Me-313-thermodynamics-11_(1).docx

8 ME 313 - THERMODYNAMICS 1

COURSE SYLLABUS/

Course Description: A course dealing with the thermodynamics properties of pure substances, ideal and real gases and the study and application of the laws of thermodynamics in the analysis of processes and cycles. It includes introduction to vapour and gas cycles. COURSE OBJECTIVES: To know the principles underlying the utilization of energy in the thermal systems; and introduction to vapour and gas cycles. COURSE OUTLINE: Topics; CHED 1. Introduction 2. Basic Principles, Concepts and definition 3. First Law of Thermodynamics 4. Ideal Gases/Ideal Gas Laws 5. Processes of Ideal Gases 6. Properties of Pure Substance 7. Processes of Pure Substance 8. Introduction to Cycle Analysis; Second law of thermodynamics 9. Introduction to Gas and Vapour cycle 10. Real Gases Textbooks and Reference Books 1. Thermodynamics 6th edition by Faires, Simman and Brewer 2. Classical Thermodynamics by Wylen and Sonntag, S I edition 3. Thermodynamics by Burghardt 4. Thermodynamics by Boyles 5. Thermodynamics by Sta. Maria 6. Power Plant Reviewer by Mandaue and Capote 7. Power Plant Reviewer by Alcorcon

INTRODUCTION In the course of our study of Thermodynamics, a number of the examples presented refer to processes that occur in such equipment as a steam power plant, a fuel cells, a vapour-compression, refrigerator, a thermoelectric cooler, a rocket engine, and an air separation plant. In this introductory chapter, a brief discussion of this equipment is given. There are at least two reasons for including such a chapter. First, many students have had limited contact with such equipment, and the solution of problems will be more significant and relevant when they have some familiarity with

the actual processes and the equipment involved. Second, this chapter will provide an introduction to Thermodynamics including the use of certain terms (which will be more formally defined in later chapters). Some of the problems for which Thermodynamics is relevant, and some accomplishments that have resulted at least in part, from the application of Thermodynamics It should be emphasised that Thermodynamics is relevant to many other processes than those cited in this chapter. It is basic to the study of materials-chemical reactions and plasmas. The student should bear in mind that this chapter is only a brief and necessary incomplete introduction to the subject of thermodynamics. Examples of Ares in Thermodynamics: a. b. c. d. e. f. g. h. i.

Simple Steam power plant Fuel cells Vapour compression refrigeration cycle Thermoelectric refrigerator Air separation Plant Combustion engines Geothermal power plant Direct energy conversion Solar energy

Founders: 1. Antoine Laurent Lavoisier, founder of modern chemistry 2. Nicolas Leonard Sadi Carnot, His observations led to the naming of a theoretical dynamic cycle after him. He wrote the famous REFLECTIONS on the MOTIVE POWER of FIRE. 3. Rudolf J Clausius, ---developed ideas about matter, and concept about entropy. 4. Lord Kelvin, developed the thermodynamic temperature scale. 5. James Prescott Joule-- made experiments that established the equivalence of heat and work. 6. J. Willard Gibbs----had a tremendous effect on the establishment of thermodynamics, statistical mechanics, chemistry and mathematics

CHAPTER 2 - Some Concepts and Definition of Thermodynamics

Definitions: By Wylen and Sonntag, it is the science that deals with heat and work and those properties of substances that bear a relation to heat and work. Basis; experimental and observation By Burghardt, it is a science that deals with energy transformation, the conversion of heat into work, or chemical energy into electrical energy. By Keenan, study of the properties of a substance and the relationship of heat and work in a system. In some textbook, it is a branch of physical sciences that treats various phenomena of energy and the related properties of matter, especially of the laws of transformation of heat into other forms of energy and vice-versa. Macroscopic versus Microscopic Point of View

The macroscopic approach to the study of thermodynamics that does not require a knowledge of the behaviour of individual particles is called CLASSICAL THERMODYNAMICS. This approach reduces the number of variables to a few that can be handled. As the word macroscopic implies, we are concerned with the gross or average effects of many molecules. Furthermore, these effects can be perceived by our senses and measured by instruments, what we really perceive and measure is the time- average influence of many molecules. From microscopic point of view, Statistical Thermodynamics is more elaborate approach to the study of thermodynamics and based on the average behaviour of large groups of individual particles.

Properties and State of a Substance Property-is a characteristic quality of the entire system and depends not on how the system changes state but only on the final state. A property can also be defined as any quantity that depends on the state of the system and is independent of the path by which the system arrived at the given state. (any characteristics of a thermodynamic system)

Two classes of property 1. Extensive property- is one that depends on the size or extent of the system, such as mass and volume, energy. Or are those properties which depends on the mass of the system. 2. Intensive property- is independent of the size of the system, example, temperature and pressure. Example of a System Or are those properties which are independent of mass of the system. Note: that an extensive property per unit mass such as specific volume is an intensive property. Sometimes it is called specific property.

Mathematically a thermodynamic property is a point function. It should be emphasized that a property of a system has significance only when the system is in equilibrium. Thermal equilibrium- uniform temperature throughout the system. Mechanical equilibrium- pressure at any point in the system does not vary with time when system is isolated. Thermodynamic equilibrium- a system condition that precludes all possible changes of state. System is defined as a collection of matter or space of fixed identity. Thermodynamic system- a quantity of matter of fixed mass and identity on which attention is focused for study. System boundary- defined as the extent of the system Control volume- a thermodynamic system that involves a flow of mass across the system boundary called control surfaces. Or a fixed region in space. Surrounding - everything external to the system. FIGURE:

HEAT Weights

LOW PRESSURE

HIGH PRESSURE Air Out

Air In

Piston WORK System Boundary y

GAS

CONTROL SURFACE

AIR COMPRESSOR MOTOR

Example of a Control Volume

Working Substance- is a substance to which heat can be stored and from which heat can be extracted. Example; pure substance and ideal gas

Properties of a working substance 1. 2. 3. 4. 5. 6. 7. 8. 9.

Pressure Temperature Specific volume, density and Specific weight Mass and Weight Volume Internal energy Flow work Enthalpy entropy

TERMS Matter

- anything that occupies space and has weight.

Mass

-Amount of matter in a substance

Volume

-Space occupied by matter.

Weight

-Force exerted by gravity on a given mass

Weight-Mass Relationship

……..

ENG

MKS

SI

W= mg

slug-ft/s2

kgm-m/s2

kgm-m/s2 , (Newton. N )

Where; m=mass

slug

g=local acceleration

lbm

kgm

kg

ft/s2

m/s2

m/s2

due to gravity If g is not specified, Then use standard gravitation (450latitude), gs=32.2 ft/s2; =9.80665 m/s2. Note that the mass of the substance remains constant with elevation but weight varies with elevation. Basic Conversion; MKS

ENG

SI

1 kg f

2.205 lbf

9.8066 N

𝒈

W=m( 𝒈𝟎 ) 𝒔

lbf Where;

kgf

go=observed or local gravity gs=standard gravity. If g0 is not specified, Then g0=gs NOTE: Within the speed of light, a given mass is constant throughout the universe. If within the speed of light , then, use; Einstein mass relativity formula. mo=

𝒎 𝟐

√𝟏−𝒗𝟐

,

where; v=velocity; c= speed of light

𝒄

Mass-Weight-Volume Relationship; Mass density=ρ=

𝒎 𝑽

in kg/m3,

Density is defined as mass per unit volume. Note: 1 slug=32.174 lb

where , V= volume in cu. m

1 lb=16 oz 1 kg= 9.80665N 1 dyne= 10-5N Multiples in SI Prefixes Factor

Prefix

Symbol

1024

Yotta

Y

1021

Zeta

Z

1018

Exa

E

1015

Peta

P

1012

Tera

T

109

Giga

G

106

Mega

M

103

Kilo

K

102

Hecto

H

101

Deka

D

10-24

yocto

y

10-21

zepto

z

10-18

atto

a

10=15

femto

f

10=12

pico

p

10-9

nano

n

10-6

micro

μ

10-3

milli

m

10-2

centi

c

10-1

deci

d

Note; 100 is unity

VOLUME CONVERSION ENG

ENG

1 ft3=

7.48 gal

1 bbl=

42 gal

1 drum=

55 gal

1 gal=

3.785 litres in mks

MKS

SI

1m3=

1 000 L

1 L=

1dm3

Specific Volume = v; in m3/kg, defined as volume per unit mass. v= V/m = 1/ρ For water at STP (600F) Density= 62.4 lb/ft3 = 1000 kgm/m3=8.33 lbm/gal; 1kg=1 L

For water other than standard temperature 𝟏

ρ=𝒗𝒇 ; refer to steam table. t

vf 𝑊 𝑉

Specific weight/ or weight density -- is defined as weight per unit volume. In symbol, ɤ= , in N/m3 or ɤ=ρg or ɤ = ρ( g0/gs). Specific gravity= it is the ratio of the weight of a substance to the weight of an equivalent volume of water. (for solids and liquids). Specific gravity- is the ratio of the mol of the gas to the mol of air. (for gases). SGg= MWg/MWair Example #01 A certain body is weight at elevation of 10,000 m (where g=9.778 m/s2) by a spring balance that was calibrated at sea level. The reading on the spring balance is 10 kg. What is the mass of this body? Solution; W = mg = 10(9.80665) = 98.0665 N m = W/g = 98.0665/9.778 = 10.0293 kg

Example#02 The specific gravity of sulphuric acid is 1.2. What is the specific weight of sulphuric acid in KN/m3? Solution; SG=specific weight of sulphuric acid/specific weight of water; then specific weight of sulphuric acid =1.2(9.80665)=11.768 KN/m3. Example. A tank contains a mixture 0f 40 kg of nitrogen and 40 kg of carbon monoxide. The total tank volume is 40 m3 . What is the density of the mixture? 𝑚𝑎𝑠𝑠 40+40 𝑘𝑔 Solution: Density = 𝑣𝑜𝑙𝑢𝑚𝑒 = 40 = 2 𝑚3 FORCE Force is defined as directly from Newton’s second law of motion which states that the force on a body is proportional to the product of the mass and the acceleration in the direction of the force. In formula; F= ma; N; Note: 1 N= 1 kg-m/s2 PRESSURE Pressure – is defined as the normal (perpendicular) force exerted by a fluid per unit area against which the force is exerted. 𝑭

P=𝑨 , in KN/m2 or Kpa. or psia.

Note: Applicable for liquid and gases

Absolute pressure- is the true pressure measured above a perfect vacuum. Or is the measure of pressure above zero. Gage pressure- is the pressure measured from the level of atmospheric pressure by most pressure recording instrument like pressure gages and open-ended manometer. Or is a measure of pressure above existing atm. Pressure. Atmospheric pressure- is the pressure obtained from barometric reading. Pabs= Pgage + Patm Where: Pabs = absolute pressure Pgage = gage pressure Patm = atmospheric pressure

= 101.325 Kpa = 14.7 psi = 1.032 or 1.033 kg/cm2 = 29.92 in. Hg =760 mm Hg =760 torr =1.013 bar =1.013 x 106dyne/cm2 = 1 atm = 0 Kpagage = 0 psig ;

NOTE : 1 bar = 100 KPa

1 ata (one technical atmosphere) = 1

𝒌𝒈 𝒇 𝒄𝒎𝟐

=98.1 kPa = 98066.5 Pa =736 torr =736 mmHg

1 cm H2O = 98.1 Pa 1 in Hg = 3390 Pa 1 bar= 1.0197 ata =14.5 psi Note: Pgage=negative (-) when the pressure is vacuum. The pressure in perfect vacuum is -101.325 Kpa. FIGURE: PRESSURE ABOVE ATMOSPHERIC PESSURE PRESSURE GAUGE, ORDINARY PRESSURE GAUGE READS DIFFERENCE BETWEEN ABSOLUTE PRESSURE AND ATMOSPHERIC PRESSURE ATMOSPHERIC PRESSURE

ABSOLUTE PRESSURE GREATER THAN ATMOSPHERIC PRESSURE

BAROMETER READS ATMOSPHERIC PRESSURE

ZERO PRESSURE E

P vacuum, -ORDINARY VACUUM GAUAGE READS DIFFERENCE BETWEEN ATMOSPHERIC PRESSURE AND ABSOLUTE PRESSURE PRESSURE LESS THAN ATMOSPHERIC PRESSURE P abs., -ABSOLUTE PRESSURE LESS THAN ATMOSPHERIC PRESSURE PERFECT VACUUM = -101.325 KPa

ILLUSTRATION OF TERMS USED IN PRESSURE MEASUREMENTS

Example Problems: A condenser vacuum gauge reads 715 mm Hg when the barometer stands at 757 mm Hg. What is the absolute pressure in the condenser in Kpa. Solution: Pabs= Patm + Pgage 101.325 𝐾𝑝𝑎

=757 + (-715) =42 mm Hg = 42mmHg( 760 𝑚𝑚 𝐻𝑔 ) = 5.60 Kpa Example: A boiler installed when the atmospheric pressure is 755 mmHg has a pressure of 12 kg/cm 2. What is the absolute pressure in Mpa. Solution: Pabs= Pg + Patm 𝟏𝟎𝟏.𝟑𝟐𝟓 ) 𝟏.𝟎𝟑𝟑

= 12(

𝟏𝟎𝟏.𝟑𝟐𝟓 ) 𝟕𝟔𝟎

+ 755(

= 1277.69 Kpa = 1.27769 Mpa

Note: Change in pressure = Δp = p of the system- p atmosphere = density x h x g Where;

h is the height of fluid in meters

TEMPERATURE Temperature is defined as the indication or degree of hotness and coldness and therefore a measure of intensity of heat. Or the degree of hotness and coldness of a substance. Zeroth law of Thermodynamics- states that when two bodies have equality in temperature with a third body, they in turn have equality of temperature with each other. The six temperature scales 1. 2. 3. 4. 5. 6.

Celsius or Centigrade Fahrenheit Kelvin Rankine Reamur Ligem

Absolute temperature is the temperature measured from absolute scale. Absolute zero is the temperature at which the molecules stop moving. Relation Between Temperature Scales 5

0

C= 9 (0F-32)

0F=9 0C

+ 32

5

Absolute Temperatures K = 0C + 273 0

R = 0F + 460

Temperature Difference ΔTk = ΔTC; 10Re =

ΔTR = ΔTF

40 C 5

ΔTC =

5 9

9 5

ΔTF = 𝛥𝑇𝑐

ΔTF;

10Re = 200L 9

Note: 1C0 =5 F0 and degree is to be written after the scale to indicate that it is temperature change. The temperature at which molecules stop moving = -2730C = -4600F FIGURE: 212⁰F

100⁰C

BOILING POINT

0⁰C

FREEZING POINT

⁰C

⁰F 32⁰F

ABSOLUTE TEMPERATURE

-460⁰F

-273⁰C ABSOLUTE ZERO

GRAPHICAL ILLUSTRATION OF TEMPERATURE SCALE

Example Problems:

#01

Convert 1000F to K. Solution: 0

5

5

C = 9 (0F-32) =9(100-32) =37.770C.

K=37.77 +273.15 = 310.920 Example #02 What is the temperature at which water freezes using the Kelvin scale.

Solution: Freezing temperature of water is 00C. Example #03

0

K=0C +273 = 0 + 273= 273 K

If the temperature inside the furnace is 700 K, what is the corresponding reading in 0F? Solution; 0

9

F =5 0C + 32

0

C = K - 273 = 700 – 273 = 427

0

9

F =5 (427) + 32 = 800.6

Processes and cycles The path of succession of states through which the system passes is called a process. 1. isothermal process – is a constant temperature process 2. isobaric or isopiestic process- is a constant pressure process 3. isochoric process - is a constant volume process 4. polytropic process; PVn =C, non- adiabatic process 5. Isentropic process; PVk =C INTERNAL ENERGY ;u, kJ/kg, heat energy due to the movement of the molecules within the substance brought about by its temperature. FLOW WORK; work due to the change in volume; kJ/kg; W=pV Enthalpy; Internal energy plus flow work; h= u+ pV Entropy; s=ʃ

𝒅𝑸 𝑻

; kJ/kg K, it is a measure of the degree of disorder. Concept of entropy; anything which is

in order eventually goes into disorder. CYCLIC PROCESS—a process which gives the same states/conditions after the system undergoes a series of processes. Quasi-Static Process-is one in which departs from equilibrium state only infinitesimally at every instant.

Problem Solving: 1. A steady force of 5 KN acts on a mass of 20 kg. What is the acceleration of this mass? Ans. 250 m/s2 2. The std. acceleration (at sea level and 450latitude) due to gravity is 9.80665 m/s2. Calculate the force due to standard gravity acting on a mass of 50 kg. Ans. 490.33 N 3. Consider a manometer containing a fluid with a density of 800 kg/m3.What pressure difference is indicated if the difference in height of the two columns is 400 mm?. What would the column height difference be if the same pressure difference were measured by a manometer containing mercury, having a density of 13 600 kg/m3? Ans. 3.14 KPa, 23.5 mm M 4. A vertical column of water will be supported to what height by standard atmospheric pressure? Ans, 10.34 m 5. Convert 225 KPa into (a) atmospheres , (b) mmHg absolute Ans. 2.22 atm, 1687.638 mmHg abs 6. A new temperature scale is desired with freezing of water at 0 0X and boiling occurring at 10000X. Derive a conversion between 0C and 0X. What is the absolute zero in 0X?. Ans. 0X= 100C, -2731.50X 7. Determine the pressure at points A and B if the density of mercury is 13590.0 kg/m3 and that of water is 1000 kg/m3.g Ans. PB=169.29 KPa, PA=158.061 KPa FIGURE:

Atmosphere

A 635 mm

H2O

510 mm

Hg

B

8. Steam at a pressure of 150 psia and a temperature of 4000F occupies a volume of 3.223 ft3/lb. a) What is its density in lb/ft3and slugs /ft3. b) Convert the pressure to in. Hg and ft water @ 600F c) Convert temperature to 0R, K,0C Ans. a) 0.31; 0.00964, b) 305.31; 346.265, c) 860; 477.4 ;204.40C 9. The piston shown below is held in equilibrium by the pressure of the gas flowing through the pipe. The piston has a mass of 21 kg: p1=600 KPa: p11= 170 KPa . Determine the pressure of the gas in the pipe; p111. Ans. FIGURE:

1

10 cm 11

20 cm

111

10. During take- off in a space ship, an 80 kg astronaut is subjected to an acceleration equal to 5 times the pull of the earth’s standard gravity. If the take-off is vertical, what force does he exert on the seat? Ans. 4704 N

ADDITIONAL PROBLEMS 1.

𝑘𝑔

200 grams of water are mixed with 100 grams of alcohol ( ρ = 790 𝑚3 ). What is the density of the mixture 𝑘𝑔

assuming the fluids mixed completely? Ans. 918. 6 𝑚3. 2. A spring scale is used to measure force and to determine the mass of a sample of moon’s rock on the 𝑚 moon’s surface. The spring were calibrated for the earth’s gravitational acceleration of 9.8 𝑠2 . The scale 𝑚

reads 4.5 kg and the moon’s gravitational attraction is 1.8 𝑠2 . determine the sample mass. What would the reading be on a beam balance scale? Ans. 24.5 kg, 24.5 kg 𝑚 3. A certain body is “weighed” at an elevation of 10 000 m ( where g= 9.778 𝑠2 by a spring balance calibrated at sea level. The reading on spring balance is 10 kg. What is the mass of this body? Ans.10.03 kg 𝑚 4. Consider an orbital space station in which an artificial gravity of 2 2 is induced by rotating the station. 𝑠

How much would a 75-kg man weigh inside? Ans. 150 N 5. The level of water in an enclosed water tank is 40 m above ground level. The pressure in the air space 𝑘𝑔

above the water is 120 kPa., and the density of the water is 1000 𝑚3. What is the water pressure at ground level? Ans. 512 kPa. 6. The reading on a pressure gage is 1.65 MPa and the local barometer reading is 94 kPa. Calculate the absolute pressure that is being measured. 7. The height of a mercury manometer column that is used to measure a vacuum is 700 mm, and the local barometer reading is 95 kPa. Calculate the absolute that is being measured assuming that the density of 𝑘𝑔

mercury is 13 600 𝑚3 . The new 8. Two liquids of different densities (ρ1= 1500 kg/m3, and ρ2 =500 kg/m3 are poured together into a 100-liter, filling it. If the resulting density of the mixture is 800 kg/m, find the respective amounts of liquids used, local g=9.675 m/s2. Ans. m1= 45 kg 9. Five masses in a region where g=30.5 fps2 are as follows: m1 is 500 gm of mass; m2 =weighs 800 gmf; m3 weighs 15 poundals; m4 weighs 3 lbf ; and m5 is 01 slug of mass. What is the total mass expressed (a) in grams, (b) in pounds, (c) in slugs Ans. (a) 4461 gm, (b) 9.84 lb, (c) 0.3057 slug 10. A Fahrenheit and a Celsius thermometer are both immersed in a fluid, (a)If the two numerical readings are identical, what is the temperature of the fluid expressed in 0R? as K ? (b) What is the fluid temperature if the Fahrenheit reading is numerically twice that of the Celsius reading? Ans. (b) 7800R, 433 K 11. An oil storage tank contains oil with specific gravity of 0.88 and depth of 20 meters. What is the hydrostatic pressure at the bottom of the tank in kg/cm2? What is the absolute pressure in kPa? Ans. 1.76 kg/cm2. 274 kPa. 12. Convert 36 0F temperature difference to 0C and to K. Ans. 200C, 20 K 13. Convert 60 lbs/cubic ft. to kg/cubic meter; to kN/m3. Ans. 960.8 kg/m3, 9.426 kN/m3 14. At what temperature are the two temperature scales 0C and 0F equal? Ans. -40 0C 15. A new temperature scale in degrees N is to be defined. The boiling and freezing points are 4000N and 1000N respectively. What will be the reading on new scale corresponding to 60 0C. Ans. 2800N 16. What pressure is column of water 100 cm high equivalent to: Ans. 9.80667kPa 17. A water temperature rise of 180F in the water cooled condenser is equivalent in 0C to: Ans. 100C 18. If the 0F scale is twice the 0C scale, what is the reading in the Fahrenheit scale? Ans. 3200F 19. A gas is contained in a vertical cylinder as shown in Figure. Atmospheric pressure outside is 1 bar, and the piston area is 500 mm2. What is the mass of the piston, if the gas pressure inside is 125 kPa? Assume standard gravitational acceleration. FIGURE:

Gas

Figure

20. Consider a manometer containing a fluid with a density of 800 kg/m 3. What pressure difference is indicated if the difference in height of the two columns is 400mm?. What would the column height difference be if the same pressure difference were measured by a manometer containing mercury having a density of 13 600 kg/m3. Ans. 3.14 kPa, 23.5 mm

CHAPTER #3 Conservation of Mass and Energy; (First law of Thermodynamics). Some definition of terms Energy - is defined as the capacity of doing work and overcoming resistance. The two most fundamental energy forms; work and heat. Work is defined as a force acting through a displacement x where the displacement being in the direction of the force. Formula; W = F x d, in N-m or Joules Work from macroscopic point of view, work done by a system, if the sole effect on the surroundings could be the raising of the weight. Work done by or produced by the system is positive, ( + ). Work done on the system or applied to a system is negative, (- ). Work is a path function and inexact differential. It is also a form of energy crossing the system boundary. 𝟐

𝟐

W=∫𝟏 𝒑𝒅𝑽, to any process or W=∫𝟏 𝜹𝑾 If the pressure must drop uniformly throughout the system, the process is called quasi-equilibrium process. Quasi- equilibrium process is defined as one in which the deviation from thermodynamic equilibrium is infinitesimal, and all the states the system passes through during a quasi- equilibrium process may be considered as equilibrium states. There are two thermodynamic systems 1. Closed system - is a system in which there is no mass flow. (fixed mass) Closed system-is a system in which there is no transfer of matter across the boundary 2. Open system - a system in which there is a flow of mass. (fixed space) Open system- a system in which there is a flow of matter through the boundary. System is defined as a collection of matter of fixed identity. Example, mass of the fluid. System is a term given to the collection of matter under consideration enclosed within a boundary. Isolated system is a system in which neither mass nor energy cross the boundaries and it is not influenced by the surroundings. Surroundings-is the region outside the boundary or the space and matter external to the system Non-flow process - is a process that takes place in a closed system. Steady flow process - a process that takes place in an open system in which the quantity of matter within the system is constant. Ex. Turbine. Non flow work-is the work for non - flow process. 𝟐

W=∫𝟏 𝒑𝒅𝑽= p(V2-V1) Steady flow work- is the work in a steady flow process. 𝟐

W=∫𝟏 𝑽𝒅𝑷 =V(P1-P2) FIGURE;

Motor

Fan

Battery

Example of Work Crossing a Boundary of a System

HEAT- is a form of energy transferred across the boundary of a system at a given temperature to another system ( or the surroundings) at a lower temperature by virtue of temperature difference between the two systems. It is a transient phenomenon. It can be identified only as it crosses the boundary. A body never contains heat. It is a path function or inexact differential. HEAT - is defined as form of energy associated with the kinetic random motion of large number of molecules. Heat transferred to the system is positive, (+).

Heat transferred from the system is negative (-). Q is a symbol of heat, in N-m or Joules Some conversion units 1 Kcal =4.187 KJ 1 Btu=778 ft-lb 1 erg= 1 dyne-cm 1 hp=42.2 Btu/min 1 hp=2545Btu/hr 1 hp=33000 ft-lb/min 1 hp=550ft-lb/sec 1hp=746 watts=0.746 Kw 1 watt= 1 J/s 1W=3.4121 Btu/hr 𝑊

𝐵𝑡𝑢

1W/m2 = 0.317 Btu/hr

1 𝑚 = 1.0403 ℎ𝑟−𝑓𝑡

1 KJ/kg = 0.4299 Btu/lbm

1𝑘𝑔0𝐶 = 0.23884 BTU/lbm0F

𝑊

𝐾𝐽

𝑊

1𝑚0𝐶 = 0.5778 Btu/hr-ft-0F

1𝑚2𝐶 = 0.1761 Btu/hr-ft2-0F

1 kg/m.s = 0.672 lbm/ft.s

1 lb/in2 = 6894.76 N/m2

1 Btu= 1.05504 KJ

1 Btu/hr =0.29307 W

1 Btu/hr.ft2 = 3.154 W/m2

1 Btu/hr.ft =0.9613 W/m

1 Btu/hr.ft3=10.35 W/m3

1 Btu/lbm =2.326 KJ/kg

1 Btu/lbm0F =4.1869 KJ/kg.0C

1 Btu/hr.ft.0F =1.7307 W/m0C

1 Btu/hrft2.0F=5.6782 W/m2-0C

1 lbm/ft.s = 1.4881 kg/m.s

1 ft2/s =0.092903 m2/s

1 KJ=0.94783 Btu

1 N/m2=1.45038 x 10-4 lbf/in2

1 kg =2.20462 lbm

1 N=0.2248 lbf

1 kg/m3 = 0.06243 lbm/ft3

1 m=3.2808 ft

1 m/s =3.2808 ft/s

1 m3 =35.3134 ft3

1 lbm =0.45359237 kg

1 lbf=4.4482 N

1 lbm/ft3 =16.018 kg/m3

1 ft/s=0.3048 m/s

1 ft3 = 0.028317 m3

1 ft2=0.092903 m2

Sensible heat- is the heat needed to change the temperature of the body without changing its phase. Qs= mCpΔT Where; Qs= sensible heat m=mass in kg Cp= specific heat Cp of water=4.187 KJ/kgK =1.0 Btu/lbF0 =1.0 cal/gC0 ΔT---- change in temperature Latent heat - is the heat needed by the body to change its phase without changing its temperature. QL=mL

Where: L = Latent heat (fusion/vaporization); Latent heat of fusion is solid to liquid. Latent vaporization is liquid to gas.

heat of

Latent heat of fusion of ice = 144 Btu/lb=334 KJ/kg=80 cal/g Latent heat of vaporization of boiling water = 970 Btu/lb=2257 KJ/kg=540 cal/g. A process in which there is no heat transfer (Q=0) is called an adiabatic process. A heat transfer per unit mass of the system is q=Q/m Example Problems Ex.#01 Let the pressure in the cylinder in Figure be the following function of volume, p=C/V. (a) If the initial pressure is 400 Kpa, the initial volume is 0.02 m3, and the final volume is 0.08 m3, find the work done. (b) Is the sign correct? Solution: Final Position 2

Initial Position I

System Boundary

Figure 3-4

AREA A

Pressure

1

5W 2

dV

Volume

Computation; 𝟐

𝟐 𝒅𝑽

W=∫𝟏 𝒑𝒅𝑽 = C∫𝟏

𝑽

𝑽

= C ln 𝑽𝟐 𝟏

𝑽

Since; C = pV; then, W = p1 V1 ln 𝑽𝟐 𝟏

0.08

W = 400(0.02) ln (0.02)) = 11.09 KJ The work is positive, indicating that the system did work on the surroundings, which corresponds to physical situation. Example#02 The pressure in the cylinder in Figure varies in the following manner with volume: p=C/V2.If the initial pressure is 500 KPa, the initial volume is 0.05m3, and the final pressure is 200 KPa, find the work done by the system.

FIGURE; Refer to Figure 3-4

Solution: 𝟐

𝟐

𝟏

𝟏

W =∫𝟏 𝒑𝒅𝑽 = C∫𝟏 𝑽-2 dV = C {𝑽𝟏 - 𝑽𝟐 };

C=

p1 V12 =p2 V2 2 ;

W= p1V1- (p2p1)1/2 V1= 500KN/m2(0.05m3)- { (500) (200) }1/2(0.05) = 9.19 KN.m = 9.19KJ. Other formula for work; 2

For compressible fluid; W=∫1 𝑝𝑑𝑉

2

For surface film; W=-∫1 𝛼 dA 2

To stretched wire; W=-∫1 𝐹 dL, Note: The minus sign indicates that a positive displacement results from work being supplied to the system. Conservation of mass The law of conservation of mass states that the total mass is a constant. For closed system a mass of fixed identity the conservation of mass is true. For open system, steady, one-dimensional flow, the continuity equation for conservation of mass is ; m= 𝐴v/ρ =Av/ѵ; in kg/sec. Examples that operates on open system; compressors, turbines, heat exchangers pumps; internal combustion engines. For steady flow system m1 = m2; or A1v1ρ1 = A2v2ρ2

The law of Conservation of mass can be also stated as mass is indestructible, provided no nuclear process is involved. Equation of Law of conservation of mass Mass entering the system = change of mass within the system+ mass leaving the system. In formula; m1 = m2 + Δm, where 𝛥m is the sum of stored mass. min = Δm +mout. If mass completely annihilated, then use Einstein’s mass energy equation, E= mc2. Where c = speed of light = 300,000 km/sec For open system; mass entering = mass leaving If two inlets and one exit, then m1+m2=m3. First Law of Thermodynamics: ( Law of conservation of Energy), it states that, Energy cannot be created nor be destroyed, it can only be transformed from one form to another. In Equation: Energy entering the system = change of energy within the system + energy leaving the system. In Formula; Ein= 𝛥E + Eout. For SSSF Process; Ein= E out

FIRST LAW OF THERMODYNAMICS 1. It also known as the conservation of energy principles. 2. Energy can neither be created nor destroyed. 3. States that one form of energy may be converted into another form or energy entering to a system is equal to energy leaving from the system. 4. In the absence of any work interaction between the system and the surroundings, the amount of heat transfer is equal to the total work of a closed system. 5. For all adiabatic processes between two specified states of a closed system, the net work done is the same regardless of the nature of the closed system and the details of the process. For First Law of Thermodynamics Equation; Energy entering the system = change of energy within the system+ energy leaving the system. In Formula; Ein = ΔE + Eout. A. If mentering = mleaving = 0 and Ein =ΔE + Eout ; then ; the system is closed and the working substance undergoes a non flow process. B. If min = mout + Δm and Δm= 0 and Ein = Eout + ΔE and ΔE = 0, then the system is open, the working substance undergoes a steady flow process

Figure:

PE₁ KE₁ H₁ Q

1 W

system 2

PE₂ KE₂ H₂

C. If min = Δm + mout and Ein = ΔE + Eout; then, the system is open, the working substance undergoes a transient process. ILLUSTRATION for Closed( Non-Flow) System Processes in a closed system are referred to as non –flow processes.

Example Problems Example #01 A gas with a density of 0.232 kg/m3 flows at a velocity of 280 m/s through a cross-sectional area of 0.0085 m2. Determine the mass rate of flow. Solution; m = ρvA = 0.232(280) (0.0085) = 0.55216 kg/s Example #02 Air is flowing in a 0.2 m-diameter pipe at a uniform velocity of 0.1m/s. The temperature and pressure are 250C and 150 KPa. Determine the mass flow rate. Solution; 𝐴𝑣 𝑅𝑇 .287 𝑥 298.2 m= 𝜈, v= 𝑃 = = 0.5705 m3/kg 150 𝛱

A = 4 (0.2)2 = 0.0314 m2 m=

0.0314 𝑥0.1 0.5705

= 0.0055 kg/s

For First Law of Thermodynamics (open system) Total Energy Entering a system = Total energy leaving a system Some terms to defined; 1. Potential energy- is a form of energy produced due to its elevation. 2. Kinetic energy- is a form of energy produced due to mass and velocity or its momentum of moving body. 3. Internal energy- heat energy due to the movement of molecules with in the substance brought about by its temperature. It is energy stored within the body. 4. Flow work- work due to the change in volume. 5. Enthalpy - internal energy + Flow work It is also defined as the heat energy transferred to a substance at a constant pressure process Formula: PE= m g h ; potential energy 𝟏 𝟐

KE = m v2;

kinetic energy

Wf = Pv; H = U + PV ; h = u + pv;

Flow work Enthalpy specific enthalpy

Note: PE, KE, u, Wf are stored energy. Transition Energy--- energy forms that are basically independent of the mass stream flow. Ex. Heat and mechanical work. Stored energy--- energy forms that may be stored or retrieved from the working substance, hence, are considered to flow with the given mass stream flow. First law open system, steady flow process. Formula: Sum energy entering = Sum energy leaving or Energy entering = Energy leaving PE1 + KE1 + H1 + Q = PE2 + KE2 + H2 + W Steady flow process is applicable to open system. An open system has a mass transfer across its boundaries, and the mass within the system is not necessarily constant.

FIGURE:

Example Problems Example #01 Steam with an enthalpy of 800 Kcal/kg enters a nozzle at a velocity of 80 m/s. Find the velocity of steam at the exit of the nozzle if its enthalpy is reduced to 750 Kcal/kg . Assuming the nozzle is horizontal and disregarding heat losses. Take g=9.81 m/s2and J=427 kg.m/kcal. Solution; Figure; Refer to figure below

Ein=Eout h1 + v12 / 2 gJ = h2 + v22 / 2gJ 800 + 802 / 2(9.81)(427) = 750 + v22 / 2(9.81)(427) V= 652.14 m/s. Example#03 A steam turbine receives steam with a flow rate of 15 kg/s and experiences a heat loss of 14 kW. Using the steam inlet and exit properties listed below, find the power produced. Inlet Exit Pressure 6205 KPa 9.859 KPa Temperature 811.1K 318.8K Velocity 30.48 m/s 274.3 m/s Spec. int. energy 3150.3 KJ/kg 2211.8 KJ/kg 3 Specific volume 0.05789m /kg 13.36 m3/kg Solution; FIGURE: m = 15.0 kg/sec h₂, KE₂

Control Volume W M = 15.0 kg/sec h₂, KE₂

Δz =3m Q= 14KW

The control Volume for Turbine

First Law: Energy In= Energy Out Q + u1 + p1v1 + KE1 + PE1 = u2 + p2v2 + KE2 + PE2 + w q=14 KW / 15 kg/s = 0.933 KJ/kg w = (u1-u2) + (p1v1-p2v2) +(

2 𝑉2 𝑉1− 2 ) 2

+ g (z1-z2) + q

w= (3150.3- 2211.8) + { (6205) (0.05789) –(9.859)(13.36) } + ( 9.8 (3.0) (1/1000) + (-0.933) = 1127.93 KJ/kg W =mw = 15 (1127.93) =16 919 KW For closed system Q= (U2-U1) + W With Paddle work

30.482− 274.32 2

1

)( 1000)+

Q=ΔU +W –Wp Example Problem; A closed gaseous system undergoes a reversible process during which 25 Btu are rejected, the volume changing from 5 ft3 to 2 ft3, and the pressure remains constant at 50 psia. Find the change of internal energy. Solution; Q= ΔU +W, where Q = -25 Btu, Then , W = P (V2- V1 )= 50 (144) ( 2-5) = - 21600 ft-lb 𝟏 𝑩𝒕𝒖 )= 𝟕𝟕𝟖 𝒇𝒕−𝒍𝒃

(

- 27.76 Btu, Then,

-25 = ΔU - 27.76; Thus ; ΔU = 2.76 Btu. THE UNFORM-STATE, UNIFORM-FLOW PROCESS

1.1 1.2

Example of USUF process Filling closed tanks with a gas or liquid Discharge from closed vessel

Basic assumptions are as follows: 1.0 2.0

The control volume remains constant relative to the coordinate frame. The state of the mass with in the control volume may change with time, but at any instant of time the state is uniform throughout the entire control volume. 3.0 The state of the mass crossing each of the areas of flow on the control surface is constant with time although the mass flow rates may be time varying. The continuity equation for the USUF process is: ( m2 – m1 ) CV + ∑ me - ∑ mi where= 0, For First Law; Qc.v. + ∑ mi ( hi +

𝑽𝒊𝟐 𝟐

+ g Zi ) =

𝒅𝑬𝒄𝒗 𝒅𝒕

+∑ me ( he +

𝑽𝟐𝒆 𝟐

+ g Ze ) + Wcv

Since at any instant of time the state within control volume is uniform we can write the First law for USUF process as, 𝑽𝟐

Qcv + ∑ mi (hi + 𝟐𝑰 + gZi ) = ∑ me (he +

𝑽𝟐𝒆 𝟐

+ gZe ) + [m2 (u2 +

𝑽𝟐𝟐 𝟐

+gZ2 ) – m1 (u1 +

𝑽𝟐𝟏 𝟐

+ gZ1 ) ]cv + Wcv.

Note: Power------- time rate of doing work or work per unit time P =Fv, where, F is force in kN and v is velocity in m/s and has a unit in kW or watts or P= 2∏Tn Note: An isolated system-a system that cannot transfer energy or mass to or from the surroundings.

Assignment 1. A nozzle is a device that converts enthalpy into kinetic energy. The kinetic energy is then usually used to drive the mechanical device such as a turbine wheel. The fluid energy is converted into mechanical work. Steam enters the nozzle at a pressure of 2700 KPa, at a velocity of 30 m/s , and with an enthalpy of 923.0 KJ/kg, and leaves with a pressure of 700 KPa, and enthalpy of 660.0 KJ/kg. a) If heat loss is 0.96 KJ/kg and if the mass flow rate is 0.2 kg/s, find the exit velocity. B) find the exit velocity for adiabatic condition. Ans. a) 725.1 m/s b) 726.5 m/s

Control Surface

m P₁

h₁ KE₁

h₁ KE₁

P₂

P₁

2. A fluid at 700 KPa, with a specific volume of 0.25m3/kg and a velocity of 175 m/s enters a device. Heat loss from the device by radiation is 23 KJ/kg. The work done by the fluid is 645 KJ/kg. The fluid exits at 136 KPa, 0.94 m3/kg and 335 m/s. Determine the change in internal energy. Ans. -481.64KJ/kg 3. A fluid enters with a steady flow of 3.7 kg/s and an initial pressure of 690 kPa, initial density of 3.2 kg/m3, initial velocity of 60 m/s/ an initial internal energy of 2000 kJ/kg. It leaves at 172 kPa, density= 0.64 kg/m 3, v= 160 m/s and u= 1950 kJ/kg. The heat loss is found to be 18.6 kJ/kg. Find the work in kilowatts. Ans. 121.1 kW 4. A mass rate of flow into a steam turbine is 1.5 kg/s, and the heat transfer from the turbine is 8.5 kW. The following data are known for the steam entering and leaving the turbine. 5. A steam condenser receives 9.47 kg/s of steam with an enthalpy of 2570 KJ/kg. The steam condenses to a liquid and leaves with an enthalpy 160.5 KJ/kg. a) Find the total heat transferred from the steam. b) Cooling water passes through the condenser with an unknown flow rate, however, the water temperature increases from 130C to 240C. Also, it is known that 1 kg of water will absorb 4.2 KJ of energy per degree temperature rise. Find the cooling water flow rate. Ans. -22817.905KW, 493.8953 kg/s 6. Steam with a flow rate of 1360 kg/h enters an adiabatic nozzle at 1378 KPa., 3.05 m/s, with a specific volume of 0.147 m3/kg and with a specific internal energy of 2510 KJ/kg. The exit conditions are p=137.8 KPa, specific volume=1.099 m3/kg and internal energy=2263 KJ/kg. Determine the exit velocity. Ans. 772.2 m/s. 7. A piston/ cylinder contains air at a pressure of 500 KPa. The piston movement is resisted by a spring and atmospheric pressure of 100 kPa. The air moves the piston and the volume changes from 0.15 m3 to 0.6m3. Determine the work when a) the force of the spring is directly proportional to the displacement, b) the force of the spring is proportional to the square root of the displacement. Ans. 562.5 KJ, 350 KJ 8. The enthalpy of air is increased by 139.586 KJ/kg in a compressor. The rate of air flow is 16.42 kg/min. The power input is 48.2 KW. Find the heat loss of the compressor in KW. Ans. -10 KW 9. If 10 lbs of water evaporated at atmospheric pressure until a volume of 288.5 ft3 is occupied, how much work is done. Ans. 610,358.11 ft-lb 10. Work done by a substance in reversible non-flow manner in accordance with V = 100/p ft3 , where; p in psia. Evaluate the work done on or by the substance as the pressure increases from 10 psia to 100 psia. Ans. -33 157.22 ft-lb 11. Gas is enclosed in a cylinder with a weighted piston as the top boundary. The gas is heated and expands from a volume of 0.04m3 to 0.10 m3 at a constant pressure of 200 Kpa. Calculate the work done by the system. Ans. 12 KJ 12. How much heat in KJ must be transferred to 20 kg of air to increase the temperature from 200C to 2800C if the pressure maintained constant. Let Cp of air=1.0 KJ/kg K . Ans. 5200 KJ

Additional Problems 1. An elastic sphere of 0.5 m diameter contains a gas at 115 kPa. Heating of the sphere causes it to increase to 0.62 m and during this process the pressure is proportional to the sphere diameter. Determine the work done by the gas. Ans. 7.7 kJ 2. Air and fuel enter a furnace used for home heating. The air has an enthalpy of 302 kJ/kg and the fuel an enthalpy of 43 027 kJ/kg. The gases leaving the furnace have an enthalpy of 616 kJ/kg. There are 17 kg air/kg fuel. Water circulates through the furnace wall receiving heat. The house requires 17.6 kW of heat. What is the fuel consumption per day? Ans. 41 kg/day 3. An internally reversible process occurs in a system during which Q = -12 kJ, ΔU = -79 kJ, and ΔH = - 111 kJ. (a) Find work if the system is non-flow, (b) Determine the shaft work and the change of flow energy if the system is steady-state, steady-flow with ΔK = 4 kJ, (c) Using the conditions stated in (b), evaluate ʃpdV and -ʃVdp in kJ. Ans. a) 67 kJ, b) 95 kJ. c)-32 kJ, d) 76 kJ 4. A closed system executes a series of processes for which two of the three quantities, W, Q and ΔU are given for each process. Find the value of the unknown quantity in each case. a) W = -35 kJ, Q= ? , ΔU = -35 kJ b) W= + 1.2 MJ, Q= + 645 kJ, ΔE = ? c) W= ?, Q = 5 Btu, ΔE= 4.22 kJ 5. A fluid changes state from p1= 15 psia, V1 = 20 ft3 to p2 =100 psia, V=2 ft3; the internal energy U is the same in both states. Find ΔH. Ans. – 18.51 Btu 6. During a reversible process executed by a non-flow system, the pressure increases from 344.74kPaa to 1378.96 kPaa in accordance with pV = C, and the internal energy increases 22,577 J; the initial volume is V1 = 85 liters. Find the heat. Ans. – 18,045 J 7. A spherical balloon contains 2 kg of water at 2500C, 400 kPa. The balloon material has an elasticity such that the pressure inside is always proportional to the balloon diameter. The water is now allowed to cool until the volume is one-half the initial volume. a) Determine the final temperature b) Calculate the work for the process considering the water as the system. 8. A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle is 5090 kJ. Heat transfer from the tank is 1500 kJ . Considering the tank and the fluid as the system, determine the change in the internal energy of the system. Ans. ΔU= 3590 kJ 9. The mass rate of flow into a steam turbine is 1.5 kg/s, and the heat transfer from the turbine is 8.5 kW. The following data are known for the steam entering and leaving the turbine.

Specific enthalpy Velocity Elevation above reference plane

Inlet conditions

Exit condition

3137 kJ/kg

2675.5 kJ/kg

50 m/s

200 m/s

6m

3m

Determine the power output of the turbine in kW. Ans. 655.7 kW 10. Steam flows through an adiabatic turbine at the rate of 100 lb/min with ΔK= 0 and Q= 0. At entry , its pressure is 175 psia, its volume is 3.16 ft3/lb and its internal energy is 1166.7 Btu/ lb. At the exit, its pressure is 0.813 psia, its volume is 328 ft3/lb and its internal energy is 854. 6 Btu/lb. a) what horsepower is developed ?. b) Same as (a) except that the turbine is not adiabatic and the heat loss from it is 10 Btu/lb of steam. Ans. a) 861 hp , b) 838 hp 11. Work done by a substance in reversible non-flow manner in accordance with V= 100/P, ft3, where P is in psia. Evaluate the work done on or by the substance as the pressure increases from 10 psia to 100 psia. Ans. -33157.22 ft-lb ,work done on the substance 12. The enthalpy of air is increased by 139.586 kJ/kg in a compressor. The rate of air flow is 16.42 kg/min. The power input is 48.2 kW. Find the heat loss from the compressor in kW? Ans. Q=-10kW 13. A closed gaseous system undergoes a reversible process during which 25 Btu are rejected, the volume changing from 5 ft3 to 2 ft3, and the pressure remains constant at 50 psia. Find the change of internal energy. Ans. ΔU=2.76 Btu 14. Steam flows into a turbine at the rate of 10 kg/s and 10kW of heat are lost from the turbine. Ignoring elevation and kinetic energy effects, calculate the power output from the turbine. Given ; h 1=2739.0 kJ/kg, h2 = 2300.5 kJ/kg. Ans. W=4375 kW

CHAPTER 4

PROPERTIES OF PURE SUBSTANCES

TWO PHASE SYSTEM (PURE SUBSTANCE) Pure Substance- is a homogeneous by nature, does not undergo chemical reactions, and is not a mechanical mixture of different species. Or it is a working substance that has homogeneous and invariable chemical composition even though there is a change of phase. Ex. Water, H2O Working substance- is a substance which energy can be stored or from which energy can be removed. A substance that receives, transport and transfers energy. Diagram: 1 atm 1 atm

1 atm

1 atm

T = 100⁰C

1 atm

VAPOR VAPOR

VAPOR LIQUID

LIQUID a Q

T = 40⁰C

b Q

T>

LIQUID c

T = 100⁰C

Q

d T = 100⁰C

Q

T = 100⁰C

e Q

Figure 4.1 The Change of a Sub-Cooled liquid to a Superheated Vapor by Constant Heat Addition

e c

d

b

T

P=C a

V

Figure 4.2 :

Constant Pressure Heat Addition to H₂O

Some definition of terms Saturation temperature - is the temperature at which liquid starts to boil and vapour start to condense. Ex. Water boils @ 2000C @ a pressure of 1.55 MPa.

Sub-cooled liquid - is a liquid whose temperature is lower than the saturation temperature corresponding to the given pressure. Ex. Liquid water @ 500C and 97.78 KPa is sub-cooled liquid since 500C is less than 990C (tsat @97.78 KPa). Compressed liquid - is a liquid whose pressure is higher than the saturation pressure corresponding to the given pressure. Ex. Liquid water at 105 kPa and 900C is a compressed liquid is 105 KPa is greater than 70.14 KPa (P 0 sat@ 90 C ). Saturated liquid - is a liquid at saturation temperature or pressure whose temperature is equal to the boiling point corresponding to the given pressure. It has no vapour content. Ex. Liquid water @500C and 12.35 KPa is a saturated liquid since, the given temperature,500C is equal to 500C (tsat@ 12.35 KPa). Vapor - is a term given to a gaseous phase that is in contact with the liquid phase. Ex. Steam. Saturated vapour – is a vapour at saturation temperature and pressure. It has no liquid or moisture content. Ex. Steam @ 3000C and 8.58 MPa. Superheated Vapour - is a vapour whose temperature is higher than the saturation temperature corresponding to the given pressure. Ex. Steam @3500C and 11.88 MPa s a superheated vapour since 3500C is greater than 3240C ( tsat@ 11.88 MPa). Degrees Superheat - is the difference between the actual superheated temperature and the saturation temperature. Ex. Find the degrees superheat of steam at 3500C and 8.58 MPa. Solution: 0

SH = actual temperature - sat. temp. Where: sat. temp. = 3000C ( tsat.@ 8.58 MPa). Thus; 0SH =350 – 300 = 500C

Degrees sub-cooled- is the difference between the saturation temperature and the actual sub-cooled temperature. Ex. Find the degrees sub-cooled of liquid water @ 500C and 19.94 Mpa. Solution; 0 SC = sat. temp. - actual temp. Where: sat. temp.= 600C ( tsat @ 19.94 KPa) ; actual temp. = 500C Thus; 0SC = 60 – 50 = 100C Wet vapour - is a mixture of saturated vapour and saturated liquid. Example; Steam @ 15 MPa and 3100C is a wet vapour since 3100C is less than 342.240C (tsat @ 15 MPa) Quality of wet vapour - is a fraction or percentage by weight that is saturated vapour. It is the ratio of the mass of saturated vapour to the total mass of the mixture. 𝑚𝑔 X = 𝑚𝑡 ; where; or x = mass of vapour / mass vapour + mass liquid X = quality of wet vapour mg = mass of vapour total mass of mixture = mass of vapour + mass of liquid or mt = mass of the mixture (wet vapour) Percentage moisture - is the fraction or percentage by weight that is saturated liquid. It is the ratio of the mass of saturated liquid to the total mass of the mixture. y = mf/mt; where; y = percentage moisture mf = mass of liquid mt = mass of the mixture (wet vapour) Critical Point - is a point that represents the pressure and temperature at which liquid and vapour can coexists in equilibrium. Properties of Mixture (WET VAPOUR/STEAM) 1. 2. 3. 4.

v = vf + x vfg u= uf + x ufg h= hf + x hfg s= sf + x sfg

5. 6. 7. 8.

vfg = vg – vf ufg= ug - uf hfg= hg - hf sfg = sg – sf

Guide to determine the phase of the working substance 1. 2. 3. 4.

If P given > Psat If T given Tsat

compressed liquid sub-cooled liquid superheated vapour superheated vapour

Where; v = specific volume u = specific internal energy h = specific enthalpy s = specific entropy

f = fluid g = vapour fg = mixture change by evaporation y = % moisture

FIGURE: The PV and TS Diagram Of PURE SUBSTANCE

Note: The properties of liquid and vapour at a given temperature and pressure can be found in Steam Tables and Mollier Chart Internal energy, u - is also a heat energy due to the movement of the molecules of a substance. Entropy - measures of randomness of the molecules of the substance. Enthalpy - also a mechanical energy of pressure transformed into energy of heat. It is a function of temp. only. EXAMPLE PROBLEMS EX. #01) Properties of saturated liquid in a tank. Problem A rigid tank contains 50 kg of saturated liquid water at 900C. Determine the pressure in the tank and volume of the tank. Solution; From Steam Tables FIGURE:

Since saturation condition exist in the tank, the pressure must be saturation pressure at 900C. Psat @ 900C = 70.14 KPa The specific volume of sat. liquid at 900C = 0.0010360 m3/kg Thus; V = mv = 50(0.0010360) = 0.0518 m3. Ex.#02 Volume and energy change during evaporation Problem;

A mass of 200 g of saturated liquid water is completely vaporized at constant pressure of 100KPa. Determine a)the volume change, b) the amount of energy transferred to the water. Solution; Figure;

For a) @ 100 Kpa; From Steam Tables Vg = 1.694 m3/kg ; vf = 0.0010432 m3/kg thus; vfg = vg - vf =1.694 -0.0010432 = 1.6929568 m3/kg then, ΔV = mvfg = 0.2 (1.6929568) = 0.33859136 m3. b) the amount of energy needed to vaporize a unit mass of a substance at a given pressure is the enthalpy of vaporization of that pressure; hfg @ 100 KPa, from steam tables; hfg=2258.0 KJ/kg, then the amount of energy transferred is; mhfg = 0.2 (2258.0) = 451.6 KJ Ex. #03 Pressure and volume at saturated mixture Problem A rigid tank contains 10 kg @ 900C.If 8 kg of the water is in the liquid form and the rest is the vapour form, a) determine the pressure in the tank, b) the volume of the tank Solution; Figure:

a) Since, the two phase coexist in equilibrium, (sat. mixture), P, pressure must be the sat. pressure at a given temp. From Steam tables, P = psat. @ 900C =70.14 kPa b) V = Vf +Vg = mf (vf) + mg(vg) Vf = 0.0010360 ; vg = 2.361 V = 8 (0.0010360) + 2(2.361) = 4.730288 m3 Or another solution X = mg/mf = 2/10 = 0.20 v = vf + x vfg = 0.0010360 + 0.20 (2.361 - 0.0010360) = 0.4730288 m3/kg. Thus; V=10(0.4730288) = 4.730288 m3. Superheated vapour is characterized by; Low pressures (P Tsat at a given P) Higher specific volumes (v>vg at a given P or T) Higher internal energies (u>ug at a given Tor P) Higher enthalpies (h>hg at a given P or T) . Superheated vapour-is a vapour whose behaviour is closer to that of the gas. A gas can never be liquid above critical temperature. Example Problem Determine the temperature of water at a state of P=0.5 MPa and h = 2890 KJ/kg Solution; From steam tables at 0.50 MPa

t = 2100C t=? t=2200C

h = 2876.7 h=2890 h=2897.9 By linear interpolation

2876.7 − 2890 210 − 𝑡 = 210 − 220 2876.9 − 2897.9

t = 216.33330C.

Compressed Liquid Compressed liquid is characterized by; 1. Higher pressures (P>Psat at a given T) 2. Lower temperatures (T
334.86−333.72 333.72

x 100 = 0.3416 %

Example: Find the enthalpy and specific volume of steam at 250 KPa and 50% quality. Solution; @ 250 KPa, From steam tables; vf=0.0010672, vg=0.7187, uf=535.10, ufg=2002.1, ug=2537.2, hf=535.37, hfg=2181.5, hg=2716.9 h=hf + x hfg= 535.37 + 0.50 (2181.5)=1626.12 KJ/kg or h= hg-(1-x)hfg= 2716.9 - (1-0.50)(2181.5)=1626.15 KJ/kg v=vf+xvfg=0.0010672 +0.5 (0.7187-0.0010672)=0.3600812 m3/kg If u is to be calculated: then u=uf+x ufg = 535.10 +0.5 (2002.1)=1536.15 KJ/kg or h=u+pv u= h - pv= 1626.12 - 250(0.3600812)=1536.0997 KJ/kg SOLID-LIQUID-VAPOR EQUILIBRIUM Triple point- a point at which all three phases may coexists. Melting- change phase from solid to liquid Latent heat of fusion-is the change of enthalpy between solid and liquid phase.

Latent heat of vaporization-is the change of enthalpy between a liquid and vapour phase. Latent heat of sublimation-is change of enthalpy in going from a solid to a vapour phase. Heat of fusion of a pure substance is the energy required to melt the substance. FIGURE:

Some Critical Point Data CRITICAL TEMP.

CRITICAL PRESSURE

CRITICAL VOL.

Water

374.140C

22.09 MPa

0.003155m3/kg

Carbon dioxide

31.05

7.39

0.002143

Oxygen

-118.35

5.08

0.002438

Hydrogen

-239.85

1.30

0.032192

Note: Triple point of water is @ 0.010C and 0.6113kPa. A substance above critical temperature exists as a gas.

Throttling Calorimeter A throttling process is one in which the pressure is adiabatically decreased by the use of a valve. This is totally irreversible process. A first law analysis across the adiabatic valve shows that the initial and final enthalpies have the same value. If there is a sufficient decrease in pressure, the steam will be superheated at a lower pressure and temperature. FIGURE:

Example Problem; A throttling calorimeter is connected to the de-superheated steam line supplying steam to the auxiliary feed pump on a ship. The line pressure measures 2.5 MPa. The calorimeter pressure is 110 KPa and the temperature is 1500C. Determine (a) the line steam quality, (b) the line steam enthalpy. Solution; From Steam Table P2 = 110 KPa,

T2 = 1500C, then, h2 = 2775.6 KJ/kg

@P1 = 2.5 MPa; hf = 962.11, hfg = 1841.0 h2 = h1 = hf + x hfg ; 2775.6 = 962.11 + x (1841) x = 0.98505 or 98.505%

Assignments 1. A 2-m3 tank contains saturated vapour at 400C. Determine the pressure and mass in the tank if the substance is a) steam, b) ammonia 2.0. Complete the following table for water. T,0C a) b) c) d) e) f) g)

p, KPa

x,%

200 300 200

h,kJ / kg

u, KJ/kg

v, m3/kg

852.4 150 800 5000 300

300 1000

1000

0.850 80 90

2. Steam has a quality of 90% at 2000C. Determine a) the enthalpy: b) the specific volume. 3. A rigid steel tank contains a mixture of vapour and liquid water at a temperature of 650C. The tank has a volume of 0.5m3, the liquid phase occupying 30% of the volume. Determine the amount of heat added to the system to raise the pressure to 3.5 MPa. 4. Steam enters an adiabatic turbine at 3000C and 400 kPa.. It exits as a saturated vapour at 30 KPa. Determine the change in enthalpy, b) the work, c) the change in internal energy 5. A rigid vessel contains 5 kg of wet steam at 0.4 Mpa. After the addition of 9585 KJ the steam has a pressure of 2.0 MPa and a temperature of 7080C. Determine the initial internal energy and the specific volume of the steam. 6. The main steam turbine of a ship is supplied with steam from two steam generators. One steam generators delivers steam at 6.0 MPa and 5000C, and the other del[vers steam at 6.0 MPa and 5500C. Determine the steam entrance enthalpy and temperature to the turbine. 7. A throttling calorimeter is connected to a main steam line where the pressure is 1750 kPa. The calorimeter pressure is 100 mmHg vacuum and 1050C. Determine the main steam quality. 8. A 1.5 kg of wet steam at a pressure of 5 bar (hf=640 KJ/kg, hfg= 2109 KJ/kg, dryness of 0.95 is blown into 70 litres of water of 120C (hf=50.4 KJ/kg). Find the final enthalpy of the mixture. Ans. h=104.80 KJ/kg 9. A vessel with a volume of 1 m3 contains liquid water and water vapour in equilibrium @ 600 KPa. The liquid water has a mass of 1 kg. Calculate the mass of water vapour. Ans. mg=3.16 kg 10. One kg of wet steam at a pressure of 8 bar and dryness 0.94 is expanded until the pressure is 4 bar. If the expansion follows the law PVn = C where n=1.12, find the dryness fraction of the steam at the lower pressure. Ans. a) 0.9072 b) 0.4197 c) 0.2260 d)0.2404 Additional Problems 1. 2.5 liters of superheated steam at 25 bar and 4000C is expanded in an engine to a pressure of 0.1 bar when its dryness fraction is 0.90. Find the final volume of the steam.

2.

3.

4. 5.

6. 7. 8.

9. 10.

a) 163.74 litres b) 263.74 litres c) 363.74 litres d) 463.74 litres A 1.5 kg of wet steam at a pressure of 5 bar dryness 0.95 is blown into 70 litre of water of 120C. Find the final enthalpy of the mixture. a) 74.80 kJ/kg b) 84.80 kJ/kg c) 94.80 kJ/kg d) 104.80 kJ/kg 0 Steam at 0.6 MPa, 200 C enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a velocity of 600 m/s. Determine the final temperature if the steam is superheated in the final state, and the quality if it is saturated. A 500-litre tank contains a saturated mixture of steam and water at 300 0C. Determine: a) the mass of each phase if their volumes are equal; b) the volume occupied by each phase if their masses are equal. A boiler produces 600 kg of steam per hour from feed water at 40 0C assuming the boiler to be steady flow system and neglecting potential and kinetic energy changes. Find the rate of which the heat is transfered. Take enthalpy of steam as 660 kcal/kg. Ans. Q=372,000 kcal/hr A vessel having a volume of 0.4 m3 contains 2 kg of a liquid water and water vapour mixture in equilibrium at a pressure of 600kPa. Calculate; a)the volume and mass of liquid, b) the volume and mass of vapour. 180 grams of saturated water of temperature 950C undergoes evaporation process until all vapour completely vaporized. Determine the change in volume. Ans. 0.3565 m3 Steam at a gage pressure of 9.5 kg/cm2 in the steam line is passed through a throttling calorimeter in which the gage pressure is 0.07 kg/cm2 and the temperature is 104.40C. Assuming a normal atm. Pressure, determine the quality of steam. Ans. x= 95.27% At 1.3 MPa, mixture and water has an entropy of 4 kJ/kgK. Find the enthalpy of the mixture. Ans. h=1627.71 kJ/kg. Steam leaves in an industrial boiler at 827.4 kPa and 171.60C. A portion of the steam is passed through a throttling calorimeter and is exhausted to the atmosphere when the barometric pressure is 101.4 kPa. How much moisture does the steam leaving the boiler contain if the temp. of the steam at the calorimeter is 115.60C. Ans. y=3.08%

Chapter 5

Processes Involving Pure Substance

Vapours (Steam) and ideal gases have similarity in their processes but ideal gas are not generally applicable to vapours. 1. Isometric; Isochoric or iso-volumic Process

It is internally reversible constant volume process of a pure substance. a. PV and TS Diagram Figure;

b. Non Flow Work 2

Wn =∫1 𝑃𝑑𝑉 = 0; since; dV = 0 at V = C c. Steady Flow Process 2

Ws = -∫1 𝑉𝑑𝑃 = V ( P1-P2) = Q - ΔKE – ΔPE – ΔH d. Change in Internal energy ΔU = m ( u2- u1) e. Change in Enthalpy ΔH = m (h2 - h1) f. Heat transferred Q = ΔU = m (u2-u1) g. Change in Entropy ΔS = m (s2-s1) 2. Isobaric Process or isopiestic process is a constant pressure process of a pure substance. a. PV and TS diagram

b. Nonflow Work 2

Wn =∫1 𝑃𝑑𝑉 = P (V2-V1) = mP (v2-v1) c. Steady Flow Work 2

Ws = -∫1 𝑉𝑑𝑃 = 0; where; dP = 0 @ P = C d. Change in internal energy ΔU = m (u2-u1) e. Change in enthalpy ΔH = m (h2-h1) f. Heat transferred Q = u2 - u1 + Wn = ( h2 - h1 ) g. Change in entropy ΔS = m ( s2 - s1 ) 3. Isothermal Process Isothermal process- is an internally reversible constant temperature process of a pure a. PV and TS diagram

b. Non Flow Process Wn = Q – ΔU c. Steady Flow Work

substance.

Ws = Q – ΔKE – ΔPE - ΔH d. Change in internal energy ΔU = m (u2 - u1) e. Change in enthalpy ΔH = m (h2 - h1) f. Heat transferred Q = TΔS = T ( s2 - s1 ) g. Change in entropy ΔS = m (s2 - s1)

4. Isentropic or Adiabatic Process Isentropic process is a constant entropy process of a pure substance. It is a process in which no heat is transferred. a. PV and TS diagram

b. Non Flow Work Wn = Q - ΔU; where ; Q = 0; then; c. Steady Flow Work Ws = - ΔKE – ΔPE – ΔH d. Change in Internal Energy ΔU = m (u2 - u1) e. Change in enthalpy ΔH = m (h2 - h1) f. Heat transferred Q=0 g. Change in entropy ΔS = 0 5. Isenthalpic or throttling process a. PV, TS and HS diagram

Wn = -ΔU

h2 = h1 = hf1 + x1 hfg1 b. Steady flow work Ws = 0 c. Change in internal energy ΔU = m (u2 - u1) d. Change in enthalpy ΔH = 0 e. Heat transferred Q=0 f. Change in entropy ΔS = m (s2 - s1) 6. Polytropic process A polytropic process is reversible process during which PVn=C, where, n is a constant. a. PV and TS diagram

b. Nonflow work Wn =

𝑃2 𝑉2 −𝑃1 𝑉1 1−𝑛

c. Steady flow work 2

Ws = -∫1 𝑉𝑑𝑃 =

𝑛(𝑃2 𝑉2 −𝑃1 𝑉1) 1−𝑛

Ws = n Wn ; = Q - ΔKE – ΔPE – ΔH d. Change in internal energy ΔU = m( u2 – u1) e. Change in enthalpy ΔH = m (h2 - h1) f. Heat transferred Q = ΔU + Wn g. Change in entropy ΔS= m (s2-s1) Example Problems Example Prob. #01 Steam with a specific volume of 0.09596 m3/k undergoes a constant pressure process at 1.7 Mpa until the specific volume becomes 0.13796m3/kg. What are (a) the final temp., b) ΔU, c) W, d) ΔS, e) Q ? Solution; PV and TS Diagram

Steam at a pressure of 1.70 MPa and with a specific volume of 0.09596 m3/kg is a wet mixture, the specific volume of saturated steam at 1.7 Mpa is 0.11673 m3/kg. At p1 = 1.7 MPa vf1 = 0.0011634 vfg1 = 0.11557 vg1 = 0.11673 x1 =

𝑣1 − 𝑣𝑓1 𝑣𝑓𝑔1

=

uf1 = 8 70.9 ufg1 = 1727.2 ug1 = 259.7

0.09596 − 0.0011634 0.11557

hf1 = 872.06 hfg1 = 1923.6 hg1 = 2795.7

sf1 = 2.3718 sfg1 = 4.0282 sg1 = 6.400

= 0.8203

u1 = uf1 + x1ufg1 = 870.09 + (0.8203) (1727.2) = 2286.9 kJ / kg h1= hf1 + x1hfg1 = 870.09 + (0.8203) (1923.6) = 2450.0 kJ / kg s1 = sf1 + x1sfg1 = 2.3718 + (0.8203)(4.0282) = 5.6761 kJ / (kg)(K) At a pressure of 1.70 MPa and a specific volume of 0.13796 m3/kg, the steam is superheated. At p2 = 1.70 MPa 260 2707.7 t2 u2 270 2725.9

0.13621 0.13796 o.13944 0.00175 0.00323

=

𝑥1 10

=

𝑥2 18.2

=

2939.3 h2 2963.0

6.6849 s2 6.7290

𝑥3 0.0441

X1 = 0.542; tu = 260 + 5.42 = 265.40C X2 = 9.86; u2 = 2707.7 + 9.86 = 2717.6 kJ/kg X3 = 12.84; h2 = 2939.3 + 12.84 = 2952.1 kJ/kh X4 = 0.02389; s2 = 6.6849 + 0.02389 = 6.7088 kJ/(kg)(K) (a) t2 = 265.4-0C (b) Δu = u2 - u1 = 2717.6 - 2286.9 = 430.7kJ/kg (c) W = p(v2-v1) = 1700 (0.13796 - 0.09596) = 71.4 kJ/kg (d) Δs = s2 - s1 = 67088 - 5.6761 = 1.0327kJ/(kg)(K) (e) Q = h2 - h1 = 2952.1 - 2450.0 = 502.1 kJ/kg

Example #02 One kg of steam at 2600C and with an enthalpy of 1861 kJ/kg is confined in a rigid container. Heat is applied until the steam becomes saturated. Determine (a) Q, (b) ) Δh, (c) Δs, and (d) the final temperature. Solution; PV and TS Diagram

The enthalpy of saturated vapour at 2600C is 2796.9 kJ/kg, point 1 is a wet mixture. At t1 = 2600C Vf1 = 0.0012755

uf1 = 1128.39

hf1 = 1134.37

sf1 = 2.8838

Vfg1 = 0.0409345

ufg1 = 1470.6

hfgi = 1662.5

sfg1 = 3.1181

Vg1 = 0.04221

ug1 = 2599.0

hg1 = 2796.9

sg1 = 6.0019

X1 =

ℎ1 − ℎ𝑓1 ℎ𝑓𝑔1

=

1861 − 1134.37 1662.5

= 0.4371

V1 = vf1 + x1vfg1 = 0.0012755 + (0.4371)(0.0409345) = 0.019168m3/kg U1 = uf1 + x1ufg1 = 1128.39 + (0.4371)(1470.6) = 1771.2kJ/kg S1 = sf1 + x1sfg1 = 2.8838 + (0.4371)(3.1181) = 4.2467kJ/(kg)(K) V2 = vg2 = v1 = 0.019168m3/kg 0.19294

307

2551.8

2734.3

5.6479

0.019168

t2

u2

h2

s2

0.018975

308

2550.1

2732.0

5.6397

0.000126 0.000319

=

𝑥1 1

=

𝑥2 1.7

=

𝑥3 2.3

=

𝑥4 0.0082

X1 = 0.4; t2 = 307 + 0.4 = 307.40C X2 = 0.7; u2 = 2551.8 - 0.7 = 2551.1kJ/kg X3 = 0.9; h2 = 2734.3 - 0.9 = 2733.4kJ/kg X4 = 0.0032; s2 = 5.6479 - 0.0032 = 5.6447 kJ/(kg)(K) (a) Q = u2 - u1 = 2551.1 - 1771.2 = 779.9kJ/kg (b) Δh = h2 - h1 = 2733.4 – 1861 = 872.4kJ/kg (c) Δs = s2 - s1 = 5.6447 - 4.2467 = 1.398kJ/(kg)(K)

Example Prob#03 There are 2.27 kg/min of steam undergoing an isothermal process from 27.5 bar, 316 0C to 6.8 bar. Determine (a)ΔS, (b)Q, (c)W for nonflow, and (d) W for steady flow with ΔP=O, ΔK=42 kJ/min. (1 bar=100kPa) At p1 = 2.75 MPa Solution;

PV and TS diagram

The initial and final conditions are superheated. The saturation temperature at a pressure of 2.75 MPa is 2290C and at 0.68 Mpa is 164 0C. 310

2774.8

3025.9

6.6321

316

u1

h1

s1

320

2793.4

3050.1

6.6733

6 10

𝑥

𝑥

𝑥

1 2 3 = 18.6 = 24.2 = 0.0412

X1 = 11.2; u1 = 2774.8 + 11.2 = 2786 kJ/kg X2 = 14.5; h1 = 3025.0 + 14.5 = 3040.4 kJ/kg X3 = 0.0247; s1 = 6.6321 + 0.0247 = 6.6568 kJ/(kg)(K) At p2 = 0.68 MPa 310

2815.6

3080.5

7.3481

316

u2

h2

s2

320

2831.6

3101.4

7.3836

6 10

=

𝑥1 16

=

𝑥2 20.9

=

𝑥3 0.0355

X1 = 9.6; u2 = 2815.6 + 9.6 = 2825.2 kJ/kg X2 = 12.5; h2 = 3080.5 + 12.5 = 3093.0 kJ/kg X3 = 0.0213; s2 = 7.3481 + 0.0213 = 7.3694 kJ/(kg)(K) (a) ΔS = m(s2 - s1) = 2.27 (7.3694 - 6.6568) = 1.6176 kJ/(min)(K) (b) Q = T (ΔS) = (316 + 273)(1.6176) =9 52.8 kJ/min (c) ΔU = m(u2 - u1) = 2.27 (2825.2 - 2786) = 88.98 kJ/min Wn = Q - ΔU = 952.8 - 88.98 = 863.8 kJ/min (d) ΔH = m(h2 - h1) = 2.27(3093 - 3040.4) = 119.4 kJ/min W = Q - ΔK - ΔH = 952.8 - 42 - 119.4 = 791.4 kJ/min

Example Problem #04 One kg of steam expands isentropically from 2.1 MPa and 374 0C to 930C. Find the final quality and the work for non flow and steady flow processes. Solution: The saturation temperature corresponding to a pressure of 2.1 MPa is 2140C, therefore, the initial condition is superheated. PV and TS diagram;

At p1 = 2.1 MPa 370

2892.7

3179.5

7.0017

374

u1

h1

s1

380

2909.8

3201.7

7.0360

4 10

=

𝑥1 17.1

=

𝑥2 22.2

=

𝑥3 0.0343

X1 = 6.8; u1 = 2892.7 + 6.8 = 2899.5 kJ/kg X2 = 8.9; h1 = 3179.5 + 8.9 = 3188.4 kJ/kg X3 = 0.0137; s1 = 7.0017 + 0.00137 = 7.0154 kJ/(kg)(K) S2 = s1 = 7.0154 kJ/(kg)(K) At t2 = 930C hf2 = 389.54

sf2 = 1.2271

uf2 = 389.46

hfg2 = 2275.4

sfg2 = 6.2138

ufg2 = 2108.7

hg2 = 2664.9

sg2 = 7.4409

ug2 = 2498.2

x2 =

𝑠2 − 𝑠𝑓2 𝑠𝑓𝑔2

=

7.0154 − 1.2271 6.2138

= 0.9315

u2 = uf2 + x2ufg2 = 389.46 + (0.9315)(2108.7) = 2353.7 kJ/kg h2 = hf2 + x2hfg2 = 389.54 + (09315)(2275.4) = 2353.7 kJ/kg Wn = u1 - u2 = 2899.5 - 2353.7 = 545.8 kJ/kg W2 = h1 - h2 = 3188.4 - 2509.1 = 679.3 kJ/kg

Example Problem #05 Expand 2 kg of steam at 15 bar, 3000C, into the wet region to 1000C in a polytropic process where pV1.21=C. determine (a)y2, (b)ΔH, (c)ΔS, (d) nonflow and steady flow work, and (e)Q. Solution: PV and TS diagram

For a pressure p1=1.5MPa and temperature t1=3000C, the steam is superheated. The saturation temperature for a pressure of 1.5 MPa is 1980C. V1 = 0.16966 m3/kg h1 = 3037.6 kJ/kg U1 = 2783.1 kJ/kg

s1 = 6.9179 kJ/(kg)(K)

At t2 = 1000C, the saturation pressure, p2 = 0.10135 MPa. P2v2n = p1v1n 𝑃

1.5

V2 = v1 [𝑃1 ]1/n = 0.16966 [0.10135]1/1.21 = 1.5730 m3/kg 2

At t2 = 1000C Vf2 = 0.0010435

uf2 = 418.94

hf2 = 419.04

sf2 = 1.3069

Vfg2 =1.6719

ufg2 = 2087.6

hfg2 = 2257.0

sfg2 = 6.0480

Vg2 = 1.6729

ug2 = 2506.5

hg2 = 2676.1

sg2 = 7.3549

X2 =

𝑣2 −𝑣𝑓2 𝑣𝑓𝑔2

=

1.5730 − 0.0010435 1.6719

= 0.9402

U2 = uf2 + x2ufg2 = 418.94 + (0.9402)(2087.6) = 2381.1 Kj/kg H2 = hf2 + x2hfg2 = 419.04 + (0.9402)(2257.0) = 2541.1 Kj/kg S2 = sf2 + x2sfg2 = 1.3069 + (0.9402)(6.0480) = 6.9932 kJ/(kg)(K)

(a) y2 = 5.98% (b) ΔH = m (h2 - h1) = 2 (2541.1 - 3037.6) = -993 kJ (c) ΔS = m(s2 - s1) =2(6.9932 - 6.9179) = 0.753 kJ/K (d) Wn =

𝑚(𝑝2 𝑣2 − 𝑝1 𝑣1 1−𝑛

=

2 [(0.10135)(1.5730) − (1.5)(0.16966] 1 − 1.21

= 905.4 kJ Wa = (n)(Wn) = 2(2381.7 - 2783.1) = -802.8 kJ (e) ΔU = m(u2 - u1) = 2(2381.7 - 2783.1) = -802.8 kJ Q = ΔU + Wn = -802.8 + 905.4 = 102.6 kJ

Example Problem #06 A throttling calorimeter receives steam from a broiler drum at 0.11 MPa and is superheated by 10 degrees. If the boiler drum pressure is 1.55 MPa, what is the quality of the steam generated by the boiler? Solution; TS and hs diagram

At p2 = 0.11 MPa,

tsat = 102.310C

T2 = tsat + 0SH = 102.31 + 10 = 112.310C h2 = 2700 kJ/kg (by interpolation) At p1 = 1.55 MPa,

hf1 = 851.92 kJ/kg and

hfg1 = 1941.2 kJ/kg

h1 = hf1 + x1hfg1 2700 = 851.92 + x1(1941.2) X1 = 95.20%

Assignments

1. A piston cylinder containing steam at 700 KPa and 250 0C undergoes a constant pressure process until the quality is 70%. Determine per kilogram (a) the work done, (b) the heat transferred, (c) the change of internal energy, and (d) the change of enthalpy. Ans. a)-101.4 KJ/kg; b) -810 KJ/kg ; c) -708.5 KJ/kg ; d) -810 KJ/kg

2. A rigid vessel contains 5 kg of wet steam at 0.4 MPa. After the addition of 9585 KJ of steam has a pressure of 2.0 MPa and a temperature of 700 0C. Determine the initial internal energy and the specific volume of the steam. Ans. 1553.9 KJ/kg, 0.2258m3/kg.

3. Water at 2.5 Mpa and 2000C is heated is heated at constant temperature up to a quality of 80%. Find (a) the quantity of heat received by the water, b) THE CHANGE in INTERNAL ENERGY, c) the work for a non-flow process. Ans. a)2025.7 KJ/kg, b) 1396.5 KJ/kg, c) 629.17 KJ/kg

4. A steam engine isentropically expands 5 kg/sec of steam from 0.68 MPa, 220 0C,the exhaust is dry and saturated. Determine a) the work of a non-flow process , b) the final temperature and c) the work of a steady flow process. Ans. a) 641.05 KJ/sec, b) 131.9 0C. c) 838.38 KJ

5. Steam at 5 MPa and 4000C expands polytropically to 1.0 MPa according to PV1.3=C. Determine the work for non flow and steady flow, the heat transferred, change of enthalpy, and change of entropy. Ans. 298.9 KJ/kg, 388.6 KJ/kg, -7.6 KJ/kg, -396.2 KJ/kg, -0.0127 KJ/kg.K

6. Wet steam at 1 Mpa flowing through a pipe is throttled to a pressure of 0.1 MPa. If the throttling temperature is 1100C, what is the quality of the steam in the pipe. Ans. 96% 7. Steam at 3.1 MPa and with 74.280C SH rejects 1973.36 kJ per kg of steam heat at constant pressure, determine; a) final temperature, b) the change in specific entropy. Ans. a) 235.7 0C; b) -3.8572 kJ/kg-K 8. Three kg of steam initially at 2.5 MPa and a constant temperature of 3700C have 2460 kJ of heat removed at constant temperature until the quality is 90%. Determine; a) the pressure when dry saturated steam exists and, b) the work. Ans. a) 21.03 Mpa; b) -4550.7 kJ 9. Steam at 173.460C and with a specific entropy of 6 kJ/kg-K undergoes a polytropic process with n=1.18 until the pressure becomes 2MPa. a) What is the final temperature of the steam?, b) What is the work and the quantity of heat of the process? Ans. a) 212.420C, b) -126.79 kJ/kg c.)-926.79 kJ/kg 10. Steam is throttled to a 0.1 MPa with a 20 degrees of superheat. What is the quality of throttled steam if its pressure is 0.75 MPa. What is the enthalpy of the process. Ans. a) 97.6%, b) 2713 kJ/kg

CHAPTER 6

IDEAL GAS AND SPECIFIC HEATS

In this Chapter the working substance is considered to be a gas. Definition: Ideal gas is a substance that has the equation of state; PV = mRT , or Pv = RT, or PV = nṜT Where; P = absolute pressure, KPa V = volume, m3, or m3/sec m = mass, kg, or kg/sec R = gas constant, KJ/kg.K or individual gas constant T = absolute temperature, K V = specific volume. m3/kg Ṝ = universal gas constant N = no. of mols Ṝ = 8.3143 KJ/kgmol.K , = 1545 ft-lbf/mol.0R = 0.821 L-atm/mol. K 8.3143 , 𝑀

R=

where; M = molecular weight in kg/kmol

Finding the value of the gas constant given the universal gas constant: Ṝ

Examples; R (O2) =𝑀𝑊(𝑂2) =

8.3143𝐾𝐽/𝑘𝑔𝑚𝑜𝑙𝐾 32𝑘𝑔/𝑘𝑚𝑜𝑙

= 0.2598 KJ/kgK = 48.281ft-lb/lbm.K

Refer to Table at the appendix of your textbook. Properties of Air M

R

Cp

28.97

0.287

1.0 or 1.003

Cv

k

0.7176

1.4

M = molecular weight in kg air/mol of air R, Cp and Cv; are in KJ/kg K R = 53.3 ft-lb/lb 0R R=

8.343 𝐾𝐽 𝑀 𝑘𝑔 𝐾

=

1545 𝑓𝑡−𝑙𝑏 𝑀 𝑙𝑏 𝑅

Cp = 0.24 Btu/lb 0 F = 0.24 kcal/kg 0C Cv = 0.171 Btu/lb0F = 0.171 kcal/kg0C Relation Between Cp, Cv, k and R Cp - Cv = R 𝑅

Cv = 𝑘 − 1 𝐶𝑝

k = 𝐶𝑣 Cp =

𝑘𝑅 , 𝑘−1

where;

Cp = specific heat at constant pressure in KJ/kg K Cv = specific heat at constant volume in KJ/kg K Boyle᾽s Law In a confined gas, If the absolute temperature is held constant, the volume is inversely proportional to the absolute pressure. P1 V1 = P2 V2 CHARLE᾽S LAW

In a confined gas, if the absolute pressure is held constant the volume is directly proportional to the absolute temperature. 𝑉1 𝑇1

𝑉2 𝑇2

=

Also; In a confined gas, if the volume is held constant, the absolute pressure is directly proportional to the absolute temperature. 𝑃1 𝑃 = 𝑇2 𝑇 1

2

GENERAL GAS LAW ( combined Charle’s Law and Boyle’s Law) 𝑃1𝑉1

=

𝑇1

𝑃2𝑉2 𝑇1

Dalton s Law of Partial Pressure The pressure exerted in a vessel by a mixture of gases is equal to the sum of the pressures that each separate gas would exert if it alone occupied the whole volume of the vessel. P = P1 + P2 + P3 + . . . + Pn , Where; Pn = the total pressure of the mixture P1 , P2, P3, . . . = PARTIAL PRESSURE OF GASES Partial pressure is the pressure exerted by each gas

Avogadro’s Law At equal volume, at the same temperature and pressure conditions the gases contain the same number of molecules. One gram mole of a substance has 6.023 x 1022 molecules. If M is the molecular mass. Mv=ṽ 𝑚1 𝑚2

=

𝑀1 𝑀2

or

𝑅1 𝑅2

=

𝑀1 𝑀2

Pṽ = MRT = ṜT nṽ = V Where; Ṝ = MR

Actual - Gas Equation of State a) Van der Waals equation of state Ṝ𝑇

𝑎

P= ṽ−𝑏 - ṽ2 , written in 1873, THE COEFFICIENT a and b compensate for the non ideal characteristics of the gas. The constant b accounts for the finite volume occupied by the gas molecules and a/ν2 term accounts for the intermolecular forces. b) Beattie- Bridgeman equation of state for a gas. Ṝ𝑇 (1−⋴ ) 𝐴 (ṽ + B ) - ṽ2 ; ṽ2 𝑎 A = A0 (1 - ), ṽ 𝑏 B = B0 ( 1 - ṽ ), 𝑐 ⋴ = ṽ𝑇 3 , The constant A0,

P=

where:

B0, a, b, and c are determined for individual gases. Refer to Table 5.1 and Table 5.2

.Thermodynamics by Burgardth.

Example Problems Ex. Prob#01 An air bubbles rises from the bottom of a well where the temperature is 250C to the surface where the temperature is 270C. Find the percent increase in the volume of the bubbles if the depth of the well is 5 m. Atmospheric pressure is 101,528 Pascals. Solution;

Figure;

% increase =

𝑉2 − 𝑉1 𝑉1

Solving for V2 P1V1/ T1 = P2V2 / T2 Where; P1 = 5 ( 9.81) + 101.528 = 150.578 KPa T1= 25 + 273= 298 K P2 = 101.528 kPa T2= 27 + 273 = 300 K Then, 150.578 V1 / 298 = 101.528 V2 / 300 V2= 1.493 V1 % increase = 1.493 V1 - V1 / V1 = 0.493 = 49.3 % Example Problem #02 An ideal gas is 45 psig at 80 0F is heated in a closed container to 130 0F. What is the final pressure? Solution; 𝑃1 𝑇1

𝑃

= 𝑇2 ;

where;

2

P1 = 45 + 14.7 = 59.7 psia T1= 80 + 460 = 540 0R T2= 130 + 460 = 590 0R 59.7 540

Then,

=

𝑃2 590

P2 = 65.23 psia

Example No.3 Air is considered to be an ideal gas with a value of R= 0.287 KJ/kg K. If there are 2 kg of air in a piston-cylinder at a temperature of 280 K a volume of 0.2 m3, find the pressure. Solution; P=

𝑚𝑅𝑇 𝑉

= (2 (0.287)280)/(0.2)= 803.6 KPa

Example No.4 A closed vessel contains air at a pressure of 140 KPag and temperature of 20 0C. Find the final gage pressure if the air is heated at constant volume to 400C. Take the atmospheric pressure as 759 mm Hg. Solution; 𝑃1 𝑇1

=

𝑃2 , 𝑇2

where; 101.325 ) 760

Patm = 759 (

= 101.19 kpa

P1= 140 + 101.325 = 241.19 kpa

T1 = 20 + 273 = 293 K T2 = 40 + 273 = 313 K , then 241.19 293

𝑃2 313

=

P2 = 257.65 KPa P2gage = 257.65 - 101.19 = 156.46 KPag

Example Prob #05 An air with mass of 0.454 kg and unknown mass of CO2 occupy an 85 liters tank at 2068.44 KPaa, If the partial pressure of CO2 is 344.74 KPaa, determine its mass. Solution; 𝑃𝑎𝑖𝑟 𝑉𝑎𝑖𝑟 1 𝑚𝑎𝑖𝑟 𝑅𝑎𝑖𝑟 𝑇𝑎𝑖𝑟

=

𝑃𝐶𝑂2 𝑉𝑐𝑜2 1 , 𝑚𝑐𝑜2 𝑅𝑐𝑜2 𝑇𝐶𝑜2

where:

Tair = Tco2

Rair = 0.287 KJ/kg K

Vair = Vco2

Rco2 = 0.188 KJ/kg 𝑃

𝑅

Then, mco2 = mair (𝑃𝑐𝑜2 ) (𝑅 𝑎𝑖𝑟 ) 𝑎𝑖𝑟

𝑐𝑜2

Solving for P air Pair = PT - Pco2 = 2068.44 – 344.74 = 1723. 7 KPa 344.74

287.08

m CO2 = 0.454 (1723.7) (188.96) = 0.138 kg

Example Problem #06 If the initial volume of an ideal gas is compressed to one-half of its original volume and to twice its original temperature the pressure is; A) Doubles B)Quadruples C. Remains constant D) Triples Solution: P1V1 / T1 = P2 V2 / T2; T2 = 2 T1

Thus;

P1 V1 / T1 = P2( V2/2) /2 T1 , then, P1 = P2/4 P2 = 4 P 1 Example # 07 A volume of 400 cc of air is measured at a pressure of 740 mm Hg and a temp. of 18 0C. What will be the volume at 760 mmHg and 00C. Solution: P1 V1/T1 = P2 V2/ T2 740 (400)/(18 + 273) = 760 (V2)/(0 + 273 ) V2 = 365.4 cc Example # 08 Calculate the pressure of 2 mols of air at 400 K with a total volume of n 0.5 m 3. Use the ideal gas law and the Van der Waals Equation of State. Solution: ṽ =

0.5 2

= 0.25 m3/kg mol

a = 135.8 b = 0.0364 8.3143 (400)

P = (0.25 − 0.0364) -

135.8 (0.25)2

= 13.4 MPa

P=

𝑛Ṝ𝑇 𝑉

=

(2.0) (8.3143)(400) (0.5)

= 13.3 MPa

Compressibility Factor The ideal gas law works well for gases at low densities. As the pressure of the gas is increased for a given temperature, the molecules are packed closer and closed together. This brings about non-ideal behaviour due to additional forces 𝑝𝑣 acting on the molecules. Equation for ideal gas, =1. For non ideal gas, it is not equal to one. Z=

𝑝𝑣 𝑅𝑇

𝑅𝑇

= compressibility factor.

The reduced coordinates are: Pr = p/pc; where pc, is the critical pressure Tr = reduced temperature = T/Tc, , where Tc = critical temperature Vr = v/vc , where; vc = critical specific volume Use Table 5.3 Critical Properties, and also used compressibility chart. Example; Determine the specific volume of ammonia at 2000 KPa and 1600C. Solution; Pr = 2000/11280 = 0.177 Tr = 433/405.5 = 1.068 From Figure 5.4, Z = 0.94 R = 0.4882 KJ/kg K 𝑅𝑇 𝑃

v=Z( )=

(0.94)(0.4882)(433) 2000

= 0.0993 m3/kg

Specific Heats 𝛿𝑢

Specific Heat at constant volume, is defined as Cv = (𝛿𝑇 ) , KJ/kg K. Cv = du/dT then’ Du = Cv dT, dU = mCv dT, Thus the first law states δQ = dU + δW, For constant volume, δW = ∫ 𝑝dV = 0 , so δQ = dU, and for constant volume process, Q = ΔU = mCv (T2 – T1 ), and the change in internal energy is, ΔU = m Cv (T2 – T1 ) = U2 – U1 𝜹𝒉

Specific Heat at constant Pressure, is defined as, Cp = (𝜹𝑻 ) KJ/kg K From h = u + pv dh = Cp dT dH = mCp dT δQ = dH for P = C, hence, δQ = dH = mCp dT, also dh = du + R dt Cp dT = Cv dT + R dT, thus

Cp – Cv = R Refer to Table 5.4 , Formulas for specific heats variation with temp. Example Problem; For a certain ideal gas, R = 0.277 KJ/kg K and k = 1.384. What are the values of Cp and Cv? Solution; Cp =

𝑘𝑅 𝑘−1

=

1.384 (0.277) 1.384 − 1

𝑅

= 0.9984 kJ/kg K

0.277

Cv = 𝑘 − 1 = 1.384 − 1 = 0.7213 KJ/kg K Example; Determine the average Cp value in kJ/kg K of a gas if 522 KJ of heat is necessary to raise the temperature from 300 K to 800K making the pressure constant. Solution; Q = mCp ΔT 522 = 1 Cp (800 - 300) Cp = 1.044 KJ/kg K A gas which satisfies the Kinetic theory of gases is known as Perfect gas. A according to kinetic theory of gases, the kinetic energy of molecules is proportional to absolute temperature. For ideal gas, the value the value of Joule-Kelvin coefficient is zero, and also the Joule-Thomson coefficient for an ideal gas having equation pv=RT is zero.

Assignments 1. A large mining company was provided with a 3000 cm3 of compressed air tank. Air pressure in the tank drops from 700 KPa to 180 KPa while the temperature remains unchanged at 280C. What percentage has the mass of air in the tank been reduced? Ans. 74.29% 2. A mountain is to be measured by finding the change in pressure at constant temperature A barometer at the base of the mountain reads 730 mm Hg, while at the top it reads 470 mm Hg. The local gravitational acceleration is 9.6 m/s2. Find the mountains height. Assume T=298 K. Ans. 3.861 km. 3. For a particular gas, the value of R=0.280 KJ/kg K and the value of k=1.375. Determine the value of Cp and Cv. Ans. 1.026 KJ/kg K , 0.7467 KJ/kg K 4. Determine the size of spherical balloon required to lift a payload of 1360 kg. The gas to be used is Helium at 101.3 Kpa and 230C. The surrounding air is 101.3 KPa and 100C. Ans. 13.387 m 5. Helium is assumed to obey the Beattie- Bridgeman equation of state. Determine the pressure for a temperature of 5000C and a specific volume of 5.2 m3/kg. Compare with the ideal gas equation of state. Ans. 308.9 KPa, 308.7 KPa 6. Two spherical balloon each 1.81 m in diameter are connected by a pipe in which there is a valve. Each sphere contains helium at temperature of 26.260C when valve closed. One sphere contains 1.1337 kg and the other 0.56689 kg. After the valve has been open long enough for equilibrium to be obtained, what is the common pressure in the sphere if there is no loss of energy. Ans. 330.18 KPa 7. The volume of air at 100 KPa and 250C is 300m3. What is the volume of the same mass of air at 400 KPa and 350C. Ans. 77.516 m3. 8. An elliptical container has a capacity of 300000 m3 at atmospheric pressure and 160C. Determine the weight in kg if the gas inside is a) Nitrogen, b) Oxygen , c) mixture of nitrogen and oxygen if the partial pressure of oxygen is 50 kPa. Ans. a) 404892.45 kg, b) 354483.34 kg ,c) 379293.73 kg. 9. A tank has a volume of 0.5m3 and contains 10 kg of ideal gas having a molecular weight of 24. The temperature is 250C. What is the pressure. Ans. 2066 KPa 10. 10. A 200 L-tank contains CO2gas at 350C, 2 MPa. a. Estimate the mass inside the tank. Ans. 7.64 kg b. Estimate the percent error in( a) if the ideal gas model is assumed. Ans. 10.72% Additional Problems 1). What is the mass of air contained in a room 6 m x 4 m if the pressure is 100 kPa and the temperature is 25 0C? Assume air to be an ideal gas. Ans. 280.5 kg 2). A tank has a volume of 0.5 m3 and contains 10 kg of an ideal gas having a molecular weight of 24. The temperature is 250C. What is the pressure? Ans. 2066 kPa. 3. A spherical balloon has a diameter of 0.3 m and contains air at a pressure of 1.5 kgf/cm2. The diameter of the balloon increases to 0.4 m due to heating and during this process the pressure is proportional to diameter. Calculate the work done during this process. Ans. 343.75 kgf-m per kg of air 4. If the two specific heats of a gas are 0.237 and 0.169 respectively, the volume of 3 kg of air at 270C under a pressure of 3 ata would be? Ans. 0.204 cu. m 5. At 2068.44 kPaa, 37.80C, 0.142m3 of methane have a total mass of 1.82 kg. Using Avogadro’s principle find the mass of Carbon dioxide contained in a 0.85-m3tank at 2068.44 kPaa, 37.80C. Ans.) 30 kg 6. For a certain ideal gas, R=0.277 kJ/kgK and k= 1.384. a) What are the values of Cp and Cv ?, b) What mass of this gas would occupy a volume of 0.425m3 at 517.11 kPaa and 26.70C? , c) If 31.65 kJ are transferred to this gas at constant volume in (b), what are the resulting temperature and pressure?. Ans. a)Cp=0.9984 kJ/kgK , b) 2.647 kg, c) 43.270C, d)545.7 kPa. 7. A 5 m3 tank contains carbon dioxide at 300K after 3 kg of carbon dioxide has been used. Determine the final pressure, if the original pressure and temperature is 337.15 kPa and 315 K, respectively. Ans. 287 kPa 8. An automobile tire is inflated to 35 psig at 700F. After being driven, the temperature rise of 250F was determined. Find the final gage pressure assuming the volume remains constant. Ans. 37.3 psig 9. Assuming compression is according to the law PV= constant. Calculate the initial volume of a gas at a pressure of 12 psig which will occupy a volume of 4 m3 when it is compressed to a pressure of 250 kPa. Ans. 5.43m3 10. Find the pressure reading in kPag of a 15 kg air with a temperature of 65.50C and occupies 1.75 m3 tank container. Ans. 731.38 kPag

CHAPTER 7

PROCESSES OF IDEAL GAS

1. Isometric or isochoric or iso-volumic Process Isometric process is an internally reversible constant volume process of a working substance. a. PV and T relation (V = C) 𝑃1 𝑇1

=

𝑃2 𝑇2

; Charle s Law

b. PV and TS Diagram

c.

Nonflow Work 2

W𝑛 = ∫1 𝑃𝑑𝑉; where; dV = 0 @ V = C Wn = 0 d. Steady flow Work 2

W s = - ∫1 𝑉𝑑𝑃 = V (P1 – P2 ) e. Change in internal energy ΔU = mCv ΔT f. Change in enthalpy ΔH = m Cp ΔT g. Heat Transferred Q = ΔU + Wn Wnx=0, thus; Q = ΔU h. Change in entropy 𝑇

ΔS = m Cv ln (𝑇2 1

2. Isobaric Process Isobaric process is an internally reversible constant pressure process of a working substance.

a. PV and T relation ( P = C ) 𝑉1 𝑇1

b.

=

𝑉2 𝑇2

, (Charle s Law )

PV and TS Diagram

c. Nonflow work 2

Wn = ∫1 𝑃𝑑𝑉 = P( V2 – V1 ) d. Steady flow work Ws = 0 e. Change in internal energy ΔU = m Cv 𝛥𝑇 f. Change bin enthalpy ΔH = mCp ΔT g. Heat transferred Q = ΔU + Wn = ΔU + ΔPV = ΔH h. Change in entropy 𝑇

ΔS = m Cp ln (𝑇2 ) 1

3. Isothermal Process Isothermal process is an internally reversible constant temperature process of a working substance. a. PV and T relation (T = C ) P1V1 = P2 V2, ( Boyle s Law ) b. PV and TS diagram

c. Nonflow Work 2

Wn =∫1 𝑃𝑑𝑉

𝐶

Where: P = 𝑉 𝑉

𝑉

Wn = P1 V1 In (𝑉2 ) = mRT In (𝑉2 ) 1

1

d. Steady Flow Work 2

Ws = -∫1 𝑉 𝑑𝑃 Ws = -P1 V1 In

𝑉 (𝑉2 )= 1

where: V = C/P 𝑉 - mRT1 In (𝑉2 ) 1

Ws = W n e. Change in Internal Energy ΔU = 0 f. Change Enthalpy ΔH = 0 g. Heat Transferred Q = ΔU + Wn Q = Wn (for ΔU = 0) h. Change in Entropy From: dS = ΔS = mR In

𝑑𝑄 𝑇 𝑉 (𝑉2 ) 1

Isentropic Process Isentropic Process is an internally reversible constant entropy process of a working substance. It is also known as a reversible adiabatic process. Adiabatic Process is a reversible process in which there is no flow of heat between a system and its surroundings (Q=O).

a. PV and T relation ( S = C or PVk = C) P1 V1k = P2 V2k

pressure - volume

𝑇1 𝑇2

= (𝑉2 )k-1

temperature - volume

𝑇1 𝑇2

𝑃1 𝑘−1 𝑘 𝑃2

𝑉

1

=( )

temperature – pressure

b. PV and TS Diagram

c. Nonflow Work 2

𝐶

Wn = ∫1 𝑃𝑑𝑉 Wn =

𝑃2 𝑉2 − 𝑃1 𝑉1 1−𝑘

where: P = 𝑉 𝑘 =

𝑚𝑅(𝑇2 − 𝑇1 ) 1−𝑘

d. Steady Flow Work 2

𝐶

Ws = -∫1 𝑉 𝑑𝑃 Ws =

where: V =𝑝𝑘

𝑘 (𝑃2 𝑉2 − 𝑃1 𝑉1 ) 𝑘𝑚𝑅(𝑇2 − 𝑇1 ) = 1−𝑘 1−𝑘

Ws = k W n e. Change in Internal Energy ΔH = m CvΔT = -Wn f. Change in Enthalpy ΔH = m CpΔT g. Heat Transferred(Adiabatic; Q=0) Q=0 Q = ΔU + Wn ΔU = -Wn h. Change in Entropy ΔS = 0 Polytropic Process Polytropic Process is an internally reversible process during which PVn=C, where; n is a constant a. PV and T relation, (PVn = C) P1 V1n = P2 V2n

pressure - volume

𝑇1 𝑇2

= (𝑉2 )n-1

temperature - volume

𝑇1 𝑇2

= (𝑃1 )

𝑉

1

𝑃 𝑛−1

temperature – pressure

𝑛

2

b. PV and TS Diagram

c. Nonflow Work 2

Wn =∫1 𝑃𝑑𝑉 Wn =

𝑃2 𝑉2 − 𝑃1 𝑉1 1−𝑛

where: P = =

𝐶 𝑉𝑛

𝑚𝑅(𝑇2 − 𝑇1 ) 1−𝑛

d. Steady Flow Work 2

Ws = -∫1 𝑉𝑑𝑃 Ws =

where: V =

𝑛(𝑃2 𝑉2 − 𝑃1 𝑉1 ) 𝑘𝑚𝑅(𝑇2 − 𝑇1 ) = 1−𝑛 1−𝑛

Ws = nWn e. Change in Internal Energy ΔU = m CvΔT f. Change in Enthalpy ΔH = m CpΔT g. Heat Transferred(Adiabatic; Q=0) Q = m CnΔT

𝐶 𝑝𝑛

𝑘−𝑛

Where; Cn = Cv (1 − 𝑛) h. Change in Entropy From: dS =

𝑑𝑄 𝑇 𝑇 𝑇1

ΔS = m Cn In ( 2 )

GENERAL EQUATION FOR THERMODYNAMIC CURVES The general equation of any process is:

P Vn = C If :

n = 0; Isobaric Process N = 1; Isothermal Process N = k; Isentropic Process N = -𝝰 to + 𝝰; Polytropic Process N = 𝝰; Isometric Process Note: PVk is steeper than PV curve;

FIGURE:

Example Problems Ex. Prob. #01 How much work is necessary to compress air in an insulated cylinder from 0.20m 3 to 0.01 m3 . Use T= 200C and P1 = 100 KPa. Solution; W=

𝑃2 𝑉2 − 𝑃1 𝑉1 𝐾−1

Solving for P2, P1 V1k = P2 V2 k (100) (0.20 )1.4 = P2 (0.01)1.4 P2 = 6628.91 KPa W=

6628.91 (0.01) − 100 (0.20) 1.4 − 1

= 115.72 kJ

Ex. Problem #02 Twenty grams of Oxygen gas are compressed at a constant temperature of 300C to 5% of their original volume. What work is done on the system? Use R = 0.0619 cal/gm-K. Solution The work done on the system is; 𝑉

W = − ∫ 𝑃𝑑𝑉 = - mRT ln ( 𝑉2 ) 1

W = - 20 (0.0619) ( 30 +273) ln (

0.05𝑉 𝑉

) = 1123.74 cal

Example#03 In an isentropic process, P1 = 200 psi, P2 = 300 psi and T1 = 7000R. Find T2 , using k = 1.4.

Solution; 𝑇2 𝑇1

𝑃 𝑘=1 𝑘 𝑃1

= ( 2)

𝑇2 700

300

1.4 = 1 1.4

= ( 200 )

T2 = 785.98 0R Example #04 A piston cylinder containing air expands at a constant pressure of 150 KPa, from a temp.of 285K to a temperature of 550 K. Find the heat, work, change in enthalpy, and change in internal energy. Solution: First law, closed system, Q = ΔU + W ; Q= ΔH for P = C ΔH = mCp (T2 - T1 ) = 0.05 (1.0047 ) (550 - 285) = 13.31 KJ ΔU= mCv (T2 – T1) = 0.05 (0.7176)(550 - 285) = 9.51 KJ W = ΔH – ΔU = 13.31 - 9.51 = 3.8 KJ or W = mR (T2 –T1 ) = 0.05 ( 0.287) ( 550 - 285) = 3.8 KJ

Example #05 A piston cylinder containing 0.25 kg of helium at 278 K receives heat at constant temp. until the pressure is one-half its initial value. Find the heat added and the work. Solution: First law, Q = ΔU + W; then, ΔU = 0 for T = C 𝑉2 𝑉1

𝑃

=𝑃1 =2, then, 2

𝑃

𝑃

Q = W = P1 V1 ln ( 𝑃1 ) = mRT1 ln (𝑃1 ) = 0.25 (2.077)(278) ln 2 = 100.0 KJ 2

2

Example #06 A closed constant volume system receives 10.5 KJ of paddle work. The system contains oxygen at 344 KPa, 278 K, and occupies 0.06 m3. Find the heat (loss or gain) if the final temperature is 400 K. Solution: For closed system, First law, Q = ΔU + W – Wp; then , W = 0 for V = C ΔU = m Cv (T2 – T1 ) ; m =

𝑃1 𝑉1 𝑅𝑇

=

344 (0.06) 0.2598 (278)

Q = ΔU – Wp = 22.9 - 10.5 = 12.4 KJ

= 22.9 kJ

Assignments 1. A rigid , perfectly insulated system contains 0.53 m3 of helium at 1000 KPa. The system receives 1000 KJ of paddle work. Determine the final pressure. Ans. 2256 KPa

2. A closed rigid container has a volume of 1 m3 and holds air at 344.8 KPa and 273 K. Heat is added until the temperature is 600 K. Determine the heat added and the final pressure. Ans. 1032.5 KJ, 757.68 KPa.

3. Air in a piston cylinder occupies 0.12 m3 at 552 KPa. The expands in a reversible adiabatic process, doing work on the piston until the volume is 0.24 m3. Determine the a) work of the system, b) the net work if the atmospheric pressure is 101 KPa. Ans. a)40.08 KJ b)27.96 KJ

4. One kilogram of air expands at a constant temperature from a pressure of 800 KPa and a volume of 2 m 3 to a pressure of 200 KPa. Determine a) Work, b) the heat, c) the change of internal energy, d) and the change of enthalpy. Ans. a)2218.1 KJ, b) 2218.1 KJ, c) 0, d) 0

5. A 2 kg mass oxygen expands at a constant pressure of 172 KPa in a piston cylinder system from a temperature of 320C to a final temperature of 1820C. Determine a) the heat required, b) the work done, c) the change of enthalpy, d) the change of internal energy. Ans. a)275.55 KJ , b) 78 KJ, c) 275.55KJ, d) 197.55 KJ

6. Air is compressed polytropically in a cylinder according to pV2 =C. The work required is 180 KJ. Determine the change of internal energy, and the heat transferred. Ans. 450 KJ, 270 KJ. 7. There are 1.36 kg of air at 137.9 kPaa stirred with internal paddles in an insulated rigid container, whose volume is 0.142 m3under the pressure becomes 689.5 kPaa. Determine the a)the work input, b) 𝛥(pV), c) ΔEf, d) Q, Ans. a) 196.2, b) 78.3 kJ

8. There are 1.36 kg of gas , for which R= 377 J/kg-K and k=1.25, that undergo a non- flow constant volume process from p1= 551.6 and t1= 600C to p2 = 1655 kPa. During the process the gas is internally stirred and there are also added 105.5 kJ of heat. Determine: a) t2 , b) the work input , c) Q, d) ΔU and , e)ΔS Ans. a) 999 K, b) -1260 kJ, e) 2.253 kJ/K 9. During a reversible process, There are abstracted 317 kJ/s from 1.134 kg/s of a certain gas while the temperature remains constant at 26.70C. For this gas, Cp=2.232 and Cv=1.713 kJ/kg K. The initial pressure is 586 kPaa. For both non-flow and steady flow (ΔP=0, ΔK=0)process, determine (a) V1 and V2 and p2, b) W and Q, c) ΔS and ΔS. Ans. a) 302, 50 l/s , 3539 kPaa, b) -317 kJ/s, c) ΔS=-1.056 kJ/K-s 10. Assume 2 lb of gas for which R=26 ft-lb/lb 0R, and k= 1.10, undergoes a polytropic process from p1=15 psia, And t1= 1000F to p2= 75 psia and V2= 3.72 ft3. For both non-flow and steady flow(ΔP=0, ΔK=0)process, determine, a) n in pVn=C, b) ΔH and ΔU, and ΔS , c) ʃpdV and -ʃVdp, d) W and Q. Ans. a) 1.25, b) 156.5, 142.2 Btu, 0.129 Btu/0R, c) -57, -71.3, d) -57, -71.3, 85.3 Btu

Problem; A mass of 200 g of saturated liquid water is completely vaporized at constant pressure of 100KPa. Determine a)the volume change, b) the amount of energy transferred to the water. ................................................................ 24 Solution;...................................................................................................................................................................... 24 Figure; ......................................................................................................................................................................... 24 For a) ........................................................................................................................................................................... 24 @ 100 Kpa; From Steam Tables .................................................................................................................................. 24 Vg = 1.694 m3/kg ; vf = 0.0010432 m3/kg ................................................................................................................ 24 thus; vfg = vg - vf =1.694 -0.0010432 = 1.6929568 m3/kg ......................................................................................... 24 then, ΔV = mvfg = 0.2 (1.6929568) = 0.33859136 m 3. ............................................................................................... 24 b) the amount of energy needed to vaporize a unit mass of a substance at a given pressure is .............................. 24 the enthalpy of vaporization of that pressure; hfg @ 100 KPa, from steam tables; hfg=2258.0 ............................... 24 KJ/kg, then the amount of energy transferred is;....................................................................................................... 24 mhfg = 0.2 (2258.0) = 451.6 KJ ................................................................................................................................... 24 Ex. #03 Pressure and volume at saturated mixture ................................................................................................... 24

CHAPTER 8 INTRODUCTION TO CYCLE ANALYSIS AND THE SECOND LAW OF THERMODYNAMICS In its broader significance the second law involves the fact that processes proceed in a certain direction but not in opposite direction. Ex. A hot cup of coffee cools by virtue of heat transfer to the surroundings, but heat will not flow from the cooler surroundings to the hotter cup of coffee. Heat engine may be defined as a device that operates in a thermodynamic cycle and does a certain amount of net positive work as a result of heat transfer from a high temperature body to a low temperature body. Example of heat engine: a simple steam power plant in which steam is the working fluid and thermo-electric power generation device. For a heat engine, we can have a system that operates in a cycle and has a net positive work and a net positive heat transfer. Heat pump or refrigerator- we can have a system that operates in a cycle and has heat transferred to it from a lowtemperature body and heat transferred from it to a high-temperature body, though work is required to do this. Figure:

The thermal efficiency of heat engine is defined as; 𝑊 (𝑒𝑛𝑒𝑟𝑔𝑦 𝑠𝑜𝑢𝑔ℎ𝑡)

Ηthermal = 𝑄ℎ(𝑒𝑛𝑒𝑟𝑔𝑦 𝑡ℎ𝑎𝑡 𝑐𝑜𝑠𝑡) =

𝑄ℎ − 𝑄𝑙 𝑄ℎ

=1-

𝑄𝑙 𝑄ℎ

Refrigerator or heat pump a device that operates in a cycle, that requires work, and that accomplishes the objective of transferring heat from a low-temperature body to a high temperature body. The efficiency of a refrigerator is expressed in terms of the coefficient of performance. Thermal reservoir- is a body to which and from which heat can be transferred indefinitely without change in temperature of the reservoir. A reservoir from which heat is transferred is called a source( high temperature reservoir). And a reservoir to which heat is transferred is called a sink(low-temperature reservoir). 𝑸𝑳 (𝒆𝒏𝒆𝒓𝒈𝒚 𝒔𝒐𝒖𝒈𝒉𝒕)

𝑸𝑳

Coefficient of performance for heat pump = COP = β = 𝑾(𝒆𝒏𝒆𝒓𝒈𝒚 𝒕𝒉𝒂𝒕 𝒄𝒐𝒔𝒕𝒔) = 𝑸𝑯 − 𝑸𝑳

FIGURE:

SECOND LAW OF THERMODYNAMICS Two classical statement of the second law 1. The Kelvin-Planck Statement: It is impossible to construct a device that will operate in a cycle and produce no effect other than the raising of a weight and the exchange of heat with a single reservoir. In effect, it states that it is impossible to construct a heat engine that’s operates in a cycle and receives a given amount of heat from a high temperature body and does an equal amount of work. This implies that it is impossible to build a heat engine that has a thermal efficiency of 100%.The only alternative is that some heat must be transferred from the working fluid at a lower temperature to a low-temperature body. Thus work can be done by the transfer of heat only if there are two temperature levels involved ,and heat is transferred from the high temperature body to the heat engine and also from the heat engine to the low temperature body. 2. The Clausius Statement. It is impossible to construct device that operates in a cycle and produces no effect other than the transfer of heat from a cooler- body to a hotter body. This statement is related to a

refrigerator or heat pump, and in effect states that it is impossible to construct a refrigerator that operates without an input of work. This also implies that the coefficient of performance is always less than infinity Observations 1. Both are negative statement 2. The two statements of the second law are equivalent 3. The second law of thermodynamics has been stated as the impossibility of constructing a perpetual motion machine of the second kind. A PERPETUAL MOTION MACHINE of the first kind would create work from nothing. Or create mass energy , thus violating the first law. A perpetual motion machine of the second kind would violate the second law. FIGURE: 01

Figure 02

FIGURE 03

Type equation here.

THE CARNOT CYCLE AND CARNOT ENGINE A thermodynamic cycle occurs when a system undergoes two or more processes and returns to its initial state. Carnot Cycle- is the most efficient thermodynamic cycle. It consist of two isothermal and two isentropic processes. For any cycle, Wnet= Qnet 𝑜𝑢𝑡𝑝𝑢𝑡

Efficiency of the cycle = 𝑖𝑛𝑝𝑢𝑡 FIGURE:

THE PV AND TS DIAGRAM Figure:

1. Process 1 t0 2; isothermal expansion process 2

𝑉

𝑉

W =∫1 𝑝𝑑𝑉 = p1V1 ln ( 𝑉2 ) = mRTh ln ( 𝑉2 ) 1

2.

1

Process 2 to 3; isentropic expansion process 3

W = ∫2 𝑝𝑑𝑉 =

𝑝3 𝑉3 − 𝑝2 𝑉2 1−𝑘

=

𝑚𝑅𝑇 𝐶 − 𝑚𝑅𝑇𝐻 1−𝑘

3. Process 3 to 4; isothermal compression process 4

𝑉 𝑉3

𝑉 𝑉3

W = ∫3 𝑝𝑑𝑉 = p3 V3 ln ( 4 ) = mRTC ln ( 4 ) 4. Process 4 to 1; isentropic compression process 1

W = ∫4 𝑝𝑑𝑉 =

𝑝1 𝑉1 − 𝑝4 𝑉4 1−𝑘

𝑚 𝑅𝑇𝐻 − 𝑚𝑅𝑇𝐶

=

1−𝑘

Where; TH = is the temperature of the high temp. heat source. TC = is the temperature of the low temperature heat 𝑉

𝑉

Wnet =m RTH ln (𝑉2 ) + mRTC ln (𝑉4 ) 1

3

For closed system; First law of Thermodynamics ; Heat added is: Q1 - 2 = U2 – U1 + W1 - 2 ; ΔU = 0 for T =C, thus; 𝑉

Q1 - 2 = W1 - 2 = m R TH ln (𝑉2 ), 1

𝑉 𝑉3

Heat rejected; Q3-4 = mRTC ln ( 4 ) Wnet = ∑Q = Qin - / Qout / Thermal efficiency; ηth =

𝑇𝐻 − 𝑇𝐶 𝑇𝐻

=

𝑊𝑛𝑒𝑡 𝑄𝑎𝑑𝑑𝑒𝑑

x 100%

Mean Effective Pressure 𝑾

𝒏𝒆𝒕 Pm = 𝒑𝒊𝒔𝒕𝒐𝒏 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕, , in KPa;

For three process cycle involves three processes undergoing a cycle. Wnet=Qnet. Example: The following cycle involves 3 kg of air: polytropic compression from 1 to 2, where p1 =150 kPa, T1 = 360 K, p2 =750 kPa and n=1.2; constant-pressure cooling from 2 to 3 ; and constant temperature heating from 3 to 1. Find the temperatures, pressures and volumes at each states and determine the net work and heat. Ans. V1 = 2.066 m3, T2 = 470.7K, V2= 0.54 m3, p3 =p2 =750 kPa, V3 =0.413 m3, T3 =360 K Q from 1 to 2 =-238.2 kJ; W from 1 to 2 =-476.5 kJ; Q from 2 to 3 =-333.6 kJ; W from 2 to 3 =-95.3 kJ Q from 3 to 1 = 499 kJ =W from 3 to 1 Qnet= Wnet = -72.8 kJ

Example Problems

1.1 An engine operates on the Carnot Cycle. It produces 50 KW while operating between temperatures of 8000C and 1000C. Determine the efficiency and the heat added. Solution: ηth = Qinv= 2.1

𝑇𝐻 − 𝑇𝐶 1073 − 333 = = 0.652 𝑇𝐻 1073 𝑊𝑛𝑒𝑡 50 = 0.652 = 76.6 KW 𝜂𝑡ℎ

A six cylinder engine with a 10- x 10-cm bore and stroke operates on the Carnot Cycle. It receives 54 KJ per cycle of heat at 833 K and rejects heat at 555 K while running at 300 rev/min. Determine the mean effective pressure and the power of the engine. Solution: ηth =

𝑇𝐻 − 𝑇𝐶 𝑇𝐻

=

833 − 555 833

= 0.333

Wnet = (0.333)Qin = (0.333)(54) = 18 KJ/cycle 6𝛱 (0.1)2 (0.1) 4 18 𝐾𝐽/𝑐𝑦𝑐𝑙𝑒

Total piston displacement = 𝑊𝑛𝑒𝑡

Pm = 𝑃𝑖𝑠𝑡𝑜𝑛 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 =

0.004712

𝑚3 𝑐𝑦𝑐𝑙𝑒

= 0.004712m3

= 3819 KPa

W = (18KJ/cycle) (5cycles/s) = 90 KW

3.1

A Carnot engine requires 35KW from the hot source. The engine produces 15 kw of power and the temperature of the sink is 26 0C. What is the temperature of the hot source?. Solution: ηth =

𝑊𝑛𝑒𝑡 𝑄𝑖𝑛

=

𝑇𝐻 − 𝑇𝐶 𝑇𝐻

=1-

𝑇𝐶 𝑇𝐻

15

= 35 =

𝑇𝐻 − 299 𝑇𝐻

:

TH = 5.23.25 K; then, tH = 250.250 C

Reversed Carnot Cycle CARNOT ENGINE- is a power producing machine that uses heat supplied as an energy source and delivers mechanical work as an energy output. When the Carnot cycle is reversed, it means that the mechanical work as supplied as an energy input and heat maybe moved from one energy level to another energy level .Example is the refrigerator. It operates in a counter clockwise direction. The purpose of the reversed Carnot engine is to remove a quantity of heat at a low temperature, Qin, by supplying work. This result in a rejection of heat at high temperature, Qout. The performance of reversed engine is denoted by the coefficient of performance (COP). Thus. 𝐂𝐎𝐏 =

𝑸𝒊𝒏 𝑾𝒏𝒆𝒕

=𝑻

𝑻𝑪 ; 𝑯− 𝑻𝑪

good for ideal gas as a working substance.

A CARNOT heat pump is used to maintain the temperature of a heated space at higher temperature than that of the environment. The measure of its performance is called Performance Factor (PF). PF = QH/Wne t =

𝑻𝑯 𝑻𝑯 − 𝑻𝑪

Reversible Process for a system is defined as a process that once having taken place can be reversed and so doing leaves no change in either system or surroundings. Irreversible Process – a process is said to be irreversible if the system and all parts of its surroundings cannot be exactly restored to their respective initial states after the process has occurred. Actual cycles are composed of irreversible process. It represent a loss of work. Common causes of Irreversibilities 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0

ELECTRIC RESISTANCE In elastic deformation Viscous flow of fluid Solid-solid friction Heat transfer across finite temp. difference Fluid flows through valves Mixing of dissimilar gases or liquids Mixing identical fluids initially at different pressures and temperatures.

Example Problems

1.0 A reversed Carnot engine removes 40 000W from the heat sink. The temperature of the heat sink is 260 K and the temperature of the heat reservoir is 320 K. Determine the power required of the engine. 𝑄 𝑇 260 COP = 𝑊 𝑖𝑛 = 𝑇 𝐶𝑇 = 320 − 260 = 4.33 𝑛𝑒𝑡

Wnet =

𝑄𝑖𝑛 𝐶𝑂𝑃

𝐻−

=

𝐶

40 000 4.33

= 9230.7 W

a. An air-source heat pump is used to provide heat in a house during the winter season. The house is to be maintained at 210C and on a typical day the heat loss from the house amounts to 75 000 KJ/hr when the outdoor air temperature is -40C. The heat pump has a performance factor of 3.7 under these conditions. Determine the power consumption of the heat pump. Supposed the heat pump is replaced with a Carnot heat pump. Determine its PF. Solution: 𝑄

PF = 𝑊 𝐻

𝑛𝑒𝑡 𝑄𝐻

Wnet = 𝑃𝐹 = PFrev Carnot =

75000 3.7 𝑇𝐻 𝑇ℎ−𝑇𝐶

= 20270.27 KJ/hr = 5.63 KW =

294 294−269

= 11.76

First Corollary of the Second Law It is impossible to construct an engine to operate between two heat reservoir, each having a fixed uniform temperature, which will exceed the efficiency of a reversible engine operating between the same reservoir. Figure:

Second Corollary of the Second Law All reversible engines have the same efficiency when working between the same two constant-temperature heat reservoirs. Figure;

The thermal efficiency of a reversible engine operating between constant-temperature reservoirs has been shown to be a function of only the temperatures of the reservoirs, thus; η1-2 = f( T1, T2) and the thermal efficiency is also; 𝑄

η 1-2 = 1 - 𝑄12 = 1 -

𝑇2 𝑇1

The Third Law of Thermodynamics The entropy of a substance of absolute zero is zero. Or at absolute zero, the entropy of a pure crystalline substance is zero. Assuming real processes, the net entropy change in the universe is positive. For any irreversible process, the net entropy change is positive. And for any reversible process, the change in entropy of the system and surroundings is zero.

Assignments 1.0

A Carnot engine operates with 0.136 kg of air as the working substance. The pressure and volume at the beginning of isothermal expansion are 2.1 MPa and 9.6 litres respectively. The air behaves as an ideal gas and the sink temperature is 500C and the heat added is 32 KJ. Determine a) the source temperature, b) cycle efficiency, c) pressure at the end of isothermal expansion, d) heat rejected to heat sink per cycle. Ans. 516.5K, 0.375, 429.4 KPa, -20 KJ/cycle

2.0

A Carnot engine uses Nitrogen as the working fluid .The heat supplied is 53 KJ and the adiabatic expansion ratio is 16 : 1 . The receiver temperature is 295 K. Determine a) the thermal efficiency, b) the heat rejected, c) the work. Ans. 0.67, -17.49 KJ, 35.51 KJ

3.0

A Carnot engine operates between temperatures of 1000 K and 300 K. The engine operates at 2000 rev/min. and develops 200 KW. The total engine displacement is such that the mean effective pressure is 300 kPa. Determine a) the cycle efficiency, b) the heat supplied, c) the total engine displacement (m 3). ANS. 0.70, 285.7 KW, 0.02m3 .

4.0

Two reversible engines operates in series between a high-temperature (Th) and the low temperature (Tc) reservoir. Engine A rejects heat to engine B, which in turn rejects heat to the low-temperature reservoir. The high temperature reservoir supplies heat to Engine A. Let Th= 1000 K , Tc=400 K , and the engine thermal efficiencies are equal. The heat receives by Engine A is 500 KJ. Determine a) the temperature of heat rejection by engine A , b) the work of Engine A and Engine B, c) the heat rejected by Engine B. Ans. 632.45 K, 183.75 KJ, 116.25 KJ, 200 KJ.

5.0

A heat pump is used for cooling in summer and heating in the winter. The house is maintained at 24 0C year round. The heat loss is 0.44 kW per degree difference between outside and inside temperature. The average outside temperature is 320C in the summer and -40C in the winter. Determine the power requirements for both heating and cooling. Ans. 1162 KW, 0.094 KW.

6.0

A Carnot heat pump is being considered for home heating in a location where the outside temperature may be as low as -35 0C. The expected COP for the heat pump is 1.50. To what temperature could this unit provide heat. Ans. 714 K

7.0

A Carnot refrigerator rejects 2500 KJ of heat at 800C while using 1100KJ of work. Find a) the cycle low temperature, b) the COP, c) the heat absorbed. Ans. 197.7 K, 1.2727, 1400 KJ

8.0

A Carnot engine operating between 678 K and 305 K produces 65 KW power. Determine the heat supplied to the engine. Ans. 118.15 KW

9.0

Calculate the thermal efficiency of a Carnot cycle heat engine operating between 5000C and 400C. Ans. 0.595

10.0

It is proposed to heat a house during the winter using a heat pump. The house is to be maintained at 20 0C at all times. It is estimated that when the ambient temperature outside drops to -100C, the rate of heat loss from the house will be 25 KW. What is the minimum electrical power required to drive the heat pump unit? Ans. 2.558 KW

Additional Problems

1) A heat pump is to be used to heat a house in the winter and then reversed to cool the house in the summer. The interior temperature is to be maintained at 200C in the winter and 250C in the summer. Heat transfer through the walls and roof is estimated to be 2400 kJ/hr per degree temperature difference between the inside and outside. A) If the outside temperature in the winter is 00C. What is the manimum power required to drive the heat pump., b) If the power input is the same as that in part (a),what is the maximum outside summer temp. for which the inside of the house can be maintained at 250C. Ans. a) 0.91 kW, b) 318.17 K 2) A Carnot refrigerator operates between temp. limits of -50C and 300C. The power consumed is 4 kW and the heat absorbed is 30 kJ/kg. Determine the COP and the refrigerant flow. Ans. a) 7.657, b) 1.0209 kg/s. 3) A substance executes a reversed Carnot cycle during which it receives 105.5 kJ/min of heat. Determine the work required if the adiabatic compression process triples the initial absolute temperature. Ans. 3.52 kJ

CHAPTER 08

Introduction to Gas and Vapour Cycle

The Rankine Cycle Figures;

Discussion; Following the cycle from state 4 in the Figures, the water enters the steam generator as a sub cooled liquid at a pressure1.The energy supplied in the steam generator raises the state of the water from that of a subcooled liquid to that of a saturated liquid and, further, to that of saturated vapour at state 1. The vapour leaves the steam generator at state 1 and enters the steam turbine, where it expands isentropically to state 2. It enters the condenser at this point and is condensed at constant pressure from state 2 to state 3. At state 3 the water is saturated liquid at the pressure in the condenser. The liquid cannot enter the steam generator, which is at higher pressure, until the pressure is raised to that of the steam generator. A pump perform this very easily, in contrast to the compressor in the Carnot vapour Cycle, and raises the pressure of the liquid to pressure 1, the steam generator pressure. The liquid is now sub-cooled liquid at state4, and the cycle is complete. The Rankine Cycle Components 1.0 2.0 3.0 4.0

Steam Turbine Condenser Pump Steam generator (boiler)

The Rankine Cycle composed of the following processes 1-2: Reversible adiabatic expansion in the turbine 2-3: Constant pressure transfer of heat in the condenser (heat rejection) 3-4; Reversible adiabatic pumping process in the pump 4-1: Constant pressure transfer of heat in the boiler ( heat addition)

ME 313 THERMODYNAMICS 1 QUIZ #01 Instructions:

1.

2.

3.

4.

A. Choose the correct answer by shading the letter in your answer sheet. B. One point for the correct choice and 9 points for the right solution. 100 grams of water are mixed with 150 grams of alcohol ( ρ=790 kg/m 3). What is the specific gravity of the resulting mixture, assuming the fluid mixed completely? A. 0.682 B. 0.782 C. 0.862 D. 0. 932 E. 0.992 A spherical tank is full of water that has a total mass of 10, 000 kg. If the outside diameter of the tank is 2722 mm, how thick is the wall of the tank? A. 21.23 mm B. 22.55 mm C. 23.34 mm D. 24.25 mm E. 25.00 mm The height of a Hg column that is used to measure a vacuum is 700 mm and the local barometer reading is 95 kPa. What is the absolute pressure that is being measured, assuming that the density of Hg is 13 600 kg/m3? A. 1.64 kPa B. 6.14 kPa C. 10.34 kPa D. 15.95 kPa E. 18.65 kPa A spring scale is used to measure force and determine the mass of a sample of moon’s rock on the moon’s surface. The spring were calibrated for the earth’s gravitational acceleration of 9.8 m/s2. The scale reads 2 4.5 kg and the moon’s gravitational acceleration is 1.8 m/s . The sample mass is--------------------.

A 14.5 kg B. 24.5 kg C. 34.5 kg D. 44.5 kg E. 54.5 kg 5. A pollution control engineer proposes a new temperature scale based on 00Y as the freezing point and 800Y as the boiling point of water. Granting linear relationship, determine the degree F and degree Y relationship, and the absolute temperature in degree R if the thermometer reads 10 degree X. 4 4 9 A.0F= 𝑍 + 32; 415.50R B. 0F= 0Z – 32; 415.50R C. 0F= Z + 32; 514.5 0R 9

9 5

9

4

D. 0F=4 0Z – 32; 514.5 0R E. 0F= 9 0Z +460; 875.50R 6. Five masses in a region where g= 30.5 fps2 are as follows: m1 is 500 gm of mass; m2 weighs 800 gmf; m3 weighs 15 poundals; m4 weighs 3 lbf; m5 is 0.1 slug of mass. What is the total mass expressed in grams? A. 4461 B. 4465 C. 4469 D. 4475 E. 4483 7. A manometer containing water (density=62.1 lb/ft3) and Hg (specific gravity=13.55) connects two pressure regions A and B as shown in the figure. The local gravitational acceleration is 32.00 fps2. If PB=50 psig, what is the value of PA ? A. 37.07 psig B. 47.07 psig C. 57.07 psig D. 67.07 psig E. 77.07 psig Figure: H2O

A

60” 20”

15” B

Hg

8. Referring to Problem #07, except that, in lieu of the Hg, the fluid is a special compound with a specific gravity of 2.00. What is the value of PA ? a. 28.75 psig B 38.75 psig C. 48.75 psig D. 58.75 psig E. 68.75 psig 9. A Fahrenheit and Celsius thermometer are both immersed in a fluid. What is the fluid temperature if the Fahrenheit reading is numerically twice that of the Celsius reading?. Expressed as 0R and 0 K. a. 3800R, 333K B. 4800R, 333K C. 5800R, 333K D. 6800R, 433K E.7800R, 433K 10. Water enters the condenser at 250C and leaves at 400C. What is the temperature difference in 0F?. a. 24 B. 25 C. 27 D. 29 E. 31

Prepared by:

Engr. Pedro G. Azores Asst. Prof. 1

,