Thermodynamics

THERMODYNAMICS

THERMODYNAMICS CONTENTS: A. Definition of Terms B. Properties of Working Substance C. Work, Heat and Power D. Laws of Thermodynamics E. Ideal Gases F. Processes Involving Ideal Gases G. Pure Substance H. Processes Involving Pure Substance I. The Carnot Cycle

Thermodynamics – the study of heat and work and those properties of substance that bear a relation to heat and work.

Definition of Terms: 1. Surroundings → all matter and space outside to a system. 2. Isolated System → is a physical system that does not interact or exchange energy with its surroundings. 3. Control Volume → the focused volume is in space from which the substance flows. (Ex. Turbine, pumps, heater, etc.) 4. Control Surface → the surface that surrounds the control volume. 5. Phase → quantity of matter having same chemical composition or homogeneous. 6. Property → a quantity which serves to describe a substance.

Two Types of Thermodynamic Properties A.Intensive Property → a property which does not depend on the mass of the substance such as temperature, pressure, density, stress and velocity. B.Extensive Property → a property which depends on the mass of the substance such as volume, momentum and energy.

Working Substance - a substance to which heat can be extracted. Two •   Types of Working Substance:

A.Pure Substance→is a working substance whose chemical composition remains the same even if there is a change in phase. Ex. Water, ammonia, Freon-12 B. Ideal Gas→is a working substance which remain in gaseous state during its operating cycle and whose equation of state is PV= mRT. Ex. Air, .

PROPERTIES OF A WORKING SUBSTANCE

•1.Mass  

and Weight

Mass→ a property of matter that constitutes one of the fundamental physical measurements or the amount of matter a body contains. Units of mass are in , slugs, or in kg. Weight→ the force acting on a body in a gravitational field, equal to the product of its mass and the gravitational acceleration of the field. Units of weight are in N or kN.

PROPERTIES OF A WORKING SUBSTANCE

•  

2. Volume Volume → the amount of space occupied by, or contained in a body and is measured by the no. of cubes a body contains. Units of volume are in , Gallons, liters, .

PROPERTIES OF A WORKING SUBSTANCE

•3. 

Pressure

force per unit area. Units of pressure are measure in psi, kg/, kN/ or kPa. →

Absolute Pressure = Gauge Pressure + Atmospheric Pressure kPaa = kPag + 101.325 Psia = Psig + 14.7

PROPERTIES OF A WORKING SUBSTANCE

•1  Atm pressure = 0 kPag, 0 psig. = 101.325 kPa = 1.033 kg/ = 29.92 in Hg = 760 mm Hg = 14.7 psia 1 bar = 100 kPa Pressure of Perfect Vacuum = - 101.325 kPag =absolute zero pressure

PROPERTIES OF A WORKING SUBSTANCE

•4. 

Temperature

Temperature→the degree of hotness or coldness of a substance. Relations of Temperature Scales, ˚ C and ˚ F: ˚C = (˚F – 32) ˚F = + 32 Temperature at which molecules stop moving = - 273 ˚C = -460 ˚F

PROPERTIES OF A WORKING SUBSTANCE

•  

Absolute Temperatures: ˚K = ˚C + 273

˚R = ˚F +460

Temperature Temperature

Change or Difference:

∆˚C = ∆˚F

∆˚K = ∆˚C

∆˚F = ∆˚C

∆˚R = ∆˚F

PROPERTIES OF A WORKING SUBSTANCE

5. Specific Volume,  Density and Specific Weight Density, ρ= Specific Volume, ν= = = Specific Weight, = kN/ Specific Gravity of a liquid = Density of Water=1000=9.81 KN/=62.4 lb/ Specific Gravity of a Gas = Density of Air = 1.2 at 101.325 kPa and 21.1˚C

PROPERTIES OF A WORKING SUBSTANCE

6. Internal Energy, u, kJ/kg Internal Energy → heat energy due to the movement of the molecules within the substance brought about its temperature. Internal Energy is zero if temperature is constant

PROPERTIES OF A WORKING SUBSTANCE

•  

7. Flow Work, W, kJ/kg Flow Work→ work due to the change in volume. W = F x L = PA x L = Pv Where: F = Force, kN L = Distance A = Area, P = Pressure, kPa v = Specific volume, /kg

PROPERTIES OF A WORKING SUBSTANCE

8. Enthalpy, h, kJ/kg Enthalpy → the total heat and heat content of a substance which is equal to the sum of the internal energy of a body and the product of pressure and specific volume. Enthalpy = Internal Energy + Flow Work h = u = Pv

PROPERTIES OF A WORKING SUBSTANCE

•9.  Entropy, s,

Entropy→measure of randomness of the

molecules of a substance or measures the fraction of the total energy of a system that is not available for doing work. The increase in entropy is known as entropy production.

Where: = entropy production, kJ/ = energy transfer, kJ/kg = constant surrounding temperature, ˚K

WORK, HEAT AND POWER WORK → the quantity of energy transferred from one system to another. Units of work are, ft-lb, J or kJ. Work = Force x Distance, ft-lb, kN-m or kJ W = F x L = Pv

WORK, HEAT AND POWER •HEAT   →form of energy due to temperature difference. Units of heat are in Btu, cal, kcal, kJ Q = mC∆Tthen C = or Where: m = mass, kg ∆T = Temperature Change (Increase), ˚C = C = Specific heat,

WORK, HEAT AND POWER •  

SPECIFIC HEAT, C→ the heat required to change the temperature of 1 kg of a substance 1˚C. = specific heat at constant pressure, or = specific heat at constant volume, or

WORK, HEAT AND POWER

CONVERSION UNITS OF HEAT: 1Btu = 778 ft – lb 1 kcal = 4.187 kJ = 252 cal (0.252 kcal) 1N–m= 1J = 1.055 kJ 1000 J = 1 kJ

WORK, HEAT AND POWER •POWER   → time rate of doing work

Power = hp, watts or kw Conversion Units of Power: 1 hp = 550 ft-lb/sec 1MHp = 0.736 kw = 33, 000 ft-lb/min 1.014 MPh = Hp = 2545 Btu/hr 1 Boiler Hp = 33, 480 Btu/hr = 42.2 Btu/min = 35, 322 kJ/hr = 0.746 kw 1 watt= J/sec

LAWS OF THERMODYNAMICS

A.FIRST LAW OF THERMODYNAMICS

Total Energy Entering a System = Total Energy Leaving a System

FIRST LAW OF THERMODYNAMICS

•   from which: W = m(-) + m(-) + m(-) – q W = m(-) neglecting KE, PE and q Where: h = Enthalpy KE = Kinetic Energy PE = Potential Energy q = Heat loss W = Turbine work

B. SECOND LAW OF THERMODYNAMICS

Kelvin Planck statement applied to the heat engine:

“It is impossible to construct a heat engine which operates in a cycle and receives a given amount of heat from a high temperature body and does an equal amount of work.”

SECOND LAW OF THERMODYNAMICS

Clausius statement applied to the heat pump: “It is impossible to construct a heat pump that operates without an input work.”

The most efficient operating cycle is the

C.THIRD LAW OF THERMODYNAMICS

“The entropy of a substance of absolute zero temperature is zero.” D. ZEROTH’s LAW “If two bodies has the same temperature as a third body they have the same temperature with each other.”

IDEAL GASES •  

An ideal gas is a substance that has the equation of state: PV = mRT Where: P = absolute pressure, kPa V = volume, or /sec m = mass, kg or kg/sec R = gas constant, T = absolute temperature,

BASIC PROPERTIES OF AN IDEAL GAS: 1. R = •1. R = 2. R 3. = k 4. = R 5. = Where: R = gas constant =specific heat at constant volume M = molecular weight k = specific heat ratio = specific heat at constant pressure

PROPERTIES OF AIR: •  

M = 28.97 kg air/mole of air k = 1.4 R = 53.3 = 0.287 = 0.24 = 0.24 = 1.0 = 0.171 = 0.171

= 0.716

PROCESS INVOLVING IDEAL GASES •   Any Process: =

= mR

- = m(-) - = m(-) - = m In - mR In

PROCESS INVOLVING IDEAL GASES •  

Reversible Process:No friction loss Adiabatic Process:No heat loss, no heat gain, that is completely insulated system Adiabatic Throttling Process:constant enthalpy or isenthalpic process, that is, = and = Constant Pressure or isobaric

PROCESS INVOLVING IDEAL GASES •  

Constant Volume or Isovolumic Process: =

Constant Temperature or Isothermal Process: = Constant Entropy or Isentropic Process:adiabatic and reversible, = Polytropic Process:non-adiabatic process

CONSTANT PRESSURE (Charles Law) •  

= = Work Done = (- ) Heat Added = m(- ) or Q = ∆H Entropy Change = m In

CONSTANT VOLUME (Charles Law) •  

= =

Work Done = 0 Heat Added = m(-)orQ = ∆U Entropy Change = m In

CONSTANT TEMPERATURE (Boyle’s Law)

•  

= =

Work Done = In Heat Added = mRIn

or Q = W

Entropy Change = mR In

CONSTANT ENTROPY •  

= C,

=

== Work Done = Heat Added = 0 Entropy Change = 0

POLYTROPIC PROCESS •   = C,

=

= = Work Done = Heat Added = Entropy Change = In

PARTICULAR VALUES OF n: Particular Values: n=0 n=1 n=k n>1 n=∞

Process: Constant Pressure or isobaric Constant Temperature or isothermal Constant entropy or isentropic (adiabatic) Polytropic process Constant volume or isochoric or isometric

MIXTURES INVOLVING IDEAL GASES

Consider a mixture of three gases, a, b, and c, at a pressure P and a temperature T, and having a volume V.

A. MASS OR GRAVIMETRIC ANALYSIS:

= + +

•  

1= + +

B. VOLUMETRIC OR MOLAL ANALYSIS: V = Va+ +

 

1= + + Where: = volume that gas a would occupy at pressure P and temperature T = volume that gas b would occupy at pressure P and temperature T = volume that gas c would occupy at pressure P and temperature T

C. DALTON’s LAW OF PARTIAL PRESSURE

•  

P= + +

= (P) = (P) = (P) Where: P = total pressure = partial pressure of gas A, that is, the pressure that gas A will exert if it alone occupies the volume occupied by the mixture, etc. = partial pressure of gas B, that is, the pressure that gas B will exert if it alone occupies the volume occupied by the mixture, etc. = partial pressure of gas C, that is, the pressure that gas C will exert if it alone occupies the volume occupied by the mixture, etc.

D. SPECIFIC HEAT OF THE MIXTURE: •   Cp = Cpa + Cpb + Cpc Cv = Cva + Cvb + Cvc Where: = specific heat at constant pressure of the mixture. = specific heat at constant volume of the mixture.

PURE SUBSTANCE is a working substance that has a homogenous and invariable chemical composition even though there is a change PHASES OFofAphase. PURE →

SUBSTANCE

1. Solid → the state of matter that does not depend on the shape of its container. 2. Subcooled liquid → liquid whose temperature is lower than the saturation temperature at the given pressure. 3. Saturated liquid → liquid water at its boiling temperature and at standard temperature.

PHASES OF A PURE SUBSTANCE 4. Liquid-vapor Mixture → substance made up of liquid and vapor portion or a two-phase liquidvapor system. 5. Saturated Vapor → vapor or steam at standard atmospheric pressure and at its boiling temperature. 6. Superheated Vapor → vapor whose temperature is higher than the saturation temperature at the given pressure. A superheated vapor absorbs more heat than is needed to vaporize.

PHASES OF A PURE SUBSTANCE Saturation Temperature - the temperature at which vaporization takes place at a given pressure, this pressure being called the saturation pressure for the given temperature. Degrees Superheat – difference between actual temperature and saturation temperature. Compressed Liquid – liquid whose pressure is higher than the saturation pressure at the given temperature. If the temperature is held constant and the pressure is increased beyond the saturation pressure.

PHASES OF A PURE SUBSTANCE Degrees Subcooling – difference between saturation temperature and actual temperature. Critical Point – is the condition of pressure and temperature at which a liquid and its vapor are indistinguishable. At the critical point the latent heat vaporization is zero. Triple Point – is the point in which the temperature and pressure at which three phases (gas, liquid and solid) of a substance may coexist in thermodynamic equilibrium.

SATURATED LIQUID AND SATURATED VAPOR

Examples of saturation temperature at various pressures for three common pure substances:

SATURATION TEMPERATURE Pressure

Water

Ammonia

Freon-12

50 kpa

81.33˚C

-46.73˚C

-45.19˚C

101.325 kpa 500kpa

100˚C

-33.52˚C

-29.79˚C

151.86˚C

4.08˚C

15.59˚C

PROPERTIES OF SATURATED LIQUID AND SATURATED VAPOR at various temperatures and pressures are found in tables (Table 1 and Table 2 for steam) with the following typical construction:

Specific Internal Volume Energy

Enthalp Entrop y y hfhfghg

TempPress = -

= -

= -

= -

MIXTURE •   x = quality or dryness factor = ratio of mass of saturated vapor to the total mass of the mixture, expressed in decimal or percent. x= = 1 – x = wetness Where: = mass of vapor = mass of liquid = total mass

PROPERTIES OF MIXTURE •  

V= +x

h= +x

u= +x

s= +x

THE T-S DIAGRAM OF A PURE SUBSTANCE

The Mollier diagram (h-s) of steam is usually useful in determining the final enthalpy of steam after an isentropic

PROCESS INVOLVING PURE SUBSTANCES:

PROCESS:

ANOTHER TERM:

Constant Pressure ( = ) Constant Temperature (=)

Isobaric Process Isothermal Process Evaporation and Condensation occur at P = C and T = C Isometric/Isovolumic/Isochronic Process Isentropic process (reversible and adiabatic Process or no friction loss that is completely insulated system) Throttling or isenthalpic Process 1. If the final state is a mixture such as in a throttling valve: +x 2. If the initial state is a mixture such as in steam calorimeter.

Constant Volume ( = ) Constant Entropy ( = )

Constant Enthalpy ()

BASIC FORMULAS: PROCESS Constant Pressure Heating or Cooling of Liquid Evaporation or Condensation ( P = C and T = C ) Constant Volume ( V = C) Constant Entropy ( S = C) Isentropic

HEAT ADDED OR REJECTED Q = mCp () For Water:Cp = 4.187 Q = m() = m() = latent heat Q = m( - ) Q = m( -)

THE CARNOT CYCLE Schematic Diagram of a heat engine:

THE T-S DIAGRAM:

PROCESS 1 → 2: ADIABATIC REVERSIBLE EXPANSION.

This piston and cylinder are completely insulated so that no heat is gained or lost during this process. The piston continues to expand with increasing volume while doing work on the surrounding.

PROCESS 2 → 3: ISOTHERMAL COMPRESSION.

The piston compresses the substance with decreasing volume and heat is transferred at constant temperature to the low temperature reservoir.

PROCESS 3 → 4: ADIABATIC REVERSIBLE COMPRESSION.

This piston and cylinder are completely insulated in no heat is gained or lost during the process. The piston compresses the working substance and causing the temperature to rise.

PROCESS 4 → 1: ISOTHERMAL EXPANSION

During this process, the piston expands with increasing volume and heat is transferred reversibly at constant temperature from the high-temperature reservoir.

BASIC FORMULAS: - = -

and

- = -

= (-) = (-)

=

W = - = (-)= = = =

or =

where: = Carnot Cycle efficiency = = highest absolute temperature = = lowest absolute temperature

BASIC WORKING CYCLES FOR VARIOUS APPLICATIONS:

APPLICATION

BASIC WORKING CYCLE

Steam Power Plant

Rankine Cycle

Gasoline Engine (Spark-Ignition) Diesel Engine (CombustionIgnition) Gas Turbine

Otto Cycle Diesel Cycle

Brayton Cycle

Refrigeration System Refrigeration Cycle

Example No. 1 A Carnot engine receives 130 Btu of heat from a hot reservoir at 700˚F and rejects 49 Btu of heat. Calculate the temperature of the cold reservoir. A. -21.9 ˚F C. -20.8 ˚F B. -24.2 ˚F D. -22.7 ˚F Example No. 2 The maximum thermal efficiency possible for a power cycle operating between 1200 ˚F and 225 ˚F is: C. 58%C. 57.54% D. 58.73% D. 57.40%

Example No. 3 An ideal gas at 45 psig and 80˚F is heated in a closed container to 130 F. What is the final pressure? A. 54 psia C. 75 psia B. 65 psia D. 43 psia Example No. 4 A Carnot engine requires 35kJ/sec from the hot source. The engine produces 15 kw of power and the temperature of the sink is 26 ˚C. What is the temperature of the hot source in ˚C? C. 245.57 C. 250.18 D. 210.10 D. 260.68

Example No. 5 An air bubble rises from the bottom of a well where the temperature is 25˚C, to the surface where the temperature is 27˚C. Find the percent increase in the volume of the bubble if the depth of the well is 5 m. Atmospheric pressure is 101,528 Pascals. A. 49.3 C. 56.7 B. 41.3 D. 38.6

Example No. 6 Steam enters a throttling calorimeter at a pressure of 1.03 MPa. The calorimeter downstream pressure and temperature are respectively 0.100 Mpa and 125˚C. What is the percentage moisture of the supply system? Properties of steam: P, MPahf hfg hg 1.032010.7 2779.25 Note: at 0.100 MPa and 125˚C, h = 2726.6 kJ/kg A. 2.62C. 3.15 B. 5.21D. 1.98

•   Example No. 7 A water temperature rise of 18˚F in the water cooled condenser is equivalent in ˚C to: A. 7.78 ˚C C. 263.56˚K B. 10 ˚C D. -9.44 ˚C Example No. 8 Steam flows into a turbine at the rate of 10 kg/s and 10 kw of heat are lost from the turbine. Ignoring elevation and kinetic energy effects, calculate the power output from the turbine. Given:= 2739.0kJ/kg and = 2300.5kJ/kg A. 4605 kw C. 4375 kw B. 4973 kw D. 4000 kw

Example No. 9 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. Which of the following values most nearly equals the heat loss from the compressor in kw? A. -10.0 C. +10.0 B. -9.95 D. +9.95

•  

Example No. 10 An iron block weighs 5 N and has a volume of . What is the density of the block? (Apr 96) A. 988 kg/cu.m C. 2550 kg/cu.m B. 1255 kg/cu.m D. 800 kg/cu.m Example No. 11 A volume of 400 cc of air is measured at a pressure of 740 mm Hg abs and a temperature of 18˚C. What will be the volume at 760 mm Hg abs and 0˚C? C. 376 ccC. 356 cc D. 326 ccD. 366 cc