M F A M F K A K Ma F: Basic Engineering Concepts

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS

BASIC ENGINEERING CONCEPTS A.

PRINCIPLES, CONCEPTS AND DEFINITIONS Thermodynamics – is that branch of physical science that treats of 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. - is the study of heat and work and those properties of substance that bear a relation to heat and work. Working Substance – a substance to which heat can be stored and from which heat can be extracted. System of Units: Newton’s Second Law states that “the acceleration of a particular body is directly proportional to the resultant force acting on it and inversely proportional to its mass”.

aα

F , m

or

a=k

F ma , therefore F = m k

Force (F) – Resultant force acting on the body. N, lbf, kgf Weight (W) – the force acting on the body in a gravitational field, equal to the product of its mass and the gravitational acceleration of the field. N, lbf, kgf

W =

mg k

Mass (m) – mass of the body. It is the property of matter that constitutes one of the fundamental physical measurements or the amount of matter a body can contain. kgm, lbm Volume (V) – the amount of space occupied by or contained in a body and is measured by the number of cubes a body 3 3 3 contained. ft , gallons, cm , m k – Proportionality constant • System of units where k is unity but not dimensionless: 2 o CGS System: 1 dyne force accelerates 1 gram mass at 1 cm/s 2 o MKS System: 1 Newton force accelerates 1 kg mass at 1 m/s 2 o FPS System: 1 lb force accelerates 1 slug mass at ft/s

k =1

g m • cm dyne • s 2

k =1

kg m • m newtons • s 2

k =1

slug • ft lb f • s 2

• System of units where k is not unity: If the same word is used for both mass and force in a given system, k is neither unity nor dimensionless. 2 o CGS System: 1 gram force accelerates 1 gram mass at 980.66 cm/s 2 o MKS System: 1 kg force accelerates 1 kg mass at 9.8066 m/s 2 o FPS System: 1 lb force accelerates 1 lb mass at 32.174 ft/s

k = 980.66

g m • cm g f • s2

k = 9.8066

kg m • m kg f • s 2

k = 32.174

Relationship between kgf and N:

k = 9.8066

kg m • m kg f • s 2

and

∴ 9.8066 So

k =1

kg m • m newtons • s 2

kg m • m kg • m =1 m 2 2 kg f • s N •s

1kg f = 9.8066 N

Relationship between lbm and slug:

k =1

slug • ft lb f • s 2

and

∴ 32.174 So

k = 32.174

lbm • ft lb f • s 2

lbm • ft slug • ft =1 2 lb f • s lb f • s 2

1 slug = 32.174lbm

Acceleration (a) – unit of force that produces unit acceleration in a body of unit mass.

lbm • ft lb f • s 2

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS

Fluid Thermodynamics Properties: Density (ρ) – is an intensive thermodynamic property that is usually used to determine the mass of the substance or system. It is the mass (not weight) per unit volume

ρ=

m V

Specific Volume (υ) – is the volume of unit mass substance, it is also defined as the reciprocal of density.

υ=

V 1 = m ρ

Specific Weight (γ) – is the force of gravity on unit volume of the substance.

γ =

W V

∴γ =

; where

W =

mg ρg = Vk k

mg k

Specific Gravity (SG) – is the ratio of the specific weight of a certain substance to that of the specific weight of water at standard condition. It is also defined as the ratio of density of substance to that of the density of water.

SG =

γS ρ = S γ W ρW

Pressure (P) – is an intensive thermodynamic property and basically defined as the force per unit area.

P=

F W = A A

Measuring Pressure o By Using a Manometer o By Using a Pressure Gage • Gauge Pressure (Pg) – is the pressure of a substance measured by a device (pressure gage). • tmospheric Pressure (PATM) – is the force exerted by the column atmospheric air per unit area. It is also the pressure of the atmosphere at a given certain location. ƒ Standard atmospheric pressure at the surface of the earth, sea level: 1 atm = 29.92” Hg = 760 mm Hg = 101.325 kPa = 14.7 psi 2 1 atm = 760 torr = 1.0332 kg/cm = 34 ft H2O Some Pressure Conversion 1 Bar = 100 kPa = 0.10 MPa Hydrostatic Pressure

F W = A A mg (ρV )g ρAhg P= = = kA kA kA ⎛ ρg ⎞ P=⎜ ⎟h = γh ⎝ k ⎠ P=

• Absolute Pressure (Pabs) – is the algebraic sum of the barometric pressure or atmospheric pressure and the gauge pressure.

Pabs = Patm ± Pg

( - ) Æ If below atmospheric pressure (Vacuum)

( + ) Æ If above atmospheric pressure Temperature (t) – it is the degree of hotness or coldness of a body. O O Temperature Relation between degrees Fahrenheit ( F) and degrees Centigrade ( C)

°C =

5 (°F − 32) = (°F − 32) 9 1.8

9 ° F = °C + 32 = 1.8°C + 32 5

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS Conversion of Temperature Difference

Δ°C =

5 Δ° F Δ° F = 9 1.8

Δ° F =

9 Δ°C = 1.8Δ°C 5

Absolute Temperature (T) – it is measured from absolute zero temperature.

° K = °C + 273

° R = ° F + 460

Absolute Zero Temperature – is the temperature at which all molecular motion ceases. Energy Concepts: o Potential Energy (PE) – energy due to its position or elevation measured with respect to a certain datum line.

PE = Wz =

mg z k

ΔPE = P2 − P1 =

mg (z 2 − z1 ) k

o Kinetic Energy (KE) – energy or stored capacity for performing work possessed by a moving body by virtue of its momentum.

KE = Ws =

m 2 V 2k

ΔKE =

(

m 2 V2 − V12 2k

)

o Internal Energy (U, u) – is the energy of a body or a substance that is the sum of the energies of all its molecules; it is also the sum of the various forms of energy that a molecule has. o Work (W) – It is defined as the energy in transmission. It is the product of the displacement of the body and the component of the force in the direction of the displacement. Non-Flow Work (WNF) It is a work done to or by the system by a non-flow closed system during a reversible process without flowing of a working substance or through a moving boundary. The work done which is represented by the area under the curve. If the process is expansion work is positive. If the process is compression it is negative.

dW = pdV 2

W NF = ∫ pdV 1

Flow Work (WF) Work or flow energy is work done in pushing a fluid across a boundary, usually into or out of a system.

WF = FL = pAL WF = pV

ΔWF = WF 2 − WF 1 ΔW = p 2V2 − p1V1 o Heat (Q) – it is defined as energy in motion, it moves from a higher temperature body into a lower temperature body. Q is positive when heat is added to the body or system and it is negative if heat is extracted. O o Specific Heat (C) – the heat required to change the temperature of 1 kg of a substance 1 C. o Enthalpy (H, h) – it is the composite property applicable to all fluids and is defined by:

h = u + pv

B.

and

H = mh = U + pV

Classification of Systems: i. Closed System – is one in which mass does not cross over. ii. Open System – is one in which mass crosses its boundaries. LAWS OF THERMODYNAMICS a. FIRST LAW OF THERMODYNAMICS – Deal with the law of Conservation of Energy. It states that “Energy can neither be created nor destroyed, it just transforms into another forms”. LAW OF CONSERVATION OF ENERGY

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS ENERGY ENTERING THE SYSTEM = ENERGY LEAVING THE SYSTEM

PE1 + KE1 + WF 1 + U 1 + Q = PE 2 + KE 2 + WF 2 + U 2 + W Q = ΔPE + ΔKE + ΔWF + ΔU + W H = U + WF = U + pV But:

h = u + pV ∴ Q = ΔPE + ΔKE + ΔH + W 1 Q = m( z 2 − z1 ) + m V22 − V12 + m(H 2 − H 1 ) + W 2

(

)

SECOND LAW OF THERMODYNAMICS – deals with the direction of flow of heat energy that is from higher temperature body to lower temperature body, and the property known as “entropy”. i. Kelvin-Planck Statement – “No cyclic process is possible whose sole result is the flow of heat from a single heat reservoir and the performance of an equivalent amount of work” ii. Clausius Statement – “it is impossible to construct a cyclic device that will cause heat to be transformed from a low-temperature reservoir to a high-temperature reservoir without the input of work. iii. Concept of Entropy – it is the measure of the microscopic ⎛ dQ ⎞ disorder of molecules of a substance. It remains constant ΔS = S 2 − S1 = ⎜ ⎟ in an adiabatic reversible process. It is given by the ⎝ T ⎠ rev following relation: c. THIRD LAW OF THERMODYNAMICS – deals with the restriction of all physical systems to the temperature regime O O that excludes absolute Zero. It states “At absolute zero, the entropy of a pure substance (in equilibrium at 0 K or R) is some ‘perfect’ crystalline form becomes zero” ZEROTH LAW – law concerning thermal equilibrium and is the basis for temperature measurement. It states that “when two bodies, isolated from other environment, and are in thermal equilibrium with a third body, the two are in thermal equilibrium with each other”. b.

∫

C.

IDEAL GASES - it is a substance that has the equation of state. Boyle’s Law - If the temperature of a given quantity of gas is held constant, the volume of the gas varies inversely with the absolute pressure during a change of state. 1 C

Vα

p

or

V =

p

pV = C

p1V1 = p 2V2

or

Charles Law - If the pressure on a particular quantity of gas is held constant, then, with any change of state, the volume will vary directly as the absolute temperature. V V V

Vα T

or

V = CT

T

=C

or

1

T1

=

2

T2

If the volume of a particular quantity of gas is held constant, then, with any change of state, the pressure will vary p p1 p2 directly as the absolute temperature.

pαT

or

p = CT

T

=C

or

T1

=

T2

Equation of State or Characteristic Equation of a Perfect Gas Combining Boyle’s and Charles’ Laws:

p1V1 p 2V2 = = C , a constant T1 T2

pV = mR T

∴ pV = mRT

Basic Properties of an Ideal Gas:

kg air kJ ft − lb ; ; MWair = 28.97 kg − ° K lb − ° R mole of air kJ ft − lb R = 8.3143 ; R = 1545 kg Mole • ° K lb Mole • ° R R=

R ; MW

Specific Heat:

C=

heat ( EnergyUnits ) (mass)(ChangeOfTemperature)

In differential quantities,

C=

dQ ; dQ = mcdT mdT

Constant Volume Specific Heat: (it is a change of the molecular internal energy for a unit mass or one mole of a substance per degree change of temperature with the volume remains constant.)

du dU dQ = = dT mdT mdT dQ = dU = mcV dT ⎯ ⎯→ Integrate

cV =

QV = ΔU = mcV ΔT

Btu lb − ° F kcal cV = 0.171 kg − °C kJ cV = 0.716 kg − °C cV = 0.171

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS

Constant Pressure Specific Heat: (is the change of enthalpy for a unit mass or one mole substance per degree change of temperature between two states without changing the pressure.)

dh dH dH dQ ; dh = ;∴ c p = = dT m mdT dT dH = dQ = dU + dW

cp =

Btu lb − ° F kcal c P = 0.24 kg − °C kJ cP = 1 kg − °C c P = 0.24

⎯→ Integrate dH = dU + pdV = mc p dT ⎯ QP = ΔH = mc p ΔT QP = ΔU + pΔV = H 2 − H 1 Ratio of specific Heat: (Relationship of cV and cP)

k=

cp cv

=

ΔH ΔU

c P = cV + R

kR k −1 R cV = k −1 cP =

PROCESSES OF IDEAL GASES 9 Constant Volume (Isometric Process) – an isometric process is a reversible constant volume process. A constant volume process may be reversible or irreversible.

Relation between p and T:

Non-flow Work

T2 p 2 = T1 p1

W NF = ∫ pdV = 0

Change of Internal Energy

Heat Transfer

ΔU = mcV (T2 − T1 )

Q = mcV (T2 − T1 ); ∴ Q = ΔU

Change of Enthalpy

Change of Entropy

ΔH = mc p (T2 − T1 )

ΔS = mcV ln

2

1

T2 dQ dQ ; dS = ; cV = T1 T mdT

Reversible Steady Flow V=C

Q = ΔPE + ΔKE + ΔU + ΔWF + W Where : Q = ΔU ; ΔPE = 0; ΔKE = 0 WSF = − ΔWNF = V ( p1 − p2 ) Irreversible Non-flow V=C

2

− ∫ Vdp = WSF + ΔKE 1

V ( p1 − p2 ) = WSF + ΔKE ; ΔKE = 0 WSF = V ( p1 − p2 )

Q = ΔU + W NF 9 Constant Pressure (Isobaric Process) – an isobaric process is an internally reversible process of a substance during which the pressure remains constant.

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS Relation between V and T:

Non-flow Work

T2 V2 = T1 V1

W NF = ∫ pdV = p (V2 − V1 )

Change of Internal Energy

Heat Transfer

ΔU = mcV (T2 − T1 )

Q = mc p (T2 − T1 ); ∴ Q = ΔH

Change of Enthalpy

Change of Entropy

ΔH = mc p (T2 − T1 )

ΔS = mc p ln

2

1

T2 dQ ; dS = T1 T

Steady Flow P=C 2

Q = ΔPE + ΔKE + ΔH + W

− ∫ Vdp = WSF + ΔKE

Where : Q = ΔH ; ΔPE = 0

0 = WSF + ΔKE

WSF = −ΔKE

WSF = − ΔKE

1

9 Constant Temperature (Isothermal Process) – an isothermal process is an internally reversible constant temperature process of a substance.

Relation between V and P:

p1 V2 = p 2 V1 Change of Internal Energy

Non-flow Work 2

W NF = ∫ pdV = p1V1 ln 1

V2 V1

Heat Transfer

ΔU = 0

Q = ΔU + W NF ; ∴ Q = W NF

Change of Enthalpy

Change of Entropy

ΔH = 0

ΔS = mR ln

p1 Q ; ΔS = p2 T

Steady Flow P=C 2

Q = ΔPE + ΔKE + ΔH + W Where : ΔPE = 0; ΔKE = 0 WSF = Q

− ∫ Vdp = WSF + ΔKE ; 1

pdV + Vdp = 0

2

− ∫ Vdp = WSF + ΔKE ; ΔKE = 0 1

WSF = p1V1 ln

V2 V1

9 Constant Entropy (Isentropic Process) – an isentropic process is a reversible adiabatic process. Adiabatic simply means that there is no heat. A reversible adiabatic is one of constant entropy.

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS Relation between p, V and T:

Non-flow Work

p1V1 = p 2V2 = C k

T2 ⎛ V1 =⎜ T 1 ⎜⎝ V2

k

⎞ ⎟⎟ ⎠

k −1

⎛p ⎞ = ⎜⎜ 2 ⎟⎟ ⎝ p1 ⎠

2

k −1 k

W NF = ∫ pdV 1

W NF =

p 2V2 − p1V1 mR(T2 − T1 ) = 1− k 1− k

Change of Internal Energy

Heat Transfer

ΔU = mcV (T2 − T1 )

Q=0

Change of Enthalpy

Change of Entropy

ΔH = mc P (T2 − T1 )

ΔS = 0

Steady Flow S=C

1

2

− ∫ Vdp = WSF + ΔKE ; C = p k V

Q = ΔPE + ΔKE + ΔH + W

1

Where : ΔPE = 0; ΔKE = 0

2

− ∫ Vdp = WSF + ΔKE

WSF = ΔH

1

2

− ∫ Vdp = 1

2 k ( p2V2 − p1V1 ) = k ∫ pdV 1 1− k

9 Polytropic Process – a polytropic process is an internally reversible process during which:

pV n = C ; and

p1V1 = p 2V2 = piVi n

n

p1V1 = p 2V2 = C T2 ⎛ V1 =⎜ T 1 ⎜⎝ V2

n

⎞ ⎟⎟ ⎠

n −1

⎛p ⎞ = ⎜⎜ 2 ⎟⎟ ⎝ p1 ⎠

where n is any constant.

Non-flow Work

Relation between p, V and T: n

n

2

n −1 n

W NF = ∫ pdV 1

W NF =

p 2V2 − p1V1 mR(T2 − T1 ) = 1− n 1− n

Change of Internal Energy

Heat Transfer

ΔU = mcV (T2 − T1 )

Q = mc n (T2 − T1 ); c n =

Change of Enthalpy

Change of Entropy

ΔH = mc P (T2 − T1 )

ΔS = mc n ln

k −n 1− n

T2 T1

Steady Flow S=C

Q = ΔPE + ΔKE + ΔH + W Where : ΔPE = 0; ΔKE = 0 WSF = Q − ΔH

2

− ∫ Vdp = WSF + ΔKE 1

2

− ∫ Vdp = WSF + ΔKE 1

2

− ∫ Vdp = 1

2 n( p2V2 − p1V1 ) = n ∫ pdV 1 1− n

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS

CURVES FOR DIFFERENT VALUES OF n Polytropic processes are all inclusive in that many of the prior equations can be obtained by choosing proper values of n.

ISOMETRIC V=C

ISOBARIC P=C

P, V, T relation

T2 p = 2 T1 p1

T2 V2 = T1 V1

0

p (V2 − V1 )

p1V1 ln

V2 V1

p 2V2 − p1V1 1− k

p 2V2 − p1V1 1− n

V ( p 2 − p1 )

0

p1V1 ln

V2 V1

k ( p 2V2 − p1V1 ) 1− k

n( p 2V2 − p1V1 ) 1− n

∫

2

1

pdV

ISOTHERMAL T=C

POLYTROPIC n pV =C

PROCESS

p1V1 = p 2V2

W NF 2

− ∫ Vdp 1

ΔU

Q

m ∫ cV dT

mcV (T2 − T1 ) mcV (T2 − T1 )

m ∫ cV dT

∞

c (specific heat)

ΔS

m ∫ c P dT

mcV (T2 − T1 ) mc P (T2 − T1 )

n

ΔH

m ∫ cV dT

m ∫ c P dT

mc P (T2 − T1 )

cV dT T T mcV ln 2 T1 m∫

mcV (T2 − T1 )

⎞ ⎟⎟ ⎠

k −1

T2 ⎛ p 2 ⎞ =⎜ ⎟ T1 ⎜⎝ p1 ⎟⎠

n −1 n

⎛V = ⎜⎜ 1 ⎝ V2

m ∫ cV dT

mcV (T2 − T1 ) m ∫ c n dT

0

1

k

-∞ to +∞

0

⎛k −n⎞ c n = cV ⎜ ⎟ ⎝ 1− n ⎠ k =C

∞

m ∫ c P dT

0

mc P (T2 − T1 )

Mixture Involving Ideal Gas

⎛V = ⎜⎜ 1 ⎝ V2

V p1V1 ln 2 V1

m ∫ c P dT

cPdT T T mc P ln 2 T1

k −1 k

m ∫ TdS

cP

m∫

T2 ⎛ p 2 ⎞ =⎜ ⎟ T1 ⎜⎝ p1 ⎟⎠

m ∫ cV dT

0

0

cV

ISENTROPIC k pV =C

mc P (T2 − T1 )

Q T mR ln

mc n (T2 − T1 )

m ∫ c P dT

mc P (T2 − T1 ) c n dT T T mc n ln 2 T1 m∫

V2 V1

0

⎞ ⎟⎟ ⎠

n −1

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS a. Mass or Gravimetric Analysis:

ma mb mc + + mT mT mT Volumetric or Molal Analysis: V V V 1= a + b + c V = Va + Vb + Vc V V V Dalton’s Law of Partial Pressure: V V V P = Pa + Pb + Pc Pa = a P; Pb = b P; Pc = c P V V V Specific Heat of Mixture ma mb mc ma mb m cP = cPa + cPb + cPc cV = cVa + cVb + c cVc mT mT mT mT mT mT mT = ma + mb + mc

b.

c. d.

D.

1=

GAS CYCLES Heat Engine or thermal engine is a closed system that exchanges only heat and work with its surrounding and that operates in cycles. Elements of a thermodynamic heat engine with a fluid as the working substance: 1. A working substance, matter that receives heat, rejects heat and does work. 2. A source of heat (Hot body, heat reservoir or just called source), from which the working substance receives heat. 3. A heat sink (receiver, cold body or just called sink), to which the working substance can reject heat. 4. An engine, where in the working substance may do work or have work done on it. BASIC WORKING CYCLE FOR VARIOUS APPLICATION Steam Power Plant Rankine Cycle Gasoline Engine (Spark-Ignition) Otto Cycle Diesel Engine (Compression) Diesel Cycle Gas Turbine Bryton Cycle Refrigeration System Refrigeration Cycle A thermodynamic cycle occurs when the working fluid of a system experiences a number of processes that eventually return the fluid to its initial state. CARNOT CYCLE - it is the most efficient thermodynamic cycle/engine that operates between two temperature levels. Components of Carnot Heat Engine

pV and TS Diagram of Carnot Cycle

Cycle Processes Process 1 – 2:

Isothermal Heat Addition process

Process 2 – 3:

Adiabatic or Isentropic Turbine Expansion

process Process 3 – 4: Isothermal Heat Rejection process Process 4 – 1: Adiabatic or Isentropic Compression process Analysis of Carnot Cycle Processes: 1. Heat Added, Consider Heater (Process 1 – 2) a. From TS Diagram: 2

2

Q A = ∫ TdS =T ∫ dS =TΔS1− 2 = TH ( S 2 − S1 ) 1

b.

1

From pV Diagram: 2

Q A = − ∫ Vdp = − p1V1 ln 1

2.

p2 p V = mRTH ln 1 = mRTH ln 2 p1 p2 V1

Heat Rejected, Consider Cooler (Process 3 – 4) a. From TS Diagram:

Prepared By: Engr. Jessie Ledesma

INDUSTRIAL & POWER PLANT ENGINEERING MODULE 1: BASIC ENGINEERING CONCEPTS-THERMODYNAMICS 4

4

3

3

QR = ∫ TdS =T ∫ dS =TΔS 3− 4 = TL ( S 4 − S 3 ) b. 3.

From pV Diagram: 4

QR = − ∫ Vdp = − p3V3 ln 3

V p4 = −mRTL ln 3 p3 V4

Net Work of the system/cycle st The 1 law states that, “When the system is undergoing a cyclic change, the net heat added is equal to the net work produce” a. From TS Diagram:

Wnet = Qnet = ∑ Q = Q A + QR = TH ( S 2 − S1 ) + TL ( S 4 − S 3 ) where : S1 = S 4 ; S 2 = S 3 so Wnet = Qnet = TH ( S 2 − S1 ) + TL ( S1 − S 2 )

∴ Wnet = Qnet = (TH − TL )( S 2 − S1 ) b.

From pV Diagram:

Wnet = Qnet = ∑ Q = QA + QR = mRTH ln

V2 V − mRTL ln 3 V1 V4

⎡ V V ⎤ Wnet = Qnet = mR ⎢TH ln 2 − TL ln 3 ⎥ V1 V4 ⎦ ⎣ ⎛V ⎞ where : ⎜⎜ 3 ⎟⎟ ⎝ V2 ⎠

k −1

⎛V ⎞ T = H = ⎜⎜ 4 ⎟⎟ TL ⎝ V1 ⎠ V V ∴ 2 = 3 V1 V4

∴Wnet = Qnet = mR ln 4.

5.

k −1

; consider processes 2 − 3 & 4 − 1

V2 [TH − TL ] V1

Cycle Thermal efficiency

V2 [TH − TL ] T − TL V1 x100% x100% = H V2 T H mRTH ln V1

mR ln

eth =

Wnet x100% = QA

eth =

Wnet (T − TL )( S 2 − S1 ) T − TL x100% = H x100% = H x100% QA TH ( S 2 − S1 ) TH

Mean Effective Pressure of the Cycle: (average constant pressure acting on the piston on one stroke that will do net work of a single cycle)

PM =

W Work per Cycle ∫ pdV = net = Volume Displacement VD VD where : VD = Vmax − Vmin = V3 − V1 mR(TH − TL ) ln

PM = 6.

Expansion Ratio =

V2 V1

V3 − V1

Ratio of Expansion, Ratio of compression

Volume at the end of exp ansion Volume at the beginning of exp ansion

Isothermal exp ansion ratio = Compressio n Ratio =

V2 V1

Isentropic exp ansion ratio =

V3 V2

Overall exp ansion ratio =

V3 V1

Volume at the beginning of compressio n Volume at the end of compressio n

Isothermal compressio n ratio =

V3 V4

Isentropic compressio n ratio , rk =

V4 V1

Overall compressio n ratio =

V3 V1