Thermodynamics 1 Instructor: Engr.
John Harvey Bequio, ME, RMP
Entropy (S) – a change of which measures the degree of heat transferred from one equilibrium state to
another. Often interpreted as the degree of disorder or randomness in the system.
Entropy (S) - unit : kJ/ K
Specific Entropy (s) – unit : kJ/kg. K
ΔS = m Cv ln (T2/T1) + m R ln (V2/V1)
ΔS = m Cp ln (T2/T1) - m R ln (P2/P1)
ΔS = m Cp ln (V2/V1) + m Cv ln (P2/P1)
Where : R = gas constant (KJ/kgm.K)
Cv = specific heat at constant volume in KJ/kg.K
Cp = specific heat at constant pressure in KJ/kg.K
Thermodynamics 1
4.0 Processes of Ideal Gas
Isometric Process is a Constant Volume Process. V=C
Relationship between pressure and temperature
P1 = P2
T1 T2
Work Non-Flow (WNF); Unit KJ
Since : V=C ; Therefore WNF = 0
Thermodynamics 1
4.0 Processes of Ideal Gas
Isometric Process is a Constant Volume Process. V=C
Change in Internal Energy (ΔU); unit kJ
ΔU = m Cv ΔT
Change in Enthalpy (ΔH); unit kJ
ΔH = m Cp ΔT
Heat (Q); unit kJ
Q = m Cv ΔT ; Q = ΔU
Thermodynamics 1
4.0 Processes of Ideal Gas
Isometric Process is a Constant Volume Process. V=C
Change in Entropy (ΔS); unit kJ/K
ΔS = m Cv ln (T2/T1)
Work steady flow (WSF); unit kJ
WSF = - V (P2 – P1)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
A perfect gas has a value of R = 58.8 ft-lbf/lbm-°R and k = 1.26. If 20 BTU are added to 5 lbm of this
gas at constant volume when the initial temperature is 90°F. Find the following:
a) Final temperature in °F
b) Change in enthalpy in BTU
c) Change in entropy in BTU/°R
d) Change in internal energy in BTU
e) Work non-flow in BTU
Given: Constant Volume (V=C)
R = 58.8 ft-lbf/lbm-°R m = 5 lbm
k = 1.26 T1= 90°F + 460 = 550 °R
Q = 20 BTU
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: a) Final temperature in °F
Given: Constant Volume (V=C)
R = 58.8 ft-lbf/lbm-°R m = 5 lbm
k = 1.26 T1= 90°F + 460 = 550 °R
Q = 20 BTU
Q = m Cv ΔT = m Cv (T2-T1)
Cv = R / k−1 = 58.8 ft-lbf/lbm-°R / (1.26-1) = 226.1538 ft-lbf/lbm-°R
Cp = Cv + R =226.1538 ft-lbf/lbm-°R + 58.8 ft-lbf/lbm-°R = 284.9538 ft-lbf/lbm-°R
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: a) Final temperature in °F
Given: Constant Volume (V=C)
R = 58.8 ft-lbf/lbm-°R m = 5 lbm
k = 1.26 T1= 90°F + 460 = 550 °R
Q = 20 BTU
Cv = 226.1538 ft-lbf/lbm-°R
Cp = 284.9538 ft-lbf/lbm-°R
Q = m Cv (T2-T1)
20 BTU x (778 ft-lbf / 1 BTU) = (5 lbm) (226.1538 ft-lbf/lbm-°R ) (T2 – 550 °R)
T2 = 563.76 °R ; °R = °F + 460
T2 = 103.76 °F (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: b) Change in enthalpy in BTU
Given: Constant Volume (V=C)
R = 58.8 ft-lbf/lbm-°R m = 5 lbm
k = 1.26 T1= 90°F + 460 = 550 °R
Q = 20 BTU T2 = 563.76 °R
Cv = 226.1538 ft-lbf/lbm-°R
Cp = 284.9538 ft-lbf/lbm-°R
ΔH = m Cp ΔT = m Cp (T2-T1)
ΔH = (5 lbm) (284.9538 ft-lbf/lbm-°R ) (563.76 °R – 550 °R)
ΔH = 19 604.82 ft-lbf x (1 BTU/778 ft-lbf)
ΔH = 25.2 BTU (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: c) Change in entropy in BTU/°R
Given: Constant Volume (V=C)
m = 5 lbm
T1= 90°F + 460 = 550 °R
T2 = 563.76 °R
Cv = 226.1538 ft-lbf/lbm-°R
Cp = 284.9538 ft-lbf/lbm-°R
ΔS = m Cv ln (T2/T1) = (5 lbm) (226.1538 ft-lbf/lbm-°R ) ln (563.76 °R/ 550 °R)
ΔS = 27.9417 ft-lbf / °R x (1 BTU/778 ft-lbf)
ΔS = 0.0359 BTU/ °R (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: d) Change in internal energy in BTU
Given: Constant Volume (V=C)
m = 5 lbm
T1= 90°F + 460 = 550 °R
T2 = 563.76 °R
Cv = 226.1538 ft-lbf/lbm-°R
Cp = 284.9538 ft-lbf/lbm-°R
ΔU = m Cv (T2-T1)
ΔU = (5 lbm) (226.1538 ft-lbf/lbm-°R ) (563.76 °R - 550 °R)
ΔU = 15 559.38 ft-lbf x (1 BTU/778 ft-lbf)
ΔU = 20 BTU (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: e) Work non-flow in BTU
Given: Constant Volume (V=C)
m = 5 lbm
T1= 90°F + 460 = 550 °R
T2 = 563.76 °R
Cv = 226.1538 ft-lbf/lbm-°R
Cp = 284.9538 ft-lbf/lbm-°R
; since V=C, Therefore WNF = 0
WNF = 0 BTU (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Isobaric Process is a Constant Pressure Process. P=C
Relationship between volume and temperature
V1 = V2
T1 T2
Work Non-Flow (WNF); Unit KJ
WNF = P (V2 –V1) also WNF = mR (T2-T1)
Thermodynamics 1
4.0 Processes of Ideal Gas
Isobaric Process is a Constant Pressure Process. P=C
Change in Internal Energy (ΔU); unit kJ
ΔU = m Cv ΔT
Change in Enthalpy (ΔH); unit kJ
ΔH = m Cp ΔT
Heat (Q); unit kJ
Q = m Cp ΔT ; Q = ΔH
Thermodynamics 1
4.0 Processes of Ideal Gas
Isobaric Process is a Constant Pressure Process. P=C
Change in Entropy (ΔS); unit kJ/K
ΔS = m Cp ln (T2/T1)
Work steady flow (WSF); unit kJ
WSF = - V (P2 – P1)
Since P=C; Therefore WSF = 0
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
3 lbm of a perfect gas with R = 38 ft-lbf/lbm-°R and k = 1.667 has 300 BTU of heat added during a
constant pressure process. The initial temperature is 100°F. Determine:
a) Final temperature in °F
b) Change in enthalpy in BTU
c) Work steady flow in BTU
d) Change in internal energy in BTU
e) Change in entropy in BTU/°R
Given: Constant Pressure (P=C)
R = 38 ft-lbf/lbm-°R m = 3 lbm Q = 300 BTU
k = 1.667 T1= 100°F + 460 = 560 °R
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: a) Final temperature in °F
Given: Constant Pressure (P=C)
R = 38 ft-lbf/lbm-°R m = 3 lbm Q = 300 BTU
k = 1.667 T1= 100°F + 460 = 560 °R
Q = m Cp ΔT = m Cp (T2-T1)
Cv = R / k−1 = 38 ft-lbf/lbm-°R / (1.667-1) = 56.9715 ft-lbf/lbm-°R
Cp = Cv + R =56.9715 ft-lbf/lbm-°R + 38 ft-lbf/lbm-°R = 94.9715 ft-lbf/lbm-°R
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: a) Final temperature in °F
Given: Constant Pressure (P=C)
R = 38 ft-lbf/lbm-°R m = 3 lbm Q = 300 BTU
k = 1.667 T1= 100°F + 460 = 560 °R
Cv = 56.9715 ft-lbf/lbm-°R
Cp = 94.9715 ft-lbf/lbm-°R
Q = m Cp (T2-T1)
300 BTU x (778 ft-lbf / 1 BTU) = (3 lbm) (94.9715 ft-lbf/lbm-°R ) (T2 – 560 °R)
T2 = 1379.193 °R ; °R = °F + 460
T2 = 919.193 °F (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: b) Change in enthalpy in BTU
Given: Constant Pressure (P=C)
m = 3 lbm
T1= 560 °R T2 = 1379.193 °R
Cv = 56.9715 ft-lbf/lbm-°R Cp = 94.9715 ft-lbf/lbm-°R
ΔH = m Cp ΔT = m Cp (T2-T1)
ΔH = (3 lbm) (94.9715 ft-lbf/lbm-°R) (1379.193 °R – 560 °R)
ΔH = 233 399.964 ft-lbf x (1 BTU/778 ft-lbf)
ΔH = 300 BTU (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: c) Work steady flow in BTU
Given: Constant Pressure (P=C)
m = 3 lbm
T1= 560 °R T2 = 1379.193 °R
Cv = 56.9715 ft-lbf/lbm-°R Cp = 94.9715 ft-lbf/lbm-°R
Since: P=C
Wsf = 0 BTU (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: d) Change in internal energy in BTU
Given: Constant Pressure (P=C)
m = 3 lbm
T1= 560 °R T2 = 1379.193 °R
Cv = 56.9715 ft-lbf/lbm-°R Cp = 94.9715 ft-lbf/lbm-°R
ΔU = m Cv (T2-T1)
ΔU = (3 lbm) (56.9715 ft-lbf/lbm-°R ) (1379.193 °R - 560 °R)
ΔU = 140 011.962 ft-lbf x (1 BTU/778 ft-lbf)
ΔU = 179.964 BTU (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Solution: e) Change in entropy in BTU/°R
Given: Constant Pressure (P=C)
m = 3 lbm
T1= 560 °R T2 = 1379.193 °R
Cv = 56.9715 ft-lbf/lbm-°R Cp = 94.9715 ft-lbf/lbm-°R
ΔS = m Cp ln (T2/T1) = (3 lbm) (94.9715 ft-lbf/lbm-°R) ln (1379.193 °R / 560 °R)
ΔS = 256.798 ft-lbf / °R x (1 BTU/778 ft-lbf)
ΔS = 0.33 BTU/ °R (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Isothermal Process is a Constant Temperature Process. T=C
Relationship between volume and pressure
P1 V1 = P2 V2
Work Non-Flow (WNF); Unit KJ
WNF = P1 V1 ln (V2 / V1)
Thermodynamics 1
4.0 Processes of Ideal Gas
Isothermal Process is a Constant Temperature Process. T=C
Change in Internal Energy (ΔU); unit kJ
ΔU = m Cv ΔT ; Since T=C ; ΔU =0
Change in Enthalpy (ΔH); unit kJ
ΔH = m Cp ΔT ; Since T=C ; ΔH =0
Heat (Q); unit kJ
Q = WNF
Q = P1 V1 ln (V2 / V1)
Thermodynamics 1
4.0 Processes of Ideal Gas
Isothermal Process is a Constant Temperature Process. T=C
Change in Entropy (ΔS); unit kJ/K
ΔS = m R ln (P1/P2)
Work steady flow (WSF); unit kJ
WSF = WNF
WSF= P1 V1 ln (V2 / V1)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Four pounds of air gain 0.491 BTU/°R of entropy during a non-flow isothermal process. If initial
pressure is 120 psia and the final volume is 42.5 ft^3, find the following:
a) Final pressure in Psia
b) Initial volume in ft^3
c) Work non-flow in BTU
d) Heat in BTU
Given: Constant Temperature (T=C)
m = 4 lbm, ΔS = 0.491 BTU/°R, P1=120 Psia, V2= 42.5 ft^3
Constant: (Air) Cpair=0.24 BTU/lbm⋅∘R, Cvair=0.1714BTU/lbm⋅∘R , Rair=53.34 lbf⋅ft/lbm⋅∘R
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Constant Temperature (T=C)
m = 4 lbm, ΔS = 0.491 BTU/°R, P1=120 Psia, V2= 42.5 ft^3
Constant: (Air) Cpair=0.24 BTU/lbm⋅∘R, Cvair=0.1714BTU/lbm⋅∘R , Rair=53.34 lbf⋅ft/lbm⋅∘R
a) Final pressure in Psia
Using:
ΔS = m R ln (P1/P2)
0.491 BTU/°R = (4 lbm) (53.34 lbf⋅ft/lbm⋅∘R x (1 BTU/778 lbf-ft)) ln (120 Psia/P2)
P2 = 20.027 Psia (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Constant Temperature (T=C)
m = 4 lbm, ΔS = 0.491 BTU/°R, P1=120 Psia, V2= 42.5 ft^3, P2 = 20.027 Psia
Constant: (Air) Cpair=0.24 BTU/lbm⋅∘R, Cvair=0.1714BTU/lbm⋅∘R , Rair=53.34 lbf⋅ft/lbm⋅∘R
b) Initial volume in ft^3
P1 V1 = P2 V2
(120 Psia) (V1) = (20.027 Psia) (42.5 ft^3)
V1 = 7.093 ft^3 (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Constant Temperature (T=C)
m = 4 lbm, ΔS = 0.491 BTU/°R, P1=120 Psia, V2= 42.5 ft^3, P2 = 20.027 Psia
Constant: (Air) Cpair=0.24 BTU/lbm⋅∘R, Cvair=0.1714BTU/lbm⋅∘R , Rair=53.34 lbf⋅ft/lbm⋅∘R
b) Initial volume in ft^3
P1 V1 = P2 V2
(120 Psia) (V1) = (20.027 Psia) (42.5 ft^3)
V1 = 7.093 ft^3 (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Constant Temperature (T=C)
m = 4 lbm, ΔS = 0.491 BTU/°R, P1=120 Psia, V2= 42.5 ft^3, P2 = 20.027 Psia, V1 = 7.093 ft^3
Constant: (Air) Cpair=0.24 BTU/lbm⋅∘R, Cvair=0.1714BTU/lbm⋅∘R , Rair=53.34 lbf⋅ft/lbm⋅∘R
c) Work non-flow in BTU
WNF = P1 V1 ln (V2 / V1)
WNF = [ (120 lbf/in^2) x (144 in^2 / 1 ft^2)] (7.093 ft^3 ) ln (42.5 ft^3/ 7.093 ft^3 )
WNF = 219 443.5 ft-lbf x (1 BTU / 778 ft-lbf) = 282.06 BTU (Answer)
d) Heat in BTU; Since Q= WNF
Q= 282.06 BTU (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Isentropic Process is a Constant Entropy Process. Also known as reversible adiabatic process. S=C
Relationship between volume, pressure and temperature.
P1 V1 = P2 V2
T1 T2
P1 V1 K = P2 V2 K Where: k –specific heat ratio of gas
T2 = (V1 / V2 ) K -1
T1
T2 = (P2 / P1 ) (K -1/K)
T1
Thermodynamics 1
4.0 Processes of Ideal Gas
Isentropic Process is a Constant Entropy Process. Also known as reversible adiabatic process. S=C
Work Non-Flow (WNF); Unit KJ
WNF = P2 V2 – P1 V1
1–k
Change in Internal Energy (ΔU); unit kJ
ΔU = m Cv ΔT
Thermodynamics 1
4.0 Processes of Ideal Gas
Isentropic Process is a Constant Entropy Process. Also known as reversible adiabatic process. S=C
Change in Enthalpy (ΔH); unit kJ
ΔH = m Cp ΔT
Heat (Q); unit kJ and Change in Entropy (ΔS); unit kJ/K
Q = 0 and ΔS = 0
Work steady flow (WSF); unit kJ
WSF = - ΔH
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Helium (k = 1.666) is compressed under reversible adiabatic process from 745 kPaa, 5.1 m3/min, 5°C
to 1765 kPaa. Find the following:
a) Final volume flow rate in m3/min.
b) Final temperature in °C.
c) Work non-flow in kJ/min.
Given: Constant Entropy (S=C)
k = 1.666, P1 = 745 kPaa, P2 = 1765 kPaa and V1 =5.1 m3/min
T1 = 5°C +273 = 278 °K
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Constant Entropy (S=C)
k = 1.666, P1 = 745 kPaa, P2 = 1765 kPaa and V1 =5.1 m3/min
T1 = 5°C +273 = 278 °K
Solution: a) Final volume flow rate in m3/min
P1 V1 K = P2 V2 K
(745 kPaa) (5.1 m3/min) (1.666) = (1765 kPaa) (V2) (1.666)
V2 =3.039 m3/min (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Constant Entropy (S=C)
k = 1.666, P1 = 745 kPaa, P2 = 1765 kPaa and V1 =5.1 m3/min
T1 = 5°C +273 = 278 °K, V2 =3.039 m3/min
Solution: b) Final temperature in °C.
T2 = (V1 / V2 ) K -1
T1
T2 = (5.1 m3/min/ 3.039 m3/min ) 1.666 -1
278 K
T2 = 392.45 °K ;°K °C +273
T2 = 119.45 °C (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Constant Entropy (S=C)
k = 1.666, P1 = 745 kPaa, P2 = 1765 kPaa and V1 =5.1 m3/min
T1 = 5°C +273 = 278 °K, V2 =3.039 m3/min , T2 = 392.45 °K
Solution: Work non-flow in kJ/min.
WNF = P2 V2 – P1 V1
1–k
WNF = (1765 kPaa )(3.039 m3/min) – (745 kPaa)(5.1 m3/min)
1 – 1.666
WNF = - 2348.85 kJ/min (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Throttling or Isenthalpic Process is a Constant Enthalpy Process. Also known as irreversible adiabatic
process. H=C
Change in Enthalpy (ΔH); unit kJ
ΔH = m Cp ΔT ; Since H=C
ΔH = 0
Thermodynamics 1
4.0 Processes of Ideal Gas
Polytropic process is a quasi-static process. Described with the condition PV n =C.
Relationship between volume, pressure and temperature.
P1 V1 = P2 V2
T1 T2
P1 V1 n = P2 V2 n Where: n – constant other than k.
T2 = (V1 / V2 ) n -1
T1
T2 = (P2 / P1 ) (n -1/n)
T1
Thermodynamics 1
4.0 Processes of Ideal Gas
Polytropic process is a quasi-static process. Described with the condition PV n =C.
Work Non-Flow (WNF); Unit KJ
WNF = P2 V2 – P1 V1
1–n
Change in Internal Energy (ΔU); unit kJ
ΔU = m Cv ΔT
Thermodynamics 1
4.0 Processes of Ideal Gas
Polytropic process is a quasi-static process. Described with the condition PV n =C.
Change in Enthalpy (ΔH); unit kJ
ΔH = m Cp ΔT
Heat (Q); unit kJ
Q = m Cn ΔT
Where: Cn= Cv [k−n / 1−n]
Thermodynamics 1
4.0 Processes of Ideal Gas
Polytropic process is a quasi-static process. Described with the condition PV n =C.
Change in Entropy (ΔS); unit kJ/K
ΔS = m Cn ln (T2/T1)
Where: Cn= Cv [k−n / 1−n]
Work steady flow (WSF); unit kJ
WSF = Q - ΔH
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
A polytropic process of air from 150 psia, 300°F, and 1 ft3 occurs to P2 = 20 psia in accordance with
PV1.3 = C. Determine the following:
a) Final volume in ft3.
b) Final temperature in °F.
c) Heat in BTU if the mass of air is 5 lb.
Given: Polytropic process
n = 1.3, P1 = 150 psia, P2 = 20 psia, V1 = 1 ft3
T1 = 300°F + 460= 760 °R;
Polytropic process; PV1.3 = C
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Polytropic process; PV1.3 = C
n = 1.3, P1 = 150 psia, P2 = 20 psia, V1 = 1 ft3
T1 = 300°F + 460= 760 °R
Solution : a) Final volume in ft3.
P1 V1 n = P2 V2 n
(150 psia) (1 ft3) 1.3 = (20 psia) (V2) 1.3
V2 =4.71 ft3 (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Polytropic process; PV1.3 = C
n = 1.3, P1 = 150 psia, P2 = 20 psia, V1 = 1 ft3 ,V2 =4.71 ft3
T1 = 300°F + 460= 760 °R
Solution : b) Final temperature in °F.
T2 = (V1 / V2 ) n -1
T1
T2 = (1 ft3 / 4.71 ft3 ) 1.3 -1
760 R
T2 = 477.427 °R ;°R = °F +460
T2 = 17.427 °F (Answer)
Thermodynamics 1
4.0 Processes of Ideal Gas
Sample Problem:
Given: Polytropic process; PV1.3 = C
n = 1.3, P1 = 150 psia, P2 = 20 psia, V1 = 1 ft3 ,V2 =4.71 ft3
T1 = 300°F + 460= 760 °R, T2 = 477.427 °R
Solution : Heat in BTU if the mass of air is 5 lb.
Q = m Cn ΔT
Where: Cn= Cv [k−n / 1−n] , k air = 1.4 and Cv air =0.1714 BTU/lbm⋅∘R
Cn= 0.1714 BTU/lbm⋅∘R [1.4−1.3 / 1−1.3] = -0.05713 BTU/lbm⋅∘R
Q = m Cn ΔT
Q = (5 lbm)(-0.05713 BTU/lbm⋅∘R) (477.427 °R - 760 °R)
Q = 80.717 BTU (Answer)
Thermodynamics 1
