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DEPARTMENT OF CHEMICAL & BIOMOLECULAR ENGINEERING
CH2008: THERMODYNAMICS
SOLUTION TO TUTORIAL ASSIGNMENT 3
1. A quantity of an ideal gas (C P = 7R/2), at 293 K and 1 bar and having a volume of 70 m3
is heated at constant pressure to 298 K by the transfer of heat from a hot reservoir at 313 K. Calculate the quantity of heat transferred to the gas, the entropy change of the gas, the entropy change of hot reservoir, and change in total entropy of the universe. What is the irreversible feature of the process?
SOLUTION:
System: 70 m3 of an ideal gas initially at 293K and 1 bar;
Process: Isobaric heating of the gas from 293 k to 313K
No. of kmol of the ideal gas: n = PV/RT = (101.33)(70)/(8.314)(293) = 2.912 kmol
Volume of gas at state 2: V2 = nRT/P = 71.2 m3
(i) Quantity of heat transferred to the gas:
From 1st law: ∆U t = Q + W = Q – n ∫ PdV = Q - nP(V2-V1)
Based on Eq. (2.47) for isobaric heating
Q = ∆H t = n ∫ Cp dT = n Cp (T2-T1) = 423.7 kJ
(ii) Change in entropy of the system:
Based on Eq. (2.49) for isobaric heating
∆S t gas = n ∫ (Cp/T) dT = nCp ln (T2/T1) = 1.433 kJ/K
(iii) Change in entropy of the reservoir (the surroundings)
∆S t res = Q surr/T res = -Q /T res = (- 423.7)/313 = -1.354 kJ/K
(iv) Total entropy change of universe:
∆Stotal = ∆S t gas + ∆S t res = 0.08 kJ/K >0
Irreversible heat transfer due to finite temperature between the hot-reservoir and the cold
gas
2. A lump of copper (specific heat: C = 0.4 kJ/kg-K) having a mass of 10 kg at a temperature of 873 K is dropped into a well-insulated bucket containing 100 kg of oil (C = 2.4
kJ/kg-K) at 293 K. (a) What is the final equilibrium temperature of copper and oil? (b) Calculate the changes in entropy of copper and oil, the surroundings, and the universe.
SOLUTION:
System: (Cu + oil) in the insulated bucket (assume negligible heat capacity of bucket); Process: Internal heat transfer between hot copper and cold oil
(i) Final temperature of the system T F:
From 1st law ∆U t = Q + W = 0
∆U t = ∆U t cu + ∆U t oil = 0
Based on Eq. (4.46) and for constant volume process:
∆U t cu + ∆U t oil = [m C (T F -T I)] Cu + [m C (T F -T I)]oil = 0 (C V = C P = C)
T F (final equilibrium temp) = 302.5K
(ii) Entropy change of Cu and oil:
∆S t Cu = [∫ (mC/T) dT] Cu = (10)(0.4) ln (T F /T I) Cu = -4.239 kJ/K
∆S t oil = [∫ (mC/T) dT] oil = (100)(2.4) ln (T F / T I)oil = 7.768 kJ/K
∆S t sys = 3.419 kJ/K
(iii) Entropy change of the surroundings
∆S t surr = Q surr/T res = 0
(iv) Entropy change of the universe
∆S univ = ∆S sys + ∆S t surr = 3.419 kJ/K >0
3. A piston/cylinder device contains 5 kmol of a monoatomic ideal gas (Cp = 5R/2 and Cv = 3R/2) at 293 K and 1 bar. This gas is compressed reversibly and adiabatically to 10 bar, where the piston is locked in position. The cylinder is then brought into thermal contact with a heat reservoir at 293 K, and heat transfer continues until the gas also reaches this temperature. Determine the entropy changes of the reservoir, the gas, and the universe.
SOLUTION:
System: 5 kmol of an ideal gas initially at 293 K and 1 bar
Process: (a) Isentropic compression followed by (b) isometric cooling
(i) Gas compression isentropically from 1 bar and T1 = 293 K to 10 bar and T2
T2 / T1= (P2 / P1) (γ-1)/ γT2 = 293 x 2.51 = 736K
(ii) Isometric heat transfer from hot gas to cold reservoir at 293K
From 1st law, ∆U t = Q + W = Q as W = - ∫ P dV t = 0
From Eq. (4.46) and for isometric process:
Q = ∆U t = n ∫ C V dT = nCv (T3 - T2) = 5(3/2R) (293 - 736) = -27,623 kJ
(iii) Entropy change of the reservoir:
∆S t res = Q res /T res = (-Q/T res) = 27,623/293 = 94.27 kJ/K
(iv) Entropy change of the gas: (Based on Eq. (2.48) for isometric process)
∆S t gas = ∆S t(a) + ∆S t(b) = ∫ (nCv/T) dT = 5(3R/2) ln (293/736) = -57.43 kJ/K
(v) Entropy change of the universe:
∆S total = ∆S t gas + ∆S t res = 36.87 kJ/K
4. One kmol of an ideal gas is compressed isothermally but irreversibly at 130 o C from 2.5 bar to 6.5 bar in a piston-cylinder device. The work required is 30% greater than the work of a reversible isothermal compression. The heat is transferred from the gas during compression flows to a cold reservoir at 25 o C. Calculate the actual quantity of heat transferred and the entropy changes of the gas and the reservoir. What is the entropy generation of the irreversible process? Solution:
System: One kmol of an ideal gas initially at 130 o C from 2.5 bar
Process: Isothermal compression from 2.5 bar to 6.5 bar at 130 o C
(i) Isothermal heat transfer to the cold reservoir (ideal gas):
∆U t = Q + W = 0
Q = -W rev = - ( - ∫ PdV) = ∫ (RT/V) dV = RT ∫ d (lnV)
= RT ln (V2/ V1) = RT ln (P1/P2) = - 3201.5 kJ/kmol
Q irrev = -W irrev = -1.3 W rev = - 4,162 kJ/kmol
(ii) Entropy changes of the gas and the reservoir:
Entropy change of gas:
From Eq. (2.49) and isothermal process
∆S gas = - ∫ (∂V/∂T)P dP
Assume ideal gas: (∂V/∂T)P = [∂(RT/P)/∂T]P = R/P
∆S gas = - ∫ (R/P) dP = - R ∫ dlnP = - Rln (P2/P1) = -7.944 kJ/kmol-K
Entropy change of reservoir: ∆S res = Q surr/T res = - Q irrev/T res =13.97 kJ/kmol-K (iii)Entropy generation:
S G = ∆S total = ∆S res + ∆S gas = 6.025 kJ/kmol-K
5. An ideal gas is changed from state 1 (P1, V1) to state 2 (P2 , V2) by the three different processes (i) isothermal (ii) isobaric followed by isometric and (iii) adiabatic followed by isometric as discussed in the example in Chapter 3. Show that the change in entropy of the three different processes are the same as follows: ∆S t = n R ln(V2 / V1).
(Hint: (ii) Substitute for C P using C P = C V + R; (iii) Based on adiabatic process using Eqs.
(3.25) and (3.26): (R/C V) = γ - 1 and (T2”/T1) =(V1/V2”)γ-1 )
Solution:
System: n kmol of an ideal gas initially at state 1
Processes:
1. Isothermal process from 1 to 2
2. Isobaric step to 2’ (P1, V2’) followed by isometric step
3. Adiabatic process to 2” (P2”, V2”) followed by isometric step
V
(i) Isothermal process
dS t = n(C v/T)dT + n(∂P/∂T)v dV
From PV= RT (∂P/∂T)v = R/V
dS t = n(C v/T)dT + n(R/V) dV = n(R/V) dV
∆S t = n R ∫ dln V = nR ln (V2 / V1).
(ii) Isobaric followed by isometric
A. Isobaric step:
dS t = n(C p/T)dT - n(∂V/∂T)p dP = n(C p/T)dT
∆S t = n ∫ (C p/T)dT = n C p ln (T2’ / T1).
B.Isometric step:
dS t = n(C v/T)dT + n(∂P/∂T)v dV = n(C v/T)dT
∆S t = n ∫ (C v/T)dT = nC V ln (T2 / T2’).
Combined Isobaric & isometric:
∆S t= n [C p ln (T2’ / T1) + C V ln (T2 / T2’)
= n [(C v + R) ln (T2’ / T1) - C V ln (T2’ / T2)] (as T1 = T2)
= n [R ln (T2’ / T1)] = nR ln (V2’ / V1) (Const – P)
= nR ln (V2 / V1). (as V2’ = V2)
(iii) adiabatic followed by isometric
Adiabatic step: dQ = 0; dS t = 0
Isometric step: ∆S t = n ∫ (C v/T)dT = nC V ln (T2 / T2’’)
From R/C V = (γ-1)
and (T2’’ / T1) = (V1 / V2’’) (γ - 1) (Adiabatic process)
∆S t= nC V ln (T2 / T2’’) = n [R /(γ-1)] ln [T2 / (T1(V1 / V2’’)(γ - 1)]
= n [R /(γ-1)] ln [ (V2’’ / V1)(γ - 1)]
= n R ln [ (V2’’ / V1)]
= nR ln (V2 / V1).
6. 1000 kmol of an ideal gas at T1 = 500K and P1 = 6 bar expands through an adiabatic
turbine and discharges at T2 = 371K and P2 = 1.2 bar. Assuming steady state operation, determine based on 1 s of operation, the following:
(a)the actual work W S produced
(b)the ideal work W ideal;
(c)the lost work W lost; and
(d)the entropy generation S g
The heat capacity of the gas at constant P may be taken to be 7R/2.
Ans: - 3.754 x 103 kJ; -5.163 x 103 kJ;; 1.409 x 103 kJ; 4.698 kJ/K
SOLUTION:
System: 1 000 kmol/s of an ideal gas initially at T1 = 500K and P1 = 6 bar
Process: adiabatic expansion through a turbine to T2 = 371K and P2 = 1.2 bar
(a) the actual work W S produced
From 1st law neglecting kinetic and potential energy changes and based on 1 s of
operation:
∆H t = Q + W S = W S
W S = ∆H t = n ∫ C p dT = nC P ln (T2 - T1) = - 3.754 x 103 kJ
(b) The ideal work W ideal;
The change in entropy of the gas during expansion:
dS = (C p/T)dT - (∂V/∂T)P dP = (C p/T)dT – (R/P) dP
dS = C p dlnT – R dln dP
∆S = C p ln (T2 / T1) – R ln (P2 / P1) = 4.698 kJ/K
The ideal work: W ideal = (∆H t – Tσ∆S t) = - 3.754 x 103 - (300)* 4.698 = -5.163 x 103 kJ (c) The lost work W lost
W lost = W S - W ideal = - 3.754 x 103 - ( -5.163 x 103) = 1.409 x 103 kJ
(d) The entropy generation S g
From W lost = Tσ S g
S g = W lost / Tσ = 4.698 kJ/K。