CENGEL'S 9TH EDITION OF THERMODYNAMICS:
CENGEL'S 9TH EDITION OF THERMODYNAMICS:
9th Edition
ISBN: 9781260917055
Author: CENGEL
Publisher: MCG
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Chapter 7.13, Problem 201RP

a)

To determine

The final temperature in each tank A and tank B.

a)

Expert Solution
Check Mark

Answer to Problem 201RP

The final temperature in tank A is 120.2°C.

The final temperature in tank B is 116.1°C.

Explanation of Solution

Write the formula to calculate the specific volume of steam from tables (v).

v=vf+x(vgvf) (I)

Here, specific volume of saturated liquid is vf, specific volume of saturated vapor is vg, and dryness fraction of water is x.

Write the formula to calculate the specific internal energy of steam from tables (u).

u=uf+x(ugf) (II)

Here, specific internal energy of saturated liquid is uf and specific internal energy of saturated liquid vapor mixture is ufg.

Write the formula to calculate the specific entropy of steam from tables (s).

s=sf+x(sgf) (III)

Here, specific entropy of saturated liquid is sf, and the specific entropy of saturated liquid vapor mixture is sfg.

Write the formula to calculate the mass of the steam (m).

m=νv (IV)

Here, volume of the steam is ν.

Write the expression for the mass balance.

minmout=Δm (V)

Here, mass of the water entering into the system is min, the mass of water leaving out of the system is mout and the change in mass is Δm.

Write the expression for the energy balance Equation for a closed system.

EinEout=ΔEsystem (VI)

Here, net energy transfer into the control volume is Ein, net energy transfer exit from the control volume is Eout and change in internal energy of the system is ΔEsystem

Conclusion:

From Table A-5, “Saturated water-Pressure table”, obtain the following properties of water at initial pressure (P1A) of 400 kPa in tank A as follows:

vf=0.001084 m3/kgvg=0.46242 m3/kguf=604.22 kJ/kg

ufg=1948.9 kJ/kgsf=1.7765 kJ/kgKsfg=5.1191 kJ/kgK

Substitute 0.001084 m3/kg for vf, 0.46242 m3/kg for vg, and 0.6 for x in Equation (I) for initial specific volume of steam in tank A (v1,A).

v1,A=0.001084 m3/kg+0.6(0.46242 m3/kg0.001084 m3/kg)v1,A=0.27788 m3/kg

Substitute 604.22 kJ/kg for uf, 1948.9 kJ/kg for ufg, and 0.6 for x in Equation (II) for initial specific volume of steam in tank A (u1,A).

u1,A=604.22 kJ/kg+0.6(1948.9 kJ/kg)u1,A=1773.6 kJ/kg

Substitute 1.7765 kJ/kgK for sf, 5.1191 kJ/kgK for sfg, and 0.6 for x in Equation (III) for initial specific entropy of steam in tank A (s1,A).

s1,A=1.7765 kJ/kgK+0.6(5.1191 kJ/kgK)s1,A=4.8479 kJ/kgK

From Table A-5, “Saturated water-Pressure table”, obtain the following properties of water at final pressure (P2A) of 200 kPa in tank A as follows:

T2A=Tsat=120.2°Cvf=0.001061 m3/kgvg=0.8858 m3/kguf=504.50 kJ/kg

ufg=2024.6 kJ/kgsf=1.5302 kJ/kgKsfg=5.5968 kJ/kgK

Here, final temperature of steam in tank A is T2A.

The steam in tank A undergoes isentropic process, Thus final specific entropy of steam in tank A (s2A) is expressed as:

s2A=s1A

Substitute 4.8479 kJ/kgK for s2A, 1.5302 kJ/kgK for sf, and 5.5968 kJ/kgK for sfg in Equation (III) for final dryness fraction of steam in tank A (x2,A).

x2,A=4.8479 kJ/kgK1.5302 kJ/kgK5.5968 kJ/kgKx2,A=0.5928

Substitute 0.001061 m3/kg for vf, 0.8858 m3/kg for vg, and 0.5928 for x2,A in Equation (I) to calculate the final specific volume of steam in tank A (v2,A).

v2,A=0.001061 m3/kg+0.5928(0.8858 m3/kg)v2,A=0.52552 m3/kg

Substitute 504.50 kJ/kg for uf, 2024.6 kJ/kg for ufg, and 0.5928 for x2,A in Equation (II) to calculate the final specific internal energy of steam in tank A (u2,A).

u2,A=504.50 kJ/kg+0.5928(2024.6 kJ/kg)u2,A=1704.7 kJ/kg

From Table A-6, “Superheated water”, note the properties for steam in tank B initially at the pressure of 200 kPa and temperature of 250°C as follows:

v1,B=1.1989 m3/kgu1,B=2731.4 kJ/kgs1,B=7.7100 kJ/kgK

Substitute 0.3 m3 for νA and 0.27788 m3/kg for v1,A in Equation (IV) to calculate the initial mass of steam in tank A (m1,A).

m1,A=0.3 m30.27788 m3/kgm1,A=1.08 kg

Substitute 0.3 m3 for νA and 0.27788 m3/kg for v1,A in Equation (IV) to calculate the final mass of steam in tank A (m2,A).

m2,A=0.3 m30.52552 m3/kgm2,A=0.5709 kg

Substitute 1.080 kg for m1,A and 0.5709 kg for m2,A using Equation (V) to calculate the mass of steam flows into tank B from tank A (ΔmA).

ΔmA=1.08 kg0.5709 kgΔmA=0.5091 kg

Rewrite the Equation (V) to calculate the final total mass of steam in tank B (m2,B).

m2,B=m1,B+ΔmA (VII)

Here, initial mass of steam in tank B is m1,B.

Substitute 2 kg for m1,B and 0.5091 kg for ΔmA in Equation (VII).

m2,B=2 kg+0.5091 kgm2,B=2.5091 kg

Substitute 2 kg for m1,B and 1.1989 m3/kg for v1,B in Equation (IV) to calculate the volume of steam in tank B (νB).

νB=2 kg(1.1989 m3/kg)νB=2.3978 m3

Substitute 2.3978 m3 for νB and 2.5091 kg for m2,B in Equation (IV) to calculate the final specific volume of steam in tank B (v2,B).

v2,B=2.3978 m32.5091 kgv2,B=0.9558 m3/kg

From first law of thermodynamics, Re-write the Equation (VI) for heat transfer (Q) as follows:

QW=ΔUQW=(ΔU)A+(ΔU)B

QW=(m2,Au2,Am1,Au1,A)+(m2,Bu2,Bm1,Bu1,B) (VIII)

Here, work done is W, total change in internal energy is ΔU, change in internal in tank A is (ΔU)A, the change in internal energy in tank B is (ΔU)B, final specific internal energy of steam in tank B is u2,B.

Substitute 300 kJ for Q, 0 for W, 0.5709 kg for m2,A, 1704.7 kJ/kg for u2,A, 1.08 kg for m1,A, 1773.6 kJ/kg for u1,A, 2.5091 kg for m2,B, 2 kg for m1,B, and 2731.4 kJ/kg for u1,B in Equation (VIII).

300 kJ0=[(0.5709 kg×1704.7 kJ/kg1.08 kg×1773.6 kJ/kg)+(2.5091 kg(u2,B)2 kg×2731.4 kJ/kg)]u2,B=2433.3 kJ/kg

From Table A-5, “Saturated water-Temperature table”, obtain the following properties of water at u2,B of 2433.3 kJ/kg and v2,B of 0.9558 m3/kg as follows:

T2,B=116.1°Cs2,B=6.9156 kJ/kgK

Here, the temperature of the steam in tank at final state is T2,B, and the specific entropy of steam in tank B finally is s2,B.

Thus, the final temperature of steam in tank A is 120.2°C, and final temperature of steam in tank B is 116.1°C.

b)

To determine

The entropy generated during the process.

b)

Expert Solution
Check Mark

Answer to Problem 201RP

The entropy generated during the process is 0.498 kJ/K.

Explanation of Solution

Write the expression for the entropy balance Equation of the system.

S˙inS˙out+S˙gen=ΔS˙system (IX)

Here, rate of net entropy in is S˙in, rate of net entropy out is S˙out, rate of entropy generation is S˙gen and change of entropy of the system is ΔS˙system

Conclusion:

Re-write the Equation (IX) for the entropy generated (Sgen) during the process.

QTsurr+Sgen=ΔSA+ΔSBSgen=ΔSA+ΔSBQTsurr

Sgen=(m2,As2,Am1,As1,A)+(m2,Bs2,Bm1,Bs1,B)QTsurr (X)

Here, temperature of the surroundings is Tsurr, entropy change for steam in tank A is ΔSA, and the entropy change for the steam in tank B is ΔSB.

Substitute 300 kJ for Q, 0.5709 kg for m2,A, 1.08 kg for m1,A, 2.5091 kg for m2,B, 2 kg for m1,B, 4.8479 kJ/kgK for s2A, 4.8479 kJ/kgK for s1A, 6.9156 kJ/kgK for s2,B, 7.7100 kJ/kgK for s1,B, and 17°C for Tsurr in Equation (X).

Sgen={[(0.5709 kg×4.8479 kJ/kgK)(1.08 kg×4.8479 kJ/kgK)]+[(2.5091 kg×6.9156 kJ/kgK)(2 kg×7.7100 kJ/kgK)](300 kJ)17°C}={[(0.5709 kg×4.8479 kJ/kgK)(1.08 kg×4.8479 kJ/kgK)]+[(2.5091 kg×6.9156 kJ/kgK)(2 kg×7.7100 kJ/kgK)]+300 kJ(17+273)K}=0.498 kJ/K

Thus, the entropy generated during this process is 0.498 kJ/K.

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Chapter 7 Solutions

CENGEL'S 9TH EDITION OF THERMODYNAMICS:

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