CONNECT FOR THERMODYNAMICS: AN ENGINEERI
CONNECT FOR THERMODYNAMICS: AN ENGINEERI
9th Edition
ISBN: 9781260048636
Author: CENGEL
Publisher: MCG
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Chapter 8.8, Problem 99RP

(a)

To determine

The final temperature of the steam in tank A.

The final temperature of the steam in tank B.

(a)

Expert Solution
Check Mark

Answer to Problem 99RP

The final temperature of the steam in tank A is 133.52°C.

The final temperature of the steam in tank B is 110.1°C.

Explanation of Solution

Write the expression to calculate the specific volume of saturated water (v).

v=vf+x(vgvf) (I)

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

Write the expression to calculate the specific internal energy of saturated water (u).

u=uf+x(ufg) (II)

Here, specific internal energy of saturated liquid is uf, and the specific internal energy of vaporization is ufg.

Write the expression to calculate the specific entropy of saturated water (s).

s=sf+x(sfg) (III)

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

Write the expression to calculate the mass from the specific volume.

m=Vv (IV)

Write the mass balance equation for the fluid flow process.

min=mout (V)

Here, mass of the steam entered is min and the mass of the steam exit out from system is mout.

Write the energy balance equation for the entire system considering it as a stationary closed system.

EinEout=ΔEsystemQout=ΔUQout=(ΔU)A+(ΔU)B

Qout=(m2u2m1u1)A+(m2u2m1u1)B (VI)

Here, net energy input to the system is Ein, net energy output to the system is Eout, the change in net energy is ΔEsystem, the amount of heat lost is Qout, change in internal energy of steam in tank A is (ΔU)A, change in internal energy of steam in tank B is (ΔUB), final mass of steam is m2, initial mass of steam is m1, final internal energy is u2, and the initial internal energy is u1.

Conclusion:

For Tank A:

Refer the Table A-5 of “Saturated water: Pressure”, obtain the properties of steam at the pressure (P1) of 400kPa as follows:

vf=0.001084m3/kgvg=0.46242m3/kguf=604.22kJ/kgufg=1948.9kJ/kgsf=1.7765kJ/kgKsfg=5.1191kJ/kgK

Substitute 0.001084m3/kg for vf, 0.46242m3/kg for vg, and 0.8 for x1 in Equation (I).

v1,A=0.001084m3/kg+0.8(0.46242m3/kg0.001084m3/kg)=0.37015m3/kg

Substitute 604.22kJ/kg for uf, 1948.9kJ/kg for ufg, and 0.8 for x1 in Equation (II).

u1,A=604.22kJ/kg+0.8(1948.9kJ/kg)=2163.3kJ/kg

Substitute 1.7765kJ/kgK for sf, 5.1191kJ/kgK for sfg, and 0.8 for x1 in Equation (III).

s1,A=1.7765kJ/kgK+0.8(5.1191kJ/kgK)=5.8717kJ/kgK

Refer the Table A-5 of “Saturated water: Pressure”, obtain the properties of steam at the pressure (P2) of 300kPa as follows:

vf=0.001073m3/kgvg=0.60582m3/kguf=561.11kJ/kgufg=1982.1kJ/kgsf=1.6717kJ/kgKsfg=5.320kJ/kgKT2,A=Tsat=133.52°C

Thus, the final temperature of the steam in tank A is 133.52°C.

Write the final specific entropy of steam in tank A from isentropic relation.

s2,A=s1,A=5.8717kJ/kgK

Substitute 5.8717kJ/kgK for s2,A, 1.6717kJ/kgK for sf, and 5.320kJ/kgK for sfg in Equation (III).

5.8717kJ/kgK=1.6717kJ/kgK+x2,A(5.320kJ/kgK)x2,A=0.7895

Substitute 0.001073m3/kg for vf, 0.60582m3/kg for vg, and 0.7895 for x2,A in Equation (I).

v2,A=0.001073m3/kg+0.7895(0.60582m3/kg0.001073m3/kg)=0.4785m3/kg

Substitute 561.11kJ/kg for uf, 1982.1kJ/kg for ufg, and 0.7895 for x2,A in Equation (II).

u2,A=561.11kJ/kg+0.7895(1982.1kJ/kg)=2125.9kJ/kg

For Tank B:

Refer to Table A-6 of “Superheated water”, obtain the properties of steam for pressure (P1) of 200kPa and temperature (T1) of 250°C as

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

Substitute 0.2m3 for VA and 0.37015m3/kg for v1,A in Equation (IV).

m1,A=0.2m30.37015m3/kg=0.5403kg

Substitute 0.2m3 for VA and 0.47850m3/kg for v2,A in Equation (IV).

m2,A=0.2m30.47850m3/kg=0.418kg

Write the expression to calculate the mass flowing into tank B (me) from Equation (V).

me=m1,Am2,A (VII)

Substitute 0.5403kg for m1,A and 0.418kg for m2,A in Equation (VII).

me=0.5403kg0.418kg=0.122kg

Calculate the final mass of steam in tank B (m2,B) from Equation (V).

m2,B=m1,B+me (VIII)

Substitute 3kg for m1,B and 0.122kg for me in Equation (VIII).

m2,B=3kg+0.122kg=3.122kg

Write the expression to calculate the final specific volume of steam in tank B from Equation (IV).

v2,B=m1,Bv1,Bm2,B (IX)

Substitute 3kg for m1,B, 1.1989m3/kg for v1,B, and 3.122kg for m2,B in Equation (IX).

v2,B=3kg×1.1989m3/kg3.122kg=1.152m3/kg

Substitute 900kJ for Qout, 0.5403kg for m1,A, 0.418kg for m2,A, 2125.9kJ/kg for u2,A, 2163.3kJ/kg for u1,A, 3kg for m1,B, 3.122kg for m2,B, and 2731.4kJ/kg for u1,B in Equation (VI).

900kJ=(0.418×2125.90.5403×2163.3)+[3.122(u2,B)3×2731.4]u2,B=2425.9kJ/kg

From the Table A-4 of “Saturated water: Temperature”, obtain the properties of water in tank B at specific volume of 1.152m3/kg and specific internal energy of 2425.9kJ/kg as

T2,B=110.1°Cs2,B=6.9772kJ/kgK

Thus, the final temperature of the steam in tank B is 110.1°C.

(b)

To determine

The amount of work potential wasted during the process.

(b)

Expert Solution
Check Mark

Answer to Problem 99RP

The amount of work potential wasted during the process is 337kJ.

Explanation of Solution

Write the entropy generation (Sgen) equation for the process from entropy balance.

SinSout+Sgen=ΔSsystem

Sgen=ΔSA+ΔSB+QoutTb,surr

Sgen=(m2s2m1s1)A+(m2s2m1s1)B+QoutTb,surr (X)

Here, entropy input to the system is Sin, entropy exiting out is Sout, change in the entropy system is ΔSsystem, and the surrounding’s temperature is Tb,surr.

Write the expression to calculate the exergy destroyed (Xdest).

Xdest=T0Sgen (XI)

Here, the surrounding’s temperature is T0.

Conclusion:

Substitute 900kJ for Qout, 0.5403kg for m1,A, 0.418kg for m2,A, 5.8717kJ/kgK for s2,A, 5.8717kJ/kgK for s1,A, 3kg for m1,B, 3.122kg for m2,B, 6.9772kJ/kgK for s2,B, 7.7100kJ/kgK for s1,B, and 273K for Tb,surr in Equation (X).

Sgen=[(0.418×5.87170.5403×5.8717)+(3.122×6.97723×7.7100)+900kJ273K]=1.234kJ/K

Substitute 273K for T0 and 1.234kJ/K for Sgen in Equation (XI).

Xdest=273K×1.234kJ/K=337kJ

Thus, the amount of work potential wasted during the process is 337kJ.

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Chapter 8 Solutions

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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