Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 5.5, Problem 173RP
To determine

The final temperature and mass of the air in the ballast tank.

Expert Solution & Answer
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Answer to Problem 173RP

The final temperature and mass of the air in the ballast tank is 386.8K and 9460kg respectively.

Explanation of Solution

Write the general mass balance equation.

m˙inm˙out=ddt(msystem)m˙inm˙out=dmsystemdt (I)

Here, the inlet mass flow rate is m˙in, the exit mass flow rate is m˙out, and the change in mass of the system is dmsystemdt.

It is given that the mass of air pumped into the tank to get the submarine to the surface. Here, there is no exit mass of air. Thus, the exit mass (m˙out) of air is negligible.

m˙out=0

Refer Equation (I).

Write the mass balance equation for air.

m˙a,inm˙a,out=dmadtm˙a,in0=dmadtm˙a,in=dmadt (II)

Here, the subscript a indicate the air.

While air entering into the ballast tank, the seawater present in the tank is released to the sea. Here, there is no inlet mass of water. Thus, the inlet mass (m˙in) of water is negligible.

m˙in=0

Refer Equation (I).

Write the mass balance equation for water.

m˙w,inm˙w,out=dmwdt0m˙w,out=dmwdtm˙w,out=dmwdt (III)

Here, the subscript w indicates the seawater.

Write the general energy rate balance equation.

E˙inE˙out=ΔE˙system (IV)

Here, the rate of total energy in is E˙in, the rate of total energy out is E˙out, and the rate of change in net energy of the system is ΔE˙system.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

ΔE˙system=0

Refer Equation (IV).

Write the energy balance equation for the system (ballast tank).

d(mu)adt+d(mu)wdt+hw,outm˙w,outha,inm˙a,in=0 (V)

Here, the mass is m, the internal energy is u, the enthalpy is h, and the subscript a indicates air, w indicates water.

Substitute dmadt for m˙a,in and dmwdt for m˙w,out in Equation (V).

d(mu)adt+d(mu)wdt+hw,out(dmwdt)ha,in(dmadt)=0d(mu)adt+d(mu)wdthw,outdmwdtha,indmadt=0d(mu)adt+d(mu)wdt=hw,outdmwdt+ha,indmadtd(mu)a+d(mu)w=hw,outdmw+ha,indma (VI)

Integrate the Equation (VI) at initial (1) and final (2) states.

12d(mu)a+12d(mu)w=12hw,outdmw+12ha,indma{[(mu)2(mu)1]a+[(mu)2(mu)1]w}=hw,out(m2m1)w+ha,in(m2m1)a (VII)

The enthalpy and internal energy is expressed as follows.

h=cpTu=cvT

The mass of air is expressed as follows.

ma=PνRT

Here, the specific volume is v.

For water the specific heat is cp=cv=cw.

At final state, the mass sea water is zero and the ballast is completely filled with air.

(m2)w=0

Rewrite the Equation (VII) as follows.

[P2ν2RT2cvT2P1ν1RT1cvT1]a+[0m1cwT1]w=cwT1,w(0m1)w+cp,aTin,a(P2ν2RT2P1ν1RT1)a[P2ν2RT2cvT2P1ν1RT1cvT1]am1,wcwT1,w=m1,wcwT1,w+cp,aTin,a(P2ν2RT2P1ν1RT1)a[P2ν2RT2cvT2P1ν1RT1cvT1]a=cp,aTin,a(P2ν2RT2P1ν1RT1)acv,a(P2ν2RP1ν1R)a=cp,aTin,a(P2ν2RT2P1ν1RT1)a

cv,acp,a(P2ν2RP1ν1R)a=Tin,a(P2ν2RT2P1ν1RT1)a (VIII)

Here,

The specific heat ratio is, k=cpcv.

The initial and final pressure of air is, P2=P1.

Rewrite the Equation (VIII) as follows.

1ka(ν2ν1)a=Tin,a(ν2T2ν1T1)a1kaTin,a(ν2ν1)a+(ν1T1)a=(ν2T2)aT2,a=ν2,a1kaTin,a(ν2ν1)a+(ν1T1)a (IX)

Write the formula for mass flow rate of air at final state.

m˙2,a=(P2ν2RT2)a (X)

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant (R) of air is 0.287kPa.m3/kgK.

Refer Table A-2a, “Ideal-gas specific heats of various common gases”.

The specific heat ratio (ka) of air at room temperature is 1.4.

Conclusion:

Substitute 700m3 for v2,a, 1.4 for ka, (20+273)K for Tin,a, 100m3 for ν1, and (15+273)K for T1 in Equation (IX).

T2,a=700m311.4[(20+273)K](700m3100m3)+100m3(15+273)K=700m31.4627m3/K+0.3472m3/K=386.7828K386.8K

Substitute 1500kPa for P2, 700m3 for v2, 0.287kPa.m3/kgK for R, and 386.8K for T2 in Equation (X).

m˙2,a=1500kPa(700m3)0.287kPa.m3/kgK×(386.8K)=1050000kPam3111.0116kPa.m3/kg=9458.47kg9460kg

Thus, the final temperature and mass of the air in the ballast tank is 386.8K and 9460kg respectively.

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Chapter 5 Solutions

Thermodynamics: An Engineering Approach

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