THERMODYNAMICS-SI ED. EBOOK >I<
THERMODYNAMICS-SI ED. EBOOK >I<
9th Edition
ISBN: 9781307573022
Author: CENGEL
Publisher: MCG/CREATE
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Chapter 4.5, Problem 138RP
To determine

The final pressure of the two rigid tanks.

The amount of heat transfer to the two rigid tanks.

Expert Solution & Answer
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Answer to Problem 138RP

The final pressure of the two rigid tanks is 3.1698kPa_.

The amount of heat transfer to the two rigid tanks is 2171kJ_.

Explanation of Solution

Write the expression for the energy balance equation.

EinEout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (I) and write energy balance two rigid tanks.

WinQout=ΔUA+ΔUB (II)

Here, the work to be done into the system is Win, the heat to be transfer by system is Qout, the change in internal energy of the tank A is ΔUA, and the change in the internal energy of the tank B is ΔUB.

Take the two rigid tanks as the system.

Substitute 0 for Win in the Equation (II).

(0)Qout=U2,A+BU1,A+U1,BQout=[U2,A+BU1,A+U1,B]=[m2,totalu2(m1u1)A(m1u1)B] (III)

Here, the total mass of the two rigid tank is m2,total, the final specific internal energy of the two rigid tank is u2, the initial mass of the tank A is m1,A, the initial specific internal energy of the tank A is u1, the initial mass of the tank B is m1,B, and the initial specific internal energy of the tank B is u1.

Determine the initial specific volume of the tank A.

v1,A=vf+x1(vgvf) (IV)

Here, the specific volume of the saturated liquid phase is vf, the initial dryness fraction is x1, and the specific volume of the saturated vapour phase is vg.

Determine the initial internal energy of the tank A.

u1,A=uf+x1ufg (V)

Here, the specific internal energy of the saturated liquid phase is uf, the initial dryness fraction is x1, and the specific internal energy change upon vaporization is vfg.

Determine the total mass of the two rigid tanks.

mt=m1,A+m1,B=(νAv1,A)+(νBv1,B) (VI)

Determine the final specific volume of the two rigid tanks.

v2=νtmt (VII)

Here, the total volume of the two tanks is νt.

Determine the final dryness fraction of the two rigid tanks.

x2=v2vfvfg=v2vfvgvf (VIII)

Here, the specific volume change upon vaporization is vfg.

Determine the final internal energy of the tanks.

u2=uf+x2ufg (IX)

Conclusion:

For Tank A:

From the Table A-5, “Saturated water-Pressure”, obtain the value of the specific volume of liquid, the specific volume of vapour, the specific internal energy of liquid, and the specific  internal energy change upon vaporization at 400 kPa of pressure and 0.80 of dryness fraction of water in tank A as:

vf=0.001084m3/kgvg=0.46242m3/kguf=604.22kJ/kgufg=1948.9kJ/kg

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

v1,A=(0.001084m3/kg)+(0.8)(0.46242m3/kg0.001084m3/kg)=(0.001084m3/kg)+(0.8)(0.461336m3/kg)=(0.001084m3/kg)+(0.369069m3/kg)=0.370153m3/kg

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

u1,A=(604.22kJ/kg)+(0.8)(1948.9kJ/kg)=(604.22kJ/kg)+(1559.12kJ/kg)=2163.34kJ/kg

For tank B:

The unit conversion of pressure from kPa to MPa.

P1=200kPa×(103MPa1kPa)=0.2MPa

From the Table A-5, “Superheated water-Pressure”, obtain the value of the initial specific volume of liquid and the initial specific internal energy of liquid at 0.2 MPa of pressure and 250°C of temperature of water in tank B as:

v1,B=1.1989m3/kgu1,B=2731.4kJ/kg

Substitute 0.2m3 for νA, 0.37015m3/kg for v1,A, 0.5m3 for νB, and 1.1989m3/kg for v1,B in Equation (VI).

mt=(0.2m30.37015m3/kg)+(0.5m31.1989m3/kg)=0.540321kg+0.417049kg=0.95737kg

Substitute 0.7m3 for νt and 0.95737kg for mt in Equation (VII).

v2=0.7m30.95737kg=0.731169m3/kg

From the Table A-4, “Saturated water-Temperature”, obtain the value of the specific volume of liquid, the specific volume of vapour, the specific internal energy of liquid, the specific  internal energy change upon vaporization, and the final pressure of the saturated mixture of liquid-vapour at 25°C of temperature and 0.731169m3/kg of final specific volume of water in tank B as:

vf=0.001003m3/kgvg=43.340m3/kguf=104.83kJ/kgufg=2304.3kJ/kgP2=3.1698kPa

Thus, the final pressure of the two rigid tanks is 3.1698kPa_.

Substitute 0.731169m3/kg for v2, 0.001003m3/kg for vf, 43.340m3/kg for vg in Equation (VIII).

x2=(0.731169m3/kg)(0.001003m3/kg)(43.340m3/kg0.001003m3/kg)=0.730166m3/kg43.339m3/kg=0.0168480.01645

Substitute 104.83kJ/kg for uf, 0.01645 for x2, and 2304.3kJ/kg for ufg in Equation (IX).

u2=(104.83kJ/kg)+(0.01645)(2304.3kJ/kg)=(104.83kJ/kg)+(37.90574kJ/kg)=142.7357kJ/kg

Substitute 0.95737kg for mt, 142.7357kJ/kg for u2, 0.540321kg for m1,A, 2163.34kJ/kg for u1,A, 0.417049kg for m1,B, and 2731.4kJ/kg for u1,B in Equation (III).

Qout=[(0.95737kg)(142.7357kJ/kg)(0.540321kg)(2163.34kJ/kg)(0.417049kg)(2731.4kJ/kg)]=[(136.6509kJ)(1168.898kJ)(1139.128kJ)]=[(2171.37kJ)]=2171kJ

Thus, the amount of heat transfer to the two rigid tanks is 2171kJ_.

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Chapter 4 Solutions

THERMODYNAMICS-SI ED. EBOOK >I<

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