Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 5, Problem 120RQ

(a)

To determine

The final pressure and quality of the piston-cylinder device.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass of the steam (m) is 0.35kg.

The pressure of the steam (P) is 3.5 MPa.

The change in temperature (ΔTsuperheated) is 7.4°C.

The final temperature of the steam (T2) is .

Calculation:

Write the unit conversion pressure from MPa to kPa for piston-cylinder device.

  P1=3.5MPa×(1000kPa1MPa)=3500kPa

From the Table A-5 “Saturated water-Pressure table”, obtain the value of saturated temperature at 3500 kPa pressure as 242.56°C.

Determine the state 1 temperature of the piston-cylinder device.

  T1=Tsat@3500kPa+ΔTsuperheated

Here, the saturated temperature at 3500 kPa is Tsat@3500kPa and the superheated of the piston-cylinder device is ΔTsuperheated.

Substitute 242.56°C for Tsat@3500kPa and 7.4°C for ΔTsuperheated in Equation (VI).

  T1=242.56°C+7.4°C=250°C

From the Table A-4 through A-6 “Saturated water”, obtain the value of steam at various states for piston-cylinder device.

At state 1 pressure and temperature of steam as 3.5MPa and 250°C.

  v1=0.05875m3/kg.

  u1=2624.0kJ/kg.

At state 1-2 pressure and quality of state of steam as 3.5MPa and 0.

  v2=0.001235m3/kg.

  u2=1045.4kJ/kg.

At state 2-3 specific volume and temperature of steam as 0.001235m3/kg and 200°C.

  x3=0.00062.

  P3=1555kPa.

  u3=851.55kJ/kg.

Thus, the final pressure of the piston-cylinder device is 1555kPa_ and quality at the final state of the piston-cylinder device is 0.00062_.

(b)

To determine

The boundary work done of the piston-cylinder device.

(b)

Expert Solution
Check Mark

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 relation of piston-cylinder device.

  Wb,inQout=ΔU        (III)

Here, the work to be done into the system is Wb,in, the heat to be transfer by the system is Qout, and the change in the internal energy is ΔU.

Simplify the Equation (III), write energy balance relation when the piston first hits the stops state(1-2).

  Wb,inQout,1-2=m(u2u1)        (IV)

Here, the mass of the piston-cylinder device is m, the specific internal energy at state 2 for steam is u2, and the specific internal energy at state 1 for steam is u1.

Similarly the Equation (IV), when the piston first hits and the final state(1-3).

  Wb,inQout,1-3=m(u3u1)        (V)

Here, the mass of the piston-cylinder device is m, the specific internal energy at state 3 for steam is u3, and the specific internal energy at state 1 for steam is u1.

Substitute 0 for Qout in Equation (III) and write energy balance relation of boundary work for piston-cylinder device.

  Wb,in=ΔU=mP1(v1v2)        (VI)

Here, the mass of piston-cylinder device is m, the pressure at state 1 is P1, the specific volume of the state 1 is v1, and the specific volume of the state 2 is v2.

Substitute 0.35kg for m, 3500kPa for P1, 0.05875m3/kg for v1, and 0.001235m3/kg in Equation (VI).

  Wb,in=(0.35kg)(3500kPa)(0.058750.001235)m3/kg=(0.35kg)(3500kPa)(0.057515)m3/kg=70.45588kPam3×(1kJ1kPam3)=70.46kJ

Thus, the boundary work done of the piston-cylinder device is 70.46kJ_.

(c)

To determine

The amount of heat transfer when the piston first hits the stops.

(c)

Expert Solution
Check Mark

Explanation of Solution

Substitute Wb,in=70.45kJ, m=0.35kg, u2=1045.4kJ/kg, and u1=2624.0kJ/kg Equation (IV).

  Qout,1-2=(70.45kJ)(0.35kg)(1045.42624.0)kJ/kg=(70.45kJ)(0.35kg)(1569.6)kJ/kg=619.81kJ620kJ

Thus, the amount of heat transfer when the piston first hits the stops is 620kJ_.

(d)

To determine

The total amount heat transfer in the piston-cylinder device.

(d)

Expert Solution
Check Mark

Explanation of Solution

Substitute Wb,in=70.45kJ, m=0.35kg, u3=851.55kJ/kg, and u2=2624.0kJ/kg in Equation (IV).

  Qout,1-2=(70.45kJ)(0.35kg)(851.552624.0)kJ/kg=(70.45kJ)(0.35kg)(1772.45)kJ/kg=690.8kJ

Thus, the total amount heat transfer in the piston-cylinder device is 690.8kJ_.

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Chapter 5 Solutions

Fundamentals of Thermal-Fluid Sciences

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