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 6, Problem 167RQ

(a)

To determine

The final temperature of air in the cylinder.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The initial mass of refrigerant (m) is 2 kg.

The initial pressure (P1) is 800 kPa.

The initial temperature (T1) is 80°C.

The final pressure (P2) is 500 kPa.

The final temperature (T1) is 20°C.

Write the equation of mass balance.

  minme=Δmsystem        (I)

Here, the inlet mass is min, the exit mass is me and the change in mass of the system is Δmsystem.

The change in mass of the system for the control volume is expressed as,

  Δmsystem=(m2m1)cv

Here, the suffixes 1 and 2 indicates the initial and final states of the system.

Consider the piston-cylinder as the control volume. Initially the cylinder is filled with air and the valve is in closed position, further no other mass is allowed to enter the cylinder. Hence, the inlet mass is neglected i.e. min=0.

Rewrite the Equation (I) as follows.

  0me=(m2m1)cvme=m2m1me=m1m2        (II)

Write the formula for initial volume of air present in the cylinder.

  ν1=m1RT1P1        (III)

Here, the mass of air is m, the pressure is P, the volume is ν, the gas constant of is R, the temperature is T and the subscript 1 indicates the initial state.

Write the formula for mass of air present in the cylinder at final state.

  m2=P2ν2RT2        (IV)

Here, the subscript 2 indicates the final state.

Write the energy balance equation.

  EinEout=ΔEsystem{[Qin+Win+min(h+ke+pe)in][Qe+We+me(h+ke+pe)e]}=[m2(u+ke+pe)2m1(u+ke+pe)1]system        (V)

Here, the heat transfer is Q, the work transfer is W, the enthalpy is h, the internal energy is u, the kinetic energy is ke, the potential energy is pe and the change in net energy of the system is ΔEsystem; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The pressure of 600kPa is required to move the piston i.e. boundary work required to move the piston. When the valve is opened and air is withdrawn at the bottom of the cylinder. The boundary work is done on the system. i.e. Win=Wb,in,We=0. The heat transfer occurs while air escapes the cylinder. Neglect the kinetic and potential energy changes i.e. (Δke=Δpe=0).

The Equation (V) reduced as follows.

  Wb,inQemehe=m2u2m1u1        (VI)

Write the formula for boundary work done on the cylinder.

  Wb,in=P2Δν=P2(ν1ν2)        (VII)

Here, the pressure required to move the piston is P2.

The enthalpy and internal energy in terms of temperature and specific heats are expressed as follows.

  h=cpTu=cvT

Rewrite the Equation (VI) as follows.

  Wb,inQemecpTe=m2cvT2m1cvT1        (VIII)

The temperature of the air while exiting the cylinder is considered as the average temperature of initial and final temperatures.

  Te=T1+T22=(200+273)K+T22=473K+T22=473+T22

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-2b, “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure (cp) of air corresponding to the anticipated average temperature of 450K is 1.020kJ/kgK and the specific heat at constant volume (cv) is 0.733kJ/kgK.

Calculation:

Substitute m1=1.2kg, R=0.287kPa.m3/kgK, T1=200°C , and  P1=700kPa in Equation (III).

  ν1=(1.2kg)(0.287kPa.m3/kgK)(200°C)700kPa=(1.2kg)(0.287kPa.m3/kgK)(20+273)K700kPa=162.9012kPa.m3700kPa=0.2327m3

It is given that the final volume is 80 % of initial volume.

  ν2=80100ν1=0.8ν1=0.8(0.2327m3)=0.1862m3

Substitute P=600kPa, ν1=0.2327m3, and ν2=0.1862m3 in Equation (VII).

  Wb,in=600kPa(0.2327m30.1862m3)=600kPa(0.0465m3)=27.9kPam3×1kJ1kPam3=27.9kJ

Substitute P2= for, ν2=0.1862m3, and R=0.287kPa.m3/kgK in Equation (IV).

  m2=(600kPa)(0.1862m3)(0.287kPa.m3/kgK)T2=389.27T2kg

Substitute m1=1.2kg and m2=389.27T2kg in Equation (I).

  me=1.2kg389.27T2kg=1.2389.27T2        (IX)

Substitute Wb,in=27.9kJ, Qe=40kJ, me=1.2389.27T2, cp=1.020kJ/kgK, Te=473+T22, m2=389.27T2kg, cv=0.733kJ/kgK, m1=1.2kg, T1=473K in Equation (VIII).

{27.9kJ40kJ[(1.2389.27T2)(1.020kJ/kgK)(473+T22)]}=[(389.27T2kg)(0.733kJ/kgK)T2(1.2kg)(0.733kJ/kgK)(473K)]12.1kJ[(1.2T2389.27T2)(1.020kJ/kgK)(473+T22)]=285.3349kJ416.0508kJ(1.2T2389.27T2)(1.020kJ/kgK)(473+T22)=(285.3349kJ416.0508kJ+12.1kJ)(1.2T2389.27T2)(1.020kJ/kgK)(473+T22)=118.6159kJ

(1.2T2389.27T2)(1.020kJ/kgK)(473+T22)=118.6159kJ        (X)

Use Engineering Equation Solver (EES) or online calculator to solve the Equation (X) and obtain the value of T2 and consider the positive root alone.

  T2=414.96659K415K

Thus, the final temperature of air in the cylinder is 415K.

(b)

To determine

The amount of mass escaped from the cylinder.

(b)

Expert Solution
Check Mark

Explanation of Solution

The amount of mass escaped from the cylinder is nothing but the mass of air vented out until final state i.e. me.

Refer Equation (II) and (IX).

Calculation:

Substitute T2=415K in Equation (IX).

  me=1.2kg389.27415Kkg=1.2kg0.938kg=0.262kg

Thus, the amount of mass escaped from the cylinder is 0.262kg.

(c)

To determine

The work done.

(c)

Expert Solution
Check Mark

Explanation of Solution

The work done is nothing but the work done on the piston to move it i.e. boundary work (Wb,in).

Refer part (a).

  Wb,in=27.9kJ

Thus, the work done is 27.9kJ.

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Chapter 6 Solutions

Fundamentals of Thermal-Fluid Sciences

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