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 75P

(a)

To determine

The temperature of the air at the exit.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The inlet pressure of Refrigerant (P3) is 20 psia.

The inlet quality of Refrigerant (x3) is 30%.

The outlet pressure of Refrigerant (P3) is 20 psia.

The outlet quality of steam (x4) is 1 (saturated vapor).

The inlet pressure of air (P1) is 14.7 psia.

The inlet temperature of air (T1) is 90°F.

The volume flow rate of air (V˙1) is 200 ft3/min.

The mass flow rate of refrigerant (m˙R) is 4 lbm/min.

Calculation:

Draw the schematic diagram for the evaporator section.

Fundamentals of Thermal-Fluid Sciences, Chapter 6, Problem 75P

Write the formula to calculate the specific enthalpy of steam (h) from tables.

  h=hf+x(hfg)        (I)

Here, specific enthalpy of saturated liquid is hf, specific enthalpy of saturated liquid-vapor mixture is hfg and dryness fraction of water is x.

Write the ideal gas equation for specific volume of air (v).

  v=RTP        (II)

Here, gas constant for air is R, temperature of air is T, and the pressure of air is P.

Calculate the volume flow rate (V˙1) of the air at the inlet.

  V˙1=m˙av1        (III)

Here, mass flow rate of air is m˙a and specific volume of air is v1.

Write the expression for the mass balance.

  minmout=Δm

  minmout=0(For steady state)        (IV)

Here, mass of the fluid entering into the system is min, mass of fluid leaving out of the system is mout and change in mass is Δm.

Write the energy rate balance equation for a control volume.

  E˙inE˙out=dE˙system/dt

  E˙inE˙out=0(For steady state)        (V)

Here, total energy rate at inlet is E˙in, total energy rate at outlet is E˙out and change in net energy rate is dE˙system/dt.

Refer Table A-1E, “Gas constant of common gases”, obtain the gas constant of air as 0.3704psiaft3/lbmR.

Refer Table A-2E, “Ideal – gas specific heats of common gases”, obtain the constant pressure specific heat of air as 0.240Btu/lbm.°F.

Refer Table A-12E, “Saturated R-134a-Pressure table”, obtain the properties of saturated R-134a at pressure (P3) of 20 psia.

  hf=11.436Btu/lbmhfg=91.302 Btu/lbm

Substitute hf=11.436Btu/lbm, x3=0.3, and hfg=91.302Btu/lbm in Equation (I).

  h3=11.436Btu/lbm+(0.3)91.302Btu/lbm=38.83Btu/lbm

Refer Table A-12E, “Saturated R-134a-Pressure table”, obtain the specific enthalpy (h4) of saturated vapor of R-134aat pressure (P4) of 20 psia as 102.74Btu/lbm.

Substitute R=0.3704psiaft3/lbmR, T1=90°F, and P1=14.7psia in Equation (II).

  v1=RT1P1=0.3704psiaft3/lbmR(90°F)14.7psia=0.3704psiaft3/lbmR(90+460)R14.7psia

  v1=13.86ft3/lbm

Substitute V˙1=200ft3/min and v1=13.86ft3/lbm in Equation (III).

  m˙a=200ft3/min13.86ft3/lbm=14.43lbm/min

Rewrite Equation (I) for the mass flow rate of air (m˙a).

  m˙1=m˙2=m˙a

Here, mass flow rate of air at inlet is m˙1 and mass flow rate of air at outlet is m˙2.

Rewrite the Equation (I) for the mass flow rate of R-134a (m˙R).

  m˙3=m˙4=m˙R

Here, mass flow rate of R-134a at inlet is m˙3, and mass flow rate of R-134a at outlet is m˙4.

Rewrite Equation (II) for the energy balance in the steam heating system.

  m˙1h1+m˙3h3=m˙2h2+m˙4h4        (V)

Substitute m˙a=m˙1, m˙2=m˙a, m˙3=m˙R, and m˙R for m˙4 in Equation (V).

  m˙ah1+m˙Rh3=m˙ah2+m˙Rh4m˙R(h3h4)=m˙a(h2h1)m˙R(h3h4)=m˙acp(T2T1)

  T2=T1+m˙R(h3h4)m˙acp

Substitute h3=38.83 Btu/lbm, h4=102.74 Btu/lbm, cp=0.240 Btu/lbm.°F, m˙R=4 lbm/min, m˙a= , and T1=90°F in above  Equation.

  T2=90°F+4 lbm/min×(38.83 Btu/lbm102.74 Btu/lbm)14.43 lbm/min×0.240 Btu/lbm.°F=16.2°F

Thus, the temperature of the air at the exit is 16.2°F.

(b)

To determine

The rate of heat transfer from the air to the refrigerant.

(b)

Expert Solution
Check Mark

Explanation of Solution

Write the formula to calculate the rate of heat transfer from the air to the refrigerant (Qair,out).

  Qair,out=m˙acp(T2T1)

Substitute m˙a=14.43lbm/min, cp=0.240Btu/lbm°F, T2=16.24°F and T1=90°F in above Equation.

  Qair,out=(14.43lbm/min)(0.240Btu/lbm°F)(16.24°F90°F)=255.6Btu/min

Thus, the rate of heat transfer from the air to the refrigerant is 255.6Btu/min.

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Chapter 6 Solutions

Fundamentals of Thermal-Fluid Sciences

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