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 74P
To determine

The mass flow rate of the refrigerant.

Expert Solution & Answer
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Explanation of Solution

Given:

The inlet pressure of Refrigerant (P1) is 1 MPa.

The inlet temperature of Refrigerant  (T1) is 90°C.

The outlet pressure of Refrigerant  (P2) is 1 MPa.

The outlet temperature of Refrigerant  (T2) is 30°C.

The inlet pressure of air (P3) is 100 kPa.

The inlet temperature of air (T3) is 27°C.

The outlet pressure of air (P4) is 95 kPa.

The outlet temperature of air (T4) is 60°C.

Calculation:

Consider the system is in steady state. Hence, the inlet and exit mass flow rates are equal.

The mass flow rate of air (m˙a) is as follows.

  m˙3=m˙4=m˙a

The mass flow rate of refrigerant (m˙R) is as follows.

  m˙1=m˙2=m˙R

Write the energy rate balance equation for one inlet and one outlet system.

  [Q˙1+W˙1+m˙(h1+V122+gz1)][Q˙2+W˙2+m˙(h2+V222+gz2)]=ΔE˙system        (I)

Here, the rate of heat transfer is Q˙, the rate of work transfer is W˙, the enthalpy is h and the velocity is V, the gravitational acceleration is g, the elevation from the datum is z and the rate of change in net energy of the system is ΔE˙system; the suffixes 1 and 2 indicates the inlet and outlet of the system.

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

  ΔE˙system=0

Neglect the work transfer, heat transfer to the surrounding, potential and kinetic energies.

The Equations (I) reduced as follows for refrigerant .

  [0+0+m˙1(h1+0+0)][0+0+m˙2(h2+0+0)]=0m˙1h1m˙2h2=0m˙1h1=m˙2h2        (II)

The Equations (I) reduced as follows for air.

  [0+0+m˙3(h3+0+0)][0+0+m˙4(h4+0+0)]=0m˙3h3m˙4h4=0m˙3h3=m˙4h4        (III)

Combining Equation (II) and (III).

  m˙1h1+m˙3h3=m˙2h2+m˙4h4m˙2h2m˙1h1=m˙3h3m˙4h4

Substitute m˙R=m˙1,m˙2 and m˙a=m˙3,m˙4 in the above equation.

  m˙ah4m˙ah3=m˙Rh1m˙Rh2m˙a(h4h3)=m˙R(h1h2)m˙R=m˙a(h4h3)h1h2        (IV)

Write the formula for change in enthalpy (h4h3) of air.

  h4h3=cp,a(T4T3)

Substitute h4h3=cp,a(T4T3) in Equation (IV).

  m˙R=m˙a[cp,a(T4T3)]h1h2=m˙acp,a(T4T3)h1h2        (V)

For refrigerant:

At inlet:

The refrigerant is at the state of superheated condition.

Refer Table A-13, “Superheated refrigerant-134a”.

Obtain the inlet enthalpy (h1) corresponding to the pressure of 1MPa and the temperature of 90°C.

  h1=324.66kJ/kg

At exit:

The refrigerant is at the state of saturated liquid.

Refer Table A-11, “Saturated refrigerant-134a-Temperature table”.

Obtain the exit enthalpy (h2) corresponding to the temperature of 30°C.

  hf=h2=93.58kJ/kg

For air:

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

The gas constant of air (R) is 0.287kPam3/kgK.

Refer Table A-2, “Ideal2gas specific heats of various common gases”.

The specific heat at constant pressure (cp@300K) at the temperature of 300K is 1.005kJ/kg°C.

Write the formula for mass flow rate of air (m˙a).

  m˙a=V˙P3RT3        (VI)

Here, the volumetric flow rate of air is V˙, the pressure is P, the gas constant of air is R and the temperature is T; the suffix one indicates the inlet condition of air.

Conclusion:

Substitute V˙=600m3/min, P3=100kPa, R=0.287kPam3/kgK and T3=27°C in Equation (VI).

  m˙air=(600m3/min)(100kPa)(0.287kPam3/kgK)(27°C)=60000kPam3/min(0.287kPam3/kgK)(27+273)K=60000kPam3/min86.1kPam3/kg=696.8641kg/min

Substitute m˙a=696.8641kg/min, cp,a=1.005kJ/kg°C, T4=60°C, T3=27°C, h1=324.66kJ/kg, h2=93.58kJ/kg in Equation (V).

  m˙R=(696.8641kg/min)(1.005kJ/kg°C)(60°C27°C)324.66kJ/kg93.58kJ/kg=(696.8641kg/min)(1.005kJ/kg°C)(33K)231.08=23111.4979kJ/min231.08kJ/kg=100.0151kg/min

  =100kg/min

Thus, the mass flow rate of the refrigerant is 100kg/min.

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Chapter 6 Solutions

Fundamentals of Thermal-Fluid Sciences

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