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 8, Problem 138P

(a)

To determine

The rate of heat transfer in the heat exchanger.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The mass flow rate of ethylene glycol (m˙) is 2 kg/s.

The specific heat of ethylene glycol at constant pressure (cp) is 2.56kJ/kgK.

The entry temperature of ethylene glycol (T1) is 80 C.

The exit temperature of ethylene glycol (T2) is 40 C.

The specific heat of water at constant pressure (cp) is 4.18kJ/kgK.

The entry temperature of water (T3) is 20 C.

The exit temperature of water (T4) is 55 C.

Calculation:

For the steady flow system, rate of change in internal energy of the system is zero.

ΔE˙system=0

Write the equation for the energy balance equation for closed system.

  E˙inE˙out=ΔE˙systemE˙inE˙out=0m˙h1+Q˙in=m˙h2Q˙in=m˙(h2h1)=m˙cp(T2T1)        (I)

Here, rate of net energy transfer into the control volume is E˙in, rate of net energy transfer exit from the control volume is E˙out and rate of change in internal energy of system is ΔE˙system,

Refer Table A-3, “properties of common liquids table”, select the specific heat at constant pressure (cp) for ethylene glycol substance as 2.56kJ/kgK.

The rate of heat transfer from the water must be equal to the rate of heat transfer to the ethylene glycol.

  Q˙in=Q˙outQ˙out=m˙cp(T1T2)

  Q˙out=(2kg/s)(2.56kJ/kgK)((80+273)K(40+273)K)=204.8kJ/s(1kW1kJ/s)=204.8kW

Thus, the rate of heat transfer in the heat exchanger is 204.8 kW_.

(b)

To determine

The rate of entropy generation in the heat exchanger.

(b)

Expert Solution
Check Mark

Explanation of Solution

Refer Table A-3, “Properties of common liquids table”, note down the specific heat at constant pressure (cp) for water substance as 4.18 kJ/kgK.

Calculate mass flow rate of water using the rate of heat from the geothermal water.

  Q˙in=m˙watercp(T4T3)204.8kJ/s=m˙water(4.18kJ/kgK)((55+273)K(20+273)K)m˙water=204.8146.3=1.4kg/s

Write the expression for the entropy balance in the heat exchanger.

  S˙inS˙out+S˙gen=ΔS˙system        (II)

Here, rate of net input entropy is S˙in, rate of net output entropy is S˙out, rate of entropy generation is S˙gen, and rate of change of entropy of the system is ΔS˙system.

Substitute S˙in=m˙1s1+m˙3s3, S˙out=m˙2s2+m˙4s4 and ΔS˙system=0 in Equation (II).

  m˙1s1+m˙3s3(m˙2s2+m˙4s4)+S˙gen=0m˙1s1+m˙3s3m˙2s2m˙4s4+S˙gen=0S˙gen=m˙glycols1m˙waters3+m˙glycols2+m˙waters4S˙gen=m˙glycol(s2s1)+m˙water(s4s3)

  S˙gen=m˙glycolcpln(T2T1)+m˙watercpln(T4T3)

  S˙gen={(2kg/s)(2.56kJ/kgK)ln((40°C)(80°C))+(1.4kg/s)(4.18kJ/kgK)ln((55°C)(20°C))}S˙gen={(2kg/s)(2.56kJ/kgK)ln((40+273)K(80+273)K)+(1.4kg/s)(4.18kJ/kgK)ln((55+273)K(20+273)K)}=0.0446kJ/sK(1kW1kJ/s)=0.0446kW/K

Thus, the rate of entropy generation in the heat exchanger is 0.0446kW/K_.

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Chapter 8 Solutions

Fundamentals of Thermal-Fluid Sciences

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