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 151P

(a)

To determine

The mass flow rate of the cold water

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The temperature of the hot water (T1) is 70°C.

The mass flow rate of hot water (m˙1) is 3.6kg/s.

The temperature of the cold water (T2) is 20°C.

The temperature of the mixture (T3) is 42°C.

The constant pressure (P) is 200 kPa.

Calculation:

Refer Table A-4, “Saturated water-Temperature table”, obtain the following properties of the water at the temperature of 70°C.

The initial enthalpy at entry 1 (h1) is 293.07kJ/kg.

The initial entropy at entry 1  (s1) is 0.9551kJ/kgK.

Refer Table A-4, "Saturated water- Temperature table”, obtain the following properties of the water at the temperature of 20°C.

The enthalpy at entry 2(h2) is 83.91kJ/kg.

The entropy at entry 2 (s2) is 0.2965kJ/kgK.

Refer Table A-4, “Saturated water- Temperature table”, for the temperature of 42°C to obtain the following properties of water using interpolation method.

Temperature in K

Enthalpy in kJ/kgEntropy in kJ/kgK
40167.530.5724
42? 
45188.440.6386

Write the formula of interpolation method of two variables.

  y2=(x2x1)(y3y1)(x3x1)+y1        (I)

Substitute x1=40°C, x2=42°C, x3=45°C, y1=167.53kJ/kg, and y3=188.44kJ/kg in Equation (I).

  y2=(4240)(167.53188.44)(4540)+188.44=175.90kJ/kgK

The enthalpy of the mixture (h3) is 175.90kJ/kgK_.

Substitute x1=40°C, x2=42°C, x3=45°C, y1=0.5724kJ/kgK, and y3=0.6386kJ/kgK in Equation (I).

  y2=(4240)(0.63860.5724)(4540)+0.5724=0.5990kJ/kgK

The entropy of the mixture (s3) is 0.5990kJ/kgK_.

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

  E˙inE˙out=ΔE˙system        (II)

Here, rate of net energy transfer in to 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.

Substitute ΔE˙system=0 in Equation (I).

  E˙inE˙out=0E˙in=E˙outm˙1h1+m˙2h2=m˙3h3m˙1h1+m˙2h2=(m˙1+m˙2)h3        (III)

Here, mass flow rate at entry 1 is m˙1, mass flow rate at entry 2 is m˙2, mass flow rate at exit is m˙3, enthalpy at entry 1 is h1, enthalpy at entry 2 is h2 and enthalpy at exit is h3.

Rewrite the Equation (III) to obtain the mass flow rate at entry 2.

  m˙2=m˙1(h1h3)(h3h2)m˙2=3.6kg/s(293.07kJ/kg175.90kJ/kg)(175.90kJ/kg83.91kJ/kg)=4.586kg/s

Thus, the mass flow rate of the superheated steam is 4.586kg/s.

(b)

To determine

The rate of heat entropy generation during the process.

(b)

Expert Solution
Check Mark

Explanation of Solution

Write the expression to calculate the mass balance of the system.

  m˙inm˙out=Δm˙system        (IV)

Here, inlet mass flow rate is m˙in and outlet mass flow rate is m˙out and change in mass flow rate is Δm˙system.

Substitute m˙in=m˙1+m˙2, m˙out=m˙3 and Δm˙system=0 in Equation (IV).

  m˙1+m˙2m˙3=0m˙3=m˙1+m˙2m˙3=(3.6kg/s)+(4.586kg/s)=8.186kg/s

Write the expression for the entropy balance during the process.

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

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 m˙1s1+m˙2s2 for S˙in, m˙3s3 for S˙out and 0 for ΔS˙system in Equation (V).

  m˙1s1+m˙2s2m˙3s3+S˙gen=0S˙gen=m˙3s3m˙1s1m˙2s2

  S˙gen={(8.186kg/s)(0.05990kJ/kgK)(4.586kg/s)(0.2965kJ/kgK)(3.6kg/s)(0.9551kJ/kgK)}=0.1054kW/K

Thus, the rate of heat entropy generation during the process is 0.1054kW/K.

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Chapter 8 Solutions

Fundamentals of Thermal-Fluid Sciences

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