Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 19, Problem 128RQ

(a)

To determine

The tube length using the appropriate Nusselt number relation for liquid metals.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The flow rate of mercury (m˙) is 0.6kg/s.

The diameter (D) is 5cm.

The inlet mean temperature (Ti) is 100°C.

The outlet mean temperature (To) is 200°C.

The maintained surface temperature of the tube surface (Ts) is 250°C.

Calculation:

Calculate Tf.

  Tf=100°C+200°C2=150°C

Refer to table A-20 “Properties of liquid metals” to obtain the following properties corresponding to the temperature of 150°C:

  cp=136.1J/kgKk=10.0778W/mKμ=1.126×103kg/msPr=0.0152

And the Prandtl number corresponding to the surface temperature of 250°C is (Pr)s=0.0119.

Calculate the Reynolds number (Re) by using the relation.

    Re=4m˙πDμ=4(0.6kg/s)π(5cm)(1.126×103kg/ms)=4(0.6kg/s)π[(5cm)(1m100cm)](1.126×103kg/ms)=13570

Calculate the Nusselt number (Nu) by using the relation.

    Nu=4.8+0.0156Re0.85(Pr)s0.93=4.8+0.0156(13570)0.85(0.0119)0.93=5.624

Calculate the convection heat transfer coefficient (h) using the relation.

    h=kNuD=(10.0778W/mK)(5.624)5cm=(10.0778W/mK)(5.624)(5cm)(1m100cm)=1134W/m2K

Calculate the length of the tube (L) by using the relation.

    L1=m˙cpπDhln(TsToTsTi)=(0.6kg/s)(136.1J/kgK)π(5cm)(1134W/m2K)ln(250°C100°C250°C200°C)=(0.6kg/s)(136.1J/kgK)π(5cm)(1m100cm)(1134W/m2K)ln(250°C100°C250°C200°C)=0.504m

Thus, the tube length is 0.504m.

(b)

To determine

The tube length using the appropriate Dittus–Boelter equation.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The velocity of air (V) is 25m/s.

The mean temperature (T) is 50°C.

Calculation:

Calculate the Nusselt number (Nu) by using the relation.

    Nu=0.023Re0.8(Pr)0.4=0.023(13570)0.8(0.0152)0.93=8.72

Calculate the convection heat transfer coefficient (h) using the relation.

    h=kNuD=(10.0778W/mK)(8.72)5cm=(10.0778W/mK)(8.72)(5cm)(1m100cm)=1758W/m2K

Calculate the length of the tube (L) by using the relation.

    L2=m˙cpπDhln(TsToTsTi)=(0.6kg/s)(136.1J/kgK)π(5cm)(1758W/m2K)ln(250°C100°C250°C200°C)=(0.6kg/s)(136.1J/kgK)π(5cm)(1m100cm)(1758W/m2K)ln(250°C100°C250°C200°C)=0.325m

Thus, the tube length is 0.325m.

(c)

To determine

The comparison.

(c)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Calculate the difference between the length (Difference) using the relation.

    Difference=L1L2=0.504m0.325m=0.179m

Thus, the length of the pipe obtained in first case (part ‘a’) is 0.179m greater as compared to the length of the pipe obtained by second process.

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Chapter 19 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

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