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 22, Problem 23P
To determine

The overall heat transfer coefficient based on the inner and the outer surface of copper tube.

Expert Solution & Answer
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Explanation of Solution

Given:

The heat capacity for hot fluid (C) is 4440W/K.

The inlet temperature for hot fluid (Thi) is 150°C.

The length of tube (L) is 1.5m.

The thermal conductivity of copper (k) is 250W/mK.

The inner diameter of inner tube (Di) is 2cm.

The outer diameter of inner tube (Do) is 2.25cm.

The mass flow rate of hot fluid (m˙h) is 2kg/s.

The exit temperature of hot fluid (Te) is 50°C.

The inlet temperature for cold fluid (Tci) is 20°C.

The outlet temperature for cold fluid (Tce) is 70°C.

The outside fouling factor (Rf)o is 0.00015m2K/W.

The outside fouling factor (Rf)i is 0.0001m2K/W.

Calculation:

Refer Table A-19 “Properties of liquid”.

Obtain the following properties of liquid corresponding to the temperature of 100°C as follows:

ρ=840kg/m3

cp=2220kJ/kgK

k=0.1367W/mK

μ=0.01718×103kg/ms

Pr=279.1

Refer Table A-15 “Properties of water”.

Obtain the following properties of liquid corresponding to the average temperature of 45°C as follows:

ρ=990.1kg/m3

cp=4180kJ/kgK

k=0.637W/mK

μ=0.596×103kg/ms

Pr=3.91

Calculate the inner surface areas of the heart exchanger.

  Ai=πDiL=π(2cm(1m100cm))(1.5m)=0.0942m2

Calculate the outer surface areas of the heart exchanger.

  Ao=πDoL=π(2.25cm(1m100cm))(1.5m)=0.106m2

Calculate the volume flow rate of the oil.

  V=4m˙hρπDi2=4(2kg/s)π(840kg/m3)(2cm(1m100cm))=7.58m/s

Calculate the Reynolds number for the oil.

  Re=ρVDiμ=(840kg/m3)(7.58m/s)(2cm(1m100cm))0.01718kg/ms=7412

Calculate the mass flow rate of cooling water.

  C(ThiThe)=m˙ccpc(TceTci)(4440W/K)(150°C50°C)=m˙c(70°C20°C)m˙c=2.124kg/s

Calculate the hydraulic diameter of the annular space on the shell side.

  Dh=DoDi=6cm(1m100cm)2.25cm(1m100cm)=0.0375m

Calculate the average velocity of cooling water.

  V=4m˙ρπDh2=4(2.124kg/s)(990.1kg/m3)π(0.0375m)2=1.942m/s

Calculate the number for the flow of water.

  Re=ρVDhμ=(990.1kg/m3)(1.942m/s)(0.0375m)0.596×102kg/ms=120980

Calculate the friction factor for oil.

  f=(0.79lnRe1.64)2=(0.79ln(7412)1.64)2=0.03429

Calculate the friction factor for water.

  f=(0.79lnRe1.64)2=(0.79ln(120980)1.64)2=001728

Calculate the Nusselt number for oil.

  Nu=(f/8)(Re1000)Pr1+12.7f/8(Pr2/31)=(0.03429/8)(74121000)279.11+12.70.03429/8(279.12/31)=214.94

Calculate the inner convective heat transfer coefficient for oil.

  hi=kDNu=(0.1367W/mK)(214.94)0.02m=1469.11W/m2K

Calculate the Nusselt number for water.

  Nu=(f/8)(Re1000)Pr1+12.7f/8(Pr2/31)=(0.01728/8)(1209801000)3.911+12.70.01728/8(3.912/31)=540.52

Calculate the inner convective heat transfer coefficient for water.

  ho=kDhNu=(0.637W/mK)(540.52)0.0375m=9188.25W/m2K

Calculate the overall heat transfer coefficient.

    1UAs=1hiAi+(Rf)iAi+ln(DoDi)2πkL+(Rf)oAo+1hoAo=[1(1469.1W/m2K)(0.0942m2)+0.0015m2K/W0.0942m2+ln(2.25cm2cm)2π(25W/mK)(1.5m)+0.0001m2K/W0.106m2+1(9188.25W/m2K)(0.106m2)]=0.01083K/W

Calculate the overall heat transfer based on the inner surface.

  1UiAi=1UAs1Ui=(0.01083K/W)(0.0942m2)Ui=979.4W/m2K

Calculate the overall heat transfer based on the outer surface.

  1UoAo=1UAs1Ui=(0.01083K/W)(0.106m2)Ui=871.09W/m2K

Thus, overall heat transfer based on the outer surface is 871.09W/m2K and overall heat transfer based on the inner surface is 979.4W/m2K.

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Chapter 22 Solutions

Fundamentals of Thermal-Fluid Sciences

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