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 17, Problem 122RQ

(a)

To determine

The thickness of the insulation needed to reduce the heat loss by 95 percent.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Thermal conductivity of the steel pipe (k) is 61 W/mK.

Heat transfer coefficient inside the pipe (h1) is 105 W/m2K.

Heat transfer coefficient outside the pipe (h2) is 14 W/m2K.

Inner diameter of the pipe (Di) is 10 cm.

Outer diameter of the pipe (Do) is 12 cm.

Calculation:

Determine the inner surface area of the pipe.

  A1=πDiL=π(0.1 m)(1 m)=0.3142 m2

Determine the outer surface area of the pipe.

  A2=πDoL=π(0.12 m)(1 m)=0.377 m2

Determine the surface area of the insulation.

  A3=πD3L=πD3(1 m)=3.1416D3 m2

Determine the individual thermal resistances.

  Ri=1hiAi=1(105 W/m2°C)(0.3142 m2)=0.03031 °C/W

  R1=Rpipe =ln(r2/r1)2πk1L=ln(6 cm5 cm)2π(61 W/m°C)(1 m)=0.00048 °C/W

  R2=Rinsulation =ln(r3/r2)2πk2L=ln(D3/0.12)2π(0.038 W/m°C)(1 m)=ln(D3/0.12)0.23876 °C/W

  Ro,steel =1hoAo=1(14 W/m2°C)(0.3770 m2)=0.18947 °C/W

  Roinsulation=1h0A0=1(14 W/m2°C)(3.1416D3 m2)=0.02274D3 °C/W

  Rtotal no insulation =Ri+R1+Ro.steel =0.03031 °C/W+0.00048 °C/W+0.18947 °C/W=0.22026 °C/W

  Rtotal insulation =Ri+R1+R2+Ro,insulation =0.03031+0.00048+ln(D3/0.12)0.23876+0.02274D3=0.03079+ln(D3/0.12)0.23876+0.02274D3 °C/W

Determine the steady rate of heat loss from the steam per meter pipe length in case of no insulation.

  Q˙=T1T2Rtotal=235°C20°C0.22026 °C/W=976.1 W

Determine the diameter of insulated material.

  Q˙insulation =T1T2Rtotal insulation (0.05×976.1)W=235°C20°C(0.03079+ln(D3/0.12)0.23876+0.02274D3) °C/WD3=0.3355 m

Determine the thickness of insulation.

  t=D3D22=33.55 cm12 cm2=10.8 cm

Thus, the thickness of insulation is 10.8 cm_.

(b)

To determine

The thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to 40°C.

(b)

Expert Solution
Check Mark

Explanation of Solution

Determine the diameter of insulated material.

  Q˙insulation =T1T2Rtotal insulation Q˙insulation =T2T2Ro, insulation (235°C20°C)(0.03079 °C/W+ln(D3/0.12)0.23876+0.02274D3)=(40°C20°C)0.02274D3 °C/WD3=0.1644 m

Determine the thickness of the insulation.

  t=D3D22=16.44 cm12 cm2=2.22 cm

Thus, the thickness of insulation is 2.22 cm_.

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Chapter 17 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

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