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Heat Transfer - Engineering

Q1:

A 10 cm dia steam pipe, carrying steam at 180°C, is covered with an insulation (conductivity = 0.6 W/m.°C). It losses heat to the surroundings at 30°C. Assume a heat transfer co-efficient of 0.8 W/m2.°C for heat transfer from surface to the surroundings. Neglect wall resistance of the pipe and film resistance of steam. If the insulation thickness is 2 cms, the rate of heat loss from this insulated pipe will be

A greater than that for uninsulated steam pipe.

B less than that of the uninsulated steam pipe.

C equal to that of the uninsulated steam pipe.

D less than the steam pipe with 5 cms insulation.

ANS:B - less than that of the uninsulated steam pipe.

To calculate the rate of heat loss from the insulated pipe, we can use the formula for heat transfer through a cylindrical wall with insulation: surroundings)Q=hins​Ains​1​+2πkins​lnrpipe​rins​​​2πL​×(Tpipe​−Tsurroundings​) Where:

  • Q is the rate of heat loss (in Watts),
  • L is the length of the pipe (in meters),
  • ℎinshins​ is the heat transfer coefficient from the insulation surface to the surroundings (in W/m².°C),
  • insAins​ is the surface area of the insulation (in m²),
  • insrins​ is the radius of the insulation (in meters),
  • piperpipe​ is the radius of the pipe (in meters),
  • inskins​ is the thermal conductivity of the insulation (in W/m.°C),
  • pipeTpipe​ is the temperature of the pipe (in °C), and
  • surroundingsTsurroundings​ is the temperature of the surroundings (in °C).
Given:
  • Diameter of the steam pipe = 10 cm = 0.1 m
  • Radius of the steam pipe (piperpipe​) = 0.05 m
  • Thickness of the insulation (L) = 2 cm = 0.02 m
  • Thermal conductivity of the insulation (inskins​) = 0.6 W/m.°C
  • Temperature of the steam pipe (pipeTpipe​) = 180°C
  • Temperature of the surroundings (surroundingsTsurroundings​) = 30°C
  • Heat transfer coefficient from insulation surface to surroundings (ℎinshins​) = 0.8 W/m².°C
First, we need to calculate the radius of the insulation (insrins​): ins=pipe+insulation thicknessrins​=rpipe​+insulation thickness ins=0.05 m+0.02 m=0.07 mrins​=0.05m+0.02m=0.07m Now, we can calculate the surface area of the insulation (insAins​): ins=2insAins​=2πrins​L ins=2××0.07×0.02Ains​=2×π×0.07×0.02 ins≈0.0088 m2Ains​≈0.0088m2 Now, we can substitute the values into the formula and calculate the rate of heat loss (Q): surroundings)Q=hins​Ains​1​+2πkins​lnrpipe​rins​​​2πL​×(Tpipe​−Tsurroundings​) =2×0.0210.8×0.0088+ln⁡0.070.052×0.6×(180−30)Q=0.8×0.00881​+2π×0.6ln0.050.07​​2π×0.02​×(180−30) =0.04113.636+ln⁡1.40.6×150Q=113.636+0.6ln1.4​0.04​×150 ≈0.04113.636+0.389×150Q≈113.636+0.3890.04​×150 ≈0.04114.025×150Q≈114.0250.04​×150 ≈0.001315×150Q≈0.001315×150 ≈0.19725 WQ≈0.19725W Therefore, the rate of heat loss from the insulated pipe is approximately 0.19725 Watts. To compare with an uninsulated steam pipe, we need to calculate the rate of heat loss without insulation. The formula for heat loss without insulation is: