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Calculation of overall heat transfer coefficient

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Problem Statement

Determine the overall heat transfer coefficient ‘U’ for heat transfer occurring from superheated steam in a steel pipe to atmosphere, with the following conditions.

Pipe nominal size = 8″
pipe schedule = STD
Average steam temperature over the pipe length = 2000C
Ambient air temperature = 220C
heat transfer coefficient on steam side = hS = 0.08 W/0Cm2
heat transfer coefficient on air side = hA = 0.04 W/0Cm2
conductivity‘k’ of steel = 60 W/mK (at given temperature range)

Solution

The sample problem can be solved by following the steps given here. First an equation is developed to determine the overall heat transfer coefficient for this pipe as a function of the individual heat transfer coefficients on both sides as well as the conductivity and then the overall heat transfer coefficient is calculated using the developed equation.

Step1

Refer to EnggCyclopedia’s article about heat transfer coefficients, for relation between heat transfer rate and the individual heat transfer coefficients on inside and outside of the pipe.

Q = hS×AS×(TS-T1) – steam side heat transfer
(TS-T1) = Q/(hS×AS) … (1)

Q = hA×AA×(T2-TA) – steam side heat transfer
(TA-T2) = Q/(hA×AA) … (2)

For conductive heat transfer across the pipe wall,

Hence,

(T1-T2) = Q × ln(r2/r1) / (2πkN) … (3)

(1) + (2) + (3) gives,

(TS-TA) = (TS-T1) + (TA-T2) + (T1-T2)

(TS-TA) = Q × (1/hS×AS + 1/hA×AA + ln(r2/r1)/2πkN)

Since Q = UA×AA×(TS-TA),

UA = 1/(AA/hSAS + 1/hA + AAln(r2/r1)/2πkN)

Consider unit length of the pipe, i.e. N=1m

Then, AA = 2πr2 and AS = 2πr1

UA = 1/(r2/hSr1 + 1/hA + r2×ln(r2/r1)/k) … (4)

Step2

where N is the tube length and ‘r’ stands for tube radius. The subscripts 1 and 2 stand for inner and outer tube wall respectively. Tube metal conductivity is expressed as ‘k’.

From EnggCyclopedia’s standard piping dimensions calculator, for 8″ STD schedule pipe,
pipe inner diameter = d1 = 202.72 mm = 0.20 m
pipe outer diameter = d2 = 219.08 mm = 0.22 m
r10 = 0.1 m and r2 = 0.11 m

Using the given data and equation (4),

UA = 1/(0.11/(0.08×0.10) + 1/0.04 + 0.11×ln(1.1)/60)

UA = 1/(13.75 + 25 + 0.00018) ≈ 1/38.75

UA =0.026 W/0Cm2

Step3

It should be noted that the terms ‘1/hS‘, ‘1/hA‘ and ‘r2×ln(r2/r1)/k’ represent the heat transfer resistance for convection inside and outside the pipe and for conduction across the pipe wall, respectively. Smaller magnitude of the heat transfer resistance indicates higher ease of heat transfer.

Thus it can be noted that heat transfer is most easy for conduction across the pipe wall and is represented by a negligible heat transfer resistance value.

On the other hand heat transfer resistance is higher for the convective heat transfer and inversely proportional to the related heat transfer coefficient. Hence it can be seen that most of the heat transfer resistance is contributed by heat transfer on air side of the pipe, which has the lowest heat transfer coefficient.

Finally it can be observed that the overall heat transfer coefficient is lower than heat transfer coefficients on both sides of the pipe. This is explained simply by the definition of the overall heat transfer coefficient which associates it with the largest temperature difference in the heat transfer system.

 

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