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- Homework Statement
- A water pipe of bore 65 mm bore and 6mm wall thickness, carrying water at 85ºC is

insulated with one layer of lagging. The lagging is made from fire glass which is 35 mm thick

and has a thermal conductivity of 0.04 W/m K. The outside air temperature is 10ºC and the

thermal conductivity of steel is 48 W/m K. Calculate the heat loss per metre length of pipe,

the total heat loss over a 30 m length of the pipe and the interface temperature (t2).

[Answers: 57.99 W/m: 1739.7 W: 84.9ºC].

- Relevant Equations
- Q = 2 x 3.142 x L (t1 – t3) / {[Ln (r2/r1)/k1] + [Ln (r3/r2)/k2]}

t2 – t3 = Q Ln (r3/r2) / k2 x 2 x 3.142 x L

Im practicing the questions in the problem book and seem to be getting different answers to the book can somebody check cheers.

[Answers: 57.99 W/m: 1739.7 W: 84.9ºC] textbook answers

A water pipe of bore 65 mm bore and 6mm wall thickness, carrying water at 85ºC is

insulated with one layer of lagging. The lagging is made from fire glass which is 35 mm thick

and has a thermal conductivity of 0.04 W/m K. The outside air temperature is 10ºC and the

thermal conductivity of steel is 48 W/m K. Calculate the heat loss per metre length of pipe,

the total heat loss over a 30 m length of the pipe and the interface temperature (t2).

Q = 2 x 3.142 x L (t1 – t3) / {[Ln (r2/r1)/k1] + [Ln (r3/r2)/k2]}

t2 = (Q Ln (r3/r2) / k2 x 2 x 3.142 x L) + t3

[Answers: 57.99 W/m: 1739.7 W: 84.9ºC] textbook answers

A water pipe of bore 65 mm bore and 6mm wall thickness, carrying water at 85ºC is

insulated with one layer of lagging. The lagging is made from fire glass which is 35 mm thick

and has a thermal conductivity of 0.04 W/m K. The outside air temperature is 10ºC and the

thermal conductivity of steel is 48 W/m K. Calculate the heat loss per metre length of pipe,

the total heat loss over a 30 m length of the pipe and the interface temperature (t2).

- r1 = 65/2 = 32.5/1000 = 0.0325m
- r2 = 0.0325 + 0.006 = 0.0385m
- r3 = 0.385 + 0.035 = 0.0735m

Q = 2 x 3.142 x L (t1 – t3) / {[Ln (r2/r1)/k1] + [Ln (r3/r2)/k2]}

- Q = (2 x pi x 1 x (85-10)) / ((Ln (0.0385/0.0325))/48) + ((Ln(0.0735/0.0385))/0.04)
- Q = 471.239 / 0.00353 + 16.166
- Q = 471.239 / 16.170
- Q = 29.143

t2 = (Q Ln (r3/r2) / k2 x 2 x 3.142 x L) + t3

- t2 = (29.143 x Ln (0.0735/0.0385)) / ( 0.04 x 2 x pi x 1 )
- t2 = 18.845 / 0.251
- t2 = 75.0797

- 29.143 x 30 = 874.29