Engineering Heat transfer from composite pipe -- Is my answer right?

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The discussion revolves around calculating heat loss from a composite pipe carrying water at 85ºC, insulated with fire glass. The user is seeking validation for their calculations, which yield a heat loss of approximately 29.143 W/m and a total heat loss of 874.29 W over 30 meters. The interface temperature (t2) is calculated to be around 75.08ºC. The textbook answers provided are 57.99 W/m, 1739.7 W, and 84.9ºC, indicating discrepancies in the user's results. The calculations involve thermal conductivity values and geometric parameters of the pipe and insulation.
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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).
  1. r1 = 65/2 = 32.5/1000 = 0.0325m
  2. r2 = 0.0325 + 0.006 = 0.0385m
  3. r3 = 0.385 + 0.035 = 0.0735m

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

  1. Q = (2 x pi x 1 x (85-10)) / ((Ln (0.0385/0.0325))/48) + ((Ln(0.0735/0.0385))/0.04)
  2. Q = 471.239 / 0.00353 + 16.166
  3. Q = 471.239 / 16.170
  4. Q = 29.143
heat loss over whole pipe

t2 = (Q Ln (r3/r2) / k2 x 2 x 3.142 x L) + t3
  1. t2 = (29.143 x Ln (0.0735/0.0385)) / ( 0.04 x 2 x pi x 1 )
  2. t2 = 18.845 / 0.251
  3. t2 = 75.0797

  1. 29.143 x 30 = 874.29
Find temperature (t2)
 
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