Question

Pressurized steam at 400 K flows through a long, thin- walled pipe of 0.6-m diameter. The pipe is enclosed in a concrete casing that is of square cross section and 1.75 m on a side. The axis of the pipe is centered in the casing, and the outer surfaces of the casing are maintained at 300 K. What is the rate of heat loss per unit length of pipe

Answer 1

Explanation:

assume steady state conditions, negligible steam side convention resistance, pipe wall resistance and contact resistance i.e $T_{1} =400k$

and constant properties of concrete (300 k); $k=1.4Wm/k$

the heat rate can be expressed as

$q=Sk(T_{1} -T_{2} )=Ak(T_{1} -T_{2} )$

the shape factor is

$S=(2\pi L/ln(1.08w/D))$

$hence \\q'=q/L\\=2\pi k(T_{1}- T_{2}) /ln(1.08w/D)$

insert values

$=(2\pi *1.4W/m.k*(400-300)k)/(ln(1.08*1.75/0.6)$

$q'=766W/m$   ( rate of heat loss per unit length)

Answer 2

Answer:

Q= 930.92W/m

Explanation:

Note that thermal conductivity of some common materials is from engineering tool box.

And the proper shape factor needed for the solution is found from the List of shape factor

Pressure of steam(T1)=400k

Diameter of the pipe(D)=0.6m

Square cross section of the concrete casing (W) =1.75m

Outer surface of casting maintained at a pressure (T2)=300k

Thermal conductivity (K) =1.7W/mK

The proper shape factor needed for the solution is found from the List of shape factor

Rate of heat loss (Q) =?

The heat rate (q) :

q=SKΔT1-T2=SK(T1-T2)

S=2πL/In(1.08W/D)

Heat loss per unit length is in terms of W/M

Note that, Q=q/L

By substitution, the full equation will be

Q=2πK(T1-T2)/In(1.08W/D)

Substituting values :

Q=2π(1.7W/mK)*(400K - 300k) / In[ (1.08(1.75m)/ 0.6m ]

Q= 930.92W/m