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
Explanation:
assume steady state conditions, negligible steam side convention resistance, pipe wall resistance and contact resistance i.e [tex]T_{1} =400k[/tex]
and constant properties of concrete (300 k); [tex]k=1.4Wm/k[/tex]
the heat rate can be expressed as
[tex]q=Sk(T_{1} -T_{2} )=Ak(T_{1} -T_{2} )[/tex]
the shape factor is
[tex]S=(2\pi L/ln(1.08w/D))[/tex]
[tex]hence \\q'=q/L\\=2\pi k(T_{1}- T_{2}) /ln(1.08w/D)[/tex]
insert values
[tex]=(2\pi *1.4W/m.k*(400-300)k)/(ln(1.08*1.75/0.6)[/tex]
[tex]q'=766W/m[/tex] ( rate of heat loss per unit length)
