Answer:
a) heat gain per unit tube length = [tex]\frac{23-6}{1.35} = 12.6W/m[/tex]
b) heat gain per unit tube length = [tex]\frac{23-6}{2.20} = 7.7W/m[/tex]
Explanation:
Assumptions:
a) What is the heat gain per unit tube length
[tex]R_{conv,i}'=\frac{1}{2\pi r_1h_i}[/tex]
[tex]d_1=36mm[/tex] Therefore [tex]r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}[/tex]
[tex]r_2=2mm=2*10^{-3}m[/tex]
[tex]k_{st}=14.2W/m.k[/tex]
[tex]h_o=6W/m^2[/tex]
[tex]h_i=400W/m^2[/tex]
[tex]R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W[/tex]
[tex]R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W[/tex]
[tex]R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W[/tex]
[tex]R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W[/tex]
heat gain per unit tube length = [tex]\frac{23-6}{1.35} = 12.6W/m[/tex]
b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube
[tex]r_3=r_1+r_2+10mm=30mm=0.03m[/tex]
[tex]R_{conv,i}'[/tex] and [tex]R_{cond,st}'[/tex] are the same, but [tex]R_{conv,o}'[/tex] changes.
Therefore:
[tex]R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W[/tex]
[tex]R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W[/tex]
The total resistance [tex]R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W[/tex]
heat gain per unit tube length = [tex]\frac{23-6}{2.20} = 7.7W/m[/tex]
Answer:
a) The heat gain per unit is 13.9 W/m
b) The heat gain per unit is 7.08 W/m
Explanation:
In the attached Word file is the explanation.
