Answer to Question #119484 in Physics for Ali

Question

A standard 1-in.schedule 40steel pipe carries saturated steam at 250^(@)F . The pipe is lagged (insulated) with a 2-in.layer of 85percent magnesia pipe covering and outside this magnesia there is a 1/2in layer of the cork. The inside temperature of the pipe wall is 249^(@)F and the outside temperature of the cork is 90^(@)F .Thermal conductivities in Btu/ft-h.F are: for steel 26; for magnesia 0.034; for cork 0.03.Calculate (a) the heat loss from 100ft of pipe in Btu per hour; (b) the temperature at the boundaries between metal and magnesia and between magnesia and cork.(For 1-in. schedule 40steel pipe outside diameter is 1.315in and inside diameter is 1.049in )

Answer 1

The heat transferred thru the steel and its thermal resistance is

$$q=\frac{2\pi Lk_s}{\text{ln}(R_\text{out}/R_\text{in})}(250-T_1),\\\space\\ R_\text{steel}=\frac{\text{ln}(R_\text{st.out}/R_\text{st.in})}{2\pi Lk_s}.$$

The heat transferred thru the magnesium is

$$q=\frac{2\pi Lk_m}{\text{ln}(R_\text{out}/R_\text{in})}(T_1-T_2),\\\space\\ R_\text{mg}=\frac{\text{ln}(R_\text{mg.out}/R_\text{mg.in})}{2\pi Lk_m}$$

The heat transferred thru the cork is

$$q=\frac{2\pi Lk_c}{\text{ln}(R_\text{out}/R_\text{in})}(T_2-90),\\\space\\ R_\text{cork}=\frac{\text{ln}(R_\text{cork.out}/R_\text{cork.in})}{2\pi Lk_c}.$$

In these equations, k is the thermal conductivity.

The total heat loss is

$$q=\frac{250-90}{R_\text{steel}+R_\text{mag}+R_\text{cork}}=\\\space\\ =\frac{250-90}{\frac{\text{ln}(R_\text{s}/R_\text{w})}{2\pi Lk_s}+\frac{\text{ln}(R_\text{m}/R_\text{s})}{2\pi Lk_m}+\frac{\text{ln}(R_\text{c}/R_\text{m})}{2\pi Lk_c}},$$

where

"R_w=1.049"\\\\\nR_s=1.315"\\\\\nR_m=1.315"+2"=3.315"\\\\\nR_c=3.315"+0.5"=3.815""

Substituting these values, we get

$$q=3152.8\text{ BTU/hr}.$$

Now, since we know the total heat and the thermal resistances, we can find temperatures $T_1$ and $T_2$ using the equation listed in the beginning of the solution:

$$T_1=249.96°\text{ F},\\ T_2=113.50°\text{ F}.$$