Question

Calculate the rate of heat loss to the surroundings and the quantity of steam that would condense per hour per meter of a 1.5 inch steel pipe (o.d.=0.04826 m, i.d.=0.04089 m, k=45 W/m.K) containing saturated steam at 130°C. The heat transfer coefficient on the steam side is 11,400 W/m² K and that on the outside of the pipe to air is 5.7 W/m²K. Ambient air averages 15°C for the year. (b) How much energy would be saved in 1 year if the pipe is insulated with 5 cm thick insulation having a thermal conductivity of 0.07 W/m.K. The heat transfer coefficients on the steam and air sides are the same as in (a).

Answer 1

(a)

$$R_{conv,i} ={1 \over h_i\pi D_i}$$

$$R_{conv,i} = {1 \over 5.7*3.14*0.04089} = 1.36640 m*K/W$$

$$R_{conv,o} ={1 \over h_o\pi D_o}$$

$$R_{conv,o} = {1 \over 11 400*3.14*0.04826} = 0.00058m*K/W$$

$$R'_W = {ln(D_o/D_i) \over 2\pi k_W}$$

$$R'_W = {0.1657 \over 2*3.14*45} = 0.00059 m*K/W$$

$$1/UA = (R_{conv,i} + R_{conv,o} + R'_W)/L$$

1.5 inch = 0.0381 m

$$1/UA = (1.36640+0.00058+0.00059)/0.0381 = 35.90971K/W$$

$$UA = 1/35.90971 = 0.02785W/K$$

Then:

$$A = 0.02785/U$$

$$1/U = 1/h_i + 1/h_o + R_{conv,o} + R_{conv,i}$$

$$1/U = 1/11 400 + 1/5.7 + 0.00058 + 1.36640 = 1.54251$$

$$U = 1/1.54251 = 0.64829W/m^2*K$$

From this:

$$A = 0.02785/0.64829 = 0.04296m^2$$

Then:

$$Q/t = {A\Delta T \over R_w}$$ $$Q = (0.04926*115)/0.00059 = 9601.52542 J/sec$$

$$с_{vap} \approx 2000 J/(kg*K)$$

$$m = Q/c\Delta T = 9601.52542/(2000*115) = 0.041746 kg$$

(b)

$$R''_W = {ln(D^+_o/D_0) \over 2\pi k}$$

$$R''_W = 0.09858/0.4396 = 0.22425m*K/W$$

$$1/UA = (R_{conv,i} + R_{conv,o} + R'_W + R''_W)/L$$

$$1/UA = 41.78005K/W$$

$$UA = 0.02393W/K$$

$$1/U = 1/11 400 + 1/5.7 + 0.00058 + 1.36640 + 0.22425 = 1.76676$$

$$U = 1/1.76676 = 0.56601W*m^2/K$$

$$A = 0.02393/0.56601 = 0.04229m^2$$

$$Q = (0.04229*115)/0.00059 = 8 242.9661 J/sec$$

$$\Delta Q = 9601.52542 - 8 242.96610 = 1358.55932J/sec$$

Per year:

$$1358.55932*365*24*60*60 =4.28*10^{10} J$$