Answer to Question #107437 in General Chemistry for Arshdeep Kaur Sidhu

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:

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