Question

1 A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. (a) Calculate the rate of heat loss per unit length for a calm day. (b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s. (c) For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K). Would the heat loss change significantly with an appreciable wind speed?

Answer 1

Answer:

a) q' = 351.22 W/m

b) q'_total = 1845.56 W / m

c) q'_loss = 254.12 W/m

Explanation:

Given:-

- The diameter of the steam line, d = 100 mm

- The surface emissivity of steam line, ε = 0.8

- The temperature of the steam, Th = 150°C

- The ambient air temperature, T∞ = 20°C

Find:-

(a) Calculate the rate of heat loss per unit length for a calm day.

Solution:-

- Assuming a calm day the heat loss per unit length from the steam line ( q ' ) is only due to the net radiation of the heat from the steam line to the surroundings.

- We will assume that the thickness "t" of the pipe is significantly small and temperature gradients in the wall thickness are negligible. Hence, the temperature of the outside surface Ts = Th = 150°C.

- The net heat loss per unit length due to radiation is given by:

                     q' = ε*σ*( π*d )* [ Ts^4 - T∞^4 ]      

Where,

          σ: the stefan boltzmann constant = 5.6703 10-8 (W/m2K4)

          Ts: The absolute pipe surface temperature = 150 + 273 = 423 K

          T∞:The absolute ambient air temperature = 20 + 273 = 293 K

Therefore,

                    q' = 0.8*(5.6703 10-8)*( π*0.1 )* [ 423^4 - 293^4 ]    

                    q' = (1.4251*10^-8)* [ 24645536240 ]    

                    q' = 351.22 W / m   ... Answer

Find:-

(b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s.

Solution:-

- We have an added heat loss due to the convection current of air with free stream velocity of U∞ = 8 m/s.

- We will first evaluate the following properties of air at T∞ = 20°C = 293 K

                  Kinematic viscosity ( v ) = 1.5111*10^-5 m^2/s

                  Thermal conductivity ( k ) = 0.025596

                  Prandtl number ( Pr ) = 0.71559

- Determine the flow conditions by evaluating the Reynold's number:

                 Re = U∞*d / v

                      = ( 8 ) * ( 0.1 ) / ( 1.5111*10^-5 )

                      = 52941.56574   ... ( Turbulent conditions )

- We will use Churchill - Bernstein equation to determine the surface averaged Nusselt number ( Nu_D ):

           $Nu_D = 0.3 + \frac{0.62*Re_D^\frac{1}{2}*Pr^\frac{1}{3} }{[ 1 + (\frac{0.4}{Pr})^\frac{2}{3} ]^\frac{1}{4} }*[ 1 + (\frac{Re_D}{282,000})^\frac{5}{8} ]^\frac{4}{5} \\\\Nu_D = 0.3 + \frac{0.62*(52941.56574)^\frac{1}{2}*(0.71559)^\frac{1}{3} }{[ 1 + (\frac{0.4}{0.71559})^\frac{2}{3} ]^\frac{1}{4} }*[ 1 + (\frac{52941.56574}{282,000})^\frac{5}{8} ]^\frac{4}{5}$

           $Nu_D = 0.3 + \frac{127.59828 }{ 1.13824 }*1.27251 = 142.95013$

- The averaged heat transfer coefficient ( h ) for the flow of air would be:

            $h = Nu_D*\frac{k}{d} \\\\h = 143*\frac{0.025596}{0.1} \\\\h = 36.58951 W/m^2K$

- The heat loss per unit length due to convection heat transfer is given by:

           q'_convec = h*( π*d )* [ Ts - T∞ ]

           q'_convec = 36.58951*( π*0.1 )* [ 150 - 20 ]

           q'_convec = 11.49493* 130

           q'_convec = 1494.3409 W / m

- The total heat loss per unit length ( q'_total ) owes to both radiation heat loss calculated in part a and convection heat loss ( q_convec ):

           q'_total = q_a + q_convec

           q'_total = 351.22 + 1494.34009

           q'_total = 1845.56 W / m  ... Answer

Find:-

For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K)

Solution:-

- To reduce the heat loss from steam line an insulation is wrapped around the line which contains a proportion of lost heat within.

- A material with thermal conductivity ( km = 0.08 W/m.K of thickness t = 20 mm ) was wrapped along the steam line.

- The heat loss through the lamination would be due to conduction " q'_t " and radiation " q_rad":

             $q'_t = 2*\pi*k \frac{T_h - T_o}{Ln ( \frac{r_2}{r_1} )}$  

             q' = ε*σ*( π*( d + 2t) )* [ Ts^4 - T∞^4 ]

             

Where,

             T_o = T∞ = 20°C

            T_s = Film temperature = ( Th + T∞ ) / 2 = ( 150 + 20 ) / 2 = 85°C

             r_2 = d/2 + t = 0.1 / 2 + 0.02 = 0.07 m

             r_1 = d/2 = 0.1 / 2 = 0.05 m

- The heat loss per unit length would be:

            q'_loss = q'_rad - q'_cond

- Compute the individual heat losses:

            $q'_t = 2*\pi*0.08 \frac{150 - 85}{Ln ( \frac{0.07}{0.05} )}\\\\q'_t = 0.50265* \frac{65}{0.33647}\\\\q'_t = 97.10 W/m$

Therefore,

             q'_loss = 351.22 - 97.10

            q'_loss = 254.12 W / m   .... Answer

- If the wind speed is appreciable the heat loss ( q'_loss ) would increase and the insulation would become ineffective.