A horizontal steel pipe having a diameter of 5 cm is maintained at a temperature of 50 o C in a large room where the air and wall temperatures are 20 o C. The surface emissivity of the pipe is 0.8.

(a) The convective coefficient, {eq}h{/eq}, can be estimated using the following empirical equation for forced convection heat transfer: {eq}Nu = 0.023 Re^{4/5} Pr^{n} {/eq} where {eq}Nu{/eq} is the Nusselt number, {eq}Re{/eq} is the Reynolds number, {eq}Pr{/eq} is the Prandtl number, and {eq}n{/eq} is an exponent that depends on the flow regime (for turbulent flow, {eq}n = 0.4{/eq}). Assuming the air flow around the pipe is turbulent, we can estimate the Reynolds number using: {eq}Re = \frac{rho v D}{mu} {/eq} where {eq}rho{/eq} is the air density, {eq}v{/eq} is the air velocity, {eq}D{/eq} is the pipe diameter, and {eq}mu{/eq} is the air viscosity. We can assume that the air velocity is equal to the free stream velocity, which we can estimate as {eq}v = 1 rm m/s{/eq}. Using the properties of air at {eq}20 ^circ rm C{/eq} (density {eq}rho = 1.204 rm kg/m^3{/eq} and viscosity {eq}mu = 1.78times 10^{-5} rm Ns/m^2{/eq}), we get: {eq}Re = \frac{1.204times 1times 0.05}{1.78times 10^{-5}} \approx 33898 {/eq} The Prandtl number for air at {eq}20 ^circ rm C{/eq} is {eq}Pr = 0.71{/eq}. Substituting these values into the Nusselt number equation, we get: {eq}Nu = 0.023times 33898^{4/5}times 0.71^{0.4} \approx 121.4 {/eq} The convective heat transfer coefficient is then given by: {eq}h = \frac{k N_u}{D} {/eq} where {eq}k{/eq} is the thermal conductivity of air at {eq}20 ^circ rm C{/eq} (0.0258 W/mK). Substituting the values we have calculated, we get: {eq}h = \frac{0.0258times 121.4}{0.05} \approx 63.3 rm W/m^2K {/eq} This is an estimate of the convective heat transfer coefficient for the given conditions. (b) The diagram with the energy fluxes on the pipe would show the following: - Heat transfer by convection from the air: {eq}q_{conv} = h A_s (T_s - T_{infty}){/eq} where {eq}A_s{/eq} is the surface area of the pipe, {eq}T_s{/eq} is the surface temperature of the pipe, and {eq}T_{infty}{/eq} is the free stream temperature of the air. - Heat transfer by radiation to the surroundings: {eq}q_{rad} = epsilon A_s sigma (T_s^4 - T_{infty}^4){/eq} where {eq}epsilon{/eq} is the surface emissivity of the pipe, {eq}sigma{/eq} is the Stefan-Boltzmann constant, and {eq}T_s{/eq} and {eq}T_{infty}{/eq} are the temperatures of the pipe and surroundings, respectively. - Total heat loss: {eq}q_{loss} = q_{conv} + q_{rad}{/eq} (c) To estimate the heat loss of the pipe per unit length, we need to know the length of the pipe. Let's assume a length of 1 meter. The surface area of the pipe is given by: {eq}A_s = pi D L {/eq} where {eq}L{/eq} is the length of the pipe. Substituting {eq}D = 5 rm cm{/eq} and {eq}L = 1 rm m{/eq}, we get: {eq}A_s = pitimes 0.05times 1 \approx 0.157 rm m^2{/eq} The temperature difference between the surface of the pipe and the free stream air is {eq}T_s - T_{infty} = 30 ^circ rm C{/eq} (the difference between 50 and 20 degrees). The heat loss per unit length of the pipe is then given by: {eq}frac{q_{loss}}{L} = q_{conv} + q_{rad} = h A_s (T_s - T_{infty}) + epsilon A_s sigma (T_s^4 - T_{infty}^4) {/eq} Substituting the values we have calculated, we get: {eq}frac{q_{loss}}{L} = 63.3times 0.157times 30 + 0.8times 0.157times 5.67times 10^{-8}times (323^4 - 293^4) \approx 81.5 rm W/m {/eq} Therefore, the estimated heat loss of the pipe per unit length is \approx imately 81.5 Watts per meter.