Solar energy falls on a black body, 30 feet by 80 feet, at a rate of 250 BTUs per hour per square foot. Determine the heat transfer rate through this body if two insulation materials are used. Outside convective heat-transfer coefficient, ho, is 6.16 BTU per hour per square foot per degree Fahrenheit.
Outside air temperature, Tb, is 30 degrees Fahrenheit. Thermal conductivity of black body is, k1, 28.9, of first insulation is k2, 0.029, and of second insulation, k3, is 0.086 BTU per hour per foot per degree Fahrenheit. Thickness of black body, L1, 0.016, of first insulation, L2, is 0.0833, and of second insulation, L3, is 0.0833 foot. Inside convective heat-transfer coefficient, hi, is 6.16 BTU per hour per square foot per degree Fahrenheit. Inside air temperature, Ta, is 68 degree Fahrenheit. Temperature of the space, Tc, is minus 4 degrees Fahrenheit.Solution:
Surface area is of the black body is 30 time 80 or 2,400 square feet. Total heat received by the body, Q, is 250 times 2400 or 6 times 105 BTUs per hour. Assuming that the outside surface temperature of the black body is T, then some of the heat is passing through the body via conduction, other by free-convection to outside bulk air at Tb, and remaining by radiation to space at Tc. Realizing that the inside surface temperature of the body is not known either, we merge, inside film coefficient in the conductive mode calculations.Thus conductive resistance is 1 over Area times (1 over hi plus L1 over k1 plus L2 over k2 plus L3 over k3). Plugging in their values, conductive resistance, R, is found to be 1.64 times 10-3 hour degrees Fahrenheit per BTU. Surface temperature, T, has to be between 490 and 528 degrees Rankin. We need to guess the value of T and check to see if the heat received by the body is equal to the sum of conductive heat term, outside-convective-heat term, and radiative heat term. Alternately, we can express this heat balance as a function, F(T) and determine the root of the function, such that F(T) is equal to zero. It can be seen that this function is non-linear in T, and we could choose Newton’s method of finding the root of the function. Newton’s method uses derivative of the function to guess the revised estimate of the root. It takes about three iterations to find the value of the outside surface temperature as 521.5 degree Rankin or 61.5 degrees Fahrenheit. Heat transfer due to conduction is found to be (Ts minus Ta) over R and is equal to 3.2 x 105 BTUs per hour.