Question

Heat Transfer

Problems 1. A window pane (1/8 inch thick) in an old house owned by a professor is 3 ft. by 4 ft. He likes to be warm in the winter so the inside temperature is kept at 70 °F even when the outside temperature is 10 •F. For this problem assume the inside and outside temperatures are for the window surface. How much heat is transferred through the window (k=.54 BTU/hr ft ° F)? 2. A small heated object is 6mm wide and 3 meters long and an emissivity of 0.85 is within a room whose walls are at a temperature of 25 C. If 2000W of heat is transferred from the object to the walls by radiation, what is the objects temperature? 3. The inside temperature of a 3 m by 10 m wall is 315 ° C. Wall is 2.5 cm thick and has a thermal conductivity of 1.4 W/m °C. The outside air temperature is 38 °C. What does the convective heat transfer coefficient (h) need to be to keep the outer wall temperature at 41 ° c. 4. A triangular steel rod is 30 cm long with each side of the cross section equal to 20 mm. One end is at a temperature of 100 °C and the other is at a temperature of - 5°C. How much heat is transferred through the steel rod? Assume adiabatic sides on the rod and ak = 73 W/m °C.

Answer 1

Solutionli) Data given; thickness (da) - lino104 tt Area = (3x4 ) ft Tinside - 70f, Toutside- -- K-54 Brulhyttur lor Fourier law of heat Conduction da 9 = kadTL - 54 & 3x4 X [ 70-10] 54.0 104 19 = 37384.615 Btu/hr Solution (2) Data given. Area = 1.00643) m² Immisivity (t) = .85 Trooma 256 = 2000W Stefan Boltzman law of Radiative heat transferi - Stefan Buitzman a = to A T T 2000 - 08585.67 * 108 * 1.096 X 3 ) { t = 125+ 2739] Constant 542 12 object

Solution (3) - Tair = 38°c Tims=3152 k = 14wl mic T Toutside = lic at Steady state - Conduction in watt Q Confection between outside wall-air - k A dan had to => ley X (2xle) [315-01) - h ( 3x15] [41-33] (9.5x102) - The 5114.67 Wilmagan

Solution (4): Ta=-5°c and K -73 wlmc 30cm Talooc a 20mm transfer - Arean crossection for heat - Area of equilateral triangle $342. 3 (102012 11.732x15 m² Contuction = KA DIL da - 73 x 1.732x lol x [ 100-(-5)] 030 a = 4.425 W