2.3 KiB
Homework 4 - Aidan Sharpe
1
The wall of a house is 7[m] wide, 6[m] high, and 0.3[m] thick with $k=0.6$[W/(m \cdot K)]. The surface temperature on the inside of the wall is 16$^\circ$C and the temperature on the outside is 6$^\circ$. Find the heat flux through the wall and the total heat loss through it.
Heat flux is given by
q''_x = -k{dT \over dx}\Delta T = 10, k = 0.6, \Delta x = 0.3
q''_x= -0.6 {10 \over 0.3} = -20\text{[W/m$^2$]}Total heat loss is the product of heat flux and the area:
-20 (6 \times 7) = -840\text{[W]}2
A 20[mm] diameter copper pipe is used to carry heated water. The external surface of the pipe is subjected to a convective heat transfer coefficient of $h = 6$[W/(m$^2 \cdot$ K)]. Find the heat loss by convection per meter length of the pipe when the external surface temperature is 80$^\circ$C and the surroundings are at 20$^\circ$C. Assuming black body radiation, what is the heat loss by radiation?
\dot{Q} = hA(T-T_f)h=6, A=2\pi 0.02 l, T = 80, T_f = 20
\dot{Q} = 6 (2 \pi 0.02 x) (80 - 20) = 45.2389xHeat loss due to convection is $45.2389$[W/m].
q'' = \sigma T_s^4 = 5.67 \times 10^{-8} (80 + 273.15) = 2 \times 10^{-5} \text{[W/m$^2$]}The total heat loss is the product of surface area and thermal flux:
2\times 10^{-5} (2\pi 0.02x) = 2.516 x \text{[W]}3
A plate 0.3[m] long and 0.1[m] wide, with a thickness of 12[mm] is made from stainless steel ($k = 16$[W/(m \cdot K)]), the top surface is exposed to a 20$^\circ$C airstream. In an experiment, the plate is heated by an electrical heater (also 0.3[m] by 0.1[m]) positioned on the underside of the plate and the temperature of the plate adjacent to the heater is maintained at 100$^\circ$C. A voltmeter and ammeter are connected to the heater and these read 200[V] and 0.25[A] respectively. Assuming that the plate is perfectly insulated on all sides except the top surface, what s the convective heat transfer coefficient?
Heat transfer from heater to plate:
P = I V = 200 \times 0.25 = 50\text{W}]q'' = {50 \over 0.3 \times 0.1} = 1666.667 \text{[W/m$^2$]}Heat transfer from the plate to the air:
q'' = 1666.667 = {d\dot{Q} \over dA}\dot{Q} = h A(T_\text{plate} - T_\text{air})h = {q'' \over T_\text{plate} - T_\text{air}} = {1666.667 \over 100 - 20} = 20.833\text{[W/(m$^2 \cdot$K)]}