The R-factor for housing insulation gives the thermal resistance in units of ft2 °F h/BTU. A good wall for harsh climates, corresponding to about 10 in of fiberglass, has R = 45 ft2 °F h/BTU.
a) Determine the thermal resistance (in m2 K/W) in SI units.
b) Find the heat flow per square meter through a wall that has insulation with an R-factor of 45, with an outside temperature of -21.1 °C and an inside temperature of 24.1°C.
dQ/dt= dTemp/R= 45.2/45 Btu/hr
a. 1 ft^2 F hr/BTu= Lord chaning units..
1m^2=10.76ft^2
1 BTU=1055 joule
1 watt= 3600Joule/hr
ok, then 1 BTU/hr=1055*3600/hr
now for the conversion
1ft^2(1m^2/10.76ft^2)*F*(9K/5F)/(1055*1036 W/hr)
and you can check that, and multiply it out.
it's wrong i think, i cannot understand
a) To determine the thermal resistance in SI units, we can use the conversion factor:
1 ft2 °F h/BTU = 0.1761 m2 K/W
Given that R = 45 ft2 °F h/BTU, we can convert it to SI units by multiplying it by the conversion factor:
Thermal resistance (in SI units) = R * conversion factor
= 45 * 0.1761
= 7.92 m2 K/W
Therefore, the thermal resistance of the wall insulation is 7.92 m2 K/W.
b) To find the heat flow per square meter through the wall, we can use the formula:
Heat flow (in watts) = (T_inside - T_outside) / thermal resistance
Convert the temperatures to Kelvin:
T_outside = -21.1°C + 273.15 = 252.05 K
T_inside = 24.1°C + 273.15 = 297.25 K
Substituting the values into the formula:
Heat flow (in watts) = (297.25 K - 252.05 K) / 7.92 m2 K/W
= 45.2 K / 7.92 m2 K/W
= 5.71 W/m2
Therefore, the heat flow per square meter through the wall is 5.71 watts/meter squared.