A carpenter builds a solid wood door with dimensions 1.95 m × 0.92 m ×5.5 cm . I

Question

A carpenter builds a solid wood door with dimensions 1.95 m × 0.92 m ×5.5 cm . Its thermal conductivity is k=0.120W/(mK). The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional 1.9 cm thickness of solid wood. The inside air temperature is 19.0 C , and the outside air temperature is -5.00 C .

By what factor is the heat flow increased if a window 0.550 m on a side is inserted in the door? The glass is 0.45 cm , and the glass has a thermal conductivity of 0.80 W/(mK). The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 cm of glass.

I keep getting 2.1 and can't find where I'm messing up in my calculations :/

Explanation / Answer

For the thermal conductivity of door ,

thermal conductivity , k = 0.120 W/m*K

rate of hat flow through door , Q1 = k * Q* delta T/t

Q1 = 0.120 * 1.95 * 0.92 * (19 + 5)/0.055

Q1 = 93.9 W

Now, when the window is added to the door ,

rate of heat flow ,

Q2 = 0.120 * (1.95 * 0.92 - 0.55^2) * (19 + 5)/0.055 + 0.80 * 0.55^2 * (19 +5 )/(0.12 + 0.055)

Q2 = 111.29 W

factor to which heat flow increased = Q2/Q1

factor to which heat flow increased = 111.29/93.9=

factor to which heat flow increased = 1.185

factor to which heat flow increased after adding window is 1.185

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.