A house has well-insulated walls 19.5cm thick (assume conductivityof air) and ar

Question

A house has well-insulated walls 19.5cm thick (assume conductivityof air) and area 410m2, a roof of wood 5.5cm thick andarea 280m2, and uncovered windows 0.65cm thick and totalarea 33m2. (A) assuming that heat is lost only byconduction calculate the rate at which heat must be supplied tothis house to maintain its inside temperature at 23 degrees C ifthe outside temperature is -15 degrees C (B) if the house isinitially at 12degrees C, estimate how much heat must be suppliedot raise the temperature to 23degrees within 30minutes. Assume that only the air needs to be heated and that is volume is750m3. (c) If natural gas costs $0.080 perkilogram and its heat of combustion is 5.4x107J/kg, howmuch is the monthly cost to maintain the house as in part (a) for24 hours each day, assuming 90% of the heat produced is used toheat the house? Take the specific heat of air to be0.24kcal/kg*degrees C.

Explanation / Answer

the rate at which heat must be supplied to this house is H = k * A * (T/x) or H = k * T * (A/x) or
H = k * T * [(A1/x1) + (A2/x2) + (A3/x3)]-------------(1)
specific heat of air = k
k = 0.24 kcal/kg·oC
     = 240 * 4.2 J/kg.oC
= 1008 J/kg.oC

T = (23 - (-15)) oC = 38 oC A1 = 410 m2 and x1 = 19.5 cm = 19.5 *10-2 m A2 = 280 m2 and x2 = 5.5 cm = 5.5 * 10-2m A3 = 33 m^2 and x3 = 0.65 cm = 0.65 * 10-2 m

(b)
From (1) we have

(E/t) = k * T * [(A1/x1) + (A2/x2) + (A3/x3)] or E = k * T * [(A1/x1) + (A2/x2) + (A3/x3)] * t T = (23 - 12) oC = 11 oC t = 30 min = 30 * 60 s = 1800 s


(c)the heat needed to heat the house E = k * T * [(A1/x1) + (A2/x2) + (A3/x3)] * t t = 24 h = 24 * 60 * 60 s = 86400 s

the percentage of heat needed to heat the house is E * (90/100) = E * 0.90 the cost needed to heat the house per month is E * 0.90 * 30t * 0.080 dollars/month
(b)
From (1) we have

(E/t) = k * T * [(A1/x1) + (A2/x2) + (A3/x3)] or E = k * T * [(A1/x1) + (A2/x2) + (A3/x3)] * t T = (23 - 12) oC = 11 oC t = 30 min = 30 * 60 s = 1800 s


(c)the heat needed to heat the house E = k * T * [(A1/x1) + (A2/x2) + (A3/x3)] * t t = 24 h = 24 * 60 * 60 s = 86400 s

the percentage of heat needed to heat the house is E * (90/100) = E * 0.90 the cost needed to heat the house per month is E * 0.90 * 30t * 0.080 dollars/month

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