Home is where the Heat is The walls of a house are made of a 10 cm layer of bric

Question

Home is where the Heat is

The walls of a house are made of a 10 cm layer of brick, 10 cm of fiberglass insulation and

1cm of drywall.

Thermal conductivity values:

Brick: 1.31 W/m*K

Fiberglass : 0.04 W/m*K

dry wall : 0.17 W/m*K

a) Calculate the thermal conductivity of a 1 m 2 section of wall.

b) Find the relationship between the thermal conductivity, k, in W/m*K and the thermal

resistance = R in (m^2 *K)/W.

In order to be energy efficient, the air intake for the home is passed through a

countercurrent heat exchanger with the indoor air going the other direction. The system

is designed to absorb heat from the outgoing air and use it to warm the incoming air. In

order to accomplish this, the outgoing air is passed through a radiator which has ethylene

glycol flowing through it. As the ethylene glycol warms, its density decreases and it flows

upward into the radiator through which the incoming air passes.

density = 1.1153-.0007*T ; T in

Explanation / Answer

a)

At h = 4050 m, density p = 1.22*e^(-4050 / 8100) = 0.74 kg/m^3

Area A = pi *r^2 = 3.14 * 5.5^2 = 94.985 m^2

dm = p*A*dh

dm = 0.74*94.985*0.5 = 35.1444 kg

New volume = A*dh - pi*h^2 /3 *(3*r_sphere - h)

= 94.985*0.5 - 3.14 *0.5 ^2 /3 *(3*5 - 0.5)

= 43.698 m^3

New density = 35.1444 / 43.698 = 0.804 kg/m^3

b)

V = pi *r^2 *h

V = 3.14 *5.5^2 *40000 = 3799400 m^3

c)

P = 1*e^(-h / 7000)

Integral P dh = Integral e^(-h / 7000) dh

= -7000 *e^(-h / 7000)

Integrating from h = 0 to 40000 m,

P_avg = -7000*[e^(-40000 / 7000) - 1] / 40000

P_avg = 0.1744 atm

d)

Adiabatic constant = 0.1744*10^5 *3799400 = 6.627*10^10

e)

Using ideal gas eqn,

PV = m (R/M) T

But m/V = p = density

Hence, P = p*(R/M)*T

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