Acold storage room is used to keep the temperature inside the room maintain at low temperature.
therefore, the special wall must be used to reduce the heat transfer rate in order to minimise the energy
consumption. a 3 m and 7 m wide wall consists of 17 cm x 24 cm cross section horizontal bricks where
thermal conductivity of the brick is given as k=0.77 w/m. k. the wall is separated by 2 cm thick layers of
cement where thermal conductivity is given as k = 0.33 w/m. k. each side of the bricks are also being covered
by the cement (2 cm thick). there are 5 cm polystyrene foam is used at the left side of the wall where k=
0.0011 w/m. k. the indoor temperature and outdoor temperatures are -5°c and 30°c respectively. the
convection heat transfer coefficients on the indoor and outdoor are h1=11 w/m²k and h2=23 w/m².k
respectively. assuming one dimensional heat transfer and disregarding radiation heat transfer, determine
the rate of the heat transfer through the wall (for one cell of bricks).

16 W

Explanation:

The sum of convective and conductive heat transfer through layers of walls can be modeled as:

Q = (T₂ − T₁) A / R

where Q is the rate of heat transfer,

T₂ − T₁ is the temperature difference,

and A is the cross sectional area.

R, the thermal resistance, is defined as:

R = ∑1/h + ∑L/k

where h is the convective heat transfer coefficient of the fluid,

L is the thickness of the wall,

and k is the thermal conductivity of the wall.

Given:

T₂ − T₁ = 35 K

A = 21 m²

h₁ = 11 W/m²/K

h₂ = 23 W/m²/K

For the polystyrene:

L₁ = 0.05 m

k₁ = 0.0011 W/m/K

For the cement:

L₂ = 0.02 m or 0.21 m

k₂ = 0.33 W/m/K

For the brick:

L₃ = 0.17 m

k₃ = 0.77 W/m/K

For the layer above and below the brick:

R = (1/11 + 1/23 + 0.05/0.0011 + 0.21/0.33)

R = 46.2 m²K/W

For the layer containing the brick:

R = (1/11 + 1/23 + 0.05/0.0011 + 0.02/0.33 + 0.17/0.77 + 0.02/0.33)

R = 45.9 m²K/W

So the average thermal resistance is:

R = (6/30) (46.2) + (24/30) (45.9)

R = 46.0 m²K/W

Therefore, the heat transfer through the wall is:

Q = (35 K) (21 m²) / (46.0 m²K/W)

Q = 16.0 W

Rounding to two significant figures, the rate of heat transfer through the wall is 16 W.

answer:

true

c

explanation: