Heat Flux Formula
The quantity of heat transmitted per unit area per unit time to or from a surface is referred to as heat flux. It is a derived quantity since it is based on the idea of two quantities: the amount of heat transfer per unit time and the region to or from which the heat transfer takes place. The joule per second or watt is the SI unit of heat rate.
Heat Flux Formula
The heat flux density is the rate of heat transfer per unit area. Heat flux density is measured in watts per meter square (W/m2) in SI units. Heat flow is a two-dimensional vector quantity with magnitude and direction. The formula for heat flux is given as,
JHc=λ × dT / dZ
Where,
JHc = conductive heat flux
T = temperature
λ = thermal conductivity constant
Heat Flow Rate Formula
The heat flow rate in a material is defined as the quantity of heat transported per unit time. The heat flow rate in a rod is determined by the rod’s cross-sectional area, the temperature differential between both ends, and the rod’s length.
Q =−k × (A/l) × (ΔT)
Where,
Q is the heat transfer per unit time
k is the thermal conductivity
A is the cross-sectional area
l is the length of the material
∆T is the temperature difference
Sample Problems
Question 1: One face of a copper plate is 10 cm thick and maintained at 500∘C, and the other face is maintained at 100∘C. Calculate the heat transferred through the plate.
Solution:
Coefficient of thermal conductivity of copper, λ = 385
dT = 500 – 100 = 400
dx = 5
Substitute the values in the given formula
JHc = λ × dT / dZ
JHc = 385 × 400 / 10
JHc = 15,400 MW
Question 2: Calculate the heat flow rate from a glass window with an area of 1.5 m x 1.0 m and a width of 3.00 mm, assuming the temperatures of the exterior and inner surfaces are 13.0 and 14.0 degrees Celsius, respectively.
Solution:
Thermal conductivity of glass λ = 0.96 W / m.K
Then,
Heat flux , JHc = λ × dT / dZ
JHc = 0.96 W/m.K × 1 K / 3.0 × 10-3 m
= 320 W / m2
Question 3: One face of a silver plate is 6 cm thick and maintained at 700∘C, and the other face is maintained at 100∘C. Calculate the heat transferred through the plate.
Solution:
Coefficient of thermal conductivity of silver, λ = 419
dT = 700 – 100= 600
dx = 6
Substitute the values in the given formula
JHc = λ × dT / dZ
JHc= 419 × 600 / 6
JHc = 41,900 MW
Question 4: Find the thickness of the copper plate which maintained 400∘ C and the other face is maintained at 200∘C where heat flux is 40900 MW.
Solution:
Coefficient of thermal conductivity of copper, λ = 385
dT = 400 – 200= 200
JHc = 40900 MW
Substitute the values in the given formula
JHc = λ × dT / dZ
40900 = 385 × 200/ dZ
dZ = 385 × 200 / 40900
= 1.88 cm
Question 5: One face of an aluminum plate is 4 cm thick and maintained at 300∘C, and the other face is maintained at 100∘C. Calculate the heat transferred through the plate.
Solution:
Coefficient of thermal conductivity of aluminium, λ = 239
dT = 300 – 100= 200
dx = 4
Substitute the values in the given formula
JHc = λ × dT / dZ
JHc= 239 × 200 / 4
JHc = 11,950MW
Question 6: One face of a lead plate is 3 cm thick and maintained at 500∘C, and the other face is maintained at 200∘C. Calculate the heat transferred through the plate.
Solution
Coefficient of thermal conductivity of aluminium, λ = 35
dT = 500 – 200= 300
dx = 3
Substitute the values in the given formula
JHc = λ x dT / dZ
JHc= 35 x 300 / 3
JHc = 3,500MW
Question 7: Find the thickness of the magnesium plate which maintained 600∘ C and the other face is maintained at 100∘C where heat flux is 45000 MW.
Solution:
Coefficient of thermal conductivity of copper, λ = 151
dT = 600 – 100= 500
JHc = 45000 MW
Substitute the values in the given formula
JHc = λ × dT / dZ
45000 = 151 × 500/ dZ
dZ = 151 × 500 / 45000
= 1.67 cm
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