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,

    JH_c=λ × dT / dZ

Where,

JH_c = 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

JH_c = λ × dT / dZ

JH_c = 385 × 400 / 10

JH_c = 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 , JH_c = λ × dT / dZ

JH_c = 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

JH_c = λ × dT / dZ

JH_c= 419 × 600 / 6

JH_c = 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

JH_c = 40900 MW

Substitute the values in the given formula

JH_c = 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

JH_c = λ × dT / dZ

JH_c= 239 × 200 / 4

JH_c = 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

JH_c = λ x dT / dZ

JH_c= 35 x 300 / 3

JH_c = 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