Heat Flux Formula (original) (raw)
Last Updated : 23 Jul, 2025
Heat flux is a fundamental concept in thermodynamics and heat transfer, describing the rate at which heat energy transfers through a surface per unit area. The heat flux formula quantifies this rate, making it a crucial tool for engineers and scientists in fields ranging from thermal engineering to environmental science.
In this article, we will learn what heat flux is also solve some important problems.
Table of Content
- Heat Flux Formula
- Heat Flow Rate Formula
- Sample Problems on Heat Flux Formula
- Heat Flux Formula Worksheet
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,
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 on Heat Flux Formula
**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
Heat Flux Formula Worksheet
**Problem 1: A surface emits 500 W/m² of heat. If the area of the surface is 3 m², calculate the total heat transfer.
**Problem 2: In a convection process, the heat transfer coefficient is 25 W/m²·K, and the temperature difference between the surface and the fluid is 40 K. Calculate the heat flux.
**Problem 3: In a steady-state heat conduction process, a wall with an area of 10 m² experiences a temperature difference of 15 K. The thermal conductivity of the wall material is 0.75 W/m·K, and the wall thickness is 0.2 m. Calculate the heat flux.
**Problem 4: A rod with varying thermal conductivity from 1 W/m·K to 2 W/m·K over a length of 2 m has a linear temperature gradient of 20 K/m. Calculate the average heat flux.
**Problem 5: During a transient heat conduction process, a surface with an area of 0.8 m² experiences a heat transfer of 2000 J in 5 seconds. Calculate the instantaneous heat flux.
**Problem 6: Calculate the heat flux through a cylindrical pipe with a length of 2 m, an inner radius of 0.1 m, and an outer radius of 0.2 m, given a temperature difference of 50 K and thermal conductivity of 45 W/m·K.
**Problem 7: A layer of liquid with a thickness of 0.02 m and thermal conductivity of 0.6 W/m·K experiences a temperature gradient of 10 K. Calculate the heat flux.
**Problem 8: A metal rod of 1 m length has its ends maintained at temperatures of 100°C and 50°C. The thermal conductivity of the rod is 200 W/m·K. Calculate the heat flux through the rod.