## Heat Flux Formula :

Heat flux formula is conductive heat flux JH_{c(w/m-1)} in watts per meter^{-1} is equal to the product of thermal conductivity costant = λ_{(wm-1/K-1)} in watts per meter^{-1} per Kelvin^{-1} and multiply the temperature dT_{(0c)} in degree Celsius and divided by the dz .Hence the Heat Flux Formula can be written as

**JH _{c(w/m-1)} = λ_{(wm-1/K-1)}* dT_{(0c)}/dZ**

**JH _{c }=Conductive heat flux in w/m^{2}**

T→ Temperature in ^{0}c

λ → thermal conductivity constant in wm^{-1 }/K^{-1}

*Heat Flow Rate Formula,*

The heat flow rate is defined as the amount of heat transferred per unit time in the material. The heat flow rate in a rod depends on the cross-sectional area of the rod, the temperature difference between both the ends and the length of the rod. Following is the formula used to calculate the heat flow rate of any material:

**Q=-K _{(wm-1/k-1)}* (A_{(m2)}/l_{(m)})*( ΔT)**

Where,

Q → heat transfer per unit time in j/s

k → thermal conductivity in wm^{-1} K^{-1}

A→ cross-sectional area in m^{2}

l → length of the material in m

∆T→ temperature difference

## Sample Problems :

### Example 1 :

Calculate the heat transferred through the plate? One face of a copper plate is 4 cm thick and maintained at 400^{∘}C, and the other face is maintained at 100^{∘}C.

### Solution :

Given parameters are

Coefficient of thermal conductivity of copper λ=385

dT=400-100=300

dx=4cm

Substitute the values in the given formula

JH_{c} = λ* dT/dZ

JH_{c}=385*300/4

JH_{c=} 28.875 Mw

### Example 2 :

Calculate the heat transferred through the plate? One face of a copper plate is 6 cm thick and maintained at 600^{∘}C, and the other face is maintained at 100^{∘}C.

### Solution :

Given parameters are

Coefficient of thermal conductivity of copper λ=385

dT=600-100=500

dx=6cm

Substitute the values in the given formula

JH_{c} = λ* dT/dZ

JH_{c}=385*600/6

JH_{c=} 38.500Mw