How to Calculate and Solve for Heat Flux | Solidification of Metals

The image above represents heat flux.

To compute for heat flux, three essential parameters are needed and these parameters are Heat Emission (h), Surface Temperature (Ts) and Initial Temperature (To).

The formula for calculating heat flux:

qx = h(Ts – To)

Where:

qx = Heat Flux (T)
h = Heat Emission
Ts = Surface Temperature
To = Initial Temperature

Let’s solve an example;
Find the heat flux when the heat emission is 21, the surface temperature is 12 and the initial temperature is 10.

This implies that;

h = Heat Emission = 21
Ts = Surface Temperature = 12
To = Initial Temperature = 10

qx = h(Ts – To)
qx = 21(12 – 10)
qx = 21(2)
qx = 42

Therefore, the heat flux is 42 W.

Calculating the Heat Emission when the Heat Flux, the Surface Temperature and the Initial Temperature is Given.

h = qx / Ts – To

Where:

h = Heat Emission
qx = Heat Flux (T)
Ts = Surface Temperature
To = Initial Temperature

Let’s solve an example;
Find the heat emission when the heat flux is 40, the surface temperature is 20 and the initial temperature is 10.

This implies that;

qx = Heat Flux (T) = 40
Ts = Surface Temperature = 20
To = Initial Temperature = 10

h = qx / Ts – To
h = 40 / 20 – 10
h = 40 / 10
h = 4

Therefore, the heat emission is 4.

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How to Calculate and Solve for Quantity of Heat Loss | Fuel and Furnaces

The image above represents quantity of heat loss.

To compute for quantity of heat loss, four essential parameters are needed and these parameters are Initial Temperature (T1), Final Temperature (T2), a and E.

The formula for calculating quantity of heat loss:

Q = a(T1 – T2)5/4 + 4.88E[(T1 + 273/100)4 – (T2 + 273/100)4]

Where:

Q = Quantity of Heat Loss
T1 = Initial Temperature
T2 = Final Temperature

Let’s solve an example;
Find the quantity of heat loss when the initial temperature is 21, the final temperature is 17, a is 14 and E is 15.

This implies that;

T1 = Initial Temperature = 21
T2 = Final Temperature = 17
a = 14
E = 15

Q = a(T1 – T2)5/4 + 4.88E[(T1 + 273/100)4 – (T2 + 273/100)4]
Q = 14(21 – 17)5/4 + 4.88(15)[(21 + 273/100)4 – (17 + 273/100)4]
Q = 14(4)5/4 + 4.88(15)[74.71 – 70.72809]
Q = 14(5.65) + 4.88(15)(3.98)
Q = 79.19 + 291.608
Q = 370.80

Therefore, the quantity of heat loss is 370.80 J/Kg K.

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