How to Calculate and Solve for Effective Heat of Fusion | Solidification of Metals

The image above represents effective heat of fusion.

To compute for effective heat of fusion, three essential parameters are needed and these parameters are Latent Heat of Fusion (Hf), Heat Capacity at Constant Pressure (cp) and Change in Temperature (ΔT).

The formula for calculating effective heat of fusion:

H = Hf + cpΔT

Where:

H = Effective Heat of Fusion
Hf = Latent Heat of Fusion
cp = Heat Capacity at Constant Pressure
ΔT = Change in Temperature

Let’s solve an example;
Find the effective heat of fusion when the latent heat of fusion is 12, the heat capacity at constant pressure is 14 and the change in temperature is 10.

This implies that;

Hf = Latent Heat of Fusion = 12
cp = Heat Capacity at Constant Pressure = 14
ΔT = Change in Temperature = 10

H = Hf + cpΔT
H = 12 + 14(10)
H = 12 + 140
H = 152

Therefore, the effective heat of fusion is 152 J/Kg.

Calculating the Latent Heat of Fusion when the Effective Heat of Fusion, the Heat Capacity at Constant Pressure and the Change in Temperature is Given.

Hf = H – cpΔT

Where:

Hf = Latent Heat of Fusion
H = Effective Heat of Fusion
cp = Heat Capacity at Constant Pressure
ΔT = Change in Temperature

Let’s solve an example;
Find the latent heat of fusion when the effective heat of fusion is 42, the heat capacity at constant pressure is 10 and the change in temperature is 2.

This implies that;

H = Effective Heat of Fusion = 42
cp = Heat Capacity at Constant Pressure = 10
ΔT = Change in Temperature = 2

Hf = H – cpΔT
Hf = 42 – (10)(2)
Hf = 42 – 20
Hf = 22

Therefore, the latent heat of fusion is 22.

Continue reading How to Calculate and Solve for Effective Heat of Fusion | Solidification of Metals

How to Calculate and Solve for %Heat Loss in Flue Gas | Fuel and Furnaces

The image above represents %heat loss in flue gas.

To compute for %heat loss in flue gas, four essential parameters are needed and these parameters are Mass (m), Specific Heat of Stock (Cp), Change in Temperature (ΔT) and Gross Calorific Value of Fuel (GCVf).

The formula for calculating %heat loss in fuel gas:

%QL = 100(mCpΔT/GCVf)

Where:

%QL = %Heat Loss in Flue Gas
m = Mass
Cp = Specific Heat of Stock
ΔT = Change in Temperature
GCVf = Gross Calorific Value of Fuel

Let’s solve an example;
Find the %heat loss in flue gas when the mass is 10, the specific heat of shock is 14, the change in temperature is 12 and the gross calorific value of fuel is 18.

This implies that;

m = Mass = 10
Cp = Specific Heat of Stock = 14
ΔT = Change in Temperature = 12
GCVf = Gross Calorific Value of Fuel = 18

%QL = 100(mCpΔT/GCVf)
%QL = 100(10(14)(12)/18)
%QL = 100(1680/18)
%QL = 100(93.33)
%QL = 9333.3

Therefore, the %heat loss of flue gas is 9333.3%.

Continue reading How to Calculate and Solve for %Heat Loss in Flue Gas | Fuel and Furnaces

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

The image above represents quantity of heat.

To compute for quantity of heat, three essential parameters are needed and these parameters are Mass of the Stock (m), Specific Heat of Stock (Cp) and Change in Temperature (ΔT).

The formula for calculating quantity of heat:

Q = mCpΔT

Where:

Q = Quantity of Heat
m = Mass of the Stock
Cp = Specific Heat of Stock
ΔT = Change in Temperature

Let’s solve an example;
Find the quantity of heat when the mass of the stock is 18, the specific heat of stock is 10 and the change in temperature is 12.

This implies that;

m = Mass of the Stock = 18
Cp = Specific Heat of Stock = 10
ΔT = Change in Temperature = 12

Q = mCpΔT
Q = 18(10)(12)
Q = 2160

Therefore, the quantity of heat is 2160 J/kg K.

Calculating the Mass of the Stock when the Quantity of Heat, the Specific Heat of Stock and the Change in Temperature is Given.

m = Q / Cp x ΔT

Where;

m = Mass of the Stock
Q = Quantity of Heat
Cp = Specific Heat of Stock
ΔT = Change in Temperature

Let’s solve an example;
Find the mass of the stock when the quantity of heat is 15, the specific heat of stock is 3 and the change in temperature is 4.

This implies that;

Q = Quantity of Heat = 15
Cp = Specific Heat of Stock = 3
ΔT = Change in Temperature = 4

m = Q / Cp x ΔT
m = 15 / 3 x 4
m = 15 / 12
m = 1.25

Therefore, the mass of the stock is 1.25 Kg.

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

How to Calculate and Solve for Quantity of Heat Conduction | Fluidization

The image above represents quantity of heat conduction.

To compute for quantity of heat conduction, three essential parameters are needed and these are Heat Transfer Coefficient (h), Area (A) and Change in Temperature (ΔT).

The formula for calculating quantity of heat conduction:

Q = hAΔT

Where:

Q = Quantity of Heat Conduction
h = Heat Transfer Coefficient
A = Area
ΔT = Change in Temperature

Let’s solve an example;
Find the quantity of heat conduction when the heat transfer coefficient is 14, the area is 7 and the change in temperature is 2.

This implies that;

h = Heat Transfer Coefficient = 14
A = Area = 7
ΔT = Change in Temperature = 2

Q = hAΔT
Q = (14)(7)(2)
Q = 196

Therefore, the quantity of heat conduction is 196 J/Kg K.

Calculating for Heat Transfer Coefficient when the Quantity of Heat Conduction, the Area, the Change in Temperature is Given.

h = Q / AΔT

Where;

h = Heat Transfer Coefficient
Q = Quantity of Heat Conduction
A = Area
ΔT = Change in Temperature

Let’s solve an example;
Find the heat transfer coefficient when the quantity of heat conduction is 22, the area is 12 and the change in temperature is 8.

This implies that;

Q = Quantity of Heat Conduction = 22
A = Area = 12
ΔT = Change in Temperature = 8

h = Q / AΔT
h = 22 / (12)(8)
h = 22 / 96
h = 0.229

Therefore, the heat transfer coefficient is 0.229.

Continue reading How to Calculate and Solve for Quantity of Heat Conduction | Fluidization

How to Calculate and Solve for Sensitivity Drift Co-efficient | System Performance Characteristics

The image above represents sensitivity drift co-efficient.

To compute for sensitivity drift co-efficient, two essential parameters are needed and these parameters are Sensitivity Drift (SD) and Change in Temperature (ΔT).

The formula for calculating for sensitivity drift co-efficient:

CSD = SD / ΔT

Where:

CSD = Sensitivity Drift Co-efficient
SD = Sensitivity Drift
ΔT = Change in Temperature

Let’s solve an example;
Find the sensitivity drift co-efficient when the sensitivity drift is 28 and the change in temperature is 12.

This implies that;

SD = Sensitivity Drift = 28
ΔT = Change in Temperature = 12

CSD = SD / ΔT
CSD = 28 / 12
CSD = 2.33

Therefore, the sensitivity drift co-efficient is 2.33.

Calculating for the Sensitivity Drift when the Sensitivity Drift Co-efficient and the Change in Temperature is Given.

SD = CSD x ΔT

Where:

SD = Sensitivity Drift
CSD = Sensitivity Drift Co-efficient
ΔT = Change in Temperature

Let’s solve an example;
Find the sensitivity drift when the sensitivity drift co-efficient is 25 and the change in temperature is 6.

This implies that;

CSD = Sensitivity Drift Co-efficient = 25
ΔT = Change in Temperature = 6

SD = CSD x ΔT
SD = 25 x 6
SD = 150

Therefore, the sensitivity drift is 150.

Continue reading How to Calculate and Solve for Sensitivity Drift Co-efficient | System Performance Characteristics

How to Calculate and Solve for Zero Drift | System Performance Characteristics

The image above represents zero drift.

To compute for zero drift, two essential parameters are needed and these parameters are Drift (D) and Change in Temperature (ΔT).

The formula for calculating zero drift:

ZD = D / ΔT

Where:

ZD = Zero Drift
D = Drift
ΔT = Change in Temperature

Let’s solve an example;
Find the zero drift when drift is 40 and the change in temperature is 22.

This implies that;

D = Drift = 40
ΔT = Change in Temperature = 22

ZD = D / ΔT
ZD = 40 / 22
ZD = 1.81

Therefore, the zero drift is 1.81.

Calculating for the Drift when the Zero Drift and the Change in Temperature is Given.

D = ZD x ΔT

Where:

D = Drift
ZD = Zero Drift
ΔT = Change in Temperature

Let’s solve an example;
Find the drift when the zero drift is 28 and the change in temperature is 14.

This implies that;

ZD = Zero Drift = 28
ΔT = Change in Temperature = 14

D = ZD x ΔT
D = 28 x 14
D = 392

Therefore, the drift is 392.

Continue reading How to Calculate and Solve for Zero Drift | System Performance Characteristics