Heat Loss Formula

Heat can be termed as the quantity of energy that flows spontaneously between two objects due to a temperature differential. During thermal systems, objects with different temperatures tend to approach thermal equilibrium. The hotter object transmits heat to the colder object until the temperatures are equal, or until the objects reach thermal equilibrium. Heat loss can be a result of any of these factors: either radiation or convection, or even conduction.

Heat Loss

Heat loss is the reduction of heat in space caused by heat transmission via walls, roofs, windows, and building surfaces. We compute heat loss by finding the product of the area values, the temperature differential between the inner and outside surfaces, and the material’s heat loss value. Convectional heat loss is the type of heat loss that is of particular relevance in the ventilation of hot processes.

None of the materials in this world can prevent heat loss, yet it can be minimized by some of them. The unit to measure heat loss is Watts (W). 

Heat Loss Formula 

The heat loss is denoted by the symbol q. Its dimensional formula is given by [M¹ L² T^-3]. It is directly proportional to the heat transmission value, area of the object, and temperature difference between the object and its surroundings. In other words, its formula equals the product of the overall transmission coefficient, area of the object, and temperature difference. Its formula is expressed as:

q = (U × A) × Δt

Where, 

• q depicts the total heat loss in Watts.
• U is the overall transmission coefficient.
• A is the area of the object/ building.
• Δt depicts the difference between the temperature inside and outside.

Sample Problems

Question 1: Calculate the amount of heat lost given that the area, coefficient of heat transfer, and Δt are 60 sq. m, 0.7, and 50°C respectively.

Solution:

Given: U = 0.7, A = 60 sq.m., Δt = 50°C

Since, q = (U × A) × Δt

⇒ q = 0.7 × 60 x 50 W

q = 2100 W

Question 2: Calculate the amount of heat lost given that the area, coefficient of heat transfer, and Δt are 70 sq. m, 0.2, and 60°C respectively.

Solution:

Given: U = 0.2, A = 70 sq. m., Δt = 60°C

Since, q = (U × A) × Δt

⇒ q = 0.2 × 70 × 60 W

q = 840 W

Question 3: Calculate the amount of heat lost given that the area, coefficient of heat transfer, and Δt are 10 sq. m, 0.3, and 10°C respectively.

Solution:

Given: U = 0.3, A = 10 sq. m., Δt = 10°C

Since, q = (U × A) × Δt

⇒ q = 0.3 × 10 x 10 W

q = 30 W

Question 4: Calculate the coefficient of heat transfer given that the area, amount of heat lost and Δt are 20 sq. m, 240 W, and 30°C respectively.

Solution:

Given: q = 240 W, A = 20 sq. m, Δt = 30°C

Since, q = (U × A) × Δt

=> U = q/AΔt

=> U = 240/(20 × 30)

U = 0.4

Question 5: Calculate the coefficient of heat transfer given that the area, amount of heat lost and Δt are 40 sq. m, 1000 W, and 50°C respectively.

Solution:

Given: q = 1000 W, A = 40 sq. m, Δt = 50°C

Since, q = (U × A) × Δt

=> U = q/AΔt

=> U = 1000/(40 × 50)

U = 0.5

Question 6: Calculate the area of the object given that the amount of heat lost, coefficient of heat transfer, and Δt are 40 W, 0.1, and 20°C respectively.

Solution:

Given: q = 40 W, U = 0.1, Δt = 20°C

Since, q = (U × A) × Δt

=> A = q/UΔt

=> A = 40/(0.1 × 20)

A = 20 sq. m

Question 7: Calculate the temperature difference given that the area, coefficient of heat transfer, and amount of heat lost are 20 sq. m, 0.9, and 216 W respectively.

Solution:

Given: U = 0.9, A = 20 sq. m, q = 216 W

Since, q = (U × A) × Δt

=> Δt = q/UA

=> Δt = 216/(0.9 × 20)

Δt = 12°C