In order to reduce the heat loss through the house wall in exercise 1.11, an insulating board with δ 3 = 6.5 cm and λ 3 = 0.040W/K m along with a facing of δ 4 = 11.5 cm and λ 4 = 0.79W/K m will...

In order to reduce the heat loss through the house wall in exercise 1.11, an insulating board with δ₃
= 6.5 cm and λ₃
= 0.040W/K m along with a facing of δ₄
= 11.5 cm and λ₄
= 0.79W/K m will replace the outer plaster wall. Calculate the heat flux ˙q and the surface temperature ϑ_W1
of the inner wall.

Derive the equations for the profile of the fluid temperatures ϑ₁
= ϑ₁(z) and ϑ₂
= ϑ₂(z) in a countercurrent heat exchanger, cf. section 1.3.3.

exercise 1.11

A house wall is made up of three layers with the following properties (inside to outside): inner plaster δ₁
= 1.5 cm, λ₁
= 0.87W/K m; wall of perforated bricks δ₂
= 17.5 cm, λ₂
= 0.68W/K m; outer plaster δ₃
= 2.0 cm, λ₃
= 0.87W/K m. The heat transfer coefficients are α₁
= 7.7W/m²K inside and α₂
= 25W/m²K outside. Calculate the heat flux, ˙q, through the wall, from inside at ϑ₁
= 22.0 ◦C to the air outside at ϑ₂
= −12.0 ◦C. What are the temperatures ϑW1 and ϑW2 of the two wall surfaces?