A metal storage tank of external area 0.5 m^2 contains water at 65 degree C. & is located inside a large enclosure with wall at 10 deg.C. the tank is painted black on the outside & its emissivity is approx. to 0.95. How much reduction in heat loss would occur if the outside surface was coated with aluminium paint of emissivity 0.55? Take radiation constant qb = 6.86*10^-8 kcal/m^2-hr-k^4

Question

A metal storage tank of external area 0.5 m^2 contains water at 65 degree C. & is located inside a large enclosure with wall at 10 deg.C. the tank is painted black on the outside & its emissivity is approx. to 0.95. How much reduction in heat loss would occur if the outside surface was coated with aluminium paint of emissivity 0.55? Take radiation constant qb = 6.86*10^-8 kcal/m^2-hr-k^4

Answer 1

To calculate the reduction in heat loss, we need to calculate the heat loss of the current situation and compare it with the heat loss of the situation with the coated surface.

1. Calculate the heat loss with the current situation:
- Calculate the radiant heat transfer from the tank's surface to the enclosure's wall using the Stefan-Boltzmann law:
Q1 = ε1 * σ * A * (T1^4 - T2^4)
- ε1 = emissivity of the current surface (0.95 for the black paint)
- σ = Stefan-Boltzmann constant (6.86 x 10^-8 kcal/m^2-hr-K^4)
- A = external area of the tank (0.5 m^2)
- T1 = temperature of the tank (65°C + 273.15 = 338.15 K)
- T2 = temperature of the enclosure's wall (10°C + 273.15 = 283.15 K)

2. Calculate the heat loss with the coated surface:
- Calculate the radiant heat transfer from the tank's surface to the enclosure's wall using the Stefan-Boltzmann law:
Q2 = ε2 * σ * A * (T1^4 - T2^4)
- ε2 = emissivity of the coated surface (0.55 for the aluminum paint)
- σ = Stefan-Boltzmann constant (6.86 x 10^-8 kcal/m^2-hr-K^4)
- A = external area of the tank (0.5 m^2)
- T1 = temperature of the tank (65°C + 273.15 = 338.15 K)
- T2 = temperature of the enclosure's wall (10°C + 273.15 = 283.15 K)

3. Find the reduction in heat loss:
Reduction = Q1 - Q2

Now, let's calculate the values:

1. Calculate the heat loss with the current situation:
Q1 = 0.95 * (6.86 x 10^-8) * 0.5 * (338.15^4 - 283.15^4)

2. Calculate the heat loss with the coated surface:
Q2 = 0.55 * (6.86 x 10^-8) * 0.5 * (338.15^4 - 283.15^4)

3. Find the reduction in heat loss:
Reduction = Q1 - Q2

By following these steps and plugging in the values into the equations, you can calculate the reduction in heat loss that would occur if the outside surface of the tank was coated with aluminum paint.