Question

In a supermarket, air at atmospheric pressure and 25 °C is supplied to an air handling unit through a rectangular duct with a cross section of 0.5 m x 0.1 m. The air handling unit will distribute 2/3 of the air to the large shopping area of the supermarket at 25 °C, and it will cool the remaining air to -5 °C and distribute it to a walk-in refrigerator. To avoid frictional pressure losses, the average air velocity in the ducts should be 5 m/s. Assume pressure changes in the system are negligible.

a. What is the mass flowrate of air into the air handling unit?
b. Find the cross-sectional areas of the two ducts leaving the air handling unit,transporting air to the shopping area and the refrigerator.

Answer 1

(a)

Considering air flow into the air handling unit-

By ideal gas equation -

PV = mRT

P = RT

101.325 = * 0.287 * 298

Density of air intake , = 1.184 kg/m³

Mass flow rate is given by the continuity equation as-

m = * A * V (density * area * velocity)

= 1.184 * 0.5 * 0.1 * 5

m = 0.296 kg/s

(b)

Considering the shopping area first-

Mass flow rate into the shopping area-

m₁ = 2/3 * m

= 2/3 * 0.296

m₁ = 0.19733 kg/s

Since the temperature and pressure conditions in the shopping floor is the same so density of air will remain the same.

Again using continuity equation -

m₁ = ₁ * A₁ * V₁

0.19733 = 1.184 * A₁ * 5

A₁ = 0.0333 m²

The rest amount of air will be cooled to -5 deg C and then fed to refrigerator.

m₂ = m - m₁

m₂ = 0.296 - 0.1973

m₂ = 0.0986 kg/s

Since the temperature is changing so the density of air will also change hence using ideal gas equation again to find the density of air-

P = ₂RT₂

101.325 = ₂ * 0.287 * 268

₂ = 1.31734 kg/m³

Again using continuity equation -

m₂ = ₂ * A₂ * V₂

0.0986 = 1.31734 * A₂ * 5

A₂ = 0.01496 m²

Note that the working pressure and velocity is same throughout the system.