Question

In: Mechanical Engineering

In a supermarket, air at atmospheric pressure and 25 °C is supplied to an air handling...

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.

Solutions

Expert Solution

(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.


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