In: Mechanical Engineering
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.
(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.