An isolated enclosure shown in the Figure has inlet P and outlet Q of 2 sqm each

Q. An isolated enclosure shown in the Figure has inlet P and outlet Q of 2 sqm each, on the opposite walls. The outdoor wind speed is 5 m/sec. If the coefficient of effectiveness is 0.6, the rate of natural ventilation in the enclosure due to wind action is                                                                              cum/hr.

Ans: 21600 to 21600

Sol:

Volume of air getting extracted from the Inlet/Outlet

Here, The area of inlet and outlet = 2 sq.m, and wind speed = 5m/sec.
Thus, volume of air getting extracted from inlet P per second = 2m² × 5m/sec = 10m³/sec.
As we know, 1 hr. = 60 min = 3600 sec.
Volume of air getting extracted inside from inlet P per hour = 10m³ × 3600/hr. = 36000 m³/hr.
If the co-efficient of effect iveness is 1 then, amount of air getting extracted frominlet = amount of air extracted from outlet.
Here, co-efficient of effectiveness = 0.6
Thus, rate of natural ventilation per hour = 0.6 × amount of air coming inside from inlet P per hour = 0.6 × 36,000 = 21,600 m³/hour.