GATE Architecture

GATE 2014

In a dance hall the indoor and outdoor temperatures are 28°C and 18°C respectively. There is an internal heat gain of 5 kW and the specific heat of air (on volume basis) is 1300 J/m³ °C, then the necessary cross sectional area (m²) of a duct with an air velocity of 2 m/s required for cooling by ventilation is ____________

Answer : 0.190 to 0.195

Soln: 5 kW of heat has to be extracted= 5000 W

Specific heat of air = 1300 J/m³ °C  ……..  it means for every m³ of air 1300 Joule  is extracted for 1 dec C temp difference.

Temp difference is 10 deg ree so for every m³ …………..1300× 10= 13,000 Joules of energy is extracted.

In order to extract only 5000 Joules amount of air to be removed will be  = 0.38 m³ per sec.

So necessary Cross section for 2m/sec air flow=   a × 2 = 0.38

Solving ‘a’= 0.19 m²