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A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed for a 20°F temperature rise? How much flow is required for 15°F rise?

User Turtle
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1 Answer

3 votes

Answer:

For
20^(\circ) - 5.556 lb/s

For
15^(\circ) - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T =
20^(\circ)

Temperature rise, T' =
15^(\circ)

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air,
C_(p) = 0.24 btu/lb ^(\circ)F

Now, for a rise of
20^(\circ) in temeprature:


\dot{m}C_(p)\Delta T = 96000


\dot{m} = (96000)/(C_(p)\Delta T) = (96000)/(0.24* 20) = 20000 lb/h = (20000)/(3600) = 5.556 lb/s

Now, for
15^(\circ):


\dot{m}C_(p)\Delta T = 96000


\dot{m} = (96000)/(C_(p)\Delta T) = (96000)/(0.24* 15) = 26666.667 lb/h = (26666.667)/(3600) = 7.4074 lb/s

User Rroche
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