Question

Moist air at 25 degrees Celsius and 1 bar enters the evaporator of an AC unit at a flow rate of 20 m^3/min. It is known that the humidity ratio is 10 gmw/kgda. Find the mass flow rates, ˙mda and ˙mwv, of dry air and water vapor respectively. If the air exits the evaporator at 18 degrees Celsius and same pressure, what is the rate of heat transfer required to cool the air.

Name: moist air at 25 degrees celsius and 1 bar enters the evaporator of an ac unit at a flow rate of 20 m3min it is known that the humidity ratio is 10 gmwkgda find the mass flow rates mda and mw 75879
Uploaded: 2022-01-21T05:36:20-08:00
Duration: 1 min 13 s
Channel: Manik Pulyani
Description: moist air at 25 degrees celsius and 1 bar enters the evaporator of an ac unit at a flow rate of 20 m3min it is known that the humidity ratio is 10 gmwkgda find the mass flow rates mda and mw 75879

          Moist air at 25 degrees Celsius and 1 bar enters the evaporator of an AC unit at a flow rate of 20 m^3/min. It is known that the humidity ratio is 10 gmw/kgda. Find the mass flow rates, ˙mda and ˙mwv, of dry air and water vapor respectively. If the air exits the evaporator at 18 degrees Celsius and same pressure, what is the rate of heat transfer required to cool the air.

Added by Amber H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Manik Pulyani

01/21/2022

Step 1

First, we need to convert the humidity ratio from grams to kilograms: ω = 10 gmw/kgda = 0.01 kgw/kgda. Using the relationship between the humidity ratio, the mass flow rate of dry air, and the density of moist air, we can find the mass flow rate of dry air. Show more…

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Moist air at 25 degrees Celsius and 1 bar enters the evaporator of an AC unit at a flow rate of 20 m^3/min. It is known that the humidity ratio is 10 gmw/kgda. Find the mass flow rates, ˙mda and ˙mwv, of dry air and water vapor respectively. If the air exits the evaporator at 18 degrees Celsius and same pressure, what is the rate of heat transfer required to cool the air.