Air is cooled and dehumidified as it flows over the coils of refrigeration system at 100 kPa from 30 ºC and relative humidityof
84%to 15 ºC and relative humidity of 100%. The mass flow rate of dry air is 0.4 kg/s.
Using the formulaand not the Psychrometric chart, determine:
a) the mass flow rate of water,
b)the heat removal from the air.
Using the formulaand not the Psychrometric chart, determine:
a) the mass flow rate of water,
b)the heat removal from the air.
1 answer:
Answer:
(a) The mass glow rate of water is 0.0032kg/s
(b) The heat removal from air is 14.17KJ/sec
Explanation:
In this question, we are asked to use the formula to calculate the mass flow rate and the heat removal from water based on the data in the question.
To answer this question, we take values of absolute humidity and enthalpy values from table from corresponding temperature
Please check attachment for complete solution and step by step explanation
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Explanation:
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Answer:
The atmospheric pressure in atm=0.885 atm
Explanation:
Given that
Local pressure (h)= 30 ft of water height ( 1 ft= 0.3048 m)
We know that pressure in given by
P=ρgh
We know that ρ of water is 1000
So pressure
P=1000(9.81)(9.144)
We know that 1000 Pa=0.00986 atm
So P=0.885 atm
The atmospheric pressure in atm=0.885 atm