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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Answer:
See attachment for chart
Explanation:
The IPO chart implements he following algorithm
The expressions in bracket are typical examples
<u>Input</u>
Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module
<u>Processing</u>
Assign variable to the input number (x)
Calculate the square (x = 5 * 5)
Display the result (25) ----> This will be passed to the output module
<u>Output</u>
Display 25
Answer:
The heater load =35 KJ/kg
Explanation:
Given that
At initial condition
Temperature= 15°C
RH=80%
At final condition
Temperature= 50°C
We know that in sensible heating process humidity ratio remain constant.
Now from chart
At temperature= 15°C and RH=80%
At temperature= 50°C
The heater load = 73 - 38 KJ/kg
The heater load =35 KJ/kg