Answer:
Temperature at 8C = 32 °C × 8 = 256
Relative humidity is = 60% × 8 of 256°C accordingly.
Explanation:
Considering the volumetric flow rate = 0.4 m³/s
Moist air delivered (Temperature T1 = 358 C)
With Relative humidity at duct outlet = 50%
Power input at steady rate = 1.7 KW
Pressure in the duct = 1 atmosphere
Temperature at 8C = ?
Relative humidity at the duct inlet = ?
Recalling that the value for specific enthalpy and specific volume is
Specific Enthalpy = h1 = 81 kj/kilogram of dry air
and Specific Volume = v1 = 0.9 m³/kg
Now, recalling the formula for mass flow rate,
We have, m = Volumetric flow rate / Specific volume
Therefore, 0.4 m³/s ÷ 0.9 m³/kg
= 0.44 kilogram / second
Recalling the enthalpy at inlet,
we have, h2 = h1 - p/m
Where h1 = Specific enthalpy
p = power input at steady rate
m = calculated mass flow rate
Now, if we substitute the values into the equation,
we have h2 = 81 kj/kilogram of dry air - 1.7 KW / 0.44 kilogram / second
h2 = 77.175 kj / kilogram of dry air.
Therefore, the properties of air at constant absolute humidity and specific enthalpy is 77.175 kj/kg.
Temperature at 8C = 32 °C × 8 = 256
Relative humidity is = 60% × 8 of 256°C accordingly.