To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.
The gas ideal law is given as,
[tex]PV=mRT[/tex]
Where,
P = Pressure
V = Volume
m = mass
R = Gas Constant
T = Temperature
Our data are given by
[tex]T_1 = 38\°C[/tex]
[tex]T_2 = 14\°C[/tex]
[tex]\eta = 97\%[/tex]
[tex]\dot{v} = 510m^3/kg[/tex]
Note that the pressure to 38°C is 0.06626 bar
PART A) Using the ideal gas equation to calculate the mass flow,
[tex]PV = mRT[/tex]
[tex]\dot{m} = \frac{PV}{RT}[/tex]
[tex]\dot{m} = \frac{0.6626*10^{5}*510}{287*311}[/tex]
[tex]\dot{m} = 37.85kg/min[/tex]
Therfore the mass flow rate at which water condenses, then
[tex]\eta = \frac{\dot{m_v}}{\dot{m}}[/tex]
Re-arrange to find [tex]\dot{m_v}[/tex]
[tex]\dot{m_v} = \eta*\dot{m}[/tex]
[tex]\dot{m_v} = 0.97*37.85[/tex]
[tex]\dot{m_v} = 36.72 kg/min[/tex]
PART B) Enthalpy is given by definition as,
[tex]H= H_a +H_v[/tex]
Where,
[tex]H_a[/tex]= Enthalpy of dry air
[tex]H_v[/tex]= Enthalpy of water vapor
Replacing with our values we have that
[tex]H=m*0.0291(38-25)+2500m_v[/tex]
[tex]H = 37.85*0.0291(38-25)-2500*36.72[/tex]
[tex]H = 91814.318kJ/min[/tex]
In the conversion system 1 ton is equal to 210kJ / min
[tex]H = 91814.318kJ/min(\frac{1ton}{210kJ/min})[/tex]
[tex]H = 437.2tons[/tex]
The cooling requeriment in tons of cooling is 437.2.
