Consider 100 m3 of moist air at 100 kPa 40 C and 80 RH Pa

Consider 100 m3 of moist air at 100 kPa , 40 C and 80 % , R.H.

Part A

Calculate the humidity ratio ( ).

Part B

Determine the mass of dry air.
Express your answer to four significant figures using the appropriate units.

Part C

Determine the enthalpy of the dry air, per unit mass of the dry air.
Express your answer in kilojoules per kilograms of dry air to four significant figures.

Solution

1) here given data as DBT=40 C ,atm.pressure= 1 bar,V=100 m3, R.H.=80%

hence here

R.H./100=Pv/Pvs

where Pvs at DBT=40 c

is Pvs=.07375 bar

hence from above equation we get

Pv=.07375*.8=.059 bar

2) specific humidity is w=.622*(Pv/P-Pv)

putting value we get

w=.03899 kg/kg of dry air

3) mass of dry air is given by

Ma=(Pa*V/Ra*T)

P=Pa+Pv

Pa=1-.059=.941bar

Ra=287 kj/kg k

we get

Ma=1.04752*10^-3 kg

and Mv=4.085*10^-5 kg

4) where DPT correspond to pressure Pv=.059 bar

Tdp=35.79 c

hence enthalpy of moist air is

H=Cpa*Tdb+w(Hg-Cpv*(Tdb-Tdp))

Cpa=1.005 kj/kg k,Cpv=1.88 kj/kg k,Hg=2565.3 at 35 c

we get

H=1.005*40+.03899(2565.3-1.88*(40-35.79)=139.9124 kj/kg k

H=139.912448 kj/kg k