Question

adiabatic container with a total volume of 10 ft3 is initially evacuated. You would like to input air from a supply line, which runs at a temperature of 100oF and a pressure of 50 psia, by opening a valve that lets the air get into this container. Calculate the amount of air that you got in the container, in lbm. Can this happen? Assume constant heat capacities.

Answer 1

Tank is evacuated and send air into the tank by supply line through a valve.

Supply line condition :

Initial temperature T1 = 100 F

Pressure P1 = 50 Psia

As we know that tank is evacuated then having no initial pressure and temperature. All inlet enthalpy of air is get converted into internal energy of air inside the tank.

1. Adiabatic operation Q = 0

2. Negligible kinetic and potential energy.

3. No boundaey work done and no shaft work.

For supply line it having flow so we use enthalpy and evacuated tank it having constant volume so we use internal energy.

Mass of the air in supply line and after getting all air inside the tank is same.

Enthalpy of air inlet = internal energy of air inside tank after getting air into the tank

H1 = U2

CpT1 = CvT2

Cp - heat capacity at constant pressure

Cv - heat capacity at constant volume

Gamma heat capacity ratio Y = Cp/Cv = 1.4

Gas constant,

R = 10.73 ft3. psia/lbmol*R

T2 = (Cp/Cv) * T1 = 1.4*100 = 140 F

Conversion of temperature : F to R

F + 459.67 = R

At 140F

R = 140+459.67 = 599.67 R

T2 = 599.67 R

Given tank volume V = 100 ft3

After getting air, volume of air V = 10 ft3, Pressure remains same because there is tank evacuated initially  P = 50 psia and T2 = 599.67 R

By using ideal gas equation

PV = nRT

n = PV/RT

n = 50 psia * 10 ft3/(10.73ft3.psia/lbmol*R)*599.67R

n = 0.0777 lbmol

Molecular weight of air M = 29 lbm/lbmol

Mass of air =0.0777lbmol * 29 lbm/lbmol = 2.253 lbm

amount of air that got the air, m = 2.253 lbm.

Yes this process happen. All flow energy get used to increase the internal energy of air after getting all air.