Oxygen is supplied to a medical facility from a 30 ft3 compressed oxygen tank. Initially, the tank is at 2000 psia and 80°F. The oxygen is removed from the tank slowly enough that the temperature in the tank remains at 80°F. After two weeks, the pressure in the tank is 100 psia. Determine the mass of oxygen used in lbm. Also determine the total heat transfer to the tank in Btu. Treat the oxygen as an ideal gas with constant specific heats at 80°F.

a) 314.81 lbm

b) 10539.84 Btu

Explanation:

We are to treat the oxygen as an ideal gas with constant specific heats at 80°F.

So;

the gas constant  (R) = 0.3353 psia.ft³/lbm.R

The specific heat at constant pressure   = 0.219 Btu/lbm.R

Specific Heat at constant volume = 0.157  Btu/lbm.R

The initial  temperature in the tank is given as 80°F.

Converting it to Rankine;  we have

 = (460 + 80) R

 = 540 R

The volume of the tank  is given as  = 30 ft³

The initial mass  in the tank from the ideal gas equation can be calculated as:

where;

P₁ = 2000 psia ( initial pressure of oxygen in the tank)

T₁ = 540 R

So; we have:

= 331.38 lbm

we can also Calculate the final mass in the tank by using the same ideal gas equation

P₂ = 100 psia (final pressure of oxygen in the tank)

= 540 R

∴  

16.57 lbm

However, we can determine the mass of oxygen used by applying the mass balance as shown below:

where;

= change in the initial mass and final mass of the oxygen in the tank

= the inlet mass of the oxygen

= the outlet mass of the oxygen

So; let replace  with

with 0

with (i.e amount of oxygen used in the system)

so; we have

 = 0 -  

=

=  331.38 - 16.57

= 314.81 lbm

Therefore, the mass of oxygen used  is 314.81 lbm

b)

To determine the total heat transfer to the tank we can use the energy balance in the system to deduce that and which is given by:

- =

- = -

where

= mass of oxygen used

= specific enthalpy of the oxygen used

= specific internal energy of the final mass

= specific internal energy of the initial mass

= total heat transfer to the tank

From above; making  the subject of the formula; we have:

=   -  

Lets replace;

 with   & with

We have:

=

Finally, we are told that the temperature remains the same throughout the tank; so we have:

where;

= 331.38 lbm

= 16.57 lbm

= 314.81 lbm

= 540 R

= 540 R

= 540 R

= 0.219 Btu/lbm.R

= 0.157 Btu/lbm.R

Substituting our parameters; we have:

= [16.57×(0.157×540)]-[331.38×(0.157×540)]+[314.81×(0.219×540)]

= 10539.84 Btu

∴ The total heat transfer to the tank = 10539.84 Btu