Answer:
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