Chemistry, 10.03.2020 06:31 Halieyrobinson3003

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.

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Answer from: cathydaves

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 $c__{p}$ = 0.219 Btu/lbm.R

Specific Heat at constant volume $c__{v}$ = 0.157 Btu/lbm.R

The initial temperature (T_1) in the tank is given as 80°F.

Converting it to Rankine; we have

(T_1) = (460 + 80) R

(T_1) = 540 R

The volume of the tank is given as = 30 ft³

The initial mass m_1 in the tank from the ideal gas equation can be calculated as:

$m_1 =\frac{P_1*V}{T_1*R}$

where;

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

T₁ = 540 R

So; we have:

$m_1 =\frac{2000*30}{540*0.3353}$

$m_1 =\frac{60000}{181.062}$

m_1 = 331.38 lbm

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

$m_2 = \frac{P_2*V}{T_2*R}$

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

T_1 =T_2 = 540 R

∴ $m_2 = \frac{100*30}{540*0.3353}$

$m_2 = \frac{3000}{181.062}$

m_2 = 16.57 lbm

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

$\delta m_{system} = m_{in}-m_{out}$

where;

$\delta m_{system}$ = change in the initial mass and final mass of the oxygen in the tank

$m_{in}$ = the inlet mass of the oxygen

$m_{out}$ = the outlet mass of the oxygen

So; let replace $\delta m_{system}$ with m_2-m_1

$m_{in}$ with 0

$m_{out}$ with m_c (i.e amount of oxygen used in the system)

so; we have

m_2-m_1 = 0 - m_c

m_c = m_1-m_2

m_c = 331.38 - 16.57

m_c = 314.81 lbm

Therefore, the mass of oxygen used is 314.81 lbm

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:

$E_{in}$ - $E_{out}$ = $E_{system}$

$Q_{in}$ - (m_c*h_c) = (m_2*u_2) - (m_1*u_1)

where

m_c = mass of oxygen used

h_c = specific enthalpy of the oxygen used

u_2 = specific internal energy of the final mass

u_1 = specific internal energy of the initial mass

$Q_{in}$ = total heat transfer to the tank

From above; making $Q_{in}$ the subject of the formula; we have:

$Q_{in}$ = (m_2*u_2) - (m_1*u_1) (m_c*h_c)

Lets replace;

u_2 with c_v*T_2 & u_1 with c_p*T_c

We have:

$Q_{in}$ = [m_2*(c_v*T_2)]-[m_1*(c_v*T_1)]+[m_c*(c_p*T_c)]

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

T_c =T_2=T_1

where;

m_1 = 331.38 lbm

m_2 = 16.57 lbm

m_c = 314.81 lbm

T_2 = 540 R

T_1 = 540 R

T_c = 540 R

c_p = 0.219 Btu/lbm.R

c_v = 0.157 Btu/lbm.R

Substituting our parameters; we have:

$Q_{In}$ = [16.57×(0.157×540)]-[331.38×(0.157×540)]+[314.81×(0.219×540)]

$Q_{In}$ = 10539.84 Btu

∴ The total heat transfer to the tank = 10539.84 Btu

Answer from: Quest

58750.08068 is the answer i think

Answer from: Quest

2.00 is the !

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