Respuesta :
Correct temperature is 80°F
Answer:
T_f = 38.83°F
Explanation:
We are given;
Volume; V = 8 ft³
Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²
Initial temperature; T_i = 80°F = 539.67 °R
Time for outlet flow; t_o = 90 s
Mass flow rate at outlet; m'_o = 0.03 lb/s
Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²
Now, from ideal gas equation,
Pv = RT
Where v is initial specific volume
R is ideal gas constant = 53.33 ft.lbf/°R
Thus;
v = RT/P
v_i = 53.33 × 539.67/(100 × 12²)
v_i = 2 ft³/lb
Formula for initial mass is;
m_i = V/v_i
m_i = 8/2
m_i = 4 lb
Now change in mass is given as;
Δm = m'_o × t_o
Δm = 0.03 × 90
Δm = 2.7 lb
Now,
m_f = m_i - Δm
Thus; m_f = 4 - 2.7
m_f = 1.3 lb
Similarly in above;
v_f = V/m_f
v_f = 8/1.3
v_f = 6.154 ft³/lb
Again;
Pv = RT
Thus;
T_f = P_f•v_f/R
T_f = (30 × 12² × 6.154)/53.33
T_f = 498.5°R
Converting to °F gives;
T_f = 38.83°F
The final temperature, in °F, of the air remaining in the tank is 38.83°F
It is given that volume V = 8 ft³
Initial Pressure Pi = 100 lbf/in² = 100 × 12² lbf/ft²
Initial temperature Ti = 80°F = 539.67 °R
Time for outlet flow; to = 90 s
Mass flow rate at outlet; m'o = 0.03 lb/s
Final pressure; Pf = 30 lbf/in² = 30 × 12² lbf/ft²
Now, from ideal gas equation,
Pv = RT
where v is initial volume, R is ideal gas constant = 53.33 ft.lbf/°R
[tex]v = RT/P\\ \\ v_i = 53.33 *539.67/(100*12^2)\\ \\ v_i = 2 ft^3/lb [/tex]
The initial mass is;
[tex]m_i = V/v_i\\ \\ m_i = 8/2\\ \\ m_i = 4 lb [/tex]
Now change in mass is given as;
Δm = [tex]m'_o*t_o[/tex]
Δm = 0.03 × 90
Δm = 2.7 lb
[tex]m_f[/tex] = [tex]m_i[/tex] - Δm
[tex]m_f[/tex] = 4 - 2.7
[tex]m_f[/tex] = 1.3 lb
now,
[tex]v_f = V/m_f\\ \\ v_f = 8/1.3\\ \\ v_f = 6.154 f^3/lb [/tex]
From the gas equation
Pv = RT
Final state:
[tex]T_f = P_fv_f/R\\\\ T_f = (30*12^2*6.154)/53.33\\\\ T_f = 498.5^oR [/tex]
Converting to °F:
[tex]T_f[/tex] = 38.83°F is the final temperature.
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