Respuesta :
Answer:
a) change in specific internal energy for the air, Δu = 675 kJ
b) Final temperature of air, [tex]T_{f} = 1017.82^{0} C[/tex]
Explanation:
The heat generated = Q
The work done = W
Time interval, Δt = 15 mins = 15 * 60
Δt = 900 s
It is stated that there is no overall change in kinetic and potential energy
[tex]\triangle KE = 0\\\triangle PE = 0[/tex]
a) The change in internal energy, ΔU = Q - W
Change in specific internal energy, Δu = (Q - W)/m
Workdone is calculated by:
[tex]W = \int {\dot{W}} \, dt \\W = \dot{W} \triangle t[/tex]
Since energy is transferred into the air, rate of energy transfer to the air is taken as negative, [tex]\dot{W} = -1 kW[/tex]
W = -1 * 900
W = - 900 kJ
Energy received by heat transfer, [tex]\dot{Q} = 0.5 kW[/tex]
[tex]Q = \int {\dot{Q}} \, dt \\Q = \dot{Q} \triangle t[/tex]
Q = 0.5 * 900
Q = 450 kJ
Mass of air, m = 2 kg
Change in specific internal energy, Δu = (Q - W)/m
Δu = (450 - (-900)/2
Δu = 675 kJ
b) By interpolation from the ideal gas property table:
At [tex]T_{i} = 200^{0} C = 473 K\\[/tex]
When T = 470K, u = 337.32 kJ/kg
When T = 480K, u = 344.70
[tex]\frac{u_{i} - 337.32 }{344.7 - 337.32} = \frac{473 - 470 }{480 -470}[/tex]
[tex]u_{i} = 339.534 kJ/kg[/tex]
[tex]\triangle u = u_{f} - u_{i}\\675 + 339.534 = u_{f}\\u_{f} = 1014.534 kJ/kg[/tex]
At [tex]u_{f} = 1014.534 kJ/kg[/tex]
When T = 1280K, u = 1004.76 kJ/kg
When T = 1300K, u = 1022.82 kJ/kg
By interpolation,
[tex]\frac{T_{f} - 1280 }{1300 - 1280} = \frac{1014.534 - 1004.76 }{1022.82 -1004.76}\\T_{f} = 1290.82 K// T_{f} = 1290.82 - 273 \\ T_{f} = 1017.82^{0} C[/tex]
