Respuesta :
Answer:
[tex]\Delta \theta = 47.57^{\circ} C[/tex]
Explanation:
given,
moles of air compressed, n = 1.70 mol
initial temperature, T₁ = 390 K
Power supply by the compressor, P = 7.5 kW
Heat removed = 1.3 kW
Angular frequency of the compressor, f = 110 rpm = 110/60 = 1.833 rps.
Time of compression = time of the hay revolution
=[tex]\dfrac{1}{2}\ T[/tex]
=[tex]\dfrac{1}{2}\times \dfrac{1}{f}[/tex]
=[tex]\dfrac{1}{2}\times \dfrac{1}{1.833}[/tex]
=0.273 s
Using first law of thermodynamics
U = Q - W
now,
[tex]\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t}- \dfrac{\Delta W}{\Delta t}[/tex]
Power supplied [tex]\dfrac{\Delta W}{\Delta t}[/tex] = 7.5 kW
heat removed [tex]\dfrac{\Delta Q}{\Delta t}[/tex] = 1.3 kW
now,
[tex]\dfrac{\Delta U}{\Delta t} = 7.5 -1.3[/tex]
[tex]\dfrac{\Delta U}{\Delta t} = 6.2 kW[/tex]
we know,
[tex]\dfrac{\Delta U}{\Delta t}=\dfrac{nC_v\Delta \theta}{\Delta t}[/tex]
C_v for air = 5 cal/° mol
= 5 x 4.186 J/mol°C = 20.93 J/mol°C
now,
[tex]\Delta \theta = \dfrac{\Delta U}{\Deta t}\times \dfrac{\Delta t}{n C_v}[/tex]
[tex]\Delta \theta = 6.2\times 10^3 \times \dfrac{0.273}{1.7\times 20.93}[/tex]
[tex]\Delta \theta = 47.57^{\circ} C[/tex]
the temperature change per compression stroke is equal to 47.57°C.
