A closed, rigid tank fitted with a paddle wheel contains 1.6 kg of air, initially at 200oC, 1 bar. During an interval of 5 minutes, the paddle wheel transfers energy to the air at a rate of 1 kW. During this time interval, the air also receives energy by heat transfer at a rate of 0.5 kW. These are the only energy transfers. Assume the ideal gas model for the air, and no overall changes in kinetic or potential energy. Do not assume specific heats are constant. Determine the change in specific internal energy for the air, in kJ/kg, and the final temperature of the air, in oC.

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Netta00 Netta00 · Answer 1 · 2019-10-24T10:46:34+02:00

Answer:

ΔU = 450 KJ

T=596.21°C

Explanation:

m= 1.6 kg

T₁= 200°C

P₁= 1 bar

Sign convention:

1. If work is done on the system then it taken as negative ans if work is done by the system then it taken as positive.

2. If heat added to the system then it taken as positive and if heat is rejected from the system then it taken as negative.

Paddle wheel transfers energy to the air at a rate of 1 kW ⇒ Work = - 1 KW

t= 5 min

W= - 1 x 5 x 60 = - 300 KJ

Air receives energy by heat transfer at a rate of 0.5 kW ⇒ Q = 0.5 KW

Q = 0.5 x 5 x 60 = 150 KJ

From first law of thermodynamics

Q= W + ΔU

150 = - 300 + ΔU

ΔU = 450 KJ

So the change in internal energy is 450 KJ.

We know that for air

ΔU= m Cv ΔT

Cv=0.71 KJ/kgk

450 = 1.6 x 0.71 x (T-200)

T=596.21°C

Final temperature is T=596.21°C.