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Stolb23 [73]
3 years ago
15

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where the

pressure is 2.5 bar, during which the pressure–volume relationship is pV = constant. Assume ideal gas behavior for the air.​​Determine the work and heat transfer, in kJ.
Engineering
1 answer:
Archy [21]3 years ago
8 0

Answer:

Work transfer is - 97.02 KJ. It means that work is given to the system.

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.

Explanation:

Given that

m= 3 kg

P₁=2 bar

T=T₁=T₂=30 °C

T=303 K

P₂=2.5 bar

PV=  Constant

This is the isothermal process .

We know that work for isothermal process  given as

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=mRT\ln \dfrac{P_1}{P_2}

For air

R= 0.287 KJ/Kg.K

Now by putting the values

W=mRT\ln \dfrac{P_1}{P_2}

W=5\times 0.287\times 303\ln \dfrac{2}{2.5}

W= - 97.02 KJ

So the work transfer is - 97.02 KJ. It means that work is given to the system.

We know that for ideal gas internal energy is the only function of temperature.The change in internal energy ΔU

ΔU = m Cv ΔT

Here ΔT= 0

So

ΔU =0

From first law of thermodynamics

Q= ΔU +W

ΔU = 0

Q= W

Q= - 97.02 KJ

Heat transfer = - 97.02 KJ . It means that heat is rejected from the system.

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Answer:

  • branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
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Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

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The power factor is cos(63.436°) ≈ 0.447 (leading)

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  It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

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The phasor diagram of the currents is attached.

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<em>Additional comment</em>

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

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