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3 years ago

What is the area enclosed by the cycle area of the Carnot cycle illustrating on a P-V diagram?

Engineering

1 answer:

Inga [223]3 years ago

7 0

Answer:

The work of the cycle.

Explanation:

The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.

The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.

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3 years ago

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1 year ago

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3 years ago

A piston-cylinder device with 15 lbm of steam is heated from a temperature and pressure of 500 °F and 350 psia to a new temperat

nevsk [136]

Answer:

The work done is 1050 Btu

Solution:

As per the question:

Mass of steam, M = 15 lbm

Temperature, T = $500^{\circ}F$

Temperature, T' = $600^{\circ}F$

Pressure, P = 350 psia

Since, the process results in the change in volume at constant pressure.

Now, work done by the steam is given by;

W = $\int_{V}^{V'} P dU$

And

U = mV

So,

W = $m\int_{V}^{V'} P dV$

W = $mP(V' - V)$ (1)

Now, using the super heated vapor table:

At 350 psia and $500^{\circ}F$ , V = $1.5 ft^{3}/lb$

At 350 psia and $600^{\circ}F$ , V' = $1.7 ft^{3}/lb$

Now, using these values in eqn(1):

W = $15\times 350(1.7 - 1.5) = 1050 Btu$

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3 years ago

Steam enters a turbine operating at steady state at 1 MPa, 200 °C and exits at 40 °C with a quality of 83%. Stray heat transfer

Andrei [34K]

Answer:

(a) Work out put=692.83 $\frac{KJ}{Kg}$

(b) Change in specific entropy=0.0044 $\frac{KJ}{Kg-K}$

Explanation:

Properties of steam at 1 MPa and 200°C

$h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}$

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

Properties of saturated steam at 40°C

$h_f= 167.5\frac{KJ}{Kg} ,h_g= 2537.4\frac{KJ}{Kg}$