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erica [24]
3 years ago
11

Steam enters a steady-flow adiabatic nozzle with a low inlet velocity (assume ~0 m/s) as a saturated vapor at 6 MPa and expands

to 1.2 MPa.
a. Under what conditions is the velocity from this nozzle maximized?
b. Determine the maximum exit velocity of the steam, in m/s.
Engineering
1 answer:
Sergio [31]3 years ago
7 0
Yea bro I don’t really know
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Waterjet cutting uses a _____. a) Very-small-diameter jet of water that exits the cutting nozzle at approximately the speed of s
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Answer:

a

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A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t
iren2701 [21]

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, P_{1}= 1 bar

v_{2} = 4 bar

pV¹°¹ = constant

From superheated propane table, at  P_{1}= 1 bar andT_{1}  = 27⁰C

v_{1} = 0.557 m³/kg

v_{2} = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1  \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557  \\\\V2 = 0.158 m^3/kg

At  = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\  \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\   \\W = -19.75KJ

(b) Heat transfer = ?

Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ

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4 years ago
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