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zavuch27 [327]
3 years ago
15

Nitrogen can be liquefied using a Joule-Thomson expansioni process. This is done by rapidlyl and adiabatically expandign cold ni

trogen gas from high pressure to a low pressure. If nitrogen at 135 K and 20 MPa undergoes a Joule-Thomson expansion to 0.4 MPa,
a. Estimate the fraction of vapor and liquid present after the expansion, and the emperature of this mixture using the pressure-enthalpy diagram for nitrogen.

b. Repeat the calculation assuming nitrogen to be an ideal gas with Cp* = 29.3 J/molK
Engineering
1 answer:
solniwko [45]3 years ago
8 0
T is desired to produced liquefied methane as illustrated in problem 1; however, theconditions are now changed so that the gas is initially available at 1 bar and 200 K, and leaving the cooler will be at 100 bar and 200 K. The flash drum is adiabatic, oerates at 1 bar, and each compressor stage canbe assumed to operate reversibly and adiabatically. A three-stage compressor will be used with the first stage compressing the gas fro 1 bar to 5 bar, each stage the gas will be isobarically cooled to 200 K.
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kirill [66]

Answer:

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3 0
2 years ago
Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.
vaieri [72.5K]

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

VT^n=c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

VT^{0.12}=V'\left ( 0.2T\right )^{0.12}

V'=\frac{V}{0.2^{0.12}}

V'=\frac{V}{0.824}=1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

6 0
3 years ago
What are two ways you can see that there is a smaller load going to the base of the transistor?
Lyrx [107]

Answer:

<h3><em>Transistor switches can be used to switch a low voltage DC device (e.g. LED’s) ON or OFF by using a transistor in its saturated or cut-off </em><em>state</em></h3>

  1. <em>. Cut-off </em><em>Region</em>

<em>Here the operating conditions of the transistor are zero input base current ( IB ), zero output collector current ( IC ) and maximum collector voltage ( VCE ) which results in a large depletion layer and no current flowing through the device. Therefore the transistor is switched “Fully-OFF”.</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>2</em><em>.</em><em>Saturation </em><em>Region</em>

<em>Here the transistor will be biased so that the maximum amount of base current is applied, resulting in maximum collector current resulting in the minimum collector emitter voltage drop which results in the depletion layer being as small as possible and maximum current flowing through the transistor. Therefore the transistor is switched “Fully-ON”.</em>

Explanation:

hope it helps

7 0
2 years ago
Answers:
frutty [35]

Answer:

the chassis ground wire

an inverter

is connected to one side of the power line

failure of a component such as the hash filter

is the hot wire

the common wire

6 0
3 years ago
2
aksik [14]

Answer:tech A

Explanation:

5 0
3 years ago
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