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4 years ago

True or False? A constricting nozzle is used

Engineering

1 answer:

marissa [1.9K]4 years ago

6 0

Answer:

True

Explanation:

Principles of plasma arc cutting, Uses a constricting nozzle to create, concentrate, and direct the high-velocity plasma. Plasma gas is always used

in plasma arc cutting When shielding gas is also used, the process is called dual flow plasma arc cutting. I hope this helps.

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Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pres

LekaFEV [45]

Answer:

44.59°c

Explanation:

Given data :

Total pressure = 105 kpa

complete combustion

A) Determine air-fuel ratio

A-F = $\frac{N_{air} }{N_{fuel} } = \frac{(Nm)_{air} }{(Nm)_{c} (Nm)_{n} }$

N = number of mole

m = molar mass

A-F = $\frac{(6.75*4.76)kmole * ( 29kg/mol)}{(3kmole)* 12kg/mol + (6kmol)*(1kg/mol)}$ = 22.2 kg air/fuel

hence the ratio of Fuel-air = 1 : 22.2

B) Determine the temperature at which water vapor in the products start condensing

First we determine the partial pressure of water vapor before using the steam table to determine the corresponding saturation temp

partial pressure of water vapor

Pv = $\frac{(N_{water vapor}) }{N_{pro} } * ( P_{ro} )$

N watervapor ( number of mole of water vapor ) = 3

N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol

Pro = 105

hence Pv = ( 3/33.53 ) * 105 = 9.39kPa

from the steam pressure table the corresponding saturation temperature to 9.39kPa = 44.59° c

Temperature at which condensing will start = 44.59°c

An equation showing the products of propylene with their mole numbers is attached below

8 0

3 years ago

. A Carnot heat pump is to be used to heat a house and maintain it at 22 °C in winter. When the outdoor temperature remains at 3

max2010maxim [7]

Answer:

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

Explanation:

The net heat daily loss of the house is:

$Q_{losses} = \left(76000\,\frac{kJ}{h}\right)\cdot (24\,h)$

$Q_{losses} = 1.824\times 10^{6}\,kJ$

In order to keep the house warm, given heat must be equal to heat losses:

$Q_{H} = Q_{losses}$

Besides, the Coefficient of Performance for a Carnot heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

Where,

$T_{L}$ - Temperature of the cold reservoir (Outdoors), measured in Kelvin.

$T_{H}$ - Temperature of the hot reservoir (House), measured in Kelvin.

Given that $T_{L} = 276.15\,K$ and $T_{H} = 295.15\,K$ , the Coefficient of Performance is:

$COP_{HP} = \frac{295.15\,K}{295.15\,K-276.15\,K}$

$COP_{HP} = 15.534$

For a real heat machine, the Coefficient of Performance is determined by the following expression:

$COP_{HP} = \frac{Q_{H}}{W}$

Where:

$Q_{H}$ - Heat received by the house, measured in kilojoules.

$W$ - Work consumed by the Carnot heat pump, measured in kilojoules.

The daily work consumed is now cleared in the previous expression:

$W = \frac{Q_{H}}{COP_{HP}}$

$W = \frac{1.824\times 10^{6}\,kJ}{15.534}$

$W = 117419.853\,kJ$

The working time is calculated by dividing this result by input power. That is:

$\Delta t = \frac{W}{\dot W}$

$\Delta t = \left(\frac{117419.853\,kJ}{9\,kW} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)$

$\Delta t = 3.624\,h$

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

8 0

3 years ago

A water jet that leaves a nozzle at 60 m/s at a certain flow rate generating power of 250 kW by striking the buckets located on

wlad13 [49]

Answer:

The mass flow rate of jet =69.44 kg/s

Explanation:

Given that

velocity of jet v= 60 m/s

Power P=250 KW

As we know that force offered by water F

$F=\rho\ A \ v^2$

Power P= F.v

So now power given as

$P=\rho\ A \ v^3$

We know that mass flow rate = ρAv

$P=mass\ flow\ rate\ \times v^2$

250 x 1000 = mass flow rate x 3600

mass flow rate = 69.44 kg/s

So the mass flow rate of jet =69.44 kg/s

4 0

4 years ago

I have five brainliest why is it only showing 2?

svp [43]

Answer: Either your computer is malfunctioning or it is glitched and still thinks you only have 2. Also good job

Explanation: Leave a brainliest it helps

4 0

3 years ago

Advance (also called Constantan) has a strain sensitivity SA=2.1 for strain as large as 8%. Determine the amount of contribution

Sholpan [36]

Answer:

$\frac{dP}{P} = 6.25$

Explanation:

Given data:

Sa = 2.1

$R = \frac{pl}{A}$

$\frac{dR}{R} =\frac{dP}{P} +\frac{dL}{L} (1_2V)$

$\frac{dR}{R} =\frac{dP}{P} +\epsilon (1_2V)$

$Sa = \frac{\frac{dR}{R}}{\epsilon} =\frac{\frac{dP}{P}}{\epsilon} +\frac{\epsilon (1_2V)}{\epsilon}$

$Sa = (1+2v) + \frac{\frac{dP}{P}}{\epsilon}$

change in specific resistance is given as $\frac{dP}{P}$

$\frac{dP}{P} = \frac{Sa -(1-2v)}{\epsilon}$ ........2

where v is elastic range = 0.30

$\epsilon = 0.08$

$\frac{dP}{P} = \frac{2.1 -(1-2\times 0.30)}{0.08}$

$\frac{dP}{P} = 6.25$

8 0

3 years ago