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Tju [1.3M]
3 years ago
11

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 21 m3/min and exits at 12 b

ar, 400 K. Heat transfer occurs at a rate of 3.5 kW from the compressor to its surroundings.
Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
Engineering
1 answer:
omeli [17]3 years ago
3 0

Answer:

- 46.5171kW

Explanation:

FIrst, the value given:

P1 = 1.05 bar (Initial pressure)

P2 = 12 bar (final pressure)

Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)

Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)

Universal gas constant, Ru = 8.3143 Kj/Kgmolk

Specific gas constant, R = 0.28699 Kj/KgK

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Finding the volume:

P1V1 = RT1

V1 = RT1 ÷ P1

= (0.28699 Kj/KgK X 300k) ÷ 105

Note convert bar to Kj/Nm by multiply it by 100

V1 =  0.81997 m3/Kg

To get the mass flow rate:

m = volumetric flow rate / V1

= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg

= 0.4268Kg/s

Using tables for the enthalpy,

hT1 = 300.19 KJ/Kg

hT2 = 400.98 KJ/Kg

The enthalpy change = hT2 - hT1

= 100.79 KJ/Kg

Power, P = Q - (m X enthalpy change)

= - 3.5 - (0.4268 X 100.79)

= - 46.5171kW

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dezoksy [38]

Answer:

<h2>698.3Kpa</h2>

Explanation:

Step one:

given data

V1=0.25m^3

T1=290k

P1=100kPa

V2=0.5m^2

T2=405k

P2=? final pressure

Step two:

The combined gas equation is given as

P1V1/T1=P2V2/T2

Substituting we have

(100*0.25)/290=P2*0.05/405

25/290=0.5P2/405

0.086=0.05P2/405

cross multiply

0.086*405=0.05P2

34.9=0.05P2

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P2=698.3Kpa

<u>Therefore the new pressure is 698.3Kpa when the gas is compressed</u>

5 0
2 years ago
The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse
sertanlavr [38]

Answer:

T = 5416.67 N

T = -2083.5 N

T = 0

Explanation:

Forward thrust has positive values and reverse thrust has negative values.

part a

Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (150 - 41.67)\\\\T = 5416.67 N

Answer: The thrust force T = 5416.67 N

part b

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:

T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (0 - 41.67)\\\\T = -2083.5 N

Answer: The thrust force T = -2083.5 N reverse direction

part c

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.

Answer: T = 0 N

5 0
3 years ago
If an object has the same number of positive and negative charges, its electrical charge is
N76 [4]

When an object has the same number of positive and negative charges, its electrical charge will become neutral.

What is an electric charge?

When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.

Now when the two equal magnitude charges with opposite natures come together they become neutral.

To know more about charges follow

brainly.com/question/24391667

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7 0
2 years ago
A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N
soldier1979 [14.2K]

Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, P_{t} = 8 kW = 8000 W

Now, the force, F is given by:

P_{t} = Force(F)\times velocity(v) = Fv               (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, F_{c} is provided by the gravitational force, F_{G}:

F_{c} = F_{G}

\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}

Thus from the above, velocity comes out to be:

v = \sqrt{\frac{GM_{e}}{R}}

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s

where

R = R_{e} + H

R = \sqrt{GM_{e}(\frac{T}{2\pi})^{2}}

where

G = Gravitational constant

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R is calculated as 42166 km

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6 0
3 years ago
The chart shows the bids provided by four engineers to test a prototype.
klasskru [66]

Answer:

D

Explanation:

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\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]

As you can see, Greg is the least cost-effective because he charges the most for the project.

8 0
3 years ago
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