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Anton [14]
3 years ago
6

An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b

y the compressor. The gas constant of air is R = 0.287 kPa·m3/kg·K. The flow work required by the compressor is
Physics
1 answer:
Mariana [72]3 years ago
6 0

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

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Is the physical phenomena arising from the force caused by magnet  objects that produce fields that attract or repel other objects while magnetic field is a region around the magnetic material or a moving electric charge within the force of magnetism acts 
 
6 0
3 years ago
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A rope is tied to a tree 4.5 feet from the ground and then run through a pulley hooked to a vehicle 33 feet from the tree. If a
Angelina_Jolie [31]

Answer:

The vehicle displacement is 9.90 feet.

Explanation:

Given that,

Height of tree = 4.5 feet

Distance = 33 feet

According to figure,

We need to calculate the value of l

Using Pythagorean theorem

l=\sqrt{(33)^2+(4.5)^2}

We need to calculate the vehicle displacement

Using horizontal component

Vehicle displacement =horizontal component of pulled rope

Vehicle\ displacement= d\cos\theta

Where, \theta is angle between rope and ground

d = pulled length of rope

Vehicle\ displacement=10\times\dfrac{33}{\sqrt{(33)^2+(4.5)^2}}

Vehicle\ displacement=9.90\ feet

Hence, The vehicle displacement is 9.90 feet.

3 0
3 years ago
A hydraulic lift has pistons with diameter 28cm and 70cm, respectively. If a force of 500 N is exerted at the input piston. What
Crazy boy [7]

Answer:

3125 N

Explanation:

diameter /2 =radius

so r1 =14cm , r2 =35cm

f1/A1 =f2/A2.

f2 = f1 × A2 / A1

=500×1225 pi cm² / 96 pi cm²

f2 =3125N

4 0
3 years ago
What is the equation: If F=10 N, a=5 m/s², m=?
Romashka-Z-Leto [24]

Answer:

2 kg

Explanation:

Remember:

F = m * a       re-arrange to

F/a   = m      substitute in the given values

10 / 5   =   2 kg

8 0
2 years ago
A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance
ArbitrLikvidat [17]

Answer:

The value is v_y  =  -48.61 \ m/s

Explanation:

From the question we are told that

   The horizontal speed is  u_x  = 252 \  m/s

    The horizontal distance is  d = 1250 \ m

Generally the time taken by the hot magma in air before landing is mathematically represented as

       t = \frac{d}{u_x}

=>    t = \frac{ 1250 }{252}

=>    t = 4.96 \  s

Generally the initial vertical velocity of the magma when it was lunched is  

    u_y = 0 \ m/ s

Then the final velocity of the magma when it hits the ground is mathematically represented s

       - v_y  =  u_y + gt

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So  

        - v_y  =  48.61 \ m/s

=>     v_y  =  -48.61 \ m/s

7 0
3 years ago
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