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VikaD [51]
3 years ago
6

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

the process is reversible and adiabatic. Use constant specific heat at 300 K to find the
Physics
1 answer:
nikitadnepr [17]3 years ago
6 0

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

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This one too dude help me
Mrac [35]

Answer:

Vas happenin!!

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3 years ago
Ety ratio
horrorfan [7]

3) The work done is D. zero

4) The kinetic energy is B. 180 J

5) The potential energy is A. 120 J

6) The work done depends on B. position

7) The example of non-renewable energy is C. coal

8) The power expended is 3\cdot 10^4 W

9) The efficiency is A. 100%

10) The velocity ratio is 5

Explanation:

3)

The work done by a force acting an object is given by:

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

When the force is applied perpendicular to the direction of motion,

\theta=90^{\circ}

Therefore, the work done is:

W=Fd(cos 90^{\circ})=0

4)

The kinetic energy of a body is given by

K=\frac{1}{2}mv^2

where

m is the mass of the body

v is its speed

For the girl in this problem, we have

m = 40 kg

v = 3 m/s

Therefore her kinetic energy is

K=\frac{1}{2}(40)(3)^2=180 J

5)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g=10 m/s^2 is the acceleration of gravity

h is the heigth of the object relative to the ground

For the ball in this problem,

m = 0.4 kg

h = 30 m

So, the potential energy is

PE=(0.4)(10)(30)=120 J

6)

A conservative field is a field for which the work done by the field on an object does not depend on the path taken, but only on the initial and final position of the object.

Gravitational and electric fields are examples of conservative fields. In fact:

  • When an object is pulled down by gravity (free fall), the work done by the gravitational field only depends on the change in height \Delta h between the two points, not on the path taken during the fall
  • When an electric charge is pushed by the electric field, the work done by the field depends only on the initial and final position of the charge in the field

For any conservative field, it is possible to define a "potential" function, which represents the energy per unit mass/charge, and depends only on the position of the object.

7.

  • Non-renewable energy sources are sources of energy whose rate of consumption is faster than the rate at which they are re-created. Examples of non-renewable sources are coal, oil, natural gas. These energy sources are consumed at a fast rate, while they take million of years to regenerate, so at the current rate they will eventually run out.
  • Renewable energy sources are sources of energy that replenish at faster rate than the rate at which it is consumed. Examples of renewable sources are solar energy, wind, hydroelectric power.

Therefore, the example of non-renewable energy in this case is

C. Coal

8.

For an object pushed by a force F and moving at a constant velocity v, the power expended is given by

P=Fv

where F is the force and v is the velocity.

for the rocket in this problem, we have:

F = 10 N is the force propelling the rocket

v = 3000 m/s is its velocity

Substituting into the equation, we find the power expended:

P=(10)(3000)=30,000 W = 3\cdot 10^4 W

9.

The efficiency of a machine is given by

\eta = \frac{W_{out}}{W_{in}}

where

W_{in} is the energy in input to the machine

W_{out} is the useful work in output from the machine

For a real machine, the useful work in output is always lower than the energy input, because part of the energy is "wasted" and converted into thermal energy due to the presence of internal frictions. However, for an ideal machine, all the input energy is converted into useful work, so

W_{out}=W_{in}

And therefore the efficiency is

\eta=1

which means 100%.

10.

The velocity ratio of a block and tackle system is the ratio between the distance moved by the effort and the distance moved by the load.

VR=\frac{d_{eff}}{d_{load}}

In a block and tackle system, the velocity ratio is also equal to the number of pulleys in the system.

For the system in the problem, there are 5 pulleys: therefore, this means that when the effort moves 5 metres, the load moves 1 metres, therefore the velocity ratio is

VR=\frac{5}{1}=5

Learn more about kinetic and potential energy:

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