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


Answer 1

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.


The exit velocity is 629.41 m/s



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((P_2)/(P_1))^{(k-1 )/(k)

k = 1.4

T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)}\n\nT_2 = 1200((80)/(150))^{(1.4-1 )/(1.4)}\n\nT_2 = 1002.714K

Work done is given as;

W = (1)/(2) *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

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

Therefore, the exit velocity is 629.41 m/s

Related Questions

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The height h (in feet) of an object shot into the air from a tall building is given by the function h(t) = 650 + 80t − 16t2, where t is the time elapsed in seconds. (a) Write a formula for the velocity of the object as a function of time t.
A ball having a mass of 500 grams is dropped from a height of 9.00 meters what is its kinetic energy when it hits the ground
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Explain why Planck’s introduction of quantization accounted for the properties of black-body radiation.

A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.


To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

m_1v_1+m_2v_2 = (m_1+m_2)v_f


m_(1,2)= Mass of each object

v_(1,2) = Initial Velocity of each object

v_f= Final velocity

Replacing we have that,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

1850*13.8+3100*0 = (1850+3100)v_f

v_f = 5.1575m/s

With the final speed obtained we can determine the acceleration through the linear motion kinematic equations, that is to say

v_f^2-v_i^2 = 2ax

Since there is no initial speed, then

v_f^2 = 2ax

5.1575^2 = 2a (1.91)

a = 6.9633m/s^2

Finally with the acceleration found it is possible to find the friction force from the balance of Forces, like this:

F_f = F_a \n\mu N = m*a \n\mu = (ma)/(N)\n\mu = (ma)/(mg)\n\mu = (a)/(g)\n\mu = (6.9633)/(9.8)\n\mu = 0.7105

Therefore the Kinetic friction coefficient is 0.7105

Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at 42.91 m/s (96.00 mph) . Assuming a pitched ball has a mass of 0.1434 kg and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?


Answer: 132.02 J


By definition, the kinetic energy is written as follows:

KE = 1/2 m v²

In our question, we know from the question, the following information:

m = 0.1434 Kg

v= 42.91 m/s

Replacing in the equation for KE, we have:

KE = 1/2 . 0.1434 Kg. (42.91)² m²/s² ⇒ KE = 132.02 N. m = 132.02 J

A concrete highway is built of slabs 18.0 m long (at 25 °C). How wide should the expansion cracks be (at 25 °C) between the slabs to prevent buckling if the annual extreme temperatures are −32 °C and 52 °C?(the coefficient of linear expansion of concrete is 1.20 × 10 − 5 °C-1) g


To solve the problem it is necessary to apply the concepts related to thermal expansion of solids. Thermodynamically the expansion is given by

\Delta L = L_0 \alpha \Delta T


L_0 = Original Length of the bar

\Delta T= Change in temperature

\alpha= Coefficient of thermal expansion

On the other hand our values are given as,

L_0 = 18m

\alpha = 12*10^(-6)/\°C

T_2 = 52\°C

T_1= 25\°C

Replacing we have,

\Delta L = L_0 \alpha (T_2-T_1)

\Delta L = (18)(12*10^(-6))(52-25)

\Delta L = 5.832*10^(-3)m

The width of the expansion of the cracks between the slabs is 0.5832cm

The width of the expansion cracks between the slabs to prevent buckling should be 0.5832cm.

How to calculate width?

According to this question, the following information are given:

  • Lo = Original length of the bar
  • ∆T = Change in temperature
  • α = Coefficient of thermal energy

The values are given as follows:

  • Lo = 18m
  • T1 = 25°C, T2 = 52°C
  • α = 12 × 10-⁶/°C

∆L = Loα (T2 - T1)

∆L = 18 × 12 × 10-⁶ (27)

∆L = 3.24 × 10-⁴ × 18

∆L = 5.832 × 10-³m

Therefore, the width of the expansion of the cracks between the slabs is 0.5832cm.

Learn more about width at:

Question 11 of 15When hydrogen is attached to the more highly electronegative oxygen atom in
a water molecule,
A. the electronegative atom becomes strongly positive
B. the hydrogen atom becomes partially positive
O C. the oxygen atom becomes partially negative

If answer is right WILL GIVE BRAINLIEST
D. the hydrogen atom becomes partially negative



I'm leaning twards A


I think B but I’m not sure

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?



The add mass = 5.465 kg


Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1  (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)


T ∝ √(m)


T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

Substituting into equation 4

m₂ = (2.07)²(0.5)/(1.18)²

m₂ = 4.285(1.392)

m₂ = 5.965 kg.

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

Thus the add mass = 5.465 kg

Ultrasonic imaging is made possible due to the fact that a sound wave is partially reflected whenever it hits a boundary between two materials with different densities within the body. the percentage of the wave reflected when traveling from material 1 into material 2 is r=(ρ1−ρ2ρ1+ρ2)2. knowing this, why does the technician apply ultrasound gel to the patient before beginning the examination?



The gel that is applied before ultrasonic imaging is a conducting material. It acts as a medium between transducer and skin. The ultrasonic waves easily transmit from the probe to the tissues because of gel. A tight bond is created between the probe and skin layer and the gel acts as a coupling agent. The density of the gel is similar to the skin layer. This reduces the attenuation of the waves.  A thin layer of gel is applied which fills the air gaps and helps in transmission of waves to the tissues. Hence, the technician apply ultrasound gel to the patient before beginning the examination

The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection.