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skelet666 [1.2K]
3 years ago
9

Plzz I need answer in 1 mint

Physics
2 answers:
garri49 [273]3 years ago
8 0

Answer:

B

Explanation:

I hope this is what you need

PLEASE MAKE ME BRAINLIEST

PilotLPTM [1.2K]3 years ago
6 0
B is the correct answer :)
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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
What is the relationship between the applied force of a hanging mass on a spring and the spring force of the spring?
zaharov [31]

Answer:

elastic force and weight are related to the acceleration of the System.

Explanation:

The relationship between these two forces can be found with Newton's second law.

        F_{e} - W = m a

        K x - m g = m a

We see that elastic force and weight are related to the acceleration of the System.

If a harmonic movement is desired, an extra force that increases the elastic force is applied, but to begin the movement this force is eliminated, in general , if the relationship between this external and elastic force is desired, the only requirement is that it be small for harmonic movement to occur

7 0
3 years ago
A rocket will move upward as long as which condition applies?
Dennis_Churaev [7]
The force of thrust is greater than the force if gravity !
Answer found on quizlet !
8 0
3 years ago
Does centripetal force change in uniform circular motion?
HACTEHA [7]
It depends on how fast you are going and in orincipal no
7 0
3 years ago
In which situation would an object weigh the LEAST? (assume all the objects have the same mass)
inessss [21]

Answer:

An object on the moon would weigh the LEAST among these. So correct answer is B.

Explanation:

  • Weight of an object on any place is given by:

W = Mass * Acceleration due to gravity(g)

  • It means when masses of different objects those are in different places are same, the weight of  those objects depends upon the 'g' of that particular place.
  • As we know, acceleration due to gravity on surface of moon (g') is 6 times weaker than the acceleration on surface of earth (g), which is due to the large M/R^2 of the earth than the moon.

i.e. g' = g/6 so W' = W/6

  • And in the space between the two, the object is weightless.
8 0
3 years ago
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