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valina [46]
3 years ago
11

A Rankine cycle with one closed feedwater heater with its drain cascaded backward has a water mass flow rate through the steam g

enerator of 140 kg/s. Superheated steam enters the turbine system at 19.3 MPa and 5408C, and saturated liquid water exits the condenser at 28 kPa absolute. The closed feedwater heater receives extracted steam from the turbine at 2.4 MPa absolute, and the temperature of the feedwater exiting the closed feedwater heater is 2108C. Determine the net power produced by the cycle and the thermal efficiency of the cycle if (a) the turbine and pump are isentropic, and (b) the turbine isentropic efficiency is 0.82 and the isentropic efficiencies of the pump is 0.65.
Physics
1 answer:
mamaluj [8]3 years ago
6 0

Answer:Draw a T-s diagram for the ideal Rankine Cycle

Explanation:

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An arcade ball is thrown with an initial speed of 7.0 m/s and follows the trajectory shown. The ball enter the basket .95 second
dmitriy555 [2]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

x  =  4.70  \  m     ,   y =  0.2803 \  m

Explanation:

From the question we are told that

     The initial velocity of the arcade ball is  u  =  7.0 \  m/s

     The time taken by the ball to enter the basket is  t =  0.95 \ s

Generally the x -component of the initial  velocity is  

            u_x  =  7 *  cos (45)    

=>         u_x  =  4.95 \  m/s

Generally the y -component of the initial  velocity is  

            u_x  =  7 *  sin(45)    

=>         u_y  =  4.95 \  m/s

  Generally the distance x is mathematically represented as

                 x  =  u_x  *  t

=>              x  =  4.95 *  0.95      

=>              x  =  4.70  \  m    

Generally from kinematic equation the distance y is mathematically represented as

               y =  u_y * t  - \frac{1}{2} *  g * t^2

=>            y = 4.95 * 0.95  - \frac{1}{2} *  9.8 * 0.95^2

=>            y =  0.2803 \  m

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Answer:

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Explanation:

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What is the net force on the purple ring in the picture above. _________
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Answer:

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Explanation:

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A 1.0-kilogram rubber ball traveling east at 4.0 meters per second hits a wall and bounces back toward the west at 2.0 meters pe
Mrrafil [7]

Answer:

8 J and 2 J

Explanation:

Given that,

Mass of the rubber ball, m = 1 kg

Initial speed of the rubber ball, u = 4 m/s (in east)

Final speed of the rubber ball, v = -2 m/s (in west)

We need to find the kinetic energy of the ball before it hits the wall, the kinetic energy of the ball after it bounces off the wall.

Initial kinetic energy,

K_i=\dfrac{1}{2}mv^2\\\\K_i=\dfrac{1}{2}\times 1\times (4)^2\\\\K_i=8\ J

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So, the initial kinetic energy is 8 J and the final kinetic energy is 2 J.

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