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blsea [12.9K]
2 years ago
13

7. 13 a turbine receives steam at 6 mpa, 600°c with an exit pressure of 600 kpa. Assume the turbine is adiabatic and neglect kin

etic energies. Find the exit temperature and the specific work
Physics
1 answer:
AnnyKZ [126]2 years ago
6 0

The work done by the turbine will be 708.2 kJ/kg. The work done by the turbine is the difference of the enthalpy at inlet and exit.

<h3 /><h3>What is temperature?</h3>

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

If the given turbine is assumed to be reversible;

\rm P_I(Initial pressure)=60 mpa = 60 bar

\rm  T_i(Initial temperature)=600° C

\rm P_e (Exit pressure)=600 kpa=6 bar

The heat balance equation is;

\rm q-W_t=h_e-h_i\\\\ q=0\\\\\ W_t=h_i-h_e

The change in the entropy is;

\rm S_2-S_1=\frac{\delta q}{dt}

The work done by the turbine is;

\rm W_t = h_i-h_e\\\\ W_t =3658.4 -29.5019\\\\  W_t =708.2 \ kJ/kg

Hence,the work done by the turbine will be 708.2 kJ/kg.

To learn more about the temperature, refer to the link;

brainly.com/question/7510619

#SPJ4

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A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0
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Answer:

Velocity = 3.25[m/s]

Explanation:

This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.

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(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]

4 0
3 years ago
A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in
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Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

\frac{GMm}{d^{2}}=\frac{mv^2}{d}

v=\sqrt{\frac{GM}{d}}

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}

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7 0
3 years ago
A 250–g piece of gold is at 19 °C. 5.192 kJ of energy is added to it by heat. The specific heat of gold is 129 J/(kg·°C). Calcul
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Answer:

A. DT is given by Q= MCs DT

m = mass of the substances

Cs= is it's specific heat capacity

Ck= <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>Q</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

Mk ×DTk

=<u>2</u><u>5</u><u>0</u><u> </u><u>×</u><u> </u><u>9</u><u> </u><u>×</u><u> </u><u>5</u><u> </u><u> </u>

129

=Dt = 180.1085271

answer is 180degree C.

Explanation:

B. = <u>2</u><u>5</u><u>×</u><u>1</u><u>0</u> ×100

1.082

=<u>2</u><u>5</u><u>0</u><u>0</u>

1.082

= 23105.360 g/kj.

7 0
3 years ago
Convert 200in/10s into m/s (1m = 39.37in)
stiv31 [10]
5.08




hope you got it right !
7 0
3 years ago
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