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Rudiy27
3 years ago
10

Steam enters an adiabatic turbine at 10 MPa and 500oC and leaves at 10 kPa with a quality of 50.027 percent. Neglecting the chan

ges in kinetic and potential energies, determine the power output in MW using 3 decimals for mass flow rate of 15.469.
Physics
1 answer:
Blizzard [7]3 years ago
3 0

Answer:

power output = 30730.71 kJ/s

Explanation:

given data

pressure  p1 = 10 MPa

temperature = 500° C

pressure p2 = 10 kPa

quality = 50.027 % = 0.50027

mass flow rate m = 15.469

to find out

power output in MW

solution

we know energy balance equation here that is

Energy (in) = Energy (out)

mh1 = P(out) + mh2  

P(out) = - m ( h2 - h1)  .................1

we know here h1 from steam table that is

h1 = 3375.1 kJ/kg and

hf = 191.81 and hfg = 2392.1 kJ/kg

h2 = hf + quality hfg

h2 = 191.81 + 0.50027 (2392.1) = 1388.50 kJ/kg

so from equation 1 we get power output

P(out) = - m ( h2 - h1)

P(out) = - 15.469 ( 1388.50 - 3375.1 )

power output = 30730.71 kJ/s

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Explanation:

Given;

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I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa

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Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.

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Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun
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Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction = u\cos\theta

and the speed in the vertical direction is = u\sin\theta upward.

For A:

The speed in the horizontal direction = u\cos75^{\circ}

and the speed in the vertical direction is = u\sin75^{\circ} upward.

For B:

The speed in the horizontal direction = u\cos15^{\circ}

and the speed in the vertical direction is = u\sin15^{\circ} upward.

Let t_A and t_B are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A = u\cos75^{\circ}\times t_A\cdots(i)

The range for the projectile B = u\cos15^{\circ}\times t_B\cdots(ii)

At the highest point, the vertical velocity is 0.

Bu using the equation of motion v^2=u^2 +2a s.

Here, the final velocity v=0, the initial velocity u = u \sin \theta , h= vertical distance up to the highest point, and a= -g (as per sign convention).

So, s= \frac{u^2\sin^2 \theta}{2g}

For projectile A: The maximum height attained.

s_A= \frac{u^2\sin^2 75^{\circ}}{2g}

For projectile B: The maximum height attained.

s_B= \frac{u^2\sin^2 15^{\circ}}{2g}

As \sin^2 75^{\circ} > \sin^2 15^{\circ}, the height of A is attained by A is more than the heigHt attained by B.

Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.

So, the total time of flight = 2 x (Time to reach the highest point)

In a similar way, by using the equation of motion v=u+at,

The time to reach the highest point =\frac {u\sin\theta}{g}

where g is the acceleration due to gravity.

So, the total time of flight

= 2 \times \frac {u\sin\theta}{g}

The total time of flight for A

=2 \times \frac {u\sin75^{\circ}}{g}

The total time of flight for A

=2 \times \frac {u\sin15^{\circ}}{g}

Now, from equations (i) and (ii),

The range for the projectile A =

u\cos75^{\circ}\times  \frac {2u\sin75^{\circ}}{g}=\frac {u^2 \sin 150^{\circ}}{g}= \frac {u^2 \sin 30^{\circ}}{g}

The range for the projectile B =

u\cos15^{\circ}\times  \frac {2u\sin15^{\circ}}{g}=\frac {u^2 \sin 30^{\circ}}{g}.

Both the projectile have the same range.

Hence, option (A) is correct.

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