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Karolina [17]
3 years ago
13

A steam turbine operates with 1.6 MPa and 350°C steam at its inlet and saturated vapor at 30°C at its exit. The mass flow rate o

f the steam is 22 kg/s, and the turbine produces 12,350 kW of power. Determine the rate at which heat is lost through the casing of this turbine.
Physics
1 answer:
WINSTONCH [101]3 years ago
3 0

Answer:

638.8kW

Explanation:

The flow rate of the steam m = 22kg/s

The Pressure of the steam at the inlet of the turbine P1 = 1.6MPa

The temperature of the steam at the inlet of the turbine T1 = 350*C

Steam quality at the exit of the turbine x2 = 1.0

The temperature of the steam at the exit of the turbine T2 = 30*C

Power produced = 12,350kW

Assuming the turbine is running on a steady state, hence we neglect the effect of kinetic and potential energy we get:

If you refer to the superheated steam table for the specific enthalpy at a pressure of 1.6MPa and temperature of 350*C, we get

h1 = 3,146kJ/kg

Refer to the steam table for saturated gas at temperature 30*C to get the specific enthalpy value h2 = 2,556.81kJ/kg

The heat that comes out from the turbine can be defined from the balance of energy in the system, and is represented as

Ein - Eout = change in system Energy = 0

Thus Ein = Eout

mh1 = mh2 + Wout + Qout

Qout = m(hi-h2) - Wout

Qout = 22 x (3146-255.6) - 12350

Qout = 638.8kW

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Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
All of the following show friction as a useful force, except _____.
prohojiy [21]

Answer: having to push a rough and heavy box across the floor to move it

Explanation:

The Friction force is any force that is in opposite direction of the motion of an object or fluid due to the contact of this object or fluid with other bodies.

In this sense, there are different types of friction force thath are useful in different situations:

-The <u>Static friction force</u> prevents surfaces from slipping across each other. For example, the friction between your feet and the floor keeping you from slipping.

-The <u>kinetic friction force</u> as the force that helps the tires in a moving vehicle to slow down and stop when necessary.

However, if you want to push a heavy box across the floor to move it, the friction force will not be useful at all.

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3 years ago
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Explanation:

i hope this helps, its not the same person but its the same equation.

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3 years ago
What kind of lens is there in our eyes? where does it form the image of an object?​
Goshia [24]

Answer:

convex lens

Explanation:

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8 0
2 years ago
A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 458 N and have a center of gravity
Step2247 [10]

Answer:

the location of the center of gravity for the entire body is 1.08 m

Explanation:

Given the data in the question;

w1 = 458 N, y1 = 1.34 m

w2 = 120 N, y2 = 0.766 m

w3 = 89.8 N, y2 = 0.204 m

The location arrangement of the body part is vertical, locate the overall centre of gravity by simply replacing the horizontal position x by the vertical position y as measured relative to the floor.

so,

Y_{centre of gravity} = (w1y1 + w2y2 + w3y3 ) / ( w1 + w2 + w3 )

so we substitute in our values

Y_{centre of gravity} = (458×1.34 + 120×0.766 + 89.8×0.204 ) / ( 458 + 120 + 89.8 )

Y_{centre of gravity} = 723.9592 / 667.8

Y_{centre of gravity} = 1.08 m

Therefore, the location of the center of gravity for the entire body is 1.08 m

5 0
2 years ago
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