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natali 33 [55]
2 years ago
5

In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an

overall turbine-generator efficiency of 85 percent, determine the flow rate of water to generate 100 kW of electricity if the irreversible head loss of the piping system between the free surfaces of the inlet and the outlet is 36 ft. Take the density of water to be rho= 62.4 lbm/ft ^3.The flow rate of water is 348 ft^3/s.
Physics
1 answer:
VashaNatasha [74]2 years ago
6 0

Answer:

V=3.475ft^3/s=3.48ft^3/s

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})

P= 73.7*10^{3}ft.lbf/s

we have other values such h=400ft and  \gamma= 62.42lb/ft^3 (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

V=\frac{P}{\gamma h \eta_0}

Where \eta_0 is the turbine efficiency, at which is,

\eta_0 = \frac{P}{\gamma Vh}

Replacing,

V=\frac{73.7*10^{3}}{62.42*400*0.85}

V=3.475ft^3/s

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

m= 217lbm/s

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3 years ago
Consider a thin rod of mass 3.2 kg, length 1.2 m and uniform density. The rod is pivoted at one end on a frictionless horizontal
notsponge [240]

Answer:

the angular acceleration is 9.7 rad/s^{2}

Explanation:

given information:

mass of thin rod, m = 3.2 kg

the length of the rod, L = 1.2

angle, θ = 38

to find the acceleration of the rod, we can use the torque's formula as below,

τ = Iα

where

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σ = acceleration

moment inertia of this rod, I

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τ = F d, d = \frac{L}{2}cosθ

τ = m g \frac{L}{2}cosθ

now we can substitute the both equation,

τ = Iα

α = τ/I

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3 0
3 years ago
1. How much energy would be required to melt 450 grams of ice at 0°C?
xenn [34]

Answer:

Explanation:

1. The amount of heat needed to melt ice at 0°C is equal to the mass of the ice times the latent heat of fusion.

q = mL

q = (450 g) (334 J/g)

q = 150,300 J

q = 150 kJ

2. The amount of heat released by the condensation of steam at 100°C is equal to the mass of the steam times the latent heat of vaporization.

q = mL

q = (325 g) (2260 J/g)

q = 734,500 J

q = 735 kJ

3. q = mL

q = (85 g) (2260 J/g)

q = 192,100 J

q = 190 kJ

4. q = mL

q = (225 g) (334 J/g)

q = 75,150 J

q = 75.2 kJ

5. Above 100°C, water is steam.  The amount of heat needed to increase the temperature of steam is equal to its mass times its specific heat times the change in temperature.

q = mCΔT

q = (20.0 g) (2.03 J/g/°C) (303.0°C − 283.0°C)

q = 812 J

6. q = mCΔT

q = (15.0 g) (2.03 J/g/°C) (250.0°C − 275.0°C)

q = -761 J

7. q = mCΔT

q = (10.0 g) (0.90 J/g/°C) (55°C − 22°C)

q = 297 J

8. q = mCΔT

198 J = (55.0 g) C (15°C)

C = 0.24 J/g/°C

9. q = mCΔT

41,840 J = m (4.184 J/g/°C) (28.5°C − 22.0°C)

m = 1540 g

10. q = mCΔT

q = (193 g) (2.46 J/g/°C) (35°C − 19°C)

q = 7600 J

11. First, the temperature of the ice must be raised to 0°C.

q = mCΔT

q = m (2.09 J/g/°C) (0°C − (-23.0°C))

q/m = 48.1 J/g

Next, the ice must be melted.

q = mL

q/m = 334 J/g

Then, the water must be heated to 100°C.

q = mCΔT

q = m (4.184 J/g/°C) (100°C − 0°C)

q/m = 418.4 J/g

The water is then vaporized.

q = mL

q/m = 2260 J/g

Finally, the steam is heated to its final temperature.

q = mCΔT

q = m (2.03 J/g/°C) (118°C − 100°C)

q/m = 36.5 J/g

So the total amount of energy needed is:

q/m = 48.1 J/g + 334 J/g + 418.4 J/g + 2260 J/g + 36.5 J/g

q/m = 3100 J/g

3 0
3 years ago
What do we call the energy that is transferred to
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Answer:

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Answer:

8.60 g/cm³

Explanation:

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This implies that:

A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} }  }

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Assuming that there is no phase change gives:

\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3}   }

  = 8.60 g/m³

3 0
3 years ago
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