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-Dominant- [34]

3 years ago

9

Assess the capabilities of a hydroelectric power plant from the following field data: Estimated water flow rate, 40 m3/s River i

nlet at 1 atm, 10 oC Discharge at 1 atm, 10.2 oC, 200 m below the intake?

1 answer:

ryzh [129]3 years ago

6 0

Answer:

This hydroelectric power plant has the capability of producing around 78.4 MW of electric energy.

Explanation:

First, we calculate the pressure the water exerts or gains when it travels 200 m below the intake. W have the data:

Density of water = ρ = 1000 kg/m^3

Acceleration due to gravity = g = 9.8 m/s²

Height lost = h = 200 m

Volume flow rate = 40 m^3/s

The pressure difference, is now given as:

ΔP = ρgh

ΔP = (1000 kg/m^3)(9.8 m/s²)(200 m)

ΔP = 1960000 Pa = 1.96 MPa

Now, we calculate the power capacity of this flow:

Power = ΔP(Volume flow rate)

Power = (1960000 N/m²)(40 m^3/s)

Power = 78400000 Watt

<u>Power = 78.4 MW</u>

Thus, this plant can produce at max 78.4 MW. The actual power produced will be slightly less than this due of the losses and efficiency of turbine system used.

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Answer:

B.197 gpm and 12.4 L/s

Explanation:

Given that

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We know that specific heat for water

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Now from energy balance

$Q=\dot{m}C_p\Delta T$

by putting the values

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$\dot{m}=12.38\ \frac{kg}{s}$ (1 Kg/s = 15.85 gal/min)

We can say that

$\dot{m}=196.31\ \frac{gal}{min}$

We know that

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12.38=1000 x volume flow rate

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3 years ago

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin

Leni [432]

Answer:

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

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$cos \theta_1 =0.77$

$\theta_1=cos^{-1}0.77$

$\theta_1=39.65 \textdegree$

And

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$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

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The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

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Answer:

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