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2 years ago

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Water flows from one reservoir to another a height, 41 m below. A turbine (η=0.77) generates power from this flow. 1 m3/s passes

through the turbine. If 12 m head loss occurs between the two reservoirs, determine the actual (i.e. useable) power generated by the turbine (in kW).

1 answer:

morpeh [17]2 years ago

5 0

Complete Question

A diagram representing this question is shown on the first uploaded image

Answer:

The value is $P = 294594.3 \ W$

Explanation:

From the question we are told that

The height is $h = 41 \ m$

The efficiency of the turbine is $\eta = 0.77$

The flow rate is $\r V = 1 m^3 / s$

The head loss is g = 2 m

Generally the head gain by the turbine is mathematically represented as

$H = h - d$

=> $H = 41 - 2$

=> $H = 39 \ m$

Generally the actual power generated by the turbine is mathematically represented as

$P = \eta *\gamma * \r V * H$

Here $\gamma$ is the specific density of water with value

$\gamma = 9810 N/m^3$

So

$P =0.77 *9810 * 1 * 39$

$P = 294594.3 \ W$

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The average velocity for this interval is :

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