Answer
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Hint: Efficiency of a turbine can be expressed as the ratio of output mechanical shaft power of the turbine to input mechanical fluid power of the turbine. That is the turbine efficiency is the ratio of actual work output of the turbine to the net input energy supplied in the form of fuel of the turbine.
Useful formula:
The power generated by the turbine generator is given by the formula of;
$P = mgh$
Where, $P$ denotes the power generated by the power hydroelectric plant, $m$ mass of the water flow, $g$ denotes the acceleration due to gravity,$h$ denotes the height of the pressurised water head.
Complete step by step solution:
The data given in the problem are;
Height of the water power head is, $h = 300\,\,m$,
Velocity of the water flow is, $v = 100\,\,{m^3}{s^{ - 1}}$,
Efficiency of the turbine generator, $\eta = 60\,\% $.
Acceleration due to gravity, $g = 9.8\,\,m{s^{ - 2}}$.
The power generated by the turbine generator is given by the formula of;
$P = mgh$
We know that $m = \rho v$;
Where, $\rho $ denotes the density of the water flow, $v$ denotes the rate of which the water flows.
That is;
$P = \rho v \times gh$
Substitute the values of gravity, height, velocity of the water on the above equation;
$
P = 1000 \times 100 \times 9.8 \times 300 \\
P = 294 \times {10^6}\,\,W \\
P = 294\,\,MW \\
$
The power generated by the turbine generator is given as $P = 294\,\,MW.$
The estimated power available for the plant is given as;
${P_a} = \eta \times P$
Where, ${P_a}$ denotes the estimated electric power available for the plant, $\eta $ denotes the efficiency of the turbine of the generator.
Substitute the value of efficiency of the turbine generator and the power of the generator;
$
{P_a} = 60\,\% \times 294 \\
{P_a} = \dfrac{{60}}{{100}} \times 294 \\
{P_a} = 0.6 \times 294 \\
{P_a} = 176.4\,\,MW \\
$
Therefore, the estimated power available for the plant is given as \[{P_a} = 176.4\,\,MW\].
Note: the estimated power available for the plant is directly proportional to the efficiency of the electric turbine motor that is if the efficiency of the electric turbine motor the power available for the plant also increases along with the efficiency.
Useful formula:
The power generated by the turbine generator is given by the formula of;
$P = mgh$
Where, $P$ denotes the power generated by the power hydroelectric plant, $m$ mass of the water flow, $g$ denotes the acceleration due to gravity,$h$ denotes the height of the pressurised water head.
Complete step by step solution:
The data given in the problem are;
Height of the water power head is, $h = 300\,\,m$,
Velocity of the water flow is, $v = 100\,\,{m^3}{s^{ - 1}}$,
Efficiency of the turbine generator, $\eta = 60\,\% $.
Acceleration due to gravity, $g = 9.8\,\,m{s^{ - 2}}$.
The power generated by the turbine generator is given by the formula of;
$P = mgh$
We know that $m = \rho v$;
Where, $\rho $ denotes the density of the water flow, $v$ denotes the rate of which the water flows.
That is;
$P = \rho v \times gh$
Substitute the values of gravity, height, velocity of the water on the above equation;
$
P = 1000 \times 100 \times 9.8 \times 300 \\
P = 294 \times {10^6}\,\,W \\
P = 294\,\,MW \\
$
The power generated by the turbine generator is given as $P = 294\,\,MW.$
The estimated power available for the plant is given as;
${P_a} = \eta \times P$
Where, ${P_a}$ denotes the estimated electric power available for the plant, $\eta $ denotes the efficiency of the turbine of the generator.
Substitute the value of efficiency of the turbine generator and the power of the generator;
$
{P_a} = 60\,\% \times 294 \\
{P_a} = \dfrac{{60}}{{100}} \times 294 \\
{P_a} = 0.6 \times 294 \\
{P_a} = 176.4\,\,MW \\
$
Therefore, the estimated power available for the plant is given as \[{P_a} = 176.4\,\,MW\].
Note: the estimated power available for the plant is directly proportional to the efficiency of the electric turbine motor that is if the efficiency of the electric turbine motor the power available for the plant also increases along with the efficiency.
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