Answer:
[tex]m = 1.74 \times 10^5 kg/s[/tex]
Explanation:
As we know that potential energy is given as
[tex]U = mgh[/tex]
here we know that
m = 1 kg
h = 33 m
now we have
[tex]U = 1(9.8)(33)[/tex]
[tex]U = 323.4 J[/tex]
Part b)
Now we need total power given as
[tex]P = 45 MW[/tex]
so we have
[tex]P = \frac{\Delta U}{t}[/tex]
here we know that it is 80% efficient so we can write
[tex]P = 0.80(\frac{\Delta U}{\Delta t})[/tex]
[tex]45 \times 10^6 = 0.80(m)(323.4)[/tex]
[tex]m = 1.74 \times 10^5 kg/s[/tex]
The change in the potential energy of the water is 323.4 J.
The mass of water that must pass through the turbine each second is 173,933.2 kg/s.
The given parameters;
The change in the potential energy of the water is calculated as follows;
ΔU = mgh
ΔU = 1 x 9.8 x 33
ΔU = 323.4 J
The efficiency of the turbine is calculated as follows;
[tex]Eff = \frac{0utput\ power }{1nput\ power } \times 100\% \\\\Eff = \frac{P_0}{P_1} \times 100\%\\\\Eff = \frac{45 \times 10^6 }{P_1} \times 100\% \\\\80\% = \frac{45 \times 10^6}{P_1} \times 100\% \\\\0.8 = \frac{45 \times 10^6}{P_1} \\\\P_1 = \frac{45 \times 10^6}{0.8} \\\\P_1 = 56,250,000 \ J/s \\\\[/tex]
The mass of water that must pass through the turbine each second is calculated as follows;
[tex]\frac{mgh}{t} = P_1\\\\\frac{m}{t} = \frac{P_1}{gh} \\\\\frac{m}{t} = \frac{56,250,000}{9.8 \times 33} \\\\\frac{m}{t} = 173,933.2 \ kg/s[/tex]
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