hydropower generation output for a dam with a given head and flow Given turbine efficiency (eta) = 0.7000; water density (rho) = 995.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; flow rate through turbine (Q) = 95.0000 m^3/s; hydraulic head (H) = 29.0000 m, determine the hydropower output (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for P:

   $$P = \eta\rho g Q H$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.7000, water density (rho) = 995.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, flow rate through turbine (Q) = 95.0000 m^3/s, hydraulic head (H) = 29.0000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.7000995.0 9.8100 95.0000 29.0000$$

5. Step 5 — Evaluate:

   $$P = 18823992\ \text{W}$$

6. Step 6 — Check: returning P = 18,823,992 W to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 37,647,984 — kept a factor of two that cancels in the correct rearrangement.
• 9,411,996 — dropped that same factor in the other direction.
• 20,706,391 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower

hydropower plant power output calculation from turbine efficiency Given turbine efficiency (eta) = 0.7200; water density (rho) = 997.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; hydraulic head (H) = 178.0 m; hydropower output (P) = 42,721,900 W, determine the flow rate through turbine (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{P}{\eta\rho g H}$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.7200, water density (rho) = 997.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, hydraulic head (H) = 178.0 m, hydropower output (P) = 42,721,900 W.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{42721900}{0.7200997.0 9.8100 178.0}$$

5. Step 5 — Evaluate:

   Q = 34.0827\ \text{m^3/s}

6. Step 6 — Check: returning Q = 34.0827 m^3/s to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 68.1654 — kept a factor of two that cancels in the correct rearrangement.
• 17.0413 — dropped that same factor in the other direction.
• 37.4910 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower

run-of-river hydropower output estimate Given turbine efficiency (eta) = 0.8600; water density (rho) = 993.5 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; flow rate through turbine (Q) = 207.0 m^3/s; hydropower output (P) = 285,047,900 W, determine the hydraulic head (H) in m.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except H is given, so isolate H symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for H:

   $$H = \dfrac{P}{\eta\rho g Q}$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.8600, water density (rho) = 993.5 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, flow rate through turbine (Q) = 207.0 m^3/s, hydropower output (P) = 285,047,900 W.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{285047900}{0.8600993.5 9.8100 207.0}$$

5. Step 5 — Evaluate:

   $$H = 164.3\ \text{m}$$

6. Step 6 — Check: returning H = 164.3 m to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 328.6 — kept a factor of two that cancels in the correct rearrangement.
• 82.1452 — dropped that same factor in the other direction.
• 180.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower

hydropower generation output for a dam with a given head and flow Given turbine efficiency (eta) = 0.8500; water density (rho) = 990.5 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; flow rate through turbine (Q) = 470.0 m^3/s; hydraulic head (H) = 40.0000 m, determine the hydropower output (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for P:

   $$P = \eta\rho g Q H$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.8500, water density (rho) = 990.5 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, flow rate through turbine (Q) = 470.0 m^3/s, hydraulic head (H) = 40.0000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.8500990.5 9.8100 470.0 40.0000$$

5. Step 5 — Evaluate:

   $$P = 155274544\ \text{W}$$

6. Step 6 — Check: returning P = 155,274,544 W to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 310,549,088 — kept a factor of two that cancels in the correct rearrangement.
• 77,637,272 — dropped that same factor in the other direction.
• 170,801,998 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower

hydropower plant power output calculation from turbine efficiency Given turbine efficiency (eta) = 0.7500; water density (rho) = 991.5 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; hydraulic head (H) = 192.0 m; hydropower output (P) = 82,389,700 W, determine the flow rate through turbine (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{P}{\eta\rho g H}$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.7500, water density (rho) = 991.5 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, hydraulic head (H) = 192.0 m, hydropower output (P) = 82,389,700 W.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{82389700}{0.7500991.5 9.8100 192.0}$$

5. Step 5 — Evaluate:

   Q = 58.8232\ \text{m^3/s}

6. Step 6 — Check: returning Q = 58.8232 m^3/s to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 117.6 — kept a factor of two that cancels in the correct rearrangement.
• 29.4116 — dropped that same factor in the other direction.
• 64.7055 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower

run-of-river hydropower output estimate Given turbine efficiency (eta) = 0.8700; water density (rho) = 998.5 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; flow rate through turbine (Q) = 271.0 m^3/s; hydropower output (P) = 51,896,400 W, determine the hydraulic head (H) in m.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except H is given, so isolate H symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for H:

   $$H = \dfrac{P}{\eta\rho g Q}$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.8700, water density (rho) = 998.5 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, flow rate through turbine (Q) = 271.0 m^3/s, hydropower output (P) = 51,896,400 W.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{51896400}{0.8700998.5 9.8100 271.0}$$

5. Step 5 — Evaluate:

   $$H = 22.4715\ \text{m}$$

6. Step 6 — Check: returning H = 22.4715 m to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 44.9430 — kept a factor of two that cancels in the correct rearrangement.
• 11.2357 — dropped that same factor in the other direction.
• 24.7186 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower

hydropower generation output for a dam with a given head and flow Given turbine efficiency (eta) = 0.7700; water density (rho) = 990.5 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; flow rate through turbine (Q) = 423.0 m^3/s; hydraulic head (H) = 154.0 m, determine the hydropower output (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for P:

   $$P = \eta\rho g Q H$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.7700, water density (rho) = 990.5 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, flow rate through turbine (Q) = 423.0 m^3/s, hydraulic head (H) = 154.0 m.

4. Step 4 — Substitute the given values:

   $$P = 0.7700990.5 9.8100 423.0 154.0$$

5. Step 5 — Evaluate:

   $$P = 487388526\ \text{W}$$

6. Step 6 — Check: returning P = 487,388,526 W to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 974,777,051 — kept a factor of two that cancels in the correct rearrangement.
• 243,694,263 — dropped that same factor in the other direction.
• 536,127,378 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower

hydropower plant power output calculation from turbine efficiency Given turbine efficiency (eta) = 0.8500; water density (rho) = 996.5 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; hydraulic head (H) = 43.0000 m; hydropower output (P) = 177,163,800 W, determine the flow rate through turbine (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{P}{\eta\rho g H}$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.8500, water density (rho) = 996.5 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, hydraulic head (H) = 43.0000 m, hydropower output (P) = 177,163,800 W.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{177163800}{0.8500996.5 9.8100 43.0000}$$

5. Step 5 — Evaluate:

   Q = 495.8\ \text{m^3/s}

6. Step 6 — Check: returning Q = 495.8 m^3/s to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 991.7 — kept a factor of two that cancels in the correct rearrangement.
• 247.9 — dropped that same factor in the other direction.
• 545.4 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower

run-of-river hydropower output estimate Given turbine efficiency (eta) = 0.8000; water density (rho) = 991.5 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; flow rate through turbine (Q) = 400.0 m^3/s; hydropower output (P) = 97,180,700 W, determine the hydraulic head (H) in m.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except H is given, so isolate H symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for H:

   $$H = \dfrac{P}{\eta\rho g Q}$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.8000, water density (rho) = 991.5 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, flow rate through turbine (Q) = 400.0 m^3/s, hydropower output (P) = 97,180,700 W.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{97180700}{0.8000991.5 9.8100 400.0}$$

5. Step 5 — Evaluate:

   $$H = 31.2225\ \text{m}$$

6. Step 6 — Check: returning H = 31.2225 m to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 62.4451 — kept a factor of two that cancels in the correct rearrangement.
• 15.6113 — dropped that same factor in the other direction.
• 34.3448 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower

hydropower generation output for a dam with a given head and flow Given turbine efficiency (eta) = 0.8800; water density (rho) = 996.5 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; flow rate through turbine (Q) = 9.0000 m^3/s; hydraulic head (H) = 87.0000 m, determine the hydropower output (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Hydropower.
• Everything except P is given, so isolate P symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Hydropower output from a dam or run-of-river plant depends on turbine efficiency, flow rate, and hydraulic head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \rho g Q H$$

2. Step 2 — Rearrange symbolically for P:

   $$P = \eta\rho g Q H$$

3. Step 3 — List the givens: turbine efficiency (eta) = 0.8800, water density (rho) = 996.5 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, flow rate through turbine (Q) = 9.0000 m^3/s, hydraulic head (H) = 87.0000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.8800996.5 9.8100 9.0000 87.0000$$

5. Step 5 — Evaluate:

   $$P = 6735824\ \text{W}$$

6. Step 6 — Check: returning P = 6,735,824 W to

   $$P = \eta \rho g Q H$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 13,471,648 — kept a factor of two that cancels in the correct rearrangement.
• 3,367,912 — dropped that same factor in the other direction.
• 7,409,407 — rounded an intermediate value before the final step.

Reference: FE Handbook — Hydropower