turbines in a hydroelectric powerhouse fed by a penstock Given turbine efficiency (eta) = 0.7000; specific weight of water (gamma) = 9,765 N/m^3; flow rate (Q) = 18.6000 m^3/s; net head (H) = 74.0000 m, determine the power output (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Turbines (hydraulic power output).
• 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.
• Turbines extract power from flowing water passing through a penstock under a net head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \gamma Q H$$

2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

3. Step 3 — List the givens: turbine efficiency (eta) = 0.7000, specific weight of water (gamma) = 9,765 N/m^3, flow rate (Q) = 18.6000 m^3/s, net head (H) = 74.0000 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$P = 9408382\ \text{W}$$

6. Step 6 — Check: returning P = 9,408,382 W to

   $$P = \eta \gamma Q H$$

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

Why the other options are there

• 18,816,764 — kept a factor of two that cancels in the correct rearrangement.
• 4,704,191 — dropped that same factor in the other direction.
• 10,349,220 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Kaplan turbine on a river diversion Given turbine efficiency (\eta) = 0.7100; water density (\rho) = 998.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 40.0000 m^3/s; net head (H) = 38.5000 m, determine the power output (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic turbine power output.
• 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.
• A reaction turbine extracts power from flow under a net head at a dam outlet works.

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) = 0.7100, water density (\rho) = 998.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 40.0000 m^3/s, net head (H) = 38.5000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.7100 998.0 9.8100 40.0000 38.5000$$

5. Step 5 — Evaluate:

   $$P = 10704801\ \text{W}$$

6. Step 6 — Check: returning P = 10,704,801 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

• 21,409,603 — kept a factor of two that cancels in the correct rearrangement.
• 5,352,401 — dropped that same factor in the other direction.
• 11,775,282 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

turbines extracting energy from a dam's reservoir head Given turbine efficiency (eta) = 0.8000; specific weight of water (gamma) = 9,760 N/m^3; net head (H) = 83.0000 m; power output (P) = 435,300 W, determine the flow rate (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Turbines (hydraulic power output).
• 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.
• Turbines extract power from flowing water passing through a penstock under a net head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \gamma Q H$$

2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

3. Step 3 — List the givens: turbine efficiency (eta) = 0.8000, specific weight of water (gamma) = 9,760 N/m^3, net head (H) = 83.0000 m, power output (P) = 435,300 W.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   Q = 0.6717\ \text{m^3/s}

6. Step 6 — Check: returning Q = 0.6717 m^3/s to

   $$P = \eta \gamma Q H$$

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

Why the other options are there

• 1.3434 — kept a factor of two that cancels in the correct rearrangement.
• 0.3358 — dropped that same factor in the other direction.
• 0.7389 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Pelton wheel fed by a penstock Given power output (P) = 2,881,508 W; turbine efficiency (\eta) = 0.9100; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; net head (H) = 78.5000 m, determine the discharge (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic turbine power output.
• 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.
• A reaction turbine extracts power from flow under a net head at a dam outlet works.

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: power output (P) = 2,881,508 W, turbine efficiency (\eta) = 0.9100, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, net head (H) = 78.5000 m.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{2881508}{0.9100 999.0 9.8100 78.5000}$$

5. Step 5 — Evaluate:

   Q = 4.1160\ \text{m^3/s}

6. Step 6 — Check: returning Q = 4.1160 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

• 8.2320 — kept a factor of two that cancels in the correct rearrangement.
• 2.0580 — dropped that same factor in the other direction.
• 4.5276 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

turbines used in a small run-of-river hydro plant Given turbine efficiency (eta) = 0.8000; specific weight of water (gamma) = 9,760 N/m^3; flow rate (Q) = 12.8000 m^3/s; power output (P) = 3,319,600 W, determine the net head (H) in m.

Start with the thinking

• The governing relation printed in this handbook section is Turbines (hydraulic power output).
• 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.
• Turbines extract power from flowing water passing through a penstock under a net head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \gamma Q H$$

2. Step 2 — Rearrange the relation so that H stands alone on the left-hand side.

3. Step 3 — List the givens: turbine efficiency (eta) = 0.8000, specific weight of water (gamma) = 9,760 N/m^3, flow rate (Q) = 12.8000 m^3/s, power output (P) = 3,319,600 W.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$H = 33.2151\ \text{m}$$

6. Step 6 — Check: returning H = 33.2151 m to

   $$P = \eta \gamma Q H$$

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

Why the other options are there

• 66.4303 — kept a factor of two that cancels in the correct rearrangement.
• 16.6076 — dropped that same factor in the other direction.
• 36.5366 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Francis turbine at a low-head dam Given power output (P) = 210,762 W; turbine efficiency (\eta) = 0.8900; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 24.5000 m^3/s, determine the net head (H) in m.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic turbine power output.
• 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.
• A reaction turbine extracts power from flow under a net head at a dam outlet works.

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: power output (P) = 210,762 W, turbine efficiency (\eta) = 0.8900, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 24.5000 m^3/s.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{210762}{0.8900 999.0 9.8100 24.5000}$$

5. Step 5 — Evaluate:

   $$H = 0.9863\ \text{m}$$

6. Step 6 — Check: returning H = 0.9863 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

• 1.9726 — kept a factor of two that cancels in the correct rearrangement.
• 0.4931 — dropped that same factor in the other direction.
• 1.0849 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

turbines in a hydroelectric powerhouse fed by a penstock Given turbine efficiency (eta) = 0.7400; specific weight of water (gamma) = 9,730 N/m^3; flow rate (Q) = 15.7000 m^3/s; net head (H) = 77.0000 m, determine the power output (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Turbines (hydraulic power output).
• 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.
• Turbines extract power from flowing water passing through a penstock under a net head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \gamma Q H$$

2. Step 2 — Rearrange the relation so that P stands alone on the left-hand side.

3. Step 3 — List the givens: turbine efficiency (eta) = 0.7400, specific weight of water (gamma) = 9,730 N/m^3, flow rate (Q) = 15.7000 m^3/s, net head (H) = 77.0000 m.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$P = 8704322\ \text{W}$$

6. Step 6 — Check: returning P = 8,704,322 W to

   $$P = \eta \gamma Q H$$

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

Why the other options are there

• 17,408,644 — kept a factor of two that cancels in the correct rearrangement.
• 4,352,161 — dropped that same factor in the other direction.
• 9,574,754 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Kaplan turbine on a river diversion Given turbine efficiency (\eta) = 0.7800; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 33.3000 m^3/s; net head (H) = 42.0000 m, determine the power output (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic turbine power output.
• 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.
• A reaction turbine extracts power from flow under a net head at a dam outlet works.

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) = 0.7800, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 33.3000 m^3/s, net head (H) = 42.0000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.7800 999.0 9.8100 33.3000 42.0000$$

5. Step 5 — Evaluate:

   $$P = 10691106\ \text{W}$$

6. Step 6 — Check: returning P = 10,691,106 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

• 21,382,211 — kept a factor of two that cancels in the correct rearrangement.
• 5,345,553 — dropped that same factor in the other direction.
• 11,760,216 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

turbines extracting energy from a dam's reservoir head Given turbine efficiency (eta) = 0.7100; specific weight of water (gamma) = 9,775 N/m^3; net head (H) = 97.0000 m; power output (P) = 4,890,300 W, determine the flow rate (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Turbines (hydraulic power output).
• 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.
• Turbines extract power from flowing water passing through a penstock under a net head.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = \eta \gamma Q H$$

2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

3. Step 3 — List the givens: turbine efficiency (eta) = 0.7100, specific weight of water (gamma) = 9,775 N/m^3, net head (H) = 97.0000 m, power output (P) = 4,890,300 W.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   Q = 7.2642\ \text{m^3/s}

6. Step 6 — Check: returning Q = 7.2642 m^3/s to

   $$P = \eta \gamma Q H$$

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

Why the other options are there

• 14.5284 — kept a factor of two that cancels in the correct rearrangement.
• 3.6321 — dropped that same factor in the other direction.
• 7.9906 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Pelton wheel fed by a penstock Given power output (P) = 2,089,642 W; turbine efficiency (\eta) = 0.7600; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; net head (H) = 86.0000 m, determine the discharge (Q) in m^3/s.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic turbine power output.
• 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.
• A reaction turbine extracts power from flow under a net head at a dam outlet works.

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: power output (P) = 2,089,642 W, turbine efficiency (\eta) = 0.7600, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, net head (H) = 86.0000 m.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{2089642}{0.7600 999.0 9.8100 86.0000}$$

5. Step 5 — Evaluate:

   Q = 3.2623\ \text{m^3/s}

6. Step 6 — Check: returning Q = 3.2623 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

• 6.5246 — kept a factor of two that cancels in the correct rearrangement.
• 1.6312 — dropped that same factor in the other direction.
• 3.5885 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines