a Francis turbine at a low-head dam Given turbine efficiency (\eta) = 0.7100; water density (\rho) = 1,000 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 1.1000 m^3/s; net head (H) = 72.0000 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) = 0.7100, water density (\rho) = 1,000 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 1.1000 m^3/s, net head (H) = 72.0000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.7100 1000 9.8100 1.1000 72.0000$$

5. Step 5 — Evaluate:

   $$P = 551636\ \text{W}$$

6. Step 6 — Check: returning P = 551,636 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

• 1,103,272 — kept a factor of two that cancels in the correct rearrangement.
• 275,818 — dropped that same factor in the other direction.
• 606,800 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Kaplan turbine on a river diversion Given power output (P) = 1,829,842 W; turbine efficiency (\eta) = 0.7100; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; net head (H) = 54.5000 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) = 1,829,842 W, turbine efficiency (\eta) = 0.7100, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, net head (H) = 54.5000 m.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{1829842}{0.7100 999.0 9.8100 54.5000}$$

5. Step 5 — Evaluate:

   Q = 4.8253\ \text{m^3/s}

6. Step 6 — Check: returning Q = 4.8253 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

• 9.6506 — kept a factor of two that cancels in the correct rearrangement.
• 2.4126 — dropped that same factor in the other direction.
• 5.3078 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Pelton wheel fed by a penstock Given power output (P) = 282,858 W; turbine efficiency (\eta) = 0.7700; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 33.3000 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) = 282,858 W, turbine efficiency (\eta) = 0.7700, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 33.3000 m^3/s.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{282858}{0.7700 999.0 9.8100 33.3000}$$

5. Step 5 — Evaluate:

   $$H = 1.1256\ \text{m}$$

6. Step 6 — Check: returning H = 1.1256 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

• 2.2513 — kept a factor of two that cancels in the correct rearrangement.
• 0.5628 — dropped that same factor in the other direction.
• 1.2382 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Francis turbine at a low-head dam Given turbine efficiency (\eta) = 0.8100; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 33.7000 m^3/s; net head (H) = 72.0000 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) = 0.8100, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 33.7000 m^3/s, net head (H) = 72.0000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.8100 999.0 9.8100 33.7000 72.0000$$

5. Step 5 — Evaluate:

   $$P = 19261137\ \text{W}$$

6. Step 6 — Check: returning P = 19,261,137 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

• 38,522,273 — kept a factor of two that cancels in the correct rearrangement.
• 9,630,568 — dropped that same factor in the other direction.
• 21,187,250 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Kaplan turbine on a river diversion Given power output (P) = 3,625,522 W; turbine efficiency (\eta) = 0.7400; water density (\rho) = 1,000 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; net head (H) = 13.0000 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) = 3,625,522 W, turbine efficiency (\eta) = 0.7400, water density (\rho) = 1,000 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, net head (H) = 13.0000 m.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{3625522}{0.7400 1000 9.8100 13.0000}$$

5. Step 5 — Evaluate:

   Q = 38.4173\ \text{m^3/s}

6. Step 6 — Check: returning Q = 38.4173 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

• 76.8345 — kept a factor of two that cancels in the correct rearrangement.
• 19.2086 — dropped that same factor in the other direction.
• 42.2590 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Pelton wheel fed by a penstock Given power output (P) = 2,859,076 W; turbine efficiency (\eta) = 0.7900; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 27.6000 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) = 2,859,076 W, turbine efficiency (\eta) = 0.7900, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 27.6000 m^3/s.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{2859076}{0.7900 999.0 9.8100 27.6000}$$

5. Step 5 — Evaluate:

   $$H = 13.3800\ \text{m}$$

6. Step 6 — Check: returning H = 13.3800 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

• 26.7599 — kept a factor of two that cancels in the correct rearrangement.
• 6.6900 — dropped that same factor in the other direction.
• 14.7180 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Francis turbine at a low-head dam Given turbine efficiency (\eta) = 0.9000; water density (\rho) = 998.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 22.2000 m^3/s; net head (H) = 56.5000 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) = 0.9000, water density (\rho) = 998.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 22.2000 m^3/s, net head (H) = 56.5000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.9000 998.0 9.8100 22.2000 56.5000$$

5. Step 5 — Evaluate:

   $$P = 11052066\ \text{W}$$

6. Step 6 — Check: returning P = 11,052,066 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

• 22,104,133 — kept a factor of two that cancels in the correct rearrangement.
• 5,526,033 — dropped that same factor in the other direction.
• 12,157,273 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Kaplan turbine on a river diversion Given power output (P) = 2,869,792 W; turbine efficiency (\eta) = 0.7900; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; net head (H) = 69.0000 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) = 2,869,792 W, turbine efficiency (\eta) = 0.7900, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, net head (H) = 69.0000 m.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{2869792}{0.7900 999.0 9.8100 69.0000}$$

5. Step 5 — Evaluate:

   Q = 5.3720\ \text{m^3/s}

6. Step 6 — Check: returning Q = 5.3720 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

• 10.7441 — kept a factor of two that cancels in the correct rearrangement.
• 2.6860 — dropped that same factor in the other direction.
• 5.9093 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Pelton wheel fed by a penstock Given power output (P) = 886,840 W; turbine efficiency (\eta) = 0.8000; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 18.3000 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) = 886,840 W, turbine efficiency (\eta) = 0.8000, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 18.3000 m^3/s.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{886840}{0.8000 999.0 9.8100 18.3000}$$

5. Step 5 — Evaluate:

   $$H = 6.1812\ \text{m}$$

6. Step 6 — Check: returning H = 6.1812 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

• 12.3623 — kept a factor of two that cancels in the correct rearrangement.
• 3.0906 — dropped that same factor in the other direction.
• 6.7993 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Francis turbine at a low-head dam Given turbine efficiency (\eta) = 0.7500; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 30.5000 m^3/s; net head (H) = 32.0000 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) = 0.7500, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 30.5000 m^3/s, net head (H) = 32.0000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.7500 999.0 9.8100 30.5000 32.0000$$

5. Step 5 — Evaluate:

   $$P = 7173739\ \text{W}$$

6. Step 6 — Check: returning P = 7,173,739 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

• 14,347,478 — kept a factor of two that cancels in the correct rearrangement.
• 3,586,870 — dropped that same factor in the other direction.
• 7,891,113 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines