a Francis turbine at a low-head dam Given turbine efficiency (\eta) = 0.8200; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 21.1000 m^3/s; net head (H) = 75.5000 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) = 0.8200, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 21.1000 m^3/s, net head (H) = 75.5000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.8200 999.0 9.8100 21.1000 75.5000$$

5. Step 5 — Evaluate:

   $$P = 12801998\ \text{W}$$

6. Step 6 — Check: returning P = 12,801,998 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

• 25,603,996 — kept a factor of two that cancels in the correct rearrangement.
• 6,400,999 — dropped that same factor in the other direction.
• 14,082,198 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Kaplan turbine on a river diversion Given power output (P) = 3,006,515 W; turbine efficiency (\eta) = 0.8300; water density (\rho) = 998.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; net head (H) = 55.0000 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) = 3,006,515 W, turbine efficiency (\eta) = 0.8300, water density (\rho) = 998.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, net head (H) = 55.0000 m.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{3006515}{0.8300 998.0 9.8100 55.0000}$$

5. Step 5 — Evaluate:

   Q = 6.7270\ \text{m^3/s}

6. Step 6 — Check: returning Q = 6.7270 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

• 13.4541 — kept a factor of two that cancels in the correct rearrangement.
• 3.3635 — dropped that same factor in the other direction.
• 7.3997 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Pelton wheel fed by a penstock Given power output (P) = 2,310,765 W; turbine efficiency (\eta) = 0.7700; water density (\rho) = 1,000 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 3.9000 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) = 2,310,765 W, turbine efficiency (\eta) = 0.7700, water density (\rho) = 1,000 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 3.9000 m^3/s.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{2310765}{0.7700 1000 9.8100 3.9000}$$

5. Step 5 — Evaluate:

   $$H = 78.4389\ \text{m}$$

6. Step 6 — Check: returning H = 78.4389 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

• 156.9 — kept a factor of two that cancels in the correct rearrangement.
• 39.2194 — dropped that same factor in the other direction.
• 86.2828 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Francis turbine at a low-head dam Given turbine efficiency (\eta) = 0.7800; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 20.6000 m^3/s; net head (H) = 65.5000 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) = 0.7800, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 20.6000 m^3/s, net head (H) = 65.5000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.7800 999.0 9.8100 20.6000 65.5000$$

5. Step 5 — Evaluate:

   $$P = 10314249\ \text{W}$$

6. Step 6 — Check: returning P = 10,314,249 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

• 20,628,498 — kept a factor of two that cancels in the correct rearrangement.
• 5,157,125 — dropped that same factor in the other direction.
• 11,345,674 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Kaplan turbine on a river diversion Given power output (P) = 1,332,116 W; turbine efficiency (\eta) = 0.7300; water density (\rho) = 998.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; net head (H) = 6.5000 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) = 1,332,116 W, turbine efficiency (\eta) = 0.7300, water density (\rho) = 998.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, net head (H) = 6.5000 m.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{1332116}{0.7300 998.0 9.8100 6.5000}$$

5. Step 5 — Evaluate:

   Q = 28.6752\ \text{m^3/s}

6. Step 6 — Check: returning Q = 28.6752 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

• 57.3504 — kept a factor of two that cancels in the correct rearrangement.
• 14.3376 — dropped that same factor in the other direction.
• 31.5427 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Pelton wheel fed by a penstock Given power output (P) = 1,402,545 W; turbine efficiency (\eta) = 0.7500; water density (\rho) = 998.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 30.0000 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) = 1,402,545 W, turbine efficiency (\eta) = 0.7500, water density (\rho) = 998.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 30.0000 m^3/s.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{1402545}{0.7500 998.0 9.8100 30.0000}$$

5. Step 5 — Evaluate:

   $$H = 6.3670\ \text{m}$$

6. Step 6 — Check: returning H = 6.3670 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

• 12.7340 — kept a factor of two that cancels in the correct rearrangement.
• 3.1835 — dropped that same factor in the other direction.
• 7.0037 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Francis turbine at a low-head dam Given turbine efficiency (\eta) = 0.7900; water density (\rho) = 998.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 23.2000 m^3/s; net head (H) = 5.5000 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) = 0.7900, water density (\rho) = 998.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 23.2000 m^3/s, net head (H) = 5.5000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.7900 998.0 9.8100 23.2000 5.5000$$

5. Step 5 — Evaluate:

   $$P = 986909\ \text{W}$$

6. Step 6 — Check: returning P = 986,909 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

• 1,973,819 — kept a factor of two that cancels in the correct rearrangement.
• 493,455 — dropped that same factor in the other direction.
• 1,085,600 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Kaplan turbine on a river diversion Given power output (P) = 2,952,299 W; turbine efficiency (\eta) = 0.9100; water density (\rho) = 998.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; net head (H) = 44.0000 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) = 2,952,299 W, turbine efficiency (\eta) = 0.9100, water density (\rho) = 998.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, net head (H) = 44.0000 m.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{2952299}{0.9100 998.0 9.8100 44.0000}$$

5. Step 5 — Evaluate:

   Q = 7.5312\ \text{m^3/s}

6. Step 6 — Check: returning Q = 7.5312 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

• 15.0625 — kept a factor of two that cancels in the correct rearrangement.
• 3.7656 — dropped that same factor in the other direction.
• 8.2844 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Pelton wheel fed by a penstock Given power output (P) = 2,268,514 W; turbine efficiency (\eta) = 0.8700; water density (\rho) = 1,000 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 17.8000 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) = 2,268,514 W, turbine efficiency (\eta) = 0.8700, water density (\rho) = 1,000 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 17.8000 m^3/s.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{2268514}{0.8700 1000 9.8100 17.8000}$$

5. Step 5 — Evaluate:

   $$H = 14.9325\ \text{m}$$

6. Step 6 — Check: returning H = 14.9325 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

• 29.8650 — kept a factor of two that cancels in the correct rearrangement.
• 7.4663 — dropped that same factor in the other direction.
• 16.4258 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines

a Francis turbine at a low-head dam Given turbine efficiency (\eta) = 0.7300; water density (\rho) = 999.0 kg/m^3; gravitational acceleration (g) = 9.8100 m/s^2; discharge (Q) = 18.7000 m^3/s; net head (H) = 48.0000 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) = 0.7300, water density (\rho) = 999.0 kg/m^3, gravitational acceleration (g) = 9.8100 m/s^2, discharge (Q) = 18.7000 m^3/s, net head (H) = 48.0000 m.

4. Step 4 — Substitute the given values:

   $$P = 0.7300 999.0 9.8100 18.7000 48.0000$$

5. Step 5 — Evaluate:

   $$P = 6421555\ \text{W}$$

6. Step 6 — Check: returning P = 6,421,555 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

• 12,843,110 — kept a factor of two that cancels in the correct rearrangement.
• 3,210,777 — dropped that same factor in the other direction.
• 7,063,710 — rounded an intermediate value before the final step.

Reference: FE Handbook — Turbines