A 75 mm jet leaves a nozzle at 32 m/s and is deflected through 60°. Compute the force on the blade when it is stationary, then when the blade moves away at 16.0 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.075)^{2}/4 = 0.00442 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00442)(32)^{2}(1 - \cos 60^{\circ}) = 2,262 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 32 - 16.0 = 16.0 m/s$$

6. Substituting

   $$F = 1000(0.00442)(16.0)^{2}(1 - \cos 60^{\circ}) = 565.5 N$$

7. Power

   $$P = F u = 565.5(16.0) = 9.05 kW$$

Why the other options are there

• 1,131 N (mixed absolute and relative velocity)
• P = 36.19 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade

A 55 mm jet leaves a nozzle at 22 m/s and is deflected through 90°. Compute the force on the blade when it is stationary, then when the blade moves away at 2.2 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.055)^{2}/4 = 0.00238 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00238)(22)^{2}(1 - \cos 90^{\circ}) = 1,150 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 22 - 2.2 = 19.8 m/s$$

6. Substituting

   $$F = 1000(0.00238)(19.8)^{2}(1 - \cos 90^{\circ}) = 931.4 N$$

7. Power

   $$P = F u = 931.4(2.2) = 2.05 kW$$

Why the other options are there

• 115.0 N (mixed absolute and relative velocity)
• P = 2.53 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade

A 30 mm jet leaves a nozzle at 21 m/s and is deflected through 180°. Compute the force on the blade when it is stationary, then when the blade moves away at 9.5 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.030)^{2}/4 = 0.00071 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00071)(21)^{2}(1 - \cos 180^{\circ}) = 623.4 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 21 - 9.5 = 11.5 m/s$$

6. Substituting

   $$F = 1000(0.00071)(11.5)^{2}(1 - \cos 180^{\circ}) = 188.6 N$$

7. Power

   $$P = F u = 188.6(9.5) = 1.78 kW$$

Why the other options are there

• 280.6 N (mixed absolute and relative velocity)
• P = 5.89 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade

A 55 mm jet leaves a nozzle at 32 m/s and is deflected through 120°. Compute the force on the blade when it is stationary, then when the blade moves away at 8.0 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.055)^{2}/4 = 0.00238 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00238)(32)^{2}(1 - \cos 120^{\circ}) = 3,649 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 32 - 8.0 = 24.0 m/s$$

6. Substituting

   $$F = 1000(0.00238)(24.0)^{2}(1 - \cos 120^{\circ}) = 2,053 N$$

7. Power

   $$P = F u = 2,053(8.0) = 16.42 kW$$

Why the other options are there

• 912.3 N (mixed absolute and relative velocity)
• P = 29.19 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade

A 80 mm jet leaves a nozzle at 43 m/s and is deflected through 180°. Compute the force on the blade when it is stationary, then when the blade moves away at 4.3 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.080)^{2}/4 = 0.00503 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00503)(43)^{2}(1 - \cos 180^{\circ}) = 18,588 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 43 - 4.3 = 38.7 m/s$$

6. Substituting

   $$F = 1000(0.00503)(38.7)^{2}(1 - \cos 180^{\circ}) = 15,056 N$$

7. Power

   $$P = F u = 15,056(4.3) = 64.74 kW$$

Why the other options are there

• 1,859 N (mixed absolute and relative velocity)
• P = 79.93 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade

A 25 mm jet leaves a nozzle at 29 m/s and is deflected through 90°. Compute the force on the blade when it is stationary, then when the blade moves away at 11.6 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.025)^{2}/4 = 0.00049 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00049)(29)^{2}(1 - \cos 90^{\circ}) = 412.8 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 29 - 11.6 = 17.4 m/s$$

6. Substituting

   $$F = 1000(0.00049)(17.4)^{2}(1 - \cos 90^{\circ}) = 148.6 N$$

7. Power

   $$P = F u = 148.6(11.6) = 1.72 kW$$

Why the other options are there

• 165.1 N (mixed absolute and relative velocity)
• P = 4.79 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade

A 45 mm jet leaves a nozzle at 20 m/s and is deflected through 90°. Compute the force on the blade when it is stationary, then when the blade moves away at 7.0 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.045)^{2}/4 = 0.00159 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00159)(20)^{2}(1 - \cos 90^{\circ}) = 636.2 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 20 - 7.0 = 13.0 m/s$$

6. Substituting

   $$F = 1000(0.00159)(13.0)^{2}(1 - \cos 90^{\circ}) = 268.8 N$$

7. Power

   $$P = F u = 268.8(7.0) = 1.88 kW$$

Why the other options are there

• 222.7 N (mixed absolute and relative velocity)
• P = 4.45 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade

A 35 mm jet leaves a nozzle at 25 m/s and is deflected through 60°. Compute the force on the blade when it is stationary, then when the blade moves away at 8.8 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.035)^{2}/4 = 0.00096 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00096)(25)^{2}(1 - \cos 60^{\circ}) = 300.7 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 25 - 8.8 = 16.3 m/s$$

6. Substituting

   $$F = 1000(0.00096)(16.3)^{2}(1 - \cos 60^{\circ}) = 127.0 N$$

7. Power

   $$P = F u = 127.0(8.8) = 1.11 kW$$

Why the other options are there

• 105.2 N (mixed absolute and relative velocity)
• P = 2.63 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade

A 55 mm jet leaves a nozzle at 18 m/s and is deflected through 180°. Compute the force on the blade when it is stationary, then when the blade moves away at 1.8 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.055)^{2}/4 = 0.00238 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00238)(18)^{2}(1 - \cos 180^{\circ}) = 1,540 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 18 - 1.8 = 16.2 m/s$$

6. Substituting

   $$F = 1000(0.00238)(16.2)^{2}(1 - \cos 180^{\circ}) = 1,247 N$$

7. Power

   $$P = F u = 1,247(1.8) = 2.24 kW$$

Why the other options are there

• 154.0 N (mixed absolute and relative velocity)
• P = 2.77 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade

A 25 mm jet leaves a nozzle at 12 m/s and is deflected through 120°. Compute the force on the blade when it is stationary, then when the blade moves away at 1.8 m/s, and find the power transferred to the moving blade.

Start with the thinking

• Momentum flux uses the velocity relative to the blade — that is the only change between the fixed and moving cases.
• A 180° bucket doubles the force because the jet reverses direction.

Step-by-step solution

1. Jet area

   $$A = \pi(0.025)^{2}/4 = 0.00049 m^{2}$$

2. Formula

   $$F_{fixed} = \rho A V_j^2\left(1-\cos\theta\right)$$

3. Substituting

   $$F = 1000(0.00049)(12)^{2}(1 - \cos 120^{\circ}) = 106.0 N$$

4. Formula

   $$F_{moving} = \rho A (V_j-u)^2(1-\cos\theta)$$

5. Relative velocity

   $$V_j - u = 12 - 1.8 = 10.2 m/s$$

6. Substituting

   $$F = 1000(0.00049)(10.2)^{2}(1 - \cos 120^{\circ}) = 77 N$$

7. Power

   $$P = F u = 77(1.8) = 0.14 kW$$

Why the other options are there

• 16 N (mixed absolute and relative velocity)
• P = 0.19 kW (fixed-blade force used)

Reference: FE Reference Handbook — Fluid Mechanics → Fixed Blade