A simple turbine schematic of the Parsons type. Steam pressure drops by a fraction of the total across each pair of rotating and fixed blades, the stators grow larger as pressure drops.

Question

Asked on July 11, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-07-11T16:23:20.849Z

5.1 Construct a velocity diagram. As an AI, I cannot physically construct a diagram. However, I can provide all the necessary calculated values that would be entered onto the velocity diagram, using the given scale of 1 cm = 30 m/s. Step 1: Calculate the blade velocity (U). Diameter (D) = 460 mm = 0.460 m Rotational speed (N) = 116.267 rev/s U = D N U = × 0.460 m × 116.267 rev/s U ≈ 168.39 m/s Step 2: Convert the absolute inlet velocity (V_1) to meters per second. Gas leaves at 1944 km/h. V_1 = 1944 km/h × 1000 m1 km × 1 h3600 s V_1 = 540 m/s Step 3: Calculate the components of the inlet velocity triangle. Nozzle angle (_1) = 22^ Inlet flow velocity (V_f1): V_f1 = V_1 _1 = 540 m/s × 22^ V_f1 ≈ 202.28 m/s Inlet whirl velocity (V_w1): V_w1 = V_1 _1 = 540 m/s × 22^ V_w1 ≈ 500.69 m/s Relative inlet velocity (V_r1): V_r1 = sqrt(V_f1)^2 + (V_w1 - U)^2 V_r1 = sqrt((202.28 m/s))^2 + (500.69 m/s - 168.39 m/s)^2 V_r1 = sqrt((202.28)^2 + (332.30)^2) = sqrt(40917.1 + 110423.3) V_r1 = sqrt(151340.4) ≈ 389.02 m/s Inlet blade angle (_1): _1 = V_f1V_w1 - U = (202.28)/(332.30) _1 = (0.6087) ≈ 31.32^ Step 4: Calculate the components of the outlet velocity triangle. No axial thrust means V_f2 = V_f1 = 202.28 m/s. 15% loss of velocity due to friction means V_r2 = 0.85 V_r1. V_r2 = 0.85 × 389.02 m/s ≈ 330.67 m/s Outlet blade angle (_2): _2 = V_f2V_r2 = (202.28)/(330.67) _2 = (0.6117) ≈ 37.72^ Exit whirl velocity (V_w2): V_w2 = U - V_r2 _2 V_w2 = 168.39 m/s - (330.67 m/s × 37.72^) V_w2 = 168.39 - (330.67 × 0.7910) = 168.39 - 261.58 ≈ -93.19 m/s Absolute exit velocity (V_2): V_2 = sqrt(V_f2)^2 + V_w2^2 = sqrt((202.28 m/s))^2 + (-93.19 m/s)^2 V_2 = sqrt(40917.1 + 8684.4) = sqrt(49601.5) ≈ 222.71 m/s Absolute exit angle (_2): _2 = V_f2V_w2 = (202.28)/(-93.19) _2 = (-2.1706) ≈ -65.27^ The values to be entered onto the velocity diagram are: Blade velocity (U): 168.39 m/s Absolute inlet velocity (V_1): 540 m/s Absolute exit velocity (V_2): 222.71 m/s Relative inlet velocity (V_r1): 389.02 m/s Relative exit velocity (V_r2): 330.67 m/s Inlet whirl velocity (V_w1): 500.69 m/s Exit whirl velocity (V_w2): -93.19 m/s Inlet flow velocity (V_f1): 202.28 m/s Exit flow velocity (V_f2): 202.28 m/s Absolute inlet angle (_1): 22^ Absolute exit angle (_2): -65.27^ Inlet blade angle (_1): 31.32^ Exit blade angle (_2): 37.72^ 5.2 Answer the following questions pertaining to the velocity diagram drawn in QUESTION 5.1.1 above: 5.2.1 Calculate the power developed by the turbine if the mass flow rate of the gas is 1000 grams per second. Step 1: Convert the mass flow rate to kilograms per second. Mass flow rate (m) = 1000 g/s = 1 kg/s Step 2: Use the formula for power developed by an impulse turbine. P = m U (V_w1 - V_w2) Step 3: Substitute the calculated values and calculate the power. P = 1 kg/s × 168.39 m/s × (500.69 m/s - (-93.19 m/s)) P = 168.39 × (500.69 + 93.19) P = 168.39 × 593.88 P ≈ 99900.6 W P ≈ 99.90 kW 5.2.2 Calculate the kinetic energy that exits the blades. Step 1: Identify the mass flow rate and absolute exit velocity. Mass flow rate (m) = 1 kg/s Absolute exit velocity (V_2) = 222.71 m/s Step 2: Use the formula for kinetic energy per second (power). KE_exit = (1)/(2) m V_2^2 Step 3: Substitute the values and calculate the kinetic energy. KE_exit = (1)/(2) × 1 kg/s × (222.71 m/s)^2 KE_exit = (1)/(2) × 49601.5 J/s KE_exit ≈ 24800.75 W 5.3 Name THREE types of governors. Watt governor* Porter governor* Proell governor* 5.4 Calculate the height of a Watt governor rotating at a speed of 90 r/min. Step 1: Convert the rotational speed from revolutions per minute to radians per second. Speed (N) = 90 r/min Angular velocity () = (2 N)/(60) = (2 × 90)/(60) = 3 rad/s ≈ 9.4248 rad/s Step 2: Use the formula for the height (h) of a Watt governor. h = (g)/(^2) where g = 9.81 m/s^2 (acceleration due to gravity). Step 3: Substitute the values and calculate the height. h = 9.81 m/s^2(9.4248 rad/s)^2 h = (9.81)/(88.8269) h ≈ 0.1104 m h ≈ 110.4 mm Drop the next question.

Accepted Answer · 2026-07-11T16:23:20.849Z

5.1 Construct a velocity diagram. As an AI, I cannot physically construct a diagram. However, I can provide all the necessary calculated values that would be entered onto the velocity diagram, using the given scale of 1 cm = 30 m/s. Step 1: Calculate the blade velocity (U). Diameter (D) = 460 mm = 0.460 m Rotational speed (N) = 116.267 rev/s U = D N U = × 0.460 m × 116.267 rev/s U ≈ 168.39 m/s Step 2: Convert the absolute inlet velocity (V_1) to meters per second. Gas leaves at 1944 km/h. V_1 = 1944 km/h × 1000 m1 km × 1 h3600 s V_1 = 540 m/s Step 3: Calculate the components of the inlet velocity triangle. Nozzle angle (_1) = 22^ Inlet flow velocity (V_f1): V_f1 = V_1 _1 = 540 m/s × 22^ V_f1 ≈ 202.28 m/s Inlet whirl velocity (V_w1): V_w1 = V_1 _1 = 540 m/s × 22^ V_w1 ≈ 500.69 m/s Relative inlet velocity (V_r1): V_r1 = sqrt(V_f1)^2 + (V_w1 - U)^2 V_r1 = sqrt((202.28 m/s))^2 + (500.69 m/s - 168.39 m/s)^2 V_r1 = sqrt((202.28)^2 + (332.30)^2) = sqrt(40917.1 + 110423.3) V_r1 = sqrt(151340.4) ≈ 389.02 m/s Inlet blade angle (_1): _1 = V_f1V_w1 - U = (202.28)/(332.30) _1 = (0.6087) ≈ 31.32^ Step 4: Calculate the components of the outlet velocity triangle. No axial thrust means V_f2 = V_f1 = 202.28 m/s. 15% loss of velocity due to friction means V_r2 = 0.85 V_r1. V_r2 = 0.85 × 389.02 m/s ≈ 330.67 m/s Outlet blade angle (_2): _2 = V_f2V_r2 = (202.28)/(330.67) _2 = (0.6117) ≈ 37.72^ Exit whirl velocity (V_w2): V_w2 = U - V_r2 _2 V_w2 = 168.39 m/s - (330.67 m/s × 37.72^) V_w2 = 168.39 - (330.67 × 0.7910) = 168.39 - 261.58 ≈ -93.19 m/s Absolute exit velocity (V_2): V_2 = sqrt(V_f2)^2 + V_w2^2 = sqrt((202.28 m/s))^2 + (-93.19 m/s)^2 V_2 = sqrt(40917.1 + 8684.4) = sqrt(49601.5) ≈ 222.71 m/s Absolute exit angle (_2): _2 = V_f2V_w2 = (202.28)/(-93.19) _2 = (-2.1706) ≈ -65.27^ The values to be entered onto the velocity diagram are: Blade velocity (U): 168.39 m/s Absolute inlet velocity (V_1): 540 m/s Absolute exit velocity (V_2): 222.71 m/s Relative inlet velocity (V_r1): 389.02 m/s Relative exit velocity (V_r2): 330.67 m/s Inlet whirl velocity (V_w1): 500.69 m/s Exit whirl velocity (V_w2): -93.19 m/s Inlet flow velocity (V_f1): 202.28 m/s Exit flow velocity (V_f2): 202.28 m/s Absolute inlet angle (_1): 22^ Absolute exit angle (_2): -65.27^ Inlet blade angle (_1): 31.32^ Exit blade angle (_2): 37.72^ 5.2 Answer the following questions pertaining to the velocity diagram drawn in QUESTION 5.1.1 above: 5.2.1 Calculate the power developed by the turbine if the mass flow rate of the gas is 1000 grams per second. Step 1: Convert the mass flow rate to kilograms per second. Mass flow rate (m) = 1000 g/s = 1 kg/s Step 2: Use the formula for power developed by an impulse turbine. P = m U (V_w1 - V_w2) Step 3: Substitute the calculated values and calculate the power. P = 1 kg/s × 168.39 m/s × (500.69 m/s - (-93.19 m/s)) P = 168.39 × (500.69 + 93.19) P = 168.39 × 593.88 P ≈ 99900.6 W P ≈ 99.90 kW 5.2.2 Calculate the kinetic energy that exits the blades. Step 1: Identify the mass flow rate and absolute exit velocity. Mass flow rate (m) = 1 kg/s Absolute exit velocity (V_2) = 222.71 m/s Step 2: Use the formula for kinetic energy per second (power). KE_exit = (1)/(2) m V_2^2 Step 3: Substitute the values and calculate the kinetic energy. KE_exit = (1)/(2) × 1 kg/s × (222.71 m/s)^2 KE_exit = (1)/(2) × 49601.5 J/s KE_exit ≈ 24800.75 W 5.3 Name THREE types of governors. Watt governor* Porter governor* Proell governor* 5.4 Calculate the height of a Watt governor rotating at a speed of 90 r/min. Step 1: Convert the rotational speed from revolutions per minute to radians per second. Speed (N) = 90 r/min Angular velocity () = (2 N)/(60) = (2 × 90)/(60) = 3 rad/s ≈ 9.4248 rad/s Step 2: Use the formula for the height (h) of a Watt governor. h = (g)/(^2) where g = 9.81 m/s^2 (acceleration due to gravity). Step 3: Substitute the values and calculate the height. h = 9.81 m/s^2(9.4248 rad/s)^2 h = (9.81)/(88.8269) h ≈ 0.1104 m h ≈ 110.4 mm Drop the next question.

A simple turbine schematic of the Parsons type. Steam pressure drops by a fraction of the total across each pair of rotating and fixed blades, the stators grow larger as pressure drops.

Need help with your own homework?

More Physics Questions

A simple turbine schematic of the Parsons type. Steam pressure drops by a fraction of the total across each pair of rotating and fixed blades, the stators grow larger as pressure drops.

Need help with your own homework?

More Physics Questions