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.

Diameter ($D$) = $460 \text{ mm} = 0.460 \text{ m}$
Rotational speed ($N$) = $116.267 \text{ rev/s}$

Gas leaves at $1944 \text{ km/h}$.

Nozzle angle ($\alpha_1$) = $22^\circ$
Inlet flow velocity ($V_{f1}$):

Inlet whirl velocity ($V_{w1}$):

Relative inlet velocity ($V_{r1}$):

Inlet blade angle ($\beta_1$):

No axial thrust means $V_{f2} = V_{f1} = 202.28 \text{ m/s}$.
15% loss of velocity due to friction means $V_{r2} = 0.85 V_{r1}$.

Outlet blade angle ($\beta_2$):

Exit whirl velocity ($V_{w2}$):

Absolute exit velocity ($V_2$):

Absolute exit angle ($\alpha_2$):

The values to be entered onto the velocity diagram are:

• Blade velocity ($U$): $168.39 \text{ m/s}$
• Absolute inlet velocity ($V_1$): $540 \text{ m/s}$
• Absolute exit velocity ($V_2$): $222.71 \text{ m/s}$
• Relative inlet velocity ($V_{r1}$): $389.02 \text{ m/s}$
• Relative exit velocity ($V_{r2}$): $330.67 \text{ m/s}$
• Inlet whirl velocity ($V_{w1}$): $500.69 \text{ m/s}$
• Exit whirl velocity ($V_{w2}$): $-93.19 \text{ m/s}$
• Inlet flow velocity ($V_{f1}$): $202.28 \text{ m/s}$
• Exit flow velocity ($V_{f2}$): $202.28 \text{ m/s}$
• Absolute inlet angle ($\alpha_1$): $22^\circ$
• Absolute exit angle ($\alpha_2$): $-65.27^\circ$
• Inlet blade angle ($\beta_1$): $31.32^\circ$
• Exit blade angle ($\beta_2$): $37.72^\circ$

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.

Mass flow rate ($\dot{m}$) = $1000 \text{ g/s} = 1 \text{ kg/s}$

5.2.2 Calculate the kinetic energy that exits the blades.

Mass flow rate ($\dot{m}$) = $1 \text{ kg/s}$
Absolute exit velocity ($V_2$) = $222.71 \text{ m/s}$

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.

Speed ($N$) = $90 \text{ r/min}$
Angular velocity ($\omega$) = $\frac{2 \pi N}{60}$

where $g = 9.81 \text{ m/s}^2$ (acceleration due to gravity).