Question 4.28: The turbine–generator system for Example Problem 4.27 is ill...

The turbine–generator system for Example Problem 4.27 is illustrated in Figure EP 4.24, where the required hydraulic power to be extracted by the reaction turbine is 800 kW. The electric power output by the generator is 700 kW, and the rotating shaft of the turbine has an angular speed of 350 rad/sec and a shaft torque of 2200 N-m. (a) Determine the shaft power delivered to the generator by the turbine. (b) Determine the turbine efficiency. (c) Determine the generator efficiency. (d) Determine the turbine–generator system efficiency.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

(a) The shaft power delivered to the generator by the turbine, (P_{t})_{out} = ωT_{shaft,out} = (P_{g})_{in} = ωT_{shaft,in} defined for Equation 4.174\eta _{generator}= \frac{electricalpoweroutputfromgenerator}{shaftpowerinputfromturbine} = \frac{\overset{\cdot }{W}_{elect,out} }{(\overset{\cdot }{W}_{shaft,in} )_{e}} = \frac{P_{out of gen}}{P_{in to gen}} = \frac{(P_{g})_{out}}{(P_{g})_{in}} is applied as follows:

w: = 350 \frac{rad}{sec}                                                T_{ shaftout }: = 2200 N.m

P_{tout} : = w. T_{ shaftout } = 770 kW                  P_{gin} : = P_{tout} = 770 kW

which is illustrated in Figure EP 4.28.
(b) The turbine efficiency defined in Equation 4.167 \eta _{turbine}= \frac{shaft power}{hydraulic power} = \frac{\overset{\cdot }{W}_{turbine} }{(\overset{\cdot }{W}_{turbine} )_{e}} = \frac{\overset{\cdot }{W}_{shaft,out} }{(\overset{\cdot }{W}_{turbine} )_{e}} = \frac{wT_shaft,out}{(\overset{\cdot }{W}_{turbine} )_{e}} = \frac{P_{out of turbine}}{P_{in of turbine}} = \frac{(P_{t})_{out}}{(P_{t})_{in}} is applied as follows:

P_{tin} : = 800 kW                            \eta _{ turbine} : = \frac{P_{tout} }{P_{tin}} = 0.963

Thus, the turbine is 96.3% efficient.
(c) The generator efficiency defined in Equation 4.174 \eta _{generator}= \frac{electricalpoweroutputfromgenerator}{shaftpowerinputfromturbine} = \frac{\overset{\cdot }{W}_{elect,out} }{(\overset{\cdot }{W}_{shaft,in} )_{e}} = \frac{P_{out of gen}}{P_{in to gen}} = \frac{(P_{g})_{out}}{(P_{g})_{in}} is applied as follows:

P_{gout}: = 700 kW                              \eta _{ generator} : = \frac{P_{gout} }{P_{gin}} = 0.909

Thus, the generator is 90.9% efficient.
(d) The turbine–generator system efficiency defined in Equation 4.176 \eta _{turbine – generator}= \frac{electricalpower}{hydraulic power} = \frac{\overset{\cdot }{W}_{elect,out} }{(\overset{\cdot }{W}_{turbine} )_{e}} = \frac{(P_{g})_{out}}{(P_{t})_{in}} = \eta _{turbine} \eta _{generator} is applied as follows:

\eta _{turbine generator} : = \frac{P_{gout} }{P_{tin}} = 0.875
\eta _{turbine generator} :\eta _{ turbine} . \eta _{ generator} = 0.875

Thus, the turbine–generator system is 87.5% efficient.

Related Answered Questions