Orlík hydroelectric plant

The Orlík hydroelectric power plant, built in 1954-1961, consists of four Kaplan turbines. For each of them, the water with a flow rate of Q = 150 m3/s is supplied with a flow rate of h = 70.5 m at full power.
a) What is the total installed power of the power plant at efficiency n = 87%?
b) For the fast charging of electric vehicles overnight, power consumption up to P0 = 11kW is required. How many would electric cars cover power plant power?
c) How many days of full power operation corresponds to the delivered energy E = 398GWh

Result

P =  361.018 MW
n =  32820
d =  45.9

Solution:

h=70.5 m Q=150 m3/s p=87/100=87100=0.87 n=4 g=9.81 m/s2 w=1000 kg/m3  E=m g h P1=Q w g h=150 1000 9.81 70.5=103740750 W P11=P1/106=103740750/106103.7408 MW P=p n P11=0.87 4 103.7408361.0178=361.018  MW h = 70.5 \ m \ \\ Q = 150 \ m^3/s \ \\ p = 87/100 = \dfrac{ 87 }{ 100 } = 0.87 \ \\ n = 4 \ \\ g = 9.81 \ m/s^2 \ \\ w = 1000 \ kg/m^3 \ \\ \ \\ E = m \cdot \ g \cdot \ h \ \\ P_{ 1 } = Q \cdot \ w \cdot \ g \cdot \ h = 150 \cdot \ 1000 \cdot \ 9.81 \cdot \ 70.5 = 103740750 \ W \ \\ P_{ 11 } = P_{ 1 }/10^6 = 103740750/10^6 \doteq 103.7408 \ MW \ \\ P = p \cdot \ n \cdot \ P_{ 11 } = 0.87 \cdot \ 4 \cdot \ 103.7408 \doteq 361.0178 = 361.018 \ \text{ MW }
P0=11/1000=111000=0.011 MW n=P/P0=361.0178/0.01132819.8182=32820P_{ 0 } = 11/1000 = \dfrac{ 11 }{ 1000 } = 0.011 \ MW \ \\ n = P/ P_{ 0 } = 361.0178/ 0.011 \doteq 32819.8182 = 32820
E=398 GWh  P2=P/1000=361.0178/10000.361 GW d=E/(24 P2)=398/(24 0.361)45.9349=45.9E = 398 \ GWh \ \\ \ \\ P_{ 2 } = P/1000 = 361.0178/1000 \doteq 0.361 \ GW \ \\ d = E/(24 \cdot \ P_{ 2 }) = 398/(24 \cdot \ 0.361) \doteq 45.9349 = 45.9



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