411344922-turbine-seatwork-docx.pdf - 1 In Ma Cristina Hydro Electric project the available head is 140m the water flow is 1m3/s efficiency of the

# 411344922-turbine-seatwork-docx.pdf - 1 In Ma Cristina...

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1. In Ma. Cristina Hydro Electric project the available head is 140m the water flow is 1m 3 /s , efficiency of the turbine is 95% ; efficiency of the generator is 95%; three-phase, 60 cycles, voltage delivered is 4160 volts, the estimated yearly output of the generator in kw-hr. Given: Hnet = 140 meters Q = 1 ? 3 ? Ƞ ? = 95% Ƞ ? = 95% Gen. spec 3 phase 60 cycles 4160 volts Generation: yearly 365days x 24 ℎ???? 1 ?𝑎? = 8760 ℎ??. EP/ year = ? Ƞ ? = ?𝑃 𝑃𝑎?𝑎?? Ƞ ? = ?𝑃 ?𝑃 Ƞ ? = ?𝑃 Ƞ 𝑔 𝑃𝑎?𝑎?? EP = Ƞ ? Ƞ ? ?????? EP = (0.95)(0,95)(9.81 KN/ ? 3 )(1 ? 3 ? ) (140m) EP = 1239.49 x 8760 hrs. EP = 10857963.06 kw-hr 2. In problem number 1, if the cost of the generated electric energy is 0.60 per kw-hr, the estimated yearly income of the plant is ? Cost = (EP)(EC) Cost = 10859763.16 kw-hr ( 0.6 ??−ℎ? ) Cost = Php 6514777.84 3. In problem number 1, if the power factor is 0.8, the generator current is ? EP= √3 (EI) P.F 1239.49 𝐾?.? ? = √3 (4.160 KV) (I) (0.8) I = 214.94 AMP Subscribe to view the full document.

4.) An impulse turbine which has a diameter of 150 cm; speed of 350 rpm, the bucket angle ? 160 ° , coefficient velocity is equal to 0.98, relative speed is equal to 0.45, generator efficiency of 90%, friction factor is 0.9 and the jet diameter from a nozzle of 15 cm. The power input kw is. Given: D= 150 cm 𝜑 = 0.45 N= 350 rpm ? ???. = 90% ? = 160 ° k= 0.9 Cv = 0.98 Solution: = 𝜋?? √2?ℎ = 0.45 = 𝜋(1.5?)(350???/ min ? 1???/60sec) √2( 9.81𝑚 ? )(ℎ) - peripheral coefficient ??? − peripheral velocity √2?ℎ - velocity of the jet h = 190.192m vj= cv √2?ℎ vj = 0.98 √2( 9.81? ? )(190.192?) vj = 59.865m/s Q = AV Q = 𝜋 4 (1.5?) 2 (59.865 m/s) Q = 1.058 ? 3 /? P = ??ℎ P = 1.058 ? 3 /? (9.82m/s)(190.192m) P = 1973.82 kw or 1974 kw 5.) In problem no. 4 the kw power output is? V = ??? V = ?( 1.5m)(350rev/min x 1min/60 sec) V = 27.49 m/s Ft = ? Q( 1 − ???𝛽 1+? )(?? − ?) Ft = 1000kg/ ? 3 (1.058 ? 3 /? )(1- ???160° 1+0.9 )(59.865m/s-27.49m/s) Ft = 51193.31 kg.m/s Ft = 51.19 KN T = Ft r , r = ? 2 T = 51.19 KN ( 150 ?? 2 ? 1? 100?? ) T = 38.3925 KN.m P = 2?𝑇? P = 2 ? (38.3925 KN.m)(350rev/min x 1min/60 sec) P = 1407.15 kw P = 1410 kw 6. In problem number 4, the turbine efficiency is? Ƞ ? = ?? ?????? ? 100% Ƞ ? = 1407.15 ?? 1973.82 ?? ? 100% Ƞ ? = 71.29% 7. In problem number 4, the annual electrical gen. in MW-hr is ? Ƞ ? = ?𝑃 𝑃 𝑎?𝑎?? ; Ƞ ? = Ƞ ? ? ) Ƞo = 0.713(0.90) Ƞo = 0.6417 0.6417 = ?𝑃 1973.82 𝐾? EP = 1266.6 KW x 8760 hr EP = 11095418.58 KW-hr x 1 ?? Subscribe to view the full document. • Summer '16
• Robel Nomorosa

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