Given P s 25 MW h 19812 m Required Power Solution P act 292 h 1000 P act 292

# Given p s 25 mw h 19812 m required power solution p

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Given: P s = 2.5 MW h = 1981.2 m Required: Power Solution: P act = 29.2 - h 1000 P act = 29.2 - 1981.2 m 1000 ( 3.281 ft 1m ) P act = 23.4196828 in Hg Power = P s ( Pact 29.92 ) Power = 2.5 MW ( 23.4196828 in Hg 29.92 ) PROBLEM 9 A 373 kW (500Hp) internal combustion engine has a brake mean effective pressure of 551.5 kPa at full load. What is the friction power if mechanical efficiency is 85%. Power = 1.956858523 MW P b = 321.1040678 kPa
Given: Brake Power = 373 kW = 500 Hp IP = 551.5 Hp ƞ m = 85% Required: Friction Power (FP) Solution: ƞ m = BP IP 0.85 = 500 Hp IP IP = 588.2352941 Hp FP = IP BP = 588.2352941 Hp 500 Hp REVIEW PROBLEM NO. 23 GAS TURBINE POWER PLANT/ HYDRO- ELECTRIC POWER PLANT Mea, Jane May M. 14-55383 Problem No. 1 Water is being heated by the exhaust gases from a gas turbine. The gases leave the gas turbine at 648 ̊ C and may be cooled to 148 ̊ C. The water enters the heater at 9 C. The rate of gas flow is 25 kg/s and the water flow is 31.5 kg/s. Assume that the mean specific FP = 88.23529412 Hp
heat of the gas and the water are respectively 1.088 and 4.27 kJ/kg- ̊ C. What is the available energy removed from the hot gases in kW? Take available sink temperature as 311 ̊ K. Given: T gt = 648 ̊ C T c = 148 ̊ C T h = 95 ̊ C m gf = 25 kg/s m w = 31.5 kg/s MEH g = 1.088kJ/kg- ̊ C MEH w = 4.27kJ/kg- ̊ C Sink Temperature= 311 ̊ C Required: Energy removed from hot gases, kW Solution: ΔQ = Q gas Q water Q gas = m g CΔT Q gas = (25 kg/s)(1.088 kJ/kg - ̊ C)(648-148) ̊ C Q gas = 73600 kW Q water = m w CΔT Q water = (31.5 kg/s)(4.27kJ/kg - ̊ C)(311-93) ̊ C
Q water = 7397.7558 kW ΔQ = 13600 kW – 7397.7758 kw ΔQ = 8345.6 kW Problem No. 2 In a gas turbine plant, the mass flow rate is 6.2 kg/s, the enthalpy at the combustor entrance is 250 kJ/kg and the enthalpy at the exit is 980 kJ/kg. What is the capacity of the combustor in kW? Given: m= 6.2 kg/s h 1 = 250 kJ/kg h 2 =980 kJ/kg Required: Capacity of the combustor, kW Solution: W= Δh = m(h 2 - h 1 ) = 6.2kg/s (980 kJ/kg - 250 kJ/kg) W= 4526 kW
Problem No. 3 A Brayton cycle has a compressor rated power of 451.3kw. If compressor efficiency is 78%, find the ideal compressor work. Given: η c = 78% W T = 451.3 kw Required: Ideal compressor work, Wc Solution: η c = W c /W t 0.78 = Wc/ 451.3 kW Wc = 352.014 kW Coro, Corinne S. 14 54200 Problem No. 4 A hydroelectric generating station is supplied from a reservoir of a capacity 6000000 m3 at a head of 170 m. Assume hydraulic efficiency of 80% and electrical efficiency of 90%. The fall in the reservoir level after a load of 15 MW has been supplied for 3 hours, if the area of the reservoir is 2.5 sq. km is closest to:
Given: h = 170 m Ƞ T = 0.80 Ƞ G = 0.90 Generating Output = 15 MW t = 3 hours A = 2.5 km 2 Required: Fall in the reservoir, H Solution: A = 2.5 km 2 A = 2.5 x 10 6 m 2 Generating Output = (w Q h) ( Ƞ T ) ( Ƞ G ) 15,000 kW = (9.81 kN m 3 x Q x 170 m) (0.80)(0.90) Q = 12.4922548 m 3 sec After 3 hours, Q = 12.4922548 m 3 sec (3 hours x 3600 sec 1 hour ) Q = 134,916.3519 m 3 Volume = Area x Height
134,916.3519 m 3 = (2.5 x 10 6 m 2 ) (H) H = 0.05396654074 m Problem No. 5 In hydro-electric plant having 50 sq. km reservoir area and 100 m head is used to generate power. The energy utilized by the consumers whose load is connected to the power plant during a five-hour period is 13.5 x 10 to the 6th power kW-hr. The overall generation efficiency is 75%. Find the fall in the height of water in the reservoir after 5- hour period.

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