Thermodynamics lecture notes, Exercises of Thermodynamics

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Ph Cran ere -D. 59,465 on) zi “Mass Flow Rate = Density x Volume flow Rate : a m= 1000Q Le Solving for Q : j 4 ‘ Py = Qh HB] 100 m 50,000 = | 9.81Q( 100) thus; , “ ; _@ Q = 50.968 m*/s = 50,968 kg/s fe oj 2, Past ME Board Problem A reaction turbine develops 500 BHP. enters at 20 fps with a 100 ft pressure head. The elev the tailwater level is 10 ft. Find the effective head. Flow through the turbine is 50 cfs. Water ation of the turbine above A. 130.2 ft C. 110.2 ft B. 120.2 ft D. 116.2 ft Ss Pp Vv 2 ay 2 hb = nad mre Vy = 10 fps wtp 2 (20)? - (0) = 100+10 + 2 (32.2) thus; ma =" 116.2 ft yet... (0.35) 9.81 2(9.81) = 28.039 m then; substituting : 7(0.55)(520/60) e———————_—_—_—_ /2(9,81)(28.039) thus; ; 7 > =0.638 ~ % Past ME Board Problem ving 50 sq. km reservoir area and 100 m head is i e energy utilized by the consumers whose !0a th five hour period is 13.5 x 10° KW-hr. - Find the fall in the height of water in C_322.10m D. 0.53 m fe ao ed by a turbine working under. t a sh ic ired from the water to the runner cal efficiency of 95%, what is the discharge throug 135i D. 1.234 wee Since the given power of kW is the shaft power, a hydraulic efficiency of ~ Say, 87% will be assumed : Shaft Power = ‘y Qhem en 45 = 9.81 Q (40)(0.95)(0.87) thus; = Q = 0.135 m/s Alternate Solution : Consider that the given energy is 350 J/kg of water: J 1k) m2 ki 45 350— Oi) 0 | islam) ° s {0 s) T0195 m thus; 3 7Q= 0135 7 Z Ss se 7 Past ME Board Problem Scope neal “il IS installed on a Francis turbine and the total head to i is 5 m/s, Th sla Casing at the inlet is 38 m and velocity of water at the inlé wane discharge is 2.1 m?/s. The hydraulic efficiency is 0.87 and over-al and 1.5 m/s re, ae velocities at the inlet and exit of the draft tube are 5 We of the spiral enh ively. The top of the draft tube is 1 m below the center ing draft tube, The Ng while the tailrace (water) level is 3 m from the top of Hi "and leakage | Te Is no velocity of whirl at either top or bottom of draft tu ? SSse$ are negligible, What is the power output of the turbine in k B. 632, C. 901.3 kw mY D. 832.6 kW y 7 el = 38+ (143) 4 ic ~(1.5) ~ 2.81) = 43.16m A substituting : R= Aa 81(2.1) ( 43. 16)(0. e) ew =P, = 746.9 kW % ‘ "Past ME Board Problem A hydro-electric generating station is supplied from a reservoir of capacity 6,000,000 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 hrs, if the area of the reservoir is 2.5 km? is closest to: A 5.39 cm C. 5.98.cm B. 4.32 cm D. 4.83 cm Leta Volume = Area x Height Solving for the Volume, (V) : 15,000 = 9.81 Q( 170 ( 0.80 )( 0.90) Q = 12.492 m/s "1h 3 hrs, volume of water consumed: then; «134,914 = (25x 108)h thus; . h =5.39cm \ -V = 12,492 (3600)(3) = 134,914 m j then; » substituting Yearly Income = 10,857,963 3(0. 60) @ Yearly Income = P 6,514,777.80 5 | | | | | | | 6,514, 777.80 6, 7,514,777 80 By , a eines ic energy Is P 0.60 per kW: fs ae Cp 8,514,777.80 ~D. P9,514,777.80 fearly Income = yearly generator ene x P0.60 iy pat for the yearly generator Output : cen = ¥ Qh & ec ( 8760 ) = 9.81(1)(140)(0.95)(0.95)(8760) = 10,857,963 kWH ’ ipgeategiay . Past ME Board Problem Water flows steadily with a velocity of 3.05 m/s in a horizontal pipe having a . diameter of 15.24 cm. At one section of the pipe, the temperature and pressure of the water are 21°C and 689.3 kPa respectively. The pressure in the other section is 516.9 kPa, and the length of the pipe is 304. Bn m. What is the friction factor? Note: Ps = 516.9 kPa ,L = 304.8 m A. 0.156 B. 0.0091 Sletten, 2 h, = fv 2gD Solving for fh, h. = Pg =Ps Y / — 689.3-516.9 9.81 = 17.574m C. 0.0185 D. 0.781 age ead on the turbine is 150 m while of 80%. The total gross h t of head du “an ig 4% of the gross head. The runaway speed jg “of head due to friction Is © 7. ecards Determine the flow of ken | hee ts | | ie [Pe | SLitane eae Generator Output = 7 Qh e& gen : Solving for h : h = hy-h : = 150 - 0.04(150) = 144m : then; substituting : 20,000 Q 9.81 Q ( 144) (0.80 ) ( 0.96 ) 18.435 m/s thus; = Q = 650.94 ft/s " 13. Past ME Board Problem driving @ From a height of 65 m, water flows at the rate of 0.85 m?/s and is P water turbine connected to an electric generator revolving at 160 rpm. calc a Power developed by the turbine in kW if the total resisting torque due : Tiction is 540 N-m and the velocity of the water leaving the turbine plades ADS mls. EN 523.42 C. 453.12 BL 543.21 D. 435.21 ting: oa er = 9.81(49.94)(25.5) er = 12,492.74 kw 0 days per year, find the annual energy in kW-hr that the the hydraulic turbi ine that will be used has an efficiency efficiency of 92%, Consider a headwork loss of 4 % of i _ C. 54,786,518 D. 48,576,185 * Output, KW:) ( 24 hrs/day ) ( 350 days/yr ) ing for the generator Output: WS Dw — Heyy = 600~4go y~ RONG Ga. rt ast 9.81 (11)(115.2)(0.80)(0.92) 9,149.39 kW Output = I (9,149.39)(24)(350) P 76,854,851 kW-hr Tae ' ale : j ” », Past ME Board Problem =) ‘ | _ Aremote community in Mountain Province plans to put a small hydroelectric - plant to service six closely-located barangays estimated to consume 52,650,000 kW-hrs per annum. Expected flow of water is 1665 m?/min. The most favorable leeation for the plant fixes the tailwater level at 480 m. The manufacturer of turbine generator set have indicated the following performance’ data: turbine is 92%, loss in head work not exceed 3.8% efficiency 87%, generator efficiency of available head. In order to pinpoint the most suitable area for the dam , determine the headwater elevation: A. 508.67 m ¢, 750.08 m 8. 605.87 m D. 875.06 m | Shin } Daw = Ag + Hew = h, + 480 i Solving for hg: _ Annual energy output = y Qh ecegen X 8760 52,650,000 = 9.81(1665/60) h (0.87)(0-92)(8760) h = 27.583™ hg = ‘h + fe io hy = 27383 + 0.038M, ees 28.67 Mm ock iS | friction in the tu draulic efficiency. 45 % ~ 650-65 = 0.8547 or 85.47 % 18, Past ME Board Problem The difference in elevation between the source of the water supply and the | centerline of the base of the nozzle of a Pelton wheel is 1300 ft. During the test the pressure at the end of the nozzle was 500 psig when the flow is 45 cfs. Inside diameter of penstock is 25 in. Compute the water horsepower at the base of the nozzle. A. 5904.76 Hp C. 7459.60 Hp B. 4509.67 Hp D. 6945.70 Hp P | Sloitinn Py = 7Qh Solving for h : | 2 ; ] ees 2) : 4\12 | ‘ | = 3.41 ft? Me Qh cy Pe : A 3.41 ft PS | ft—lbp ; : Wy 25 = 3247,62048 44, ‘eat “Py = 5,904.76 Hp het " le 42.71 = 1156.56 ft sf] (1156.56 ft) Past ME Board Problem 2112.34 Hp . 2506.34 Hp > A pelton wheel runs at a constant speed under a head of 650 sectional area of the jet is 0.50 ft? and the nozzle friction loss is to Suppose the needle of the nozzle is to be adjusted as to reduce the area jet from 0.50 ft? to 0.20 ft? . Under these conditions the efficiency of the WM _|s known to be 70% . Find the power output of the wheel. C. 3017.62 Hp D. 3462.74 Hp ft. The cross neglected: be neg! of the Re d total efficiencies of a turbine are 94%, anical volumetric an (Cine cHEREVE RS Sa a tively. Calculate the total head if the et oo €. 35.44m By D. 53.44 m ~ Total head = eh ‘ Solving for ey: ans * ah ae ea i 0.82 = 0.94(e,)(0.96) & = 0.9087 or 90.87 % thus; : Total Head = 0.9087 (50) : =, Total Head = 45.43 m i 22. Past ME Board Problem | a A proposed hydro-electric Power plant has the following data: Elevation of normal headwater surface 194m _ Elevation of normal tailwater surface 60m Loss of head due to friction 6.5m Turbine discharge at full gate opening 5 m/s Turbine efficiency at rated Capacity 90 % . . ‘ © turbine is to be connected to a 60 cycle AC generator. Calculate turbine brakepower output. é A. 6852.5 kW C. 5,268.5 kw B, gost kW D. 5,628.5 kW z turbine Output = 5,628.49 kW me * e past ME Board Problem mA hydro-electric power plant consumes 60,000,000 kW-hr per year . What is the net head if the expected flow is 1500 m?/min and see ainannctneh is 63%. A. 34.34m ' C. 44.33.m B. 43.43 m : 2. D. 33.44m Py = yQh Solving for water power, Pw : _ GeneratorO utput Oa) oa pa Water Power Be. 001000,0008768 Pw P, = 10,871.93 kW then; substituting : 10,871.93 = ai 1) h 9.81 60 thus; Fh = 44.33m Velocity Head 1.28 ft H,0 Total Dynamic Head = 129.72 +9,73+1.28 = 140.73 ft thus; 5 aula Bireconer Developed = ©2-A44-SX140.73N0192). 1 ogi . 550 ve ; ra HPturbine = 650 Hp .Y & | ] 4, Past ME Board Problem te ‘t River flows at 30 m?/s and develops a total brake power of 5 MW. What is the ee ] available head if it is proposed to install two turbines with 85% efficiency each? ae ‘A. 19.99 m C. 16.75 m Stet B. 25.26 m D. 23.18 m Rey Py = yQh Solving for water power Py : : | a) p. . 5000 = 2 = 5,882.35 kW Boss °° * Wi, = 15.MWor1,500 KW a3 8 Past ME Board Problem The over-all-efficiency of a 10 MW hydro -electric plant is 85%, What js the secondary power could this Plant deliver if the actual Power received by the | _ Customer for that day is 100,000 kw-hr? | A. 110,000 kW-hrs C. 124,000 kW-hrs B. 104,000 kW-hrs . _D. 130,000 kW-hrs bee 10,000 ( 0.85 24 ) 204,000 kW-hrs 204,000 ~ 100, 099 u I 7] 104,000 kW-hrs