Pipe Problems (ALCORCON) - University of Cebu - Main Campus BSME ME415

PIPE PROBLEMS Find the work posses for a Helium gas at 200C *A. 609 KJ/kg B. 168 KJ/kg C. 229 KJ/kg D. 339 KJ/kg Solution: W = m R T = m (8.314 / M) T For helium, M = 4 W/m = (8.314/4)(20 + 273) = 609 KJ/kg Two k ...

PIPE PROBLEMS Find the work posses for a Helium gas at 200C *A. 609 KJ/kg B. 168 KJ/kg C. 229 KJ/kg D. 339 KJ/kg Solution: W = m R T = m (8.314 / M) T For helium, M = 4 W/m = (8.314/4)(20 + 273) = 609 KJ/kg Two kilogram of gas is confined in a 1 m3 tank at 200 kpa and 880C. What type of gas is in the tank? A. Helium *B. Ethane C. Methane D. Ethene Solution: P V = m R T 200 (1) = 2 (8.314/M)(88+273) M = 30 Therefore: the gas is Ethane (C2 H6) Find the enthalpy of Helium if its internal energy is 200 KJ/kg A.144 KJ/kg B. 223.42 KJ/kg *C. 333.42 KJ/kg D. 168 KJ/kg Solution: R = 8.314/4 = 2.0785 K = 1.667 for helium Cp = k R/(k - 1) = 1.667(2.0785)/(1.667 – 1) = 5.195 KJ/kg-K Cv = R/(k – 1) = 2.0785/(1.667 – 1) = 3.116 KJ/kg – K ∆h/∆U = Cp/Cv ∆h/200 = 5.195/3.116 ∆h = 333.42 KJ/kg Compute the mass of a 2 m3 propane at 280 kpa and 40˚C. A. 6.47 kg B. 5.1 kg C. 10.20 kg *D. 9.47 kg Solution: Propane is C3 H3--------------M = 12(3) + 8(1) = 44 PV = m R T 280(2) = m (8.314/44)(40 + 273) m = 9.47 kg Compute the air flow in ft3/min of mechanical ventilation required to exhaust an accumulation of refrigerant due to leaks of the system capable of revolving air from the machinery room for a mass of 4 lbs refrigerant. *A. 200 B. 210 C. 220 D. 230 Solution: Q = 100 x G0.5 ft3/min Q = 100 x (4)0.5 = 200 ft3/min Compute the free-aperture cross section in m2 for the ventilation of a machinery room if the mass of refrigerant is 9 kg. A. 0.314 *B. 0.414 C. 0.514 D. 0.614 Solution: F = 0.138 G0.5 m2 F = 0.138 (9)0.5 = 0.414 m2 A 29.53” x 39.37” pressure vessel contains ammonia with f = 0.041. Compute the minimum required discharge capacity of the relief device in kg/hr. A. 106.71 kg/hr B. 108.71 kg/hr *C. 110.71 kg/hr D. 112.71 kg/hr Solution: C = f D L, kg/s C = 0.041(29.53/39.37)(39.37/39.37) = 0.03075 kg/s (3600) = 110.71 kg/hr Compute the maximum length of the discharge pipe installed on the outlet of a pressure-relief device in feet for internal pipe diameter of 0.5 inch and rated discharge capacity is 8 lb/min of air. The rated pressure of relief valve is 16 psig. *A. 0.286 ft B. 0.386 ft C. 0.486 ft D. 0.586 ft Solution: P = Pg + Patm = 16 x 1.1 + 14.7 = 32.3 psia L = 9P2d5/16Cr2 = 9(32.3)2(0.5)5/16(8)2 = 0.286 ft A thermal power plant has a heat rate of 11,363 Btu/kw-hr. Find the thermal efficiency of the plant. A. 28% *B. 30% C. 34% D. 40% Solution: eth = 3412/Heat rate = 3412/11,363 = 30% What is the hydraulic gradient of a 1 mile, 17 inches inside diameter pipe when 3300 gal/min of water flow with f = 0.03. *A. 0.00714 B. 0.00614 C. 0.00234 D. 0.0018 Solution: v = (3300/7.481)/(π/4)(17/12)2(60) = 4.66 ft/s L = 1 mile = 5280 ft hL = fLv2/2_D = 0.03(5280)(4.66)2/2(32.2)(17/12) = 37.7 ft Hydraulic gradient = 37.7/5280 = 0.007.14 Find the loss of head in the pipe entrance if speed of flow is 10 m/s. A. 5.10 m B. 10.2 m C. 17.4 m *D. 2.55 m Solution: Loss at entrance = 0.5 (v2/2g) = 0.5 [102 / 2(9.81)] = 2.55 m Wet material, containing 220% moisture (dry basis) is to be dried at the rate of 1.5 kg/s in a continuous dryer to give a product containing 10% (dry basis) . Find the moisture removed, kg/hr *A. 3543.75 kg/hr B. 3513.75 kg/hr C. 3563.75 kg/hr D. 3593.75 kg/hr Solution: Solid in wet feed = solid in dried product [1/(1 + 2.2)](1.5) = [1/(1 + 0.1)](x) x = 0.5156 kg/s (total dried product) Moisture removed = 1.5 – 0.5156 = 0.984 kg/s = 3543.75 kg/hr Copra enters a dryer containing 70% moisture and leaves at 7% moisture. Find the moisture removed on each pound on solid in final product. A. 6.258 lb B. 1.258 lb C. 4.258 lb *D. 2.258 lb Solution: Solid in wet feed = solid in dried product 0.3x = 1 x = 3.333 lbs 1 = 0.93y y = 1.07527 lb Moisture removed = x – y = 3.333 – 1.07527 = 2.258 lb A 1 m x 1.5 m cylindrical tank is full of oil with SG = 0.92. Find the force acting at the bottom of the tank in dynes. A. 106.33 x 103 dynes B. 106.33 x 104 dynes C. 106.33 x 105 dynes *D. 106.33 x 106 dynes Solution: P = w h = (0.92 x 9.81) (1.5) = 13.5378 kpa F = PA = 13.5378(π/4 x 12) = 10.632 KN = 10,632.56 N x 10,000 dynes/N F = 106.33 x 106 dynes Find the pressure at the 100 fathom depth of water in kpag. *A. 1,793.96 kpag B. 1,893.96 kpag C. 1,993.96 kpag D. 1,693.96 kpag Solution: H = 100 fathom x 6 = 600 ft P = w h = (600/3.281)(9.81) = 1,793.96 kpag Find the depth in furlong of the ocean (SG = 1.03) if the pressure at the sea bed is 2,032.56 kpag. *A. 1 B. 2 C. 3 D. 4 Solution: P = w h 2,032.56 = (1.03 x 9.81) h h = 201.158 m x 3.281 ft/m x 1 yd/3ft x 1 furlong/220yd = 1 furlong Find the mass of 10 quartz of water. A. 10.46 kg *B. 9.46 kg C. 11.46 kg D. 8.46 kg Solution: V = 10 quartz x 1gal/4quartz x 3.785li/1gal x 1m3/1000li V = 0.0094625 x 10-3m3 w = m/V 1000 = m/0.0094625 x 10-3 m = 9.46 kg Find the mass of carbon dioxide having a pressure of 20 psia at 200˚F with 10 ft3 volume. A. 1.04 lbs B. 1.14 lbs *C. 1.24 lbs D. 1.34 lbs Solution: PV = m R T (20 x 144)(10) = m (1545/44)(200 + 460) m = 1.24 lbs Find the heat needed to raise the temperature of water from 30˚C to 100˚C with 60% quality. Consider an atmospheric pressure of 101.325 kpa. Use the approximate enthalpy formula of liquid. A. 293.09 KJ/kg B. 1,772.90 KJ/kg C. 1,547.90 KJ/kg *D. 1,647.29 KJ/kg Solution: At 100˚C ht = cp t = 4.187 (100) = 418.7 KJ/kg htg = 2257 KJ/kg h2 = h1 + x htg = 418.7 + 0.60(2257) = 1,772.9 KJ/kg Q = 1(4.187)(100 – 30) + 1(1772.9 – 418.7) = 1,647.29 KJ/kg Find the enthalpy of water at 212˚F and 14.7 psi if the dryness factor is 30%. Use the approximate enthalpy formula of liquid. A. 461 Btu/lb *B. 471 Btu/lb C. 481 Btu/lb D. 491 Btu/lb Solution: ht = (˚F – 32) = (212 – 32) = 180 Btu/lb htg = 970 Btu/lb h = ht + x htg h = 180 + 0.3(970) = 471 Btu/lb An air compressor consumed 1200 kw-hr per day of energy. The electric motor driving the compressor has an efficiency of 80%. If indicated power of the compressor is 34 kw, find the mechanical efficiency of the compressor. A. 117.65 % B. 75 % *C. 85 % D. 90 % Solution: P/m = 1200kw-hr/24 hrs = 50 kw BP = 50(0.80) = 40 kw em = 34/40 = 85 % A refrigeration system consumed 28,000 kw-hr per month of energy. There are 20 % of energy is lost due to cooling system of compressor and motor efficiency is 90 %. If COP of the system is 6, find the tons of refrigeration of the system. A. 43.15 TR B. 46.15 TR *C. 49.15 TR D. 41.15 TR Solution: P/m = 28,800/(24 x 30) = 40 kw BP = 40(0.90) = 36 kw Wc = 36(1 – 0.20) = 28.80 kw COP = RE/Wc 6 = RE/28.80 RE = 172.8/3.516 = 49.15 TR A 23 tons refrigeration system has a heat rejected of 100 kw. Find the energy efficiency ratio of the system. A. 13.42 *B. 14.42 C. 15.42 D. 16.42 Solution: QR = RE + Wc 100 = 23(3.516) + Wc Wc = 19.132 kw COP = RE/Wc = (23 x 3.516) / 19.132 = 4.32 EER = 3.412 COP = 3.412(4.23) = 14.42 A 200 mm x 250 mm, 8-cylinder, 4-stroke diesel engine has a brake power of 150 kw. The mechanical efficiency is 80%. If two of the cylinders were accidentally cut off, what will be the new friction power? A. 31.50 kw B. 33.50 kw C. 35.50 kw *D. 37.50 kw Solution: em = BP/IP 0.8 = 150/IP IP = 187.5 kw FP1 = IP – BP = 187.5 – 150 = 37.50 kw FP1 = FP2 = 37.50 kw If the energy efficiency ratio of the refrigeration system is 12.6, what is the COP of the system? *A. 3.69 B. 4.23 C. 5.92 D. 6.83 Solution: EER = 3.412 COP 12.6 = 3.412 COP COP = 3.69 An air compressor has a power of 40 kw at 4 % clearance. If clearance will increase to 70 %, what is the new power? A. 70 kw *B. 40 kw C. 53kw D. 60 kw Solution: The power of compressor will not be affected with the changes in clearance. Therefore the power will still be 40 kw. What is the approximate value of temperature of water having enthalpy of 208 Btu/lb? A. 138.67 ˚C *B. 115.55 ˚C C. 258.67 ˚C D. 68.67 ˚C Solution: h = ˚F – 32 Find the work posses for a Helium gas at 20°C. *A. 609 KJ/Kg B. 168 KJ/Kg C. 229 KJ/Kg D. 339 KJ/Kg Solution: W = m R T = m (8.314/M) T For Helium, M = 4 W/m = (8.314/4)(20 + 273) = 609 KJ/Kg Two kilogram of gas is confined in a 1 m3 tank at 200 kpa and 88˚C. What type of gas is in the tank? A. Helium *B. Ethane C. Methane D. Ethene Solution: P V = m R T 200 (1) = 2 (8.314/m)(88 + 273) M = 30 Therefore: the gas is Ethane (C2H6) Find the enthalpy of Helium if its internal energy is 200 KJ/kg. A. 144 KJ/kg B. 223.42 KJ/kg *C. 333.42 KJ/kg D. 168 KJ/kg Solution: R = 8.314/4 = 2.0785 K = 1.667 for helium cp = k R/(k – 1) = 1.667(2.0785)/(1.667 – 1) = 5.195 KJ/kg-K cv = R/(k – 1) = 2.0785/(1.667 – 1) = 3.116 KJ/kg-K ∆h/∆U = cp/cv ∆h/200 = 5.195/3.116 ∆h = 333.42 KJ/kg Compute the mass of a 2 m3 propane at 280 kpa and 40˚C. A. 6.47 kg B. 5.1 kg C. 10.20 kg *D. 9.47 kg Solution: Propane is C3H6------------------------M = 12(3) + 8(1) = 44 PV = m R T 280(2) = m (8.314/44)(40 + 273) m = 9.47 kg Compute the air flow in ft3/min of mechanical ventilation required to exhaust an accumulation of refrigerant due to leaks of the system capable of revolving air from the machinery room for a mass of 4 lbs refrigerant. *A. 200 B. 210 C. 220 D. 230 Solution: Q = 100 x G0.5 ft3/min Q = 100 x (4)0.5 = 200 ft3/min Compute the free-aperture cross section in m2 for the ventilation of a machinery room if the mass of refrigerant is 9 kg. A. 0.314 *B. 0.414 C. 0.514 D. 0.613 Solution: F = 0.138 G0.5 , m2 F = 0.138(9)0.5 = 0.414 m2 A 29.53” x 39.37” pressure vessel contains ammonia with f = 0.041. Compute the minimum required discharge capacity of the relief device in kg/hr. A. 106.71 kg/hr B. 108.71 kg/hr *C. 110.71 kg/hr D. 112.71 kg/hr Solution: C = f D L, kg/s C = 0.041(29.53/39.37)(39.37/39.37) = 0.03075 kg/s (3600) = 110.71 kg/hr Compute the maximum length of the discharge pipe installed on the outlet of a pressure-relief device in feet for internal pipe diameter of 0.5 inch and rated discharge capacity is 8lb/min of air. The rated pressure of relief valve is 16 psig. *A. 0.286 ft B. 0.386 ft C. 0.486 ft D. 0.586 ft Solution: P = Pg + Patm = 16 x 1.1 + 14.7 = 32.3 psia L = 9P2d5/16Cr2 = 9(32.3)2(0.5)5/16(8)2 = 0.286 ft A thermal power plant has a heat rate of 11,363 Btu/kw-hr. Find the thermal efficiency of the plant. A. 28 % *B. 30 % C. 34 % D. 40 % Solution: eth = 3412/heat rate = 3412/11,363 = 30% What is the hydraulic gradient of a 1 mile, 17 inches inside diameter pipe when 3300 gal/min of water flow with f…. 0.03 *A. 0.00714 B. 0.00614 C. 0.00234 D. 0.00187 Solution: v = (3300/7.481)/(π/4)(17/12)2(60) = 4.66 ft/s L = 1 mile = 5280 ft hL =fLv2/2gD = 0.03(5280)(4.66)2/2(32.2)(17/12) = 37.7 ft Hydraulic gradient = 37.7/5280 = 0.00714 Find the loss of head in the pipe entrance if speed of flow is 10 m/s. A. 5.10 m B. 10.2 m C. 17.4 m *D. 2.55 m Solution: Loss at entrance = 0.5 (v2/2g) = 0.5[102/ 2(9.81)] = 2.55 m Wet material, containing 220 % moisture (dry basis) is to be dried at the rate of 1.5 kg/s in a continuous dryer to give a product containing 10% (dry basis). Find the moisture removed, kg/hr. *A. 3543.75 kg/hr B. 3513.75 kg/hr C. 3563.75 kg/hr D. 3593.75 kg/hr Solution: Solid in wet feed = solid in dried product [1/(1 + 2.2)](1.5) = [1/(1 + 0.1)](x) x = 0.5156 kg/s (total dried product) Moisture removed = 1.5 – 0.5156 = 0.984 kg/s = 3543.75 kg/hr Copra enters a dryer containing 70% moisture and leaves at 7% moisture. Find the moisture removed on each pound of solid in final product. A. 6.258 lb B. 1.258 lb C. 4.258 lb *D. 2.258 lb Solution: Solid in wet feed = solid in dried product 0.3x = 1 x = 3.333 lbs 1 = 0.93y y = 1.07527 lb Moisture removed = x – y = 3.333 – 1.07527 = 2.258 lbs A 1 m x 1.5 m cylindrical tank is full of oil with SG = 0.92. Find the force acting at the bottom of the tank in dynes. A. 106. 33 x 103 dynes B. 106.33 x 104 dynes C. 106.33 x 105 dynes *D. 106.33 x 106 dynes Solution: P = w h = (0.92 x 9.81)(1.5) = 13.5378 kpa F = PA = 13.5378(π/4 x 12) = 10.632 KN = 10,632.56 N x 10,000 dynes/N F = 106.33 x 106 dynes Find the pressure at the 100 fathom depth of water in kpag. *A. 1,793.96 kpag B. 1,893.96 kpag C. 1,993.96 kpag D. 1,693.96 kpag Solution: H = 100 fathom x 6 = 600 ft P = w h = (600/3.281)(9.81) = 1,793.96 kpag Find the depth in furlong of the ocean (SG = 1.03) if the pressure at the sea bed is 2,032.56 kpag. *A. 1 B. 2 C. 3 D. 4 Solution: P = w h 2,032.56 = (1.03 x 9.81) h h = 201.158 m x 3.281 ft/m x 1 yd/3ft x 1 furlong/220yd = 1 furlong Find the mass of 10 quartz of water. A. 10.46 kg *B. 9.46 kg C. 11.46 kg D. 8.46 kg Solution: V = 10 quartz x 1gal/4quartz x 3,785 li/1gal x 1m3/1000 li V = 0.0094625 x 10-3m3 w = m/V 1000 = m/0.0094625 x 10-3 m = 9.46 kg Find the mass of carbon dioxide having a pressure of 20 psia at 200°F with 10 ft3 volume. A. 1.04 lbs B. 1.14 lbs *C. 1.24 lbs D. 1.34 lbs Solution: PV = m R T (20 x 144)(10) = m (1545/44)(200 + 460) m = 1.24 lbs Find the heat needed to raise the temperature of water from 30°C to 100°C with 60% quality. Consider and atmospheric pressure of 101.325 kpa. Use the approximate enthalpy formula of liquid. A. 293.09 KJ/kg B. 1,772.90 KJ/kg C. 1,547.90 KJ/kg *D. 1,647.29 KJ/kg Solution: At 100°C hf = Cp t = 4.187(100) 418.7 KJ/kg hfg = 2257 KJ.kg h2 = hf + xhfg = 418.7 + 0.60(2257) = 1,772.9 KJ/kg Q = 1(4.187)(100-30) + 1(1772.9 – 418.7) = 1,647.20 KJ/kg Find the enthalpy of water at 212˚F and 14.7 psi if the dryness factor is 30%. Use the approximate enthalpy formula of liquid. A. 461 Btu/lb *B. 471 Btu/lb C. 481 Btu/lb D. 491 Btu/lb Solution: hf = (˚F – 32) = (212 – 32) = 180 Btu/lb hfg = 970 Btu/lb h = hf + x hfg h = 180 + 0.3(970) = 471 Btu/lb An air compressor consumed 1200 kw-hr per day of energy. The electric motor driving the compressor has an efficiency of 80 %. If indicated power of the compressor is 34 kw, find the mechanical efficiency of the compressor. A. 117.65 % B. 75% *C. 85% D. 90% Solution: Pim = 1200kw-hr/24 hrs = 50 kw BP = 50(0.80) = 40 kw em = 34/40 = 85% A refrigeration system consumed 28,800 kw-hr per month of energy. There are 20 % of energy is lost due to cooling system of compressor and motor efficiency is 90%. If COP of the system is 6, find the tons of refrigeration of the system. A. 43.15 TR B. 46.15 TR *C. 49.15 TR D. 41.15 TR Solution: Pim = 28,800/(24 x 30) = 40 kw BP = 40 (0.90) = 36 kw Wc = 36(1 – 0.20) = 28.80 kw COP = RE/Wc 6 = RE/28.80 RE = 172.8/3.516 = 49.15 TR A 23 tons refrigeration system has a heat rejected of 100 kw. Find the energy efficiency ratio of the system. A. 13.42 *B. 14.42 C. 15.42 D.16.42 Solution: QR = RE + Wc 100 = 23(3.516) + Wc Wc = 19.132 kw COP = RE/Wc = (23 x 3.516) / 19.132 = 4.23 EER = 3.412 COP = 3.412(4.23) = 14.42 A 200 mm x 250 mm, 8-cylinder, 4-stroke diesel engine has a brake power of 150 kw. The mechanical efficiency is 80 %. If two of the cylinders were accidentally cut off, what will be the new friction power? A. 31.50 kw B. 33.50 kw C. 35.50 kw *D. 37.50 kw Solution: em = BP/IP 0.8 = 150/IP IP = 187.5 kw FP1 = IP – BP = 187.5 – 150 = 37.50 kw FP1 = FP2 = 37.50 kw If the energy efficiency ratio of the refrigeration system is 12.6, what is the COP of the system? *A. 3.69 B. 4.23 C. 5.92 D. 60 kw Solution: EER = 3.412 COP 12.6 = 3.412 COP COP = 3.69 An air compressor has a power of 40 kw at 4% clearance. If clearance will increase to 7%, what is the new power? A. 70 kw *B. 40 kw C. 53 kw D. 60 kw Solution: The power of compressor will not be affected with the changes in clearance. Therefore power will still be 40 kw. What is the approximate value of temperature of water having enthalpy of 208 Btu/lb? A. 138.67˚C *B. 115.67˚C C. 258.67˚C D. 68.67˚C Solution: h = ˚F – 32 208 = F – 32 ˚F = 240 ˚F = 115.55 ˚C Convert 750˚R to ˚K A. 390.33 ˚K B. 395.33 ˚K C. 410.33 ˚K *D. 416.33 ˚K Solution: ˚R = 1.8 ˚K 750 = 1.8 ˚K ˚K = 416.667 An otto cycle has a compression ratio of 8. Find the pressure ratio during compression. *A. 18.38 B. 16.38 C. 14.38 D. 12.38 Solution: P1V1k = P2V22 (V1/V2)k = (P2/P1) rkk = rp rp = (8)1.4 = 18.38 A diesel cycle has a cut off ratio of 2.5 and expansion ratio of 4. Find the clearance of the cycle. A. 9.11 % B. 5.55 % *C. 11.11 % D. 15.15 % Solution: rk = rc re rk = 2.5(4) = 10 rk = (1 + c)/c 10 = (1 + c)/c c = 11.11 % A dual cycle has an initial temperature of 30 ˚C. The compression ratio is 6 and the heat addition at constant volume process is 600 KJ/kg. If cut-off ratio is 2.5, find the maximum temperature of the cycle. A. 3638.50 ˚C *B. 3365.50 ˚C C. 3565.50 ˚C D. 3965.50 ˚C Solution: T2 = T1 rkk-1 = (30 +273)(6)1.4-1 = 620.44 ˚K QAV = m cv (T3 – T2) 600 = 1(0.7186)(T3 – 620.44) T3 = 1455.396 ˚K rc = T4/T3 2.5 = T4/1455.396 T4 = 3638.49 ˚K = 3365.50 ˚C A three stages air compressor compresses air from 100 kpa to 1000 kpa. Find the intercooler pressure between the first and second stage. A. 505.44 kpa B. 108.44 kpa C. 316.23 kpa *D. 215.44 kpa Solution: Px = (P12P2)1/3 Px = [(100)2(1000)]1/3 = 215.44 kpa A 10-stages air compressor compresses air from 100 kpa to 800 kpa. Find the intercooler pressure between 1st and 2nd stage. A. 282.84 kpa B. 113.21 kpa *C. 123.11 kpa D. 333.51 kpa Solution: Px = (P1s-1P2)1/s Px = [(100)10-1(1000)]1/10 = 123.11 kpa A 3-stages air compressor compresses air from 100 kpa to 700 kpa. Find the intercooler pressure between the 2nd and 3rd stage. *A. 365.88 kpa B. 375.88 kpa C. 385.88 kpa D. 395.88 kpa Solution: Px = (P12P2)1/3 Px = [(100)2(700)]1/3 = 191.28 kpa Px/P1 = Py/Px Py = Px2/P1 = (191.28)2/100 = 365.88 kpa Carnot cycle A, B and C are connected in series so that the heat rejected from A will be the heat added to B and heat rejected from B will be added to C, each cycle operates between 30 ˚C and 400 ˚C. If heat added to A is 1000 kw, find the work output of C. *A. 111.44 kw B. 549.78 kw C. 247.53 kw D. 141.89 kw Solution: e1 = e2 = e3 = (400 – 30)/(400 + 273) = 54.98 % e1 = W1/QA1 = (QA1 – QR1)/QA1 0.5498 = (1000 – QR1)/1000 QR1 = 450.22 = QA2 0.5498 = (450.22 – QR2)/450.22 By heat balance: Qgain = Qloss mw cp (tb- ta) = mg cpg (t2-t1) (0.30)(4.187)( tb – 15) = (0.5)(1.0717)(150 – 80) tb = 44.86 oC A 350 mm X 450 mm steam engine running at 280 rpm has an entrance steam condition of 2 Mpa and 230 oC and exit at 0.1 Mpa. The steam consumption is 2,000 kg/hr and mechanical efficiency is 85%. If indicated mean effective pressure is 600 Kpa, determine brake thermal efficiency. At 2 Mpa and 230 oC (Table 3): h1 = 2849.6 s1 = 6. 4423 At 0.1 Mpa: sf = 1.3026 hf = 417.46 sfg = 6.0568 hfg = 2258 hf2 = 417.46 KJ/kg A. 23.34% *B. 15.25% C. 14.16% D. 27.34% SOLUTION: VD = 2[3.1416/4 (0.35) 2 (0.45)(280/60)]= 0.4041 m3/sec Indicated Power = Pmi x VD = 600 x 0.4041 =242.45 KW Brake Power = IP (em) = 242.45 (0.85) = 206.08KW Brake Power 206.08 etb = BP/ ms (h1-hf2) =206.08/ (2000/3600)(2849.6 – 417.46)=15.25% A steam turbine receives 5,000 kg/hr of steam at 5 Mpa and 4000oc and velocity of 30 m/sec. It leaves the turbine at 0.06 Mpa and 85% quality and velocity of 15 m/sec. Radiation loss is 10,000 KJ/hr. Find he KW developed. At 5 Mpa and 400oC: h1 = 3195.7 KJ/kg s1 =6.6459 At 0.006 Mpa: hf = 151.53 hfg = 2415.9 A. 1273.29 B. 2173.29 *C. 1373.60 D. 7231.29 SOLUTION: h2 = hf + xhfg = 151.53 + 0.85(2415.9) = 2205.045 KJ/ kg KE1 = ½ m v2 = ½ (5,000/3600)(30)2 = 625 W = 0.625 KW KE2 = ½ m v2 = ½ (5,000/3600)(15)2 = 156.25 W = 0.15625 KW By energy balance: KE1 + mh1 = KE2 + mh2 + Q + W W = (KE1 – KE2) + m(h1-h2) – Q 5000 10,000 W = (0.625 – 0.156) + (3600)(3195.7 – 2205.045) – 3600 = 1373.60 KW A steam turbine with 85% stage efficiency receives steam at 7 Mpa and 550oC and exhausts as 20 Kpa. Determine the turbine work. At 7 Mpa and 550oC: h1 = 3530.9 Kj/kg s1 = 6.9486 At 20 Kpa (0.020 Mpa): sf = 0.8320 hf = 251.4 sfg = 7.0766 hfg = 2358.3 A. 1,117 KJ/kg B. 1,132 KJ/ kg C. 1,123.34 KJ/ kg *D. 1,054.95 KJ/kg SOLUTION: s1 = s2 = sf + xsfg 6.9486 = 0.8320 + x(7.0766) x = 0.8643 h2 = 251.40 + 0.8643(2358.3) = 2289.78 KJ/kg h1 - h2a eST = h1 – h2 0.85 =3530.9 – h2a 3530.9 – 2289.78 h2a = 2475.95 KJ/kg WT = h1 = h2a = 3530.9 – 2475.95 = 1,054.95 KJ/kg A steam turbine with 80% stage efficiency receives steam at 7 Mpa and 550oC and exhaust as 20 Kpa. Determine the quality at exhaust. At 7 Mpa and 550oC: h1 = 3530.9 Kj/kg s1 = 6.9486 At 20 Kpa (0.020 Mpa): sf = 0.8320 hf = 251.4 *A. 96.96% B. 76.34% C. 82.34% D. 91.69% SOLUTION: sfg = 7.0766 hfg = 2358.3 s1 = s2 = sf + sfg 6.9486 = 0.8320 + x(7.0766) x = 0.8643 h2 = 251.40 + 0.8643(2358.3) = 2289.78 KJ/kg nST = h1 – h2a h1 – h2 0.80 = 3530.9 – h2a 3530.9 – 2289.78 h2a = 2538.004 KJ/kg h2a = hf + x hfg 2538.004 = 251.40 + x (2358.3) x = 96.96% A 16,000KW geothermal plant has a generator efficiency and turbine efficiency of 90% and 80%., respectively if the quality after throttling is 20% and each well discharges 400, 000 kg/hr, determine the number of wells are required to produce if the charge of enthalpy if the change of enthalpy at entrance and exit of turbine is 500KJ/kg. A. 4 wells *B. 2 wells C. 6 wells D. 8 wells SOLUTION: WT = ms(h3 – h4) 16,000 = ms (500) 0.9(0.8) ms = 44.44 kg/sec ms = 160,000 kg/hr 160,000 = 0.20 mg mg = 800,000 kg/hr No. of wells = 800,000/400,000 = 2 wells A liquid dominated geothermal plant with a single flash separator receives water at 204oC. The separator pressure is 1.04 Mpa. A direct contact condenser operates at 0.034 Mpa. The turbine has a polytropic efficiency of 0.75. For a cycle output of 60 MW, what is the mass flow rate of the well-water in kg/s? At 204oC: hf = 870.51 KJ/kg At 1.04 Mpa: hf = 770.38 hfg = 2009.2 hg = 2779.6 sg = 6.5729 At 0.034 MPa: hf = 301.40 hfg = 2328.8 sf = 0.9793 sfg = 6.7463 *A. 2,933 B. 2,100 C. 1,860 D. 2,444 SOLUTION: h3 = hg at 1.04 MPa = 2779.6 KJ/kg Solving for h4: s3 = s4 = sf + xsfg 6.5729 = 0.9793 + x4(6.7463) x4 = 0.829 h4 = 301.4 + 0.829(2328.8) = 2232.3 KJ/kg WT = ms (h3 – h4) 60,000 = ms (2779.6 – 2232.3) 0.75 ms = 146.17 kg/sec Solving for x2: (h1 = h2) h1 = h2 = hf + xhfg 870.51 = 770.38 + x2(2009.2) x2 = 0..049836 ms = x mg 146.17 = 0.049836 mg mg = 2,933.06 kg/sec An engine-generator rated 9000 KVA at 80% power factor, 3 phase, 4160 V has an efficiency of 90%. If overall plant efficiency is 28%, what is the heat generated by the fuel. A. 18,800 KW B. 28,800 KW C. 7500 KW *D. 25,714 KW SOLUTION: Gen. Output = pf x KVA = 0.8 x 9000 = 7200 KW eoverall= Gen. Output Qg 0.28 = 7200/Qg Qg = 25,714.28 KW The indicated thermal efficiency of a two stroke diesel engine is 60%. If friction power is 15% of heat generated, determine the brake thermal efficiency of the engine. A. 43% *B. 45 % C. 36% D. 37% SOLUTION: ne = IP/ Qg 0.60 = IP/Qg IP = 0.60 Qg BP = IP- FP = 0.60Qg – 0.15Qg = 0.45Qg etb = BP/Qg = 0.45Qg/Qg = 45% A 305 mm x 457 mm four stroke single acting diesel engine is rated at 150 KW at 260 rpm. Fuel consumption at rated load is 0.56 kg/KW-hr with a heating value of 43,912 KJ/kg. Calculate brake thermal efficiency A. 10.53% B. 27.45% *C. 14.64% D. 18.23% SOLUTION: mf = 0.56 kg/KW-hr x 150 KW = 84 kg/hr = 0.0233 kg/sec Brake thermal efficiency = A waste heat recovery boiler produces 4.8 Mpa(dry saturated) steam from 104°C feedwater. The boiler receives energy from 7 kg/sec of 954°C dry air. After passing through a waste heat boiler, the temperature of the air is has been reduce to 343°C. How much steam in kg is produced per second? Note: At 4.80 Mpa dry saturated, h = 2796. A. 1.30 B. 0.92 *C. 1.81 D. 3.43 SOLUTION: hf = approximate enthalpy of feedwater hf = Cpt hf = 4.187(104) hf = 435.45 KJ/kg Heat loss = Heat gain m gc p(t 1 - t 2) = m s(h - h f) 7(1.0)(954 – 343) = ms(2796.0 – 436.45) m s = 1.81 kg/sec A diesel electric plant supplies energy for Meralco. During a 24-hour period, the plant consumed 240 gallons of fuel at 28°C and produced 3930 KW-hr. Industrial fuel used is 28°API and was purchased at P30 per liter at 15.6°C. What is the cost of the fuel be to produce one KW-hr? *A. P6.87 B. P1.10 C. P41.07 D. P5.00 SOLUTION: SG 15.6C = 141.5/(131.5 + 28) = 0.887 Density at 15.6°C = 0.887(1kg/li) = 0.887 kg/li SG 28C = 0.887[1-.0007(1 – 15.6)] = .879 Density at 28°C = 0.879(1 kg/li) = 0.879 kg/li V28C / V15.6C = SG15.6C / SG28C 240 / V15.6C = 0.887 / 0.879 V15.6C = 237.835 gallons x 3.785 li/gal = 900.21 li Cost = [(30)(900.21)] / 3930 = P6.87/KW-hr In a gas turbine unit, air enters the combustion chamber at 550 kpa, 277°C and 43 m/s. The products of combustion leave the combustor at 511 kpa, 1004°C and 180 m/s. Liquid fuel enters with a heating value of 43,000 KJ/kg. For fuel-air ratio of 0.0229, what is the combustor efficiency of the unit in percent? A. 70.38% B. 79.385% *C. 75.38% D. 82.38% SOLUTION: Heat supplied by fuel = mfQh = 0.0229(43,000) = 984.7 KJ/kg air Q = heat absorbed by fuel Q/m = Cp(T2 – T1) + ½(V22 – V12) Q/m = (1.0)(1004 – 277) + ½[(180) 2 –(43) 2]/1000 =742.28 KJ/kg air Combustor Efficiency = = 75.38% The specific speed of turbine is 85 rpm and running at 450 rpm. If the head is 20 m and generator efficiency is 90%, what is the maximum power delivered by the generator. A. 450.51 KW B. 354.52 KW C. 650.53 KW*D. 835.57 KW SOLUTION: NS = (N√HP)/h5/4 85 = (450√HP)/(20 x 3.281) 5/4 Hp = 1244.52 Generator Output = (1244.52 x 0.746)(0.9) = 835.57 KW In Francis turbine, the pressure gage leading to the turbine casing reads 380 Kpa. The velocity of water entering the turbine is 8 m/sec, if net head of the turbine is 45 m, find the distance from center of spiral casing to the tailrace. *A. 3.0 m B. 3.5 m C. 4.0 m D. 4.5m SOLUTION : h = V2/2g 45 = (380/9.81) + z + [82/(2 x 9.81)] z = 3 m A turbine has a mechanical efficiency of 93%, volumetric efficiency of 95% and total efficiency of 82%. If effective head is 40 m, find the total head. A. 48.72 m B. 40.72 m *C. 36.22 m D. 34.72 m SOLUTION: eT = emehev 0.8 = 0.93(eh)(.95) ηh = 0.9055 Total head = h eh = (40)(0.9055) = 36.22 m A Pelton type turbine has 25 m head friction loss of 4.5 m. The coefficient of friction head loss (from Moorse) is 0.00093 and penstock length of 80 m. What is the penstock diameter? *A. 1,355.73 mm B. 3,476.12 mm C. 6771.23 mm D. 1686.73 mm SOLUTION: h =25- 4.5 = 20.5 v = √(2gh) = [(2 x 9.81 x 20.5)1/2] = 20.55 m/sec hL = (2fLv2)/gD 4.5 = (2)(0.00093)(80)(20.055)2 / 9.81D D = 1,355,730 m = 1,355.73 mm In an 9,000 KW hydro-electric plant the over-all efficiency is 88% and the actual power received by the customer is 110,000 KW-hrs for that day. What is the secondary power could this plant deliver during the entire day? A. 58,960 KW-hrs *B. 80,080 KW-hrs C. 65,960 KW-hrs D. 70,960 KW-hrs SOLUTION: Plant Capacity = 9,000(0.88)(24) = 190,080 KW-hrs Secondary Power = 190,080 – 110,000 = 80,080 KW-hrs A Pelton type turbine was installed 30 m below the gate of the penstock. The head loss due to friction is 12 percent of the given elevation. The length of penstock is 100 m and coefficient of friction is 0.00093. Determine the power output in KW. ( Use Moorse equation) A. 22,273 B. 23,234 C. 32,345 *D. 34,452 SOLUTION: hL = 0.12(30) = 3.6 m h = 30 – 3.6 = 26.40 m v = (2gh)1/2 = [(2)(9.81)(26.4)]1/2 = 22.759 m/sec hL= (2fLv2)/gD 3.6 = (2 x .00093 x 100 x 22.759) / (9.81D) D = 2.728 m Q = A x v = [ 2](22.759) = 133.03 m3/sec Power = w Q h = 9.81(133.03)(26.4) = 34,452 KW Water flows steadily with a velocity of 3.05 m/s in a horizontal pipe having a diameter of 25.4 cm. At one section of the pipe, the temperature and pressure of the water are 21C and 689.3 Kpa, respectively. At a distance of 304.8 m downstream A hydro electric plant having 30 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 106 kwh. The overall generation efficiency is 75%. Find the fall in the height of wa