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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. 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. ∆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. Solution: F = 0.138 G0.5 m F = 0.138 (9)0.5 = 0.414 m 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. 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. 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/(1 + 0.1) 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. 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. 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 10-3m w = m/V 1000 = m/0.0094625 x 10- 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. 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. Solution: QR = RE + Wc 100 = 23(3.516) + Wc Wc = 19.132 kw COP = RE/Wc = (23 x 3.516) / 19.132 = 4. EER = 3.412 COP = 3.412(4.23) = 14. 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
hL =fLv2/2gD = 0.03(5280)(4.66)2/2(32.2)(17/12) = 37.7 ft Hydraulic gradient = 37.7/5280 = 0. 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/(1 + 0.1) 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. 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 10-3m w = m/V 1000 = m/0.0094625 x 10- 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. 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. Solution: QR = RE + Wc 100 = 23(3.516) + Wc Wc = 19.132 kw COP = RE/Wc = (23 x 3.516) / 19.132 = 4. EER = 3.412 COP = 3.412(4.23) = 14. 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. 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. 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. Solution: P1V1k = P2V (V1/V2)k = (P2/P1) rkk = rp rp = (8)1.4 = 18. 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/T 2.5 = T4/1455. 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/ 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
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. At 0.034 MPa: hf = 301.40 hfg = 2328.8 sf = 0.9793 sfg = 6. *A. 2,933 B. 2,100 C. 1,860 D. 2, 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. h4 = 301.4 + 0.829(2328.8) = 2232.3 KJ/kg WT = ms (h3 – h4) 60,000 = ms (2779.6 – 2232.3) 0. ms = 146.17 kg/sec Solving for x2: (h1 = h2) h1 = h2 = hf + xhfg 870.51 = 770.38 + x2(2009.2) x2 = 0.. 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, 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. 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. SOLUTION: SG 15.6C = 141.5/(131.5 + 28) = 0. Density at 15.6°C = 0.887(1kg/li) = 0.887 kg/li SG 28C = 0.887[1-.0007(1 – 15.6)] =. 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. 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(T 2 – T 1 ) + ½(V 22 – V 12 ) 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 KWD. 835.57 KW SOLUTION: NS = (N√HP)/h5/ 85 = (450√HP)/(20 x 3.281) 5/ Hp = 1244. 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. 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. 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, 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, 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 = 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. cm. At one section of the pipe, the temperature and pressure of the water are 21C and 689. 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 water in the reservoir after the 5-hour period. A. 5.13 m B. 1.32 m C. 3.21 *D. 2.20 m SOLUTION Energy Output = Power x time = (w Q h) x time 13.5 x 106 = 9.81(Q)(100)(0.75)(5) Q = 3669.725 m3/s Volume after 5 hrs = 3669.725(5 x 3600) = 66,055,050 m Volume = A x height 66,055,050 = (30 x 106) h H =2.202 m The gas density of chimney is 0.75 kg/m3 and air density of 1.15 kg/m3. Find the driving pressure if the height of chimney is 63.71 m. A. 0.15 kpa *B. 0.25 kpa C. 0.35 kpa D. 0.45 kpa SOLUTION: hw = H(da – dg) = 63.71(1.15 – 0.75) (0.00981) = 0.25 kpa The actual velocity of gas entering in a chimney is 8 m/sec. The gas temperature is 25C with a gas constant of 0.287 KJ/kg-K. Determine the gas pressure for a mass of gas is 50,000 kg/hr and chimney diameter of 1.39m. A. 95 kpa *B. 98 kpa C. 101 kpa D. 92 kpa SOLUTION:
SOLUTION:
W = [(P2/P1)n-1/n – 1] 19.57 = 1.35(P1)(0.1)/(1.35-1)[(5)1.35-1/1.35 – 1] P1 = 98 KPa The initial condition of air in an air compressor is 98 KPa and 27C and discharge air at 450 KPa. The bpre and stroke are 355 mm and 381 mm, respectively with percent cleared of 8% running at 300 rpm. Find the volume of air at suction. A. 541.62 m3/hr B. 551.62 m3/hr C. 561.62 m3/hr *D. 571.62 m3/hr SOLUTION: ev = 1 + c – c(P2/P1)1/n = 1 + 0.08 - 0.08(450/98)1/1.4 = 0. VD = D2 LN = (0.355)2 (0.381)(300/60) = 0.1885 m3/sec V1 = 0.1885(0.842) = 0.15878 m3/sec = 571.62 m3/hr An air compressor has a suction volume of 0.35 m3/sec t 97 KPa and discharges to 650 KPa. How much power saved by the compressor of there are two stages? A. 18.27 KW B. 16.54 KW C. 13.86 KW *D. 11.58 KW SOLUTION: W = [(P2/P1)n-1/n – 1] = (1.4 x 97 x 0.35)/(1.4 -1) [(650/97)1.4-1/1.4 – 1] = 85.79 KW For two stages : Px = (P1P2)1/2 = (97 x 650)1/2 = 251.097 KPa W = [(Px/P1)n-1/n – 1] = 2(1.4)(97)(0.35)/(1.4 – 1) [(251.0.97/97)1.4-1/1.4 – 1] = 74.208 KW POWER SAVED = 85.79 – 74.208 = 11.582 KW A twop stage air compressor has an intercooler pressure of 4 kg/cm2. What is the discharge pressure if suction pressure is 1 kg/cm2? A. 3 kg/cm2 B. 9 kg/cm2 C. 12 kg/cm2 *D. 16 kg/cm SOLUTION: Px = (P1P2)1/ Px2 = P1(P2) 42 = 16 kg/cm A two stage air compressor compresses air at 100 KPa and 22C discharges to 750 KPa. If intercooler intake is 105C. Determine the value of n. A. 1.400 *B. 1.325 C. 1.345 D. 1. SOLUTION: Px = (100 x 750)1/2 =273.86 KPa Tx/T1 = (Px/P1)n-1/n (105 + 273)/(22 + 273) = (273.86/100)n-1/n 1.281 = (2.6268)n-1/n n = 1. A single acting compressor has a volumetric efficiency of 89%, operates at 500 rpm. It takes in air at 900 KPa and 30C and discharges it at 600 KPa. The air handled is 8 m3/min measured at discharge condition. If compression is isentropic, find mean effective pressure in KPa.
*A. 233.34 B. 973.17 C. 198.34 D. 204.
SOLUTION:
P1V1K = P2V2K
100(V11.4) = 600(6)1.
V1 = 28.768 m3/min VD = 28.768/0.89 = 32.32 m3/min W = n P1V1/n-1 x [(P2 / P1)n-1/n – 1] = [(1.4 x 100 x 32.32)/(1.4 – 1)] x [(600/100)1.4-1/1.4 – 1] W = 7562.19 KJ/min W = Pm x Vd 7562.19 = Pm x 32. Pm = 233.34 KPa A water-jacketed air compressed handles 0.343 m3/s of air entering at 96.5 KPa and 21C and leaving at 460 KPa and 132C; 10.9 kg/h of cooling water enters the jacket at 15C and leaves at 21C. Determine the compressor brake power. A. 26.163 KW *B. 62.650 KW C. 34.44 KW D. 19.33 KW SOLUTION: T2/T1 = (P2/P1) n-1/n (132+273) / (21+273) = (480/96.5)n-1/n n = 1. W = (1.249 x 96.5 x 0.343) / (1.249-1) [(480 / 96.5)1.249-1/1.249 – 1] W = 62.57 KW Q = heat loss = mcp(t2 – t1) = (10.9/3600)(4.187)(21 – 15) 0.075 KW Brake power = W + Q = 62.57 + 0.076 = 62.65 KW A double suction centrifugal pumps delivers 20 ft3/sec of water at a head of 12 m and running at 650 rpm. What is the specific speed of the pump? A. 5014.12 rpm B. 6453.12 rpm *C. 2770.73 rpm D. 9966.73 rpm SOLUTION: N = N(Q)1/2 / h3/ Q = 20/2 ft3/sec x 7.481 gal/ft3 x 60 sec/1min = 4,488.6 gal/min h = 12 x 3.281 = 39.37 ft N = (650 x (4,488.6)1/2)/(39.37)3/ N = 2,770.73 rpm Determine the number of stages needed for a centrifugal pump if it is used to deliver 400 gal/min of water and pump power of 15 Hp. Each impeller develops a head of 30 ft. A. 6 B. 4 *C. 5 D. 7 SOLUTION: Wp = w Q h 15 x 0.746 = 9.81(400 gal/min x 0.00785m3/gal x 1/60)h h = 45.20 m x 3.281 ft/m = 148.317 ft Number of stages = 148.317/40 = 4.94 stages = 5 stages The suction pressure of a pump reads 3 in. of mercury vacuum and discharge pressure reads 140 psi is use to deliver 120 gpm of water with specific volume of 0.0163 ft3/lb. Determine the pump work. A. 4.6 KW B. 5.7 KW *C. 7.4 KW D. 8.4 KW SOLUTION: P1 = -3 in Hg x 101.325/29.92 = -10.16 KPa P2 = 140 psi x 101.325/14.7 = 965 KPa w = 1/v = 1/0.163 = 61.35 lb/ft3 x 9.81/62.3 = 9.645 KN/m h = (P2 – P1)/w = (965 +10.16)/9.645 = 101.105 m
Q = 120 gal/min x 3.785/1gal x 1m3/1000li x 1/60 = 0.00757 m3/sec P = w Q h = 9.645(0.00757)(101.105) = 7.38 KW A submersible pump delivers 350 gpm of water to a height of 5 ft from the ground. The pump were installed 150 ft below the ground level and draw down of 8 ft during the operation. If water level is 25 ft above the pump, determine the pump power. A. 7.13 KW B. 4.86 KW C. 7.24 KW *D. 9.27 KW SOLUTION: h = 5 + 150 – (25 – 8) = 138/3.281 = 42.06 m Q = 350 gal/min x 0.003785 m3/gal x 1 min/60sec = 0.02246 m3/sec Wp = w Q h = 9.81(0.02246)(42.06) = 9.27 KW A vacuum pump is used to drain a flooded mine shaft of 20 water. The pump pressure of water at this temperature is 2.34 KPa. The pump is incapable of lifting the water higher than 16 m. What is the atmospheric pressure? *A. 159.30 B. 32.33 C. 196.22 D. 171. SOLUTION: Using Bernoulli’s Theorem: P1/w + V12/2g + z1 = P2/w + V2/2g + z P1/w = P2/w + (V22 - V12)/2g + (z2 - z1) P1/9.81 = 2.34/9.81 + 0 + 16 P1 = 159.30 KPa A submersible, multi-stage, centrifugal deep well pump 260 gpm capacity is installed in a well 27 feet below the static water level and running at 3000 rpm. Drawdown when pumping at rated capacity is 10 feet. The pump delivers the water into a 25,000 gallons capacity overhead storage tank. Total discharge head developed by pump, including friction in piping is 243 feet. Calculate the diameter of the impeller of this pump in inches if each impeller diameter developed a head of 38 ft. A. 3.28 B. 5.33 *C. 3.71 D. 6. SOLUTION: V = D N V = D (3000/60) = (2(32.2)(38))1/ D = 0.315 ft = 3.708 inches A fan pressure of 2.54 cm of water t 1.42 m3 per second of air at static pressure of 2.54 cm of water through a duct 300 mm diameter and discharges it through a duct 275 mm diameter. Determine the static fan efficiency if total fan mechanical is 75% and air measured at 25 and 60 mm Hg. A. 50.11% *B. 53.69% C. 65.67% D. 45.34% SOLUTION: wA = P/RT = 101.325/(0.287)(25 + 273) = 1.18 kg/m hA = hwww/wA = (0.0254)(1000)/1.18 = 21.52 m vA = 1.42/( /4)(0.3)2 = 20.09 m/s Vd = 1.42/( /4)(0.275)2 = 23.9 m/s hv = (23.9)2 – (20.09)2 / 2(9.81) = 8.54 m h = ha + hv = 21.52 + 8.54 = 30.06 m eT = wa Q h/BP 0.75 = (1.18 x 0.00981)(1.42)(30.06) / BP BP = 0.6588 KW ep = wa Q hs/BP = (1.18 x 0.00981)(1.42)(21.52) / 0.6588 = 53.69% A water cooler uses 50 lb/hr of melting ice to cool running water from 80 to 42. Based on te inside coil area, U1 = 110 Btu/hr-ft2-. Find the gpm of water cooled. A. 0.10 GPM B. 0.21 GPM *C. 0.38 GPM D. 0.45 GPM SOLUTION: Q = mf L = mwcpw(t1 – t2) 50 (144) = mW(1)(80-42) mw = 189.474 lb/hr V = (189.474/62.4) (7.48/60) = 0.38 GPM The charge in a Diesel engine consists of 18.34 grams of fuel, with lower heating value of 42,571 KJ/kg, and 409 grams of fuel and products of combustion. At the beginning of compression, t1 = 60. Let rk = 14. For constant cP = 1.11 KJ/kg-C, what should be the cut- off ratio in the corresponding ideal cycle? A. 2.05 B. 2.34 C. 5.34 *D. 2. SOLUTION: QA = mfQh = 0.01283(42,571) = 780,752 KJ T2/T1 = rkk- T2 = (60 + 273)1.4-1 = 956.964K mt + mg = 409 mt + ma + mf = 409 ma = 409 – 2(18.34) = 372.32 grams QA = macp(t3 – t2) 780.752 = 0.37232(1.11)(T3 – 956.964) T3 = 2846, rC = T3/T2 = 2846.146/956.964 = 2. The gain of entropy during isothermal nonflow process of 5 lb of air at 60 is 0.462 Btu/R. Find the V1/V2. A. 3.85 *B. 0.259 C. 1.0 D. 0. SOLUTION: s = m R T ln(V2/V1) 0.462 = 5 (53.33/778) ln (V2/V1) V2/V1 = 3. V1/V2 = 1/3.85 = 0. An auditorium seating 1500 people is to be maintained at 80 dry bulb and 85 wet bulb temperature when outdoor air is at 91 dry bulb and 75 wet bulb. Solar heat load is 110,000 Btu/hr and supply air at 60 determine the amount of supply air. *A. 93,229.17 lb/hr B. 83,229.17 lb/hr C. 73,229.17 D. 63,229.17 lb/hr SOLUTION: Sensible heat per person = 225 Btu/hr Qa = 225(1500) + 110,000 = 447,500 Btu/hr Qa = m cp(t1 – t2) 447,500 = ma(0.24)(80 – 60) ma = 93,229.17 lb/hr
Since it is greater than 2000 then it is turbulent flow What is the force is exerted by water jet 60 mm diameter if it strikes a wall at the rate of 15 m/s? *A. 636.17 N B.442.62 N C. 764.23 N D. 563.34 N SOLUTION: F = w Q v Q = A v = = 0.0424 m3/s F = (1000)(0.0424)(15) = 636.17 N A 300 mm diameter pipe discharges water at the rate of 200 li/s. Point 1 on the pipe has a pressure of 260 kpa and 3.4 m below point 1 is point 2 with a pressure of 300 kpa. Compute the head loss between points 1 and 2. A. 4.29 m B. 2.59 m C. 6.32 m *D. 1.87 m SOLUTION: hL hL = Water flowing at the rate of 10 m/s from an orifice at the bottom of a reservoir. Find the pressure at the bottom of the reservoir. A. 30 kpag B. 40 kpag *C. 50 kpag D. 60 kpag SOLUTION: h = V2/ 2g = 102/ 2(9.81) = 5.0968 m P = w h = 9.81(5.0968) = 50 kpag Steam flows through a nozzle at 400oC and 1 Mpa (h = 3263.9 KJ/kg) with velocity of 300 m/s. Find the stagnation enthalpy. A. 3300 KJ/kg B. 3290 KJ/kg *C. 3320 KJ/kg *D. 3309 KJ/kg SOLUTION: ho = h + v2/2000 = 3263.9 + 3002/2000 = 3309 KJ/kg Air flows through a nozzle at a speed of 350 m/s. Find the stagnation temperature if entrance temperature is 200oC. A. 241.25oC B. 251.25oC *C. 261.25oC D. 271.25oC SOLUTION: To = T1 + v2/2000Cp = (2000 + 273) + 3502/2000(1) To = 534.25oK = 261. Carbon dioxide flows through a nozzle with a speed of 400 m/s. Compute the dynamic temperature. A. 92.56oK *B. 94.56oK C. 96.56oK D. 98.56oK SOLUTION: For CO2: Cp = 0.846 KJ/kg-K Dynamic temperature = v2/2000Cp = 4002/2000(0.846) = 94.56oK Carbon dioxide flows through a nozzle with a speed of 380 m/s. The entrance condition of nozzle is 250oC and 1200 kpa. Find the stagnation pressure. *A. 2,136.34 kpa B. 2,146.34 kpa C. 2,156.34 kpa D. 2,166. kpa
SOLUTION:
T1 = 250 + 273 = 523oK To = T1 + v2/2000 = 523 = 3802/2000 = 595.2oK P1 = 1200 kpa T1/To = (P1/Po)k-1/k For CO2: k = 1. 523/595.2 = (1200/Po)1.289-1/1. P0 = 2,136.34 kpa Air enters a diffuser with a velocity of 200 m/s. Determine the velocity of sound if air temperature is 30oC. *A. 349 m/s B. 359 m/s C. 369 m/s D. 379 m/s SOLUTION: C = Air flows through a nozzle with temperature of entrance of 420oK stagnation temperature of 468oK. Find the mach number. A. 0.744 *B. 0.754 C. 0.764 D. 0. SOLUTION: To = T1 + v2/2000Cp 468 = 420 + v2/ v = 309.838 m/s C = M = v/C = 309.838/410.8 = 0. Air at 300oK and 200 kpa is heated at constant pressure to 600oK. Determine the change of internal energy. A. 245.58 KJ/kg B. 235.58 KJ/kg C. 225.58 KJ/kg *D. 215. KJ/kg SOLUTION: ΔU = mCv (T2 – T1) = 1(0.7186)(600 -300) = 215.58 KJ/kg An insulated rigid tank initially contains 1.5 lb of helium at 80oF and 50 psia. A paddle wheel with power rating of 0.02 hp is operated within the tank for 30 min. Determine the final temperature. A. 159.22oF B. 169.22oF *C. 179.22oF D. 189.22 oF SOLUTION: W = ΔU = m Cv (T2 – T1) 0.02 hp (0.50hr)(2545Btu/hr/hp) = 1.5(0.171)(t2 – 80) t2 = 179.22oF A 4m2 asphalt pavement with emissivity of 0.85 has a surface temperature of 50oC. Find the maximum rate of radiation that can be emitted from the surface. A. 2,068.32 watts B. 2,078.32 watts C. 2,088.32 watts D. 2.098. watts SOLUTION: Qr = e kev A Ts Kev = 5.67 x 10-8 ( Stefan Boltzman constant) Qr = 0.85(5.67 z 10-8)(4)50 +273)4 = 2,098.32 watts
Air at 10oC and 90 kpa enters a diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area diffuser is 0.40 m2. Determine the mass flow rate of air. A. 72.79 kg/s B. 74.79 kg/s C. 76.79 kg/s *D. 78.79 kg/s SOLUTION: W = P/RT = 80/0.287(10 + 273) = 0.985 kg/m m = w v A = 0.985(200)(0.40) = 78.79 kg/s Consider a refrigeration whose 40 watts light bulb remains on continuously as a result of a malfunction of the switch. If the refrigerator has a COP of 1.3 and the cost of electricity is 8 cents per kw-hr., determine the increase in the energy consumption of the refrigerator and its cost per year if the switch is not fixed. *A. P49.59 B. P47.59 C. P45.59 D. P43. SOLUTION: COP = RE/Wref 1.3 = 40/Wref Wref = 30.769 watts W = Wb + Wref = 40 + 30.769 = 70.77 watts W = 0.07077 Kw Cost = 0.07077(8760)(P0.08) = P49. A 75 hp motor that has an efficiency of 91% is worn out and is replaced by a high-efficiency motor that has an efficiency of 95.4%. Determine the reduction in heat gain of the room due to higher efficiency under full-load conditions. A. 2.24 KW *B. 2.44 KW C. 2.64 KW D. 2.84 KW SOLUTION: P01 = (75 x 0.746)(0.91) = 50.91 KW P02 = (75 x 0.746)(0.954) = 53.376 KW Qreduced = 53.376 – 50.91 = 2.44 KW A household refrigerator that has a power input of 450 watts and a COP of 2.5 is to cool five large watermelons, 10 kg each, to 8oC. If the watermelons are initially at 20oC, determine how long will take for the refrigerator cool them. The watermelons can be treated as a water whose specific heat is 4.2 KJ/kg-oK. A. 2220 seconds B. 2230 seconds *C.2240 seconds D. 2250 seconds SOLUTION: COP = RE/Wc 2.5 = RE/ RE = 1,125 watts RE = m cp (t2 – t1) 450 t = (10 x 5)(4.2)(20-8) t = 2240 seconds When a man returns to his wall-sealed house on a summer day, he finds that the house is at 32oC. He returns on the air conditioner which cools the entire house to 20oC in 15 minutes, if COP is 2.5, determine the power drawn by the airconditioner. Assume the entire mass within the house is 800 kg of air for which cv = 0.72 KJ/kg-K, cp = 1.0KJ/kg-K. A. 1.072 KW B. 2.072 KW *C. 3.072 KW D. 4.072 KW SOLUTION: RE = m cv (T2 –T1) = (800/15x60)(0.72)(32-20) RE = 7.66 KW Wc = 7.68/2.5 = 3.072 KW A heat source at 8000K losses 2000 KJ of heat to a sink at 500oK. Determine the entropy generated during this process. *A. 1.5 KJ/K B. 2.5 KJ/K C. -2.5 KJ/K D. 4 KJ/K SOLUTION: ΔSsource = -2000/800 = -2. ΔSsink = 2000/500 = 4 ΔSgen. = -2.5 + 4 = 1.5 KJ/K Helium gas is compressed in an adiabatic compressor from an initial state of 14 psia and 50oF to a final temperature of 320oF in a reversible manner. Determine the exit pressure of Helium. A. 38.5 psia *B. 40.5 psia C. 42.5 psia D. 44.5 psia SOLUTION: T2/T1 = (P2/P1)n-1/n (320 + 460)/(50 +460) = (P2/14)1.587-1/1. P2 = 40.5 psia Air pass thru a nozzle with efficiency of 90%. The velocity of air at the exit is 600 m/s. Find the actual velocity at the exit. A. 382 m/s B. 540 m/s C. 458 m/s *D. 568 m/s SOLUTION: e = (v2/v3) 0.9 = (v2/600) v2 = 568.21 m/s A 50 kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 258oK. The iron block eventually reaches thermal equilibrium with the lake water. Assuming average specific hear of 0.45 KJ/kg-K for the iron, determine the entropy generated during this process. *A. -12.65 KJ/k B. 16.97KJ/K C. 4.32 KJ/K D. 6.32 KJ/K SOLUTION: ΔSiron = m c ln (T2/T1) = 50(0.45)ln(285/500) = -12.65 KJ/K ΔSlake = Q/T = [50(0.45)(500-285)]/285 = 16.97 KJ/K ΔSgen. = -12.65 + 16.97 = 4.32 KJ/K A windmill with a 12 m diameter rotor is to be installed at a location where the wind is blowing at an average velocity of 10 /s. Using standard conditions of air (1 atm, 25oC), determine the maximum that can be generated by the windmill. A. 68 KW *B. 70 KW C. 72 KW D. 74 KW SOLUTION: w = P/RT = 101.325/(0.28)(25+ 273) = 1.1847 kg/m m = w A v = 1.1847(π/4 x 122)(10) = 1,1339.895 kg/s KE = v2/2000 = 102/2000 = 0.05 KJ/kg Power = m KE = 1,1339.895(0.05) = 70 KW Consider a large furnace that can supply heat at a temperature of 2000oR at a steady rate of 3000Btu/s. Determine the energy of this energy. Assume an environment temperature of 77oF.
Consider a person standing in a breezy room at 20oC. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 m2 and 29oC, respectively, and the convection heat transfer coefficient is 6 W/m2 with emissivity factor of 0.95. A. 86.40 watts B. 61.70 watts C. 198.1 watts *D. 168.1 watts SOLUTION: Qc = h A (t2 – t1) = (6)(1.6)(29.20) = 86.40 watts Qr = (0.95)(5.67 x 10-6)[(1.6)(29 + 273)4 – (20 + 273)4] = 81.7 watts Q = Qc + Qr = 86.40 + 81.7 = 168.1 watts Water is boiler in a pan on a stove at sea level. During 10 minutes of boiling, it is observed that 200 grams of water has evaporated. Then the rate of heat transfer to the water is: A. 0.84 KJ/min *B. 45.1 KJ/min C. 41.8 KJ/min D. 53.5 KJ/min SOLUTION: Q = m L = (0.2 / 10) (2257) = 45.1 KJ/min An aluminum pan whose thermal conductivity is 237 W/m-C has a flat bottom whose diameter is 20 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 500 watts. If the inner surface of the bottom of the pan is 105oC, determine the temperature of the surface of the bottom of the pan. A. 95.27 oC *B. 105.27oC C. 115.27oC D. 125.27oC SOLUTION: A = π / 4 ( 0.20)2 = 0.0314 m Q = 500 = T2 = 105.27oC For heat transfer purposes, a standing man can be modeled as a 30 cm diameter, 170 cm long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 34oC. For a convection heat transfer coefficient of 15 W/m2- oC, determine the rate of heat loss from this man by convection in an environment at 20oC. A. 316.46 watts B. 326.46 watts *C. 336.46 watts D. 346.46 watts SOLUTION: Qc = k A (t2 – t1) = 15 (π x 0.30 x 1.7) (34 – 20) = 336.46 watts A 5cm diameter spherical ball whose surface is maintained at a temperature of 70oC is suspended in the middle of a room at 20oC. If the convection heat transfer coefficient is 15 W/m2 – C and the emissivity of the surface is 0.8, determine the total heat transfer from the ball. A. 23.56 watts *B. 32.77 watts C. 9.22 watts D. 43.45 watts SOLUTION: A = 4 π r2 = 4 π (0.05)2 = 0.0314 m Qc = h A (t2 – t1) = 15 (0.0314)(70 – 20) = 23.56 watts Qr = (0.80)(5.67 x 10-6)(0.0314)[(70 + 273)4 – (50 + 273)4] = 9.22 watts Q = Qr + Qc = 23.56 + 9.22 = 32.77 watts A frictionless piston-cylinder device and rigid tank contain 1.2 kmol of ideal gas at the same temperature, pressure, and volume. Now heat is transferred, and the temperature of both system is raised by 15oC. The amount of extra heat that must be supplied to the gas in the cylinder that is maintained at constant pressure. SOLUTION: A. 0 B. 50 KJ C. 100 KJ *D. 150 KJ Q = m cp (t2 – t1) = (1.2 x 8.314)(1)(15) = 150 KJ A supply of 50 kg of chicken needs at 6oC contained in a box is to be frozen to -18oC in a freezer. Determine the amount of heat that needs to be removed. The latent heat of chicken is 247 KJ/kg, and its specific heat is 3.32 KJ/kg-oC above freezing and 1.77 KJ/kg-oC below freezing. The container box is 1.5 kg, and the specific heat of the box material is 1.4 Kj/kg-oC. Also the freezing temperature of chicken is -2.8oC. *A. 15,206.4 KJ B. 50.4 KJ C. 15,156 KJ D. 1,863 KJ SOLUTION: Qchicken = 50 [3.32(6 + 2.8) = 247 1.77(-2.8 + 18)] = 15,156 KJ Qbox = 1.5(1.4)(6 + 8) = 50.4 KJ Q = 15,156 + 50.4 = 15, 206.4 KJ Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 KJ of heat is transferred to the water, and 5 KJ of heat is lost to the surrounding air. The paddle wheel work amounts to 500 N-m. Determine the final energy of the system if its initial energy is 10 KJ. *A. 35.5 KJ B. 45.5 KJ C. 25.5 KJ D. 14.5 KJ SOLUTION: Final energy = Qa + ΔU – Qloss + W = 30 + 10 – 5 + 0.50 = 35.5 KJ A classroom that normally contains 40 people is to be air- conditioned with window air- conditioning units of 5 KW cooling capacity. A person at rest may be assumed to dissipate heat at a rate of about 360 KJ/hr. There are 10 light bulbs in the room, each with a rating of 100 watts. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,00 KJ/hr. If the room air is to be maintained at a constant temperature of 21oC, determine the number of window air- conditioning units required. A. 1 unit *B. 2 units C. 3 units D. 4 units SOLUTION: Q = total head load = 40(360/3600) + 10(0.100) + 15,000/3600 = 9.167 KW No. of air-conditioning = 9.167/5 = 1.833 = 2 units A 4m x 5m x 6m room is to be heated by a baseboard resistance heater. It is desired that the resistance heater be able to raise the air temperature in the room from 7 to 23oC within 15 minutes. Assuming no heat losses from the room and an atmospheric pressure of 100 kpa, determine the required power of the resistance heater. Assume constant specific heats at room temperature. A. 2.34 KW *B. 1.91 KW C. 4.56 KW D. 6.34 KW SOLUTION: w = P / R T = 100 / (0.287)(7 +273) = 1.244 kg / m m = 1.244 (4 x 5 x 6) = 149.28 kg Q = m cv (t2 – t1 ) = 149.28 (0.7186)(23 – 7) = 1,716.36 KJ Power = 1,716.36 / (15 x 60) = 1.91 KW A student in a 4m x 6m x 6m dormitory room turns on her 150 watts fan before she leaves the room on a summer day, hoping that the room will be cooler when she comes back in the
evening. Assuming all the doors and windows are tightly closed and disregarding any heat transfer through the walls and the windows, determine the temperature in the room when she comes back 10 hours later. Use specific heat values at room temperature, and assume the room to be at 100 kpa and 15oC in the morning when she leaves A. 28.13oC B. 38.13oC C. 48.13oC *D. 58.13oC SOLUTION: w = P / R T = 100 / (0.287)(15 + 273) = 1.2098 kg / m m = 1.2098(4 x 6 x 6) = 174.216 kg Q = m cv (t2 – t1) 0.15(10 x 3600) = 174.216 (0.7186)(t2 – 15) t2 = 58.13oC A piston cylinder device whose piston is resting on top of a set stops initially contains 0.50 kg of helium gas at 100 kpa and 25oC. The mass of the piston is such that 500 kpa of pressure is required to raise it. How much heat must be transferred to the helium before the piston starts rising? A. 1557.13 KJ B. 1657.13 KJ C. 1757.13 KJ *D. 1857. KJ SOLUTION: For helium: cv = R / (k-1) = (8.314 / 4) (1.667 – 1) = 3,116 KJ/ kg-K T2 = (25 + 273)(500 / 100) = 1,490oK T1 = 25 + 273 = 298oK Q = m cv (T2 – T1) = 0.50(3.116)(1490 – 298) = 1857.13 KJ In order to cool 1 ton (100kg) of water at 20oC in an insulated tank, a person pours 80 kg of ice at -5oC into the water. Determine the final equilibrium temperature in the tank. The melting temperature and the hat of fusion of ice at atmospheric pressure are 0oC and 333. kJ/kg, respectively. *A. 12.43oC B. 14.43oC C. 16.43oC D. 18.43oC SOLUTION: Qwater =Qice 1000(4.187)(20 –te) = 80(2.09)(0 + 5) + 80(333.7) + 80(4.187)(te – 0) te = 14.43oC A fan is powered by a 0.5 hp motor and delivers air at a rate of 85 m3/min. Determine the highest value for the average velocity of air mobilized by the fan. The density of air to 1. kg/m3. A. 18.23 m/s *B. 21.12 m/s C. 25.34 m/s D. 32.23 m/s SOLUTION: P = w Q h 0.50(0.746) = ( 1.18 x 0.00981)( 85 / 60) (h) h = 22.74 m v = = 21.12 m/s An Ocean – Thermal Energy Conversion power plant generates 10,000 KW using a warm surface water inlet temperature of 26oC and a cold deep- water temperature of 15oC. ON the basis of a 3oC drop in the temperature of the warm water and a 3oC rise in the temperature of the cold water due to removal and addition of heat, calculate the power required in KW to pump the cold- deep water to the surface and through the system heat.Assume a Carnot cycle efficiency and density of cold water to be 1000 kg/m3. A. 108 *B. 250 C. 146 D. 160 SOLUTION: e = (Th – TL)/ TH = [(26 + 273) – (15 + 273)] / (26 + 273) = 0. e =W / Qa 0.03676 = 10, 000 / QA QA = 271, 612. 99 KW QR = Qa – W = 271,812.99 – 10, 000 = 261,813 KW QR = m cp (Δt) 261, 813 = m (4.187)(3) M = 20, 643.32 kg/s Q = 20,843.32 kg/s or 20,843.32 li/s = 20.843 m3/s H = P / w = 12 / 9.81 = 1.223 m Wp = w Q h = 9.81(20.843)(1.223) = 250.12 KW A plate – type solar energy collector with an absorbing surface covered by a glass plate is to receive an incident radiation of 800 W/m2. The glass plate has a reflective of 0.12 and a transmissivity of 0.85. The absorbing surface has an absorptivity of 0.90. The area of the collector is 5m2. How much solar energy in watts is absorbing by the collector? A. 2500 B. 2880 C. 3510 *D. 3060 SOLUTION: Q = heat absorbed from sun Q = (800 W/m2)(5 m2)(0.85)(0.9) = 3,060 watts A tank contains liquid nitrogen at -190oC is suspended in a vacuum shell by three stainless steel rods 0.80 cm in diameter and 3 meters long with a thermal conductivity of 16.3 W/m2-C. If the ambient outside the vacuum shell is 15oC, calculate the magnitude of the conductive heat flow in watts along the support rods. *A. 0.168 B. 0.0587 C. 0.182 D. 0. SOLUTION: Q = h A (Δt) = 16.3 (π/4 x 0.0082)(15 – (-190)) = 0.168 watts An elastic sphere containing gas at 120 kpa has a diameter of 1.0 m. Heating the sphere causes it to expand to a diameter of 1.3 m. During the process the pressure is proportional to the sphere diameter. Calculate the work done by the gas in KJ. A. 41.8 B. 50.6 *C. 87.5 D. 35. SOLUTION: P α D P = k D 120 = k(1) K = 120 P = 120 D V = 4/3 π(D/d)2 = 4/24 π D dV = (12/24) π D2 dD W = D3dD W = 87.47 KJ An ideal gas with a molecular weight of 7.1 kg/kg mol is compressed from 600 kpa and 280oK to a final specific volume of 0.5 m3/kg. During the process the pressure varies according to p = 620 + 150v + 95v2 where p is in kpa and v in m3/kg. Calculate the work of compression in KJ/kg?
An ideal gas mixture consists of 2 kmol of N2 and 6 kmol of CO2. The apparent gas constant of mixture is: *A. 0.208 B. 0.231 C. 0.531 D. 0. Solution: M= (2/8)(28) + (6/8)(44) = 40 R= 8.314/M = 8.314/40 = 0.208 KJ/kg-K A Carnot cycle operates between the temperature limits of 300OK and 1500OK, and produces 600 KW of net power. The rate of entropy changes of the working fluid during the heat addition process is: A. 0 B. 0.4KW/K *C. 0.5KW/K D.2.0KW/K Solution: W= ( s) (TH – TL) 600 = ( s) (1500 – 300) ( s)= 0.50 KW/K Air in an ideal Diesel cycle is compressed from 3L to 0.15L and then it expands during the constant pressure heat addition process to 0.3L. Under cold air standard conditions, the thermal efficiency of this cycle is: rk = 3/0.15= 20 rc= 0.3/0.15= 2 1 rck- 1 1 21.2 - 1 e=1- = 1 - = 0. rkk-1 k(rc – 1) 201.4-1 1.4 (2-
=64.67% Helium gas is an ideal Otto cycle is compressed from 20oC and 2L to 0.25L and its temperature increases by an additional 800oC during the addition process. The temperature of helium before the expansion process is: *A. 1700oC B. 1440oC C. 1240oC D.880oC Solution: rk =2/0 25 = 8 T2 = (20+273) (8)1.667-1 =1,172K T3 = T2 + 800 = 1172 + 800 = 1972oK t3 = 1699oC = 1700oK In an ideal Brayton cycle has a net work output of 150KJ/kg and backwork ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 80%, the net work output of the cycle would be. A. 50KJ/kg *B. 75KJ/kg C. 98KJ/kg D.120KJ/kg Solution: Backwork ratio = WO/WT 0.40 = WO/WT WO = 0.40 WT Wnet = WT - WO 150 = WT – 0.4 WT WT =250 KJ/kg WT ‘=250(0.8) = 200KJ/kg WP = 0.40(200) =100KJ/kg WP’ = 100/0.80 =125 KJ/kg Wnet= WT’ – WC’= 200 – 125 = 75 KJ/kg Air entered a turbojet engine at 200 m/s at a rate of 20 kg/s, and exists at 800 m/s relative to the aircraft. the thrust developed by the engine is: A. 6KN *B. 12KN C.16KN D. 20KN Solution: Thrust developed = m(v2 – v1) = 20(800 – 200) = 12,000N = 12KN A thermal power has a net power 10MW. The backwork ratio of the plant is 0.005. Determine the compressor wor. A. 50.15KW B. 50.35KW *C.50.25KW D. 50.45KW Solution: Wnet= WT + WP BW= WP / WT 0.005 =WP / WT WP= 0.005WT Wnet= WT - WP 10,000 = WT – 0.005WT WT= 10,050.25 KW WC= 0.005(10,050.25) = 50.25KW A heat engine receives heat from a source at 1200oK at a rate of 500KJ/s and rejects the waste heat to a sink at 300oK. If the power output of the engine is 200KW, the second law efficiency of the heat engine is: A.35% B.40% *C.53% D.75% Solution: ea= 200/500 = 0. et= (TH – TL)/TH = (1200 – 300)/1200 = 0. es= 0.40/0.75= 53.33% A water reservoir contains 100,000 kg of water at an average elevation of 60 m. The maximum amount of electric power that can be generated from this water is: A.8KWh *B.16KWh C.1630KWh D.58, 800KWh Solution: P= m h = (100,000 x 0.00981)(60)= 58,860 KJ P= 58,860 KJ x KWh/3600 KJ = 16.35KWh A house is maintained at 22oC in winter by electric resistance heaters. If the outdoor temperature is 5oC, the second law of efficiency of the resistance heaters is: A.0% *B.5.8% C.34% D.77% Solution: ea= 100% of resistance heaters et= (22 – 5)/(22 + 273) = 5.8% es= 5.8/100 = 5.8% A thermoelectric refrigerator that resembles a small ice chest is powered by a car battery, and has a COP of 0.10. If the refrigerator cools at 0.350L canned drink from 20OC to 4OC in 30 min. determine the average electric power consumed by the thermoelectric refrigerator. *A.130 watts B.110 watts C.120 watts D.140 watts Solution: (1 x 0.35) Q= m cp(t2 – t1) = (4.187)(20 – 4) = 13 watts 30 x 60 COP= RE/Wc
0.10= 13/0.10= 130 watts A Carnot refrigerator operates in a room in which the temperature is 25OC and consumes 2 kW of power when operating. If the food compartment of the refrigerator is to be maintained at 3OC, determine the rate of heat removal from the food compartment. *A.1504.8 kJ/min B.12.86 kJ/min C.1625 kJ/min D.9.57 kJ/min Solution: COP= TL /TH– TL = (3 + 273)/ (25 + 273) – (3+273) =12. QL= COP x W = 12.54 x 2(60) = 1504.8 kJ/min A household refrigerator with EER 8.0 removes heat from the refrigerated space at a rate of 90 kJ/min. Determine the rate of heat transfer to the kitchen air. A.101.25 kJ/min B.63.05 kJ/min *C.128.46 kJ/min D.80 kJ/min Solution: COP= EER /3.412 = 8/ 3.412 = 2. COP= QL /QH – QL = 2.34 = 90 / QH – 90 QH=128.46Kj/min An air-conditioning system is used to maintain a house at 75OF when the temperature outside is 95OF. The house is gaining heat through the walls and windows at a rate of 1250 Btu/min, and the heat generation rate within the house from people, lights and appliances amounts to 350 Btu/min. Determine the minimum power input required for this air- conditioning system. A.10.06 hp B.1.36 hp *C.1.41 hp D.7.94 hp Solution: QL= 1250 + 350 = 1600 Btu/min COP= TL / TH – TL= (75 + 460)/(95 + 460) - (75+460) = 26. W= QL /COP = (1600 / 26.75) / 42.4 = 1.41 hp A refrigeration system is to cool bread loaves with an average mass of 450 g from 22OC to -10OC at a rate of 500 loaves per hour by refrigerated air. Taking the average specific and latent heats of bread to be 2.93 kJ/kg, OC and 109.3 kJ/kg, respectively, determine the product load. A.541.7 kJ/min B.351.6 Kj/min *C.761.5 kJ/min D.409.9 kJ/min Solution: Mbread= (500 breads/h) (0.45 kg / bread) = 225 kg/h Qtotal= Qbread + Qfreezing = (mcp∆T)breadl (mhlatent)bread = (225)(2.93)[22-(-10)] l (225) (109.3) Qtotal= 45,688.5 kJ/h = 761.5 kJ/min A house that was heated by electric resistance heaters consumed 1200 kWh of electric energy in a winter month. If this house were heated instead by a heat pump that has an average performance factor, PF of 2.4, determine how much money the homeowner would be saved that month. Assume a price of 0.085$/kWh for electricity. A. $42.5 *B. $59.50 C.$109 D.$97. Solution: W= QH/PF = 1200kWh / 2.4 = 500 kWh $ Savings per month = (1200 – 500) (0.085) = $59. An ammonia simple saturation cycle operates with a suction pressure of 291.6 kPa and a condenser pressure of 1204 kPa develops 15 tons of refrigeration. Determine the theoretical horsepower of the compressor. The following enthalpies have been found: condenser entrance = 1653 kJ/kg, exit =346.6 kJ/kg, compressor entrance = 1450.2 kJ/kg, exit= 1653kJ/kg. A.7.23 hp *B.13 hp C. 15 hp D.8.23 hp Solution: m= Qe/ (h1 – h4) = (15 x 3.52) / (1450.2 – 346.6) = 0.0478 kg/s W= m (h2 – h1) = (0.0478)(1653 – 1450.2) / 0.746 = 13 hp An ammonia ice plant operates between a condenser temperature of 35OC and evaporator of -15OC. It produces 10 metric tons of ice per day from water at 30OC to ice at -5OC. Assuming simple saturation cycle, determine the horsepower of the motor if the adiabatic efficiency of the compressor ηc=0.85 and mechanical efficiency ηm=0.95. The specific heat of ice is 2,094kJ/kg. OC and the latent heat is 335kJ/kg. From the table for ammonia the following enthalpies are: condenser entrance = 1703 kJ/kg, exit= 366.1 kJ/kg; compressor entrance= 1443.9kJ/kg, exit = 1703kJ/kg A.17.68 hp B.18.61 hp C.15.5 hp *D.21.9 hp Solution: qe=cpa (te – tf) lhlatentlcph(tf – ts) = (4.187) (30 – 0) + 335 + (2.094) [(0- (-5)] = 471.08 kJ/kg Qe= (10 x 1000) (471.08) / 24 = 196,283.33 kJ/hr= 54.523kJ/s m= Qe / (h2 – h4) = (54.523)/ (1443.9 – 366.1) = 0.05059kg/s W= m (h2 – h1) = (0.0509) (1703 – 1443.9)/ 0.746 = 17.68 hp Wmotor= 17.68 / (0.85) (0.95) =21.9 hp A Freon 22 air conditioning under standard operating conditions of 35OC is condensing and 5OC evaporating temperatures. The volume flow rate entering the compressor is 23.72 L/s. determine the refrigerating capacity if the refrigerating effect is 164 kJ/kg. From the table for R22 the specific volume at the compressor entrance is 40.36L/kg. A.393.3 TR B.79.3 TR C.96.4 TR *D.27.4 TR Solution: m= V1/v1= 23.72 / 40.36 = 0.5877 kg/s Qe= m (qe) = 0.5877 (164)/ 3.52 = 27.4 TR The refrigerant volume flow rate at the entrance of compressor were obtained from a test on a twin cylinder single acting 15 cm x 20 cm, 320 rpm compressor ammonia refrigerating plant 33 L/s. Determine the volumetric efficiency of the compressor. A.77.65% *B.87.6% C.97.6 TR D.65.65% Solution: VD = (π D2 L/4) N= (π /4) (0.15)2(0.2) (320) (2) = 2.26 m3/min nv =V1/VD = 33/2.26 (1000/60) = 0.876 or 87.6 %
