131.5 API 131.5 38 Solution Manual for Design of Fluid Thermal Systems SI Edition 4th Edition Janna Chapter : FLUID PROPERTIES AND BASIC EQUATIONS. Density, Specific Gravity, Specific Weight 1. What is the specific gravity of 38 API oil? 38 API oil sp.gr. = 141.5 = 141.5 + + 141.5 sp. gr. = 169.5 = 0.835. The specific gravity of manometer gage oil is 0.86. What is its density and its API rating? sp. gr. = 0.86;ρ = 1000(0.86) = 86 kg/m 3 ρ = 6.4(0.86) = 51.54 lbm/ft 3 131.5 API 0.86 sp. gr. = 141.5 131.5 + API = 141.5 API = + 171.3 ; = 131.5 API 39.8 API 40 API 3. What is the difference in density between a 50 API oil and a 40 API oil? sp. gr. = 141.5 141.5 131.5 API = 131.5 50 = 0.7796 for a 50 oil + + 141.5 141.5 sp. gr. = 131.5 = = 0.86 for a 40 oil + API 131.5 40 + 0.85 0.7796 = 0.0455 density difference 4. A 35 API oil has a viscosity of 0.85 N s/m. Express its viscosity in Saybolt Universal Seconds (SUS). 35 AFI oil sp. gr. = 141.5 141.5 131.5 API + = +35 = 0.850 μ = 0.85 N s/m ν = μg c = 0.85 = 10 10 4 ρ 0.850(1000) Highly viscous; try ν = 0.158 10 6 (SUS) if SUS > 15 10 10 4 SUS = 0.158 10 6 = 4633 SUS Link download of Solution Manual for Design of Fluid Thermal Systems SI Edition 4th Edition Janna https://digitalcontentmarket.org/download/solution-manual-for-design-of-fluid-thermal-systemssi-edition-4th-edition-janna/ https://digitalcontentmarket.org/download/solution-manual-for-design-of-fluid-thermal-systems-si-edition-4thedition-janna/
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5. Air is collected in a 1. m 3 container and weighed using a balance as indicated in Figure P.5. On the other end of the balance arm is 1. m 3 of CO. The air and the CO are at 7 C and atmospheric pressure. What is the difference in weight between these two volumes? air CO FIGURE P.5. Air at 7 C = 300 K has ρ = 1.177 kg/m 3 CO at 7 C = 300 K has ρ = 1.797 kg/m 3 For a volume of 1. m 3, the weight of air is (1.177 kg/m 3 )(1. m 3 )(9.81 m/s ) = 13.86 N For CO (1.797 kg/m 3 )(1. m 3 )(9.81 m/s ) = 1.15 N Weight difference is 1.15 13.86 = 7.9N 6. A container of castor oil is used to measure the density of a solid. The solid is cubical in shape, 30 mm 30 mm 30 mm, and weighs 9 N in air. While submerged, the object weighs 7 N. What is the density of the liquid? Castor Oil ρ = 960 kg/m 3 mg mg buoyant force in air submerged volume = V 9 7 1 ρ = = 7 551 kg/m 3 (0.03) 3 9.81 = ρg 7. A brass cylinder (Sp. Gr. = 8.5) has a diameter of 5.4 mm and a length of 101.6 mm. It is submerged in a liquid of unknown density, as indicated in Figure P.7. While submerged, the weight of the cylinder is measured as 3.56 N. Determine the density of the liquid. submerged object weight FIGURE P.7. Buoyant force = mg in air mg submerged = mg 0.8 buoyant force mg 0.8 ρg V π D h π (0.054) (0.1016) 5.15 10 5 m 3 volume = V = = 4 = 4 = mg = ρ b V g = 8500(5.15 10 5 )(9.81) = 4.9 N ρ mg 0.8 4.9 3.56 = gv = 9.81(5.15 10 5 ) ρ = 1454 kg/m 3 - https://digitalcontentmarket.org/download/solution-manual-for-design-of-fluid-thermal-systems-si-edition-4thedition-janna/
Viscosity 8. Actual tests on Vaseline yielded the following data: τ in N/m 0 00 600 1 000 d V /d y in 1/s 0 500 1 000 1 00 Determine the fluid type and the proper descriptive equation. 1 1 shear stress 8 0 0 6 0 0 4 0 0 0 0 0 0 5 0 0 1 0 0 0 1 5 0 0 s t r a i n r a t e τ = K d V dy n Can be done instantly with spreadsheet; hand calculations follow for comparison purposes: dv/dy ln(dv/dy) τ ln τ ln(τ) ln(dv/dy) 0 0 500 6.15 00 5.98 3.93 1000 6.908 600 6.397 44.19 100 7.090 1000 6.908 48.98 Sum 0.1 18.60 16.1 m = 3 Summation (ln(dv/dy)) = 136.6 b 1 = 3(16.1) 0.1(18.60) = 1.766 3(136.6) 0.1 18.60 0.1 b 0 = 3 1.766 3 = 5.697 https://digitalcontentmarket.org/download/solution-manual-for-design-of-fluid-thermal-systems-si-edition-4thedition-janna/
K = exp(b 0 ) = 0.00336; n = b 1 = 1.766 dv n dv 1.766 τ = τ o + K dy = 0.00336 dy -3 https://digitalcontentmarket.org/download/solution-manual-for-design-of-fluid-thermal-systems-si-edition-4thedition-janna/
9. A popular mayonnaise is tested with a viscometer and the following data were obtained: τ in g/cm 40 100 140 180 d V /d y in rev/s 0 3 7 15 Determine the fluid type and the proper descriptive equation. The topmost line is the given data, but to curve fit, we subtract 40 from all shear stress readings. 0 0 1 8 0 1 6 0 shear stress 1 4 0 1 0 1 0 0 8 0 6 0 4 0 0 0 0 5 10 15 0 s t r a i n r a t e dv n τ = τ o + K dy which becomes τ = τ τ o = K dy Can be done instantly with spreadsheet; hand calculations: dv n dv/dy ln(dv/dy) τ τ ln τ ln(τ ) ln(dv/dy) 0 40 0 3 1.099 100 60 4.094 4.499 7 1.946 140 100 4.605 8.961 15.708 180 140 4.94 13.38 Sum 5.753 13.64 6.84 m = 3 Summation (ln(dv/dy)) = 1.33 b 3(6.84) 5.753(13.64) 1 = 3(1.33) 5.753 = 0.56 13.64 5.753 b 0 = 3 0.56 3 = 3.537 K = exp(b 0 ) = 34.37; n = b 1 = 0.56 τ = τ o + K dv dy n = 40 + 34.37dy dv 0.56 where dv/dy is in rev/s and τ in g/cm ; these are not standard units. -4
10. A cod-liver oil emulsion is tested with a viscometer and the following data were obtained: τ in 0 40 60 80 10 d V /d y in rev/s 0 0.5 1.7 3 6 Graph the data and determine the fluid type. Derive the descriptive equation. Cod liver oil; graph excludes the first data point. 1 4 0 1 0 shear stress 1 0 0 8 0 6 0 4 0 0 0 0 4 6 8 s t r a i nr a t e τ = K dv dy n Can be done instantly with spreadsheet; hand calculations: dv/dy ln(dv/dy) τ ln τ ln(τ) ln(dv/dy) 0.5 0.6931 40 3.689.557 1.7 0.5306 60 4.094.17 3 1.099 80 4.38 4.816 6 1.79 10 4.787 8.578 Sum.79 16.95 13.01 m = 4 Summation (ln(dv/dy)) = 5.181 b 1 = 4(13.01).79(16.95) = 0.4356 4(5.181).79 b 0 = 16.95.79 4 0.4356 4 = 3.537 K = exp(b 0 ) = 51.43; n = b 1 = 0.4356 dv n dv 0.4356 τ = τ o + K dy = 51.43 dy where dv/dy is in rev/s and τ in lbf/ft ; these are not standard units. -5
11. A rotating cup viscometer has an inner cylinder diameter of 50.8 mm and the gap between cups is 5.08 mm. The inner cylinder length is 63.5 mm. The viscometer is used to obtain viscosity data on a Newtonian liquid. When the inner cylinder rotates at 10 rev/min, the torque on the inner cylinder is measured to be 0.0143 mn-m. Calculate the viscosity of the fluid. If the fluid density is 850 kg/m 3, calculate the kinematic viscosity. Rotating cup viscometer R = 5.4 mm δ = 5.08 mm L = 63.5 mm ω = (10 rev/min) (π rad/rev)(1 min /60 s) = 1.047 rad/s = dv dy T = 0.0143 10 3 N-m ρ = 850 kg/m 3 μ = μ = Tδ π R (R + δ)lω 1.43 10 5 5.08 10 3 π (0.054) (0.054 + 5.08 10 3 )(0.0635)(1.047) μ = 7.7 10 3 Pa s v 7.7 10 3 9.76 10 5 ft /s 9.06 10 6 m /s = 850 = 1. A rotating cup viscometer has an inner cylinder whose diameter is 38 mm and whose length is 80 mm. The outer cylinder has a diameter of 4 mm. The viscometer is used to measure the viscosity of a liquid. When the outer cylinder rotates at 1 rev/min, the torque on the inner cylinder is measured to be 4 10 6 N m. Determine the kinematic viscosity of the fluid if its density is 1 000 kg/m 3. R = 38/ = 0.019 m; L = 0.08 m Routside = 4/ = 1 mm δ = 1 19 = mm = 0.00 m ω = (1 rev/min)(π/60) = 1.6 rad/s T = 3.8 10 6 N m ρ = 1 000 kg/m 3 μ Tδ = π R (R + δ)(lω) μ = 1.58 10 3 N s/m = 3.8 10 6 (0.00) π (0.019) (0.019 + 0.00)(0.08) (1.6) v = ρ = 1.58 10 3 = 1.58 10 6 m /s 1 000-6
13. A rotating cup viscometer has an inner cylinder diameter of 57.15 mm and an outer cylinder diameter of 6.5 mm. The inner cylinder length is 76. mm. When the inner cylinder rotates at 15 rev/min, what is the expected torque reading if the fluid is propylene glycol? D = 57.15 mm R = 8.58 mm (R + δ) = 6.5 mm R + δ = 31.15 δ =.545 mm ρ = 968 kg/m3 μ = 0.041 Pa s ω = (15 rev/ min)(π/60) = 1.57 rad/s T = π R (R + δ)(l ω)μ = π (0.0858) (0.03115)(0.076)(1.571)(0.041) δ 0.00545 T = 3.16 10 4 N-m 14. A capillary tube viscometer is used to measure the viscosity of water (density is 1000 kg/m 3, viscosity is 0.89 10 3 N s/ m ) for calibration purposes. The capillary tube inside diameter must be selected so that laminar flow conditions (i.e., VD/v < 100) exist during the test. For values of L = 76. mm and z = 54 mm, determine the maximum tube size permissible. Capillary tube viscometer μ = 0.89 10 3 N s/m z = 0.54 m L = 0.076 m V t V = ρg z π R 4 t L 8μ ρ = 1000 kg/m 3 = Volume flow rate = AV = πr V; substituting into the equation, πr z π R 4 V = ρg L 8μ z R Rearrange and solve for V, V = ρg L 8μ The limiting value is Re < 100; using equality, ρv(r V(R) ) ν = 100; μ = 100 or V = 100μ z R ρr = ρg L 8μ Rearrange and solve for R 3 R 3 = 100μ (8)(L) 100(0.89 10 3 ) (8)(0.076) ρ gz = (1000) (9.81)(0.54) R 3 =.035 10 10 or R = 5.88 10 4 m = 0.588 mm Any larger, flow no longer laminar -7
15. A Saybolt viscometer is used to measure oil viscosity and the time required for 6 10 5 m 3 of oil to pass through a standard orifice is 180 SUS. The specific gravity of the oil is found as 44 API. Determine the absolute viscosity of the oil. For 180 SUS, ν = 0.3 10 6(180) 155 10 6 = 3.98 10 5 m/s 180 44 API oil; sp.gr. 141.5 0.8063 ρ 806.3 kg/m 3 + = 131.5 44 = ; = μ = ρν = 806.3(3.98 10 ) = 3.167 10 N s/m 16. A 10 4 m 3 capillary tube viscometer is used to measure the viscosity of a liquid. For values of L = 40 mm, z = 50 mm, and D = 0.8 mm, determine the viscosity of the liquid. The time recorded for the experiment is 1 seconds. ν z R g t (9 6 (1) π 4 0.5π (0.000 8/) 4.81) = 8LV = 8(0.04)(10 10 ) ν = 7.39 10 7 m /s 17. A Saybolt viscometer is used to obtain oil viscosity data. The time required for 60 ml of oil to pass through the orifice is 70 SUS. Calculate the kinematic viscosity of the oil. If the specific gravity of the oil is 35 API, find also its absolute viscosity. For 70 SUS, 185 10 6 ν = 0.4 10 6(70) 70 ν = 1.304 10 5 m /s 35 API oil 141.5 sp. gr. = 131.5 35 = 0.8498 ρ = 849.8 kg/m 3 + μ = ρv = 849.8(1.304 10 5 ) g c μ = 1.108 10 N s/m -8
18. A -mm diameter ball bearing is dropped into a container of glycerine. How long will it take the bearing to fall a distance of 1 m? μ = ρ D L s ρ 1 ρg 18V V = t L = 1 m D = mm = 0.00 m ρ s = 7900 kg/m 3 V = ρ = 1 63 μ = 950 10 3 Pa s ρ D 7.9 1 s ρ 1 ρg 18μ = 1.63 1 (1 63)(9.81)(0.00 ) 18(950 10 3 ) V = 0.015m/s ρvd 1 63(0.015 )(0.00) Check on Re = μ = 950 10 3 L 1 t = 0.015 ; t = 0.015 = 0.04 < 1 OK t = 65.8 s 19. A 3.175 mm diameter ball bearing is dropped into a viscous oil. The terminal velocity of the sphere is measured as 40.6 mm/s. What is the kinematic viscosity of the oil if its density is 800 kg/m 3? ρ s D L μ = ρ 1 ρg 18V V = t = 40.6 10 3 m/s D = 0.003175 m ρ s = 7900 kg/m 3 ν = ρ = ρs 1 18V = 800 1 18(40.6 10 3 ) μ ρ gd 7900 (9.81)(0.003175) ν = 1.04 10 3 m /s Check on Re V D = ν = 40.6 10 3 (0.003175) 1.04 10 3 = 0.107 < 1 OK Pressure and Its Measurement 0. A mercury manometer is used to measure pressure at the bottom of a tank containing acetone, as shown in Figure P.0. The manometer is to be replaced with a gage. What is the expected reading in psig if h = 17 mm and x = 50.8 mm? -9
d Acetone ρ a = 787 kg/m3 3 open to Hg ρ = 13600 kg/m atmosphere p A + ρ a gx = p atm + ρg h )(0.0508)(/1) p A + 787(9.81 5 = hh 1.0135 10 + 13600(9.81)(0.17) x p A + 39. = 1.0135 10 5 + 16943.8 FIGURE P.0. p A = 1.18 10 5 Pa 1. Referring to Figure P.1, determine the pressure of the water at the point where the manometer attaches to the vessel. All dimensions are in inches and the problem is to be worked using Engineering or British Gravitational units. 10 oil (sp gr. = 0.85) ρ w g 10 ρ air g 5 ρ H g g 7 ρ c g 17 p W = + + = p air g c 1 g c 1 g c 1 g c 1 atm water FIGURE P.1. 5 mercury 10 7 p w 1.94(3.)(10/1) + 13.6(1.94)(3.)(7/1) 0.85(1.94)(3.)(17/1) = 14.7(144) p W 5.06 + 495.6 75. = 117 p W = 1749 psf = 1.14 psia. Figure P. shows a portion of a pipeline that conveys benzene. A gage attached to the line reads 150 kpa. It is desired to check the gage reading with a benzene-over-mercury U-tube manometer. Determine the expected reading h on the manometer. open to atmosphere D A pressure gage pipeline hh 30 mm C B mercury FIGURE P.. -10
p D + ρ Hg g h ρ B g(0.03) = p A p D = p atm = 0 0 + 13.6(1 000)(9.81)Δh 876(9.81)(0.03) = 150 000 (which is a gage reading) 0 + 133 400 h 57.8 = 150 000 h = 150 000 + 57.8 133 400 h = 1.16 m 3. An unknown fluid is in the manometer of Figure P.3. The pressure difference between the two air chambers is 700 kpa and the manometer reading h is 60 mm. Determine the density and specific gravity of the unknown fluid. air Because ρ air ρ liquid, then air hh p A p B = ρg h;h = 60 mm = 0.06 m, and FIGURE P.3. p A p B = 700 N/m given; so p A p B 700 ρ = 1 190 kg/m 3 = g h 9.81(0.06) = 4. A U-tube manometer is used to measure the pressure difference between two air chambers, as shown in Figure P.4. If the reading h is 15.4 mm, determine the pressure difference. air Because ρ air ρ liquid, then air p A p B = ρg h; h = 15.4 10 3 hh m FIGURE P.4. p A p B = 1000 kg/m 3 (9.81)(0.154) p A p B = 1495 Pa 5. A manometer containing mercury is used to measure the pressure increase experienced by a water pump as shown in Figure P.5. Calculate the pressure rise if h is 70 mm of mercury (as shown). All dimensions are in mm. -11
outlet p ρg 600 + 40 + 70 ρ g(0.07) H g outlet + 1000 water ρg(0.04) = p inlet 600 p outlet + 1 000(9.81)(0.71) 13.6(1 000)(9.81)(0.07) 40 pump motor 70 1 000(9.81)(0.04) = p inlet mercury inlet FIGURE P.5. p outlet + 6965 9339 39.4 = p inlet p outlet p inlet = 7 66 Pa =.77 kpa 6. Determine the pressure difference between the linseed and castor oils of Figure P.6. (All dimensions are in mm.) air linseed oil 76. 101.6 castor oil 17 114.3 water p A ρ L O g(0.076) + ρ air g(0.1016) + ρ HO g(0.17) ρ C O g(0.1143) = p B ρ L O = 930 kg/m 3 ; 3 ρ C O = 960 kg/m 3 ρ HO = 1000 kg/m ρ air negligible p A p B = ρ L O g(0.076) + ρ HO g(0.17) ρ C O g(0.1143) p A p B = 930(9.81)(0.076) 1000(9.81)(0.17) 960(9.81)(0.1143) p A p B = 695. 146.3 + 1076.8 p A p B = 56 Pa FIGURE P.6. 7. For the system of Figure P.7, determine the pressure of the air in the tank. open to atmosphere air oil 17 mm 50.8 mm p air + ρ oi l g(0.0508 + 0.154) ρg(0.17 + 0.0508 + 0.154) = p atm 03) 1000(9.81)(0.330) pair + 86(9.81)(0. 5 ρ = 86 kg/m 3 15.4 mm = 1.0135 10 5 p air + 1647 340 = 1.0135 10 sp. gr. = 1.0 p air = 1.03 10 5 Pa FIGURE P.7. -1
Continuity Equation 8. Figure P.8 shows a reducing bushing. A liquid leaves the bushing at a velocity of 4 m/s. Calculate the inlet velocity. What effect does the fluid density have? D 1 = 100 mm = 0.1 m; D = 40 mm = 0.04 m V = 4 m/s Density has no effect Q = A 1 V 1 = A V 0.5 m/s 40 mm π D 100mm 1 π D water V 1 = V 4 4 V 1 V 4 0. D 04 FIGURE P.8, P.9. = D 1 = 0.1 V 1 = 0.64 m/s 9. Figure P.9 shows a reducing bushing. Liquid enters the bushing at a velocity of 0.5 m/s. Calculate the outlet velocity. D 1 = 100 mm = 0.1 m; D = 40 mm = 0.04 m V 1 = 0.5 m/s 0.5 m/s 40 mm water 100mm Q = A 1 V 1 = A V π D 1 π D V 1 = V 4 4 V V1 0.50.1 D1 FIGURE P.8, P.9. = D = 0.04 V = 3.13 m/s 30. Water enters the tank of Figure P.30 @ 0.00189 m 3 /s. The inlet line is 63.5 mm in diameter. The air vent is 38 mm in diameter. Determine the air exit velocity at the instant shown. For low pressures and temperatures, air can be treated as incompressible. water inlet 0.00189 m 3 /s air exit QHO in = Q air out Q HO in = 0.00189 m /s 3 P HO = 1 000 kg/m 3 ρ air = 1.19 kg/m 3 03. mm Q AV π D π air out = = 4 V = 4 [(0.038)] = 457. mm 1.14 10 3 V FIGURE P.30. So 0.00189 = 1.14 10 3 V V air = 1.66 m/s -13
10 00 31. An air compressor is used to pressurize a tank of volume 3 m 3. Simultaneously, air leaves the tank and is used for some process downstream. At the inlet, the pressure is 350 kpa, the temperature is 0 C, and the velocity is m/s. At the outlet, the temperature is 0 C, the velocity is 0.5 m/s, and the pressure is the same as that in the tank. Both flow lines (inlet and outlet) have internal diameters of.7 cm. The temperature of the air in the tank is a constant at 0 C. If the initial tank pressure is 00 kpa, what is the pressure in the tank after 5 minutes? 0 = m t + (ρ AV ) out (ρ AV ) in m = ( p AV ) out ( p AV ) in = Substituting, V d p p out p out R T out pv R T A V out out p in R T A V 0 = R T d t + out out out R T in For constant T, all R T products cancel d p V = p A V out out out + p A V d t in in in A V in in m t p in R T in p out = p V d p = R T d t A V in in A in = d p 4 = 5.76 10 4 m = A out Areas are equal π (0.07) 3 d t d p = p(5.76 10 4 )(0.5) + 350 000(5.76 10 4 )() d p 3 d t = 400.8.863 10 4 p or d t = 133.6 9.543 10 5 p Separating variables, 133.6 = d t 00 000 9.543 10 5 p 300 d p p 0 ln (133.6 9.543 10 5 p) 9.543 5 000 p = 300 0 ln (133.6 9.543 10 5 p) ln (133.6 9.543 10 5 (00 000)) = 300( 9.543 10 5 ) ln (133.6 9.543 10 5 p) 4.741 =.863 10 ln (133.6 9.543 10 5 p) = 4.71 Exponentiating, 133.6 9.543 10 5 p = 1.113 10 or 9.543 10 5 p =.3 p =.34 kpa -14
3. Figure P.3 shows a cross-flow heat exchanger used to condense Freon-1. Freon-1 vapor enters the unit at a flow rate of 0.065 kg/s. Freon-1 leaves the exchanger as a liquid (Sp. Gr. = 1.915) at room temperature and pressure. Determine the exit velocity of the liquid. m in = ρ oυ τ A out V out vapor inlet 1/4 in. ID tubing fins m in = 0.065 kg/s 3 ρ = 1.915(1 000)kg/ m π D π (0.5/1) A = 4 = 4 = 3.41 10 4 ft A = 3.41 10 4 (9.9 10 ) = liquid 3.17 10 5 m outlet Substituting, FIGURE P.3. 0.065 = 1.915(1 000)3.17 10 5 )V out V out = 1.07 m/s 33. Nitrogen enters a pipe at a flow rate of 90.7 g/s. The pipe has an inside diameter of 101.6 mm. At the inlet, the nitrogen temperature is 6.7 C (ρ = 1.17 kg/m 3 ) and at the outlet, the nitrogen temperature is 77 C (ρ = 0.34 kg/m 3 ). Calculate the inlet and outlet velocities of the nitrogen. Are they equal? Should they be? m = 0.0907 kg D = 0.1016 m ρ 1 = 1.17 kg/m 3 ρ = 0.34 kg/m 3 A π D π (0.1016) 8.11 10 3 m m ρ AV = 4 = 4 = = V m 0.0907 ρ 1 1 = A = 1.17(8.11 10 3 ) 0.0907 V 1 = 9.56 m/s V = 0.34(8.11 10 3 ) V = 3.8 m/s Momentum Equation 34. A garden hose is used to squirt water at someone who is protecting herself with a garbage can lid. Figure P.34 shows the jet in the vicinity of the lid. Determine the restraining force F for the conditions shown. -15
Σ F = m (V out V in ) m in = m out frictionless flow magnitude of V in = magnitude of V out F = [ρ AV ]inlet ( V in V in ) 0 mm diameter F F = ρ AV ρ = 1 000 kg/m 3 3 m/s velocity π(0.0) A = 4 = 3.14 10 4 m V = 3 m/s F = (1 000)(3.14 10 4 )(3) FIGURE P.34. F = 5.65 N 35. A two-dimensional, liquid jet strikes a concave semicircular object, as shown in Figure P.35. Calculate the restraining force F. A, V FIGURE P.35. F Σ F = m (V out V in ) m in = m out frictionless flow magnitude of V in = magnitude of V out F [ρ AV ] ( = inlet Vin V ) in F = ρ AV 36. A two-dimensional, liquid jet strikes a concave semicircular object, as shown in Figure P.36. Calculate the restraining force F. Σ F m (V V ) = gc out in m in = m out frictionless flow F magnitude of V in = magnitude of V out A, V F = FIGURE P.36. [ρ AV ] inlet ( V in V in ) g c g c = 1 in SI units F = ρ AV g c 37. A two-dimensional liquid jet is turned through an angle θ (0 < θ < 90 ) by a curved vane, as shown in Figure P.37. The forces are related by F = 3F 1. Determine the angle θ through which the liquid jet is turned. -16
Σ F = m (V out V in ); m in = m out frictionless flow g c magnitude of V in = magnitude of V out F 1 = [ρ AV ] inlet (V V ) g c outx i n x F 1 V outx = V cos θ ; V i n x = V F F 1 = F 1 = [ρ AV ] inlet ρ AV g c (V cos θ V ) = g c (cos θ 1) ρ AV [ρ AV ] inlet (V V ) (1 cos θ ) F = outy i ny g c g c A, V FIGURE P.37. V out y = V sin θ ; ρ AV V i ny = 0 [ρ AV ] inlet F = g c (V sin θ ) = g c (sin θ ) F = 3F 1 ; sin θ = 3(1 cos θ ) 1 3 sin θ = 1 cos θ T&E solution is quickest θ (1/3) sin θ1 cos θ 45 0.357 0.99 50 0.553 0.357 40 0.143 0.34 35 0.191 0.1808 37 0.006 0.014 36.8 0.1997 0.1993 θ = 36.8 38. A two-dimensional liquid jet is turned through an angle θ (0 < θ < 90 ) by a curved vane as shown in Figure P.38. The forces are related by F 1 = F. Determine the angle θ through which the liquid jet is turned. θ Σ F = m (V out V in ); m in = m out frictionless flow g c magnitude of V in = magnitude of V out F 1 = [ρ AV ] inlet (Voutx g c V ) i n x θ V outx = V cos θ ;V i n x = V A, V F 1 F 1 g c [ρ AV ] inlet = ( V cos θ V ) = ρ AV ρ AV g c (cos θ + 1) F FIGURE P.39. F 1 = g c (1 + cos θ ) -17
F = [ρ AV ] inlet g c (V V ) outy i ny V out y = V sin θ ; V i ny = 0 [ρ AV ] inlet F = F 1 = F ; g c ρ AV (V sin θ ) == g c (sin θ ) 1 + cos θ = sin θ T & E solution is quickest θ 1 cos θ sin θ 45 1.707 1.414 50 1.643 1.53 55 1.574 1.638 53 1.60 1.597 54 1.588 1.618 53.5 1.595 1.608 53.4 1.596 1.606 53. 1.599 1.601 53.1 1.600 1.599 θ = 53.1 Energy Equation 39. Figure P.39 shows a water turbine located in a dam. The volume flow rate through the system is 0.315 m 3 /s. The exit pipe diameter is 1. m. Calculate the work done by (or power received from) the water as it flows through the dam. (Compare to the results of the example problem in this chapter.) 36.6 m p = p 1 = p atm We apply the energy equation between any two sections. Section 1 = the free surface upstream, and Section = the outlet downstream. 1. m V 1 0 (reservoir surface velocity) 1.83 m z = 1.83 m; z 1 = 36.6 m A = π D π (1.) 4 = 4 = 1.169 m FIGURE P.39. 3 V = Q 0.315 A = 1.169 = 0.7 m/s ρ = 1000 kg/m 3 Q = 0.315 m ρ V A = m = 1000(1.169)(0.7) = 315.5 kg/s evaluated at outlet /s -18
Substituting, t ρ gz ρ gzρ V A W = p + V + p +V + 1 t = (0 + + 9.81(1.83) (0 + 0 + 9.81(36.6)) 315.5 W 0.7 W = {(0 + 0.03645 + 17.95 0 0 359.05}315.5 t + W = 1.075 10 5 W t 40. Air flows through a compressor at a mass flow rate of 0.0438 kg/s. At the inlet, the air velocity is negligible. At the outlet, air leaves through an exit pipe of diameter 50.8 mm. The inlet properties are 101.3 kpa and 3.9 C. The outlet pressure is 87 kpa. For an isentropic (reversible and adiabatic) compression process, we have T = p (γ 1)/γ T 1 P 1 Determine the outlet temperature of the air and the power required. Assume that air behaves as an ideal gas (dh = c p dt, du = c v dt, and ρ = p/ R T ). (γ 1)/γ T = p T 1 P 1 Determine the outlet temperature of the air and the power required. Assume that air behaves as an ideal gas (dh = c p dt, du = c v dt, and ρ = p/ R T ). Solution: m = 0.0438 kg Vin = 0 Vout = unknown pin = 101.3 103 Pa pout = 87 103 Pa Dout = 0.0508 m Aou t = π D/4 = 0.0003 m γ = 1.4 Rai r = 8.314 J/K mole c pai r = 1004 J/kg K T out = p out T in T out (γ 1)/γ p in (73 + 3.9) = 87 10 3 (1.4 1)/1.4 3 = 1.8 T out = 96.9(1.8) 101.3 10 T out = 540.95 k = 68 C -19
p 87 10 3 9 ρ 5.33 kg/m 3 out R T = 8314(540.95) = t (h + V + gz) ou t (h + V V m 0.0438 4.05 m/s V out = 4.05 = 8. ρ A = out = out 5.33(0.0003) = W = + gz) in ρ V A V W out t =(h out h in + )ρ V AP E = 0 (h out h in ) = c p (T out T in ) = 1004(68 3.9) =.45 10 5 J/kg (h out h in ) =.45 10 5 J/kg W = (.45 10 5 + 8.)(0.0438) = 10735 W t or W = 14.4 HP Assuming no losses t 41. An air turbine is used with a generator to generate electricity. Air at the turbine inlet is at 700 kpa and 5 C. The turbine discharges air to the atmosphere at a temperature of 11 C. Inlet and outlet air velocities are 100 m/s and m/s, respectively. Determine the work per unit mass delivered to the turbine from the air. p in = 700 kpa T in = 5 C V in = 100 m/s p out = 101.3 kpa T out = 11 C V out = m/s c p = 1005.7 J/(kg K) W t V = h + g c + gz g h c out + V gz g g c + c in ρ V A W / t (h out h in ) V V out in g (z out z in ) m = + 3 + g +g (h out h in ) = c p (T out T in ) c c z z out = in W / t = 1 005.7(5 11) + 100 = 1.4 104 5 103 m + W / t = 9 103 J/kg m c -0
4. A pump moving hexane is illustrated in Figure P.4. The flow rate is 0.0 m 3 /s; inlet and outlet gage pressure readings are 4 kpa and 190 kpa, respectively. Determine the required power input to the fluid as it flows through the pump. We apply the energy equation between any two sections. Section 1 = inlet pressure gage (actually the centerline of the pipe where the pressure gage is attached), and Section = outlet pressure gage. p 75 mm p = 190 000 Pa p 1 = 4000 Pa z = 1.5 m z 1 = 1.0 m 1.5 m p 1 pump motor AV = 0.0 m 3 /s 1.0 m A 1 = π D 1 π(0.10) 4 = 4 = 7.854 10 3 m π D π(0.075) A = 4 = 4 = 4.4 10 3 m Q 0.0 Q V 1 = A = 7.854 10 3 =.55 m/sv = A = 1 ρ = 0.657(1 000) for hexane W pv gz pv gz t = ρg + g ρg + g ρ V A + c c + c c 1 FIGURE P.4. 0.0 100 mm = 4.5 m/s 4.4 10 3 W 190 000 4.5 4 000.55 1.5(9.81) 1.0(9.81) 657(0.0) t = 657 + + 657 + + W = {89. + 10. + 14.7 + 6.088 3.5 9.81} (13.14) t W = 4.04 10 3 N m/s = 4.0 kw t Bernoulli Equation 43. Figure.15 shows a venturi meter. Show that the Bernoulli and continuity equations when applied combine to become Q = A g h 1 ( D 4 / D1 4 ) Hydrostatic equation for manometer; all measurements are from the centerline p 1 ρ 1 g x ρ 1 g h = p ρ 1 g x ρ g h or p 1 p = ρ 1 g h -1
m 1 = m ρ 1 A 1 V 1 = ρ A V or A 1 V 1 = A V π D 1 π D In terms of diameter, 4 V 1 = 4 Bernoulli Equation p 1 V 1 p V V = Q ρ 1 g + g + z 1 = ρ 1 g + g + z With z 1 = z, (p 1 p ) = 1 (V V1 ) Substitute for V in terms of Q ( p 1 p ) g ρ 1 g gρ1g =Q (1/ A 1 1/ A ) 1 ρ g hq 1 A Q D 1 4 4 ρ g 1 c A A A = 1 = D 1 4 4 A g h = Q 1 D / D1 or finally, Q = A g h 1 D 4 4 / D 1 44. A jet of water issues from a kitchen faucet and falls vertically downward at a flow rate of 4.44 10 5 m 3 /s. At the faucet, which is 355.6 mm above the sink bottom, the jet diameter is 15.88 mm. Determine the diameter of the jet where it strikes the sink. Q = 4.44 10 5 m 3 /s D 1 = 0.01588 m A 1 = 1.98 10 4 m 1 Q V 1 = A 1 = 0. m/s h = z 1 = 0.3556 m D 1 Bernoulli Equation p1 V 1 p V ρ 1 g + g + z 1 = ρ 1 g + g + z h p 1 = p z 1 = 0.3556 m z = 0 Substituting, 0. 1 D V 0 + (9.81) + 9.81 0.3556= 0 + (9.81) + 0 which becomes -
(.51 10 3 + 0.3556)((9.81)) = V or V =.65 m/s A Q 4.44 10 5 1.675 10 5 m = V =.65 = π D 1.675 10 5 D 4 (1.675 10 5 ) or 4 = = π D = 4.6 10 3 m = 4.6 mm 45. A jet of water issues from a valve and falls vertically downward at a flow rate of 3 10 5 m 3 /s. The valve exit is 50 mm above the ground; the jet diameter at the ground is 5 mm. Determine the diameter of the jet at the valve exit. Section 1 is the exit; section is the ground. p 1 V 1 p V ρ 1 g + g + z 1 = ρ 1 g + g + z D 1 1 Q = 3 10 5 m 3 /s; p 1 = p = p atm ; z = 0; z 1 = 0.05 m h π(0.005) D = 5 mm; A = 4 = 1.963 10 5 m V Q 30 10 6 1.53 m/s = A = 1.963 10 5 = ; Bernoulli Equation becomes V 1 g + z 1 = V g V 1 1.53 (9.81) = (9.81) 0.05 = 0.06931 D V 1 = 1.36; V1 = 1.17 m/s π D Q A 1 1 V 1 V 1 D 1 = = 4 = D 1 4(30 10 6 ) = π (1.17) D 1 = 5.7 mm = 5.7 4Q π V 1 10 3 m -3
46. A garden hose is used as a siphon to drain a pool, as shown in Figure P.46. The garden hose has a 19 mm inside diameter. Assuming no friction, calculate the flow rate of water through the hose if the hose is 5 ft long. 1. m 19 mm ID 7.6 m long FIGURE P.46. Section 1 is the free surface; section is the hose outlet. p 1 V 1 ρ 1 g p V ρ 1 + g + z 1 = g + Substituting, V g 0 + 0 + (1.) = 0 + (9.81) + 0 V = (9.81)(1.) = 4.89 m/s D = 19 mm A = Q = AV =.835 10 4 (4.89); Q = 1.386 10 3 m 3 /s π D =.835 10 4 m 4 + z p 1 = p = p atm ; V 1 = 0; z 1 = 1. m Miscellaneous Problems 47. A pump draws castor oil from a tank, as shown in Figure P.47. A venturi meter with a throat diameter 50.8 mm is located in the discharge line. For the conditions shown, calculate the expected reading on the manometer of the meter. Assume that frictional effects are negligible and that the pump delivers 186.5 W to the liquid. If all that is available is a 1.83 m tall manometer, can it be used in the config-uration shown? If not, suggest an alternative way to measure pressure difference. (All measurements are in mm.) -4
4 air air hh outlet 76 559 76. ID 50.8 throat inside diameter 178 76. ID pump motor FIGURE P.47. p 1 ρg x ρ air g(0.559) + ρg(0.559 + x 0.178) = p ρ air is negligible x terms cancel; ρ = 960 kg/m 3 p p 1 = ρg(0.559 0.178) = 960(9.81)(0.381) = 3588 Pa Energy equation, 1 to : W p V gz p V gzρ V A t ρ ρ = ++ ++ 1 D 1 = D = 0.076 m A 1 V 1 = A V so V 1 = V z 1 = 0 z = 0.178 m The power was given as ρ AV =ρ Q W = 186.5 W Substituting, t = ρ + = 960 + 186.5 ρ Q ( p p 1 ) gz 960Q 3588 9.81(0.178) Solving for Q Q = 0.0354 m 3 /s Now for the venturi meter, the throat diameter is D th = 0.0508 m D = 0.076 m A th = π D th =.03 10 3 m 4 Q = Ath 1 Dth 4 / D 4 g h 0.0354.03 10 3 (9.81)Δh = 1 (0.0508/0.076) -5
h = 1.44 m of castor oil A 1.83 m tall air-over-oil manometer is not tall enough. A Hg manometer will work; pressure transducers will also work. 48. A 4 mm ID pipe is used to drain a tank, as shown in Figure P.48. Simultaneously, a 5 mm ID inlet line fills the tank. The velocity in the inlet line is 1.5 m/s. Determine the equilibrium height h of the liquid in the tank if it is octane. How does the height change if the liquid is ethyl alcohol? Assume in both cases that frictional effects are negligible, and that z is 40 mm. inlet h exit z FIGURE P.48. Q in = AVA = π (0.05) =.14 10 3 m 4 Q in =.14 10 3 (1.5) = 3.19 10 3 m 3 /s Section 1 is the free surface in the tank, and is at the exit of the pipe. Apply the Bernoulli equation, 1 to : p 1 V 1 p V + z ρ 1 g + g + z 1 = ρ 1 g + g p 1 = p = p atm ; V h = g + z ; At equilibrium, Q out V 1 = 0; z 1 = h; z = 0.04 m; the Bernoulli equation becomes = Q in = 3.19 10 3 m 3 /s A out π(0.04) 1.39 10 3 m ; and V = 4 = V.3 h = g + z = (9.81) + 0.04 Q 3.19 10 3.3 m/s = Aout = 1.39 10 3 = h = 0.309 m which is independent of fluid properties, and with no friction -6
θ Computer Problems 49. One of the examples in this chapter dealt with the following impact problem, with the result that the ratio of forces is given by: F x (cos θ 1 cos θ ) F y = (sin θ + sin θ ) For an angle of θ 1 = 0, produce a graph of the force ratio as a function of the angle θ. 14.00 1.00 A, V 10.00 y x θ 1 F y F x Force ratio 8.00 6.00 4.00.00 0.00 0 50 100 150 00 theta in degrees FIGURE P.49. 50. One of the examples in this chapter involved calculations made to determine the power output of a turbine in a dam (see Figure P.50). When the flow through the turbine was 3.15 m 3 /s, and the upstream height is 36.6 m, the power was found to be 1.06 kw. The relationship between the flow through the turbine and the upstream height is linear. Calculate the work done by (or power received from) the water as it flows through the dam for upstream heights that range from 18.3 to 36.6 m. D 1 1 36.6 m 1. 80 mm m 1.83 m FIGURE P.50. FIGURE P.51. 0D.3 cm -7
inkw dw/dt 1194 1044 895 746 597 448 98 149 0 0 15 30 45 h in m Diameter in mm 3.5 3.5 1.5 1 0.5 0 0 0.1 0. 0.3 0.4 Volume in m 3 x 10 3 51. One of the examples in this chapter dealt with a water jet issuing from a faucet. The water flow rate was 3.15 10 5 m 3 /s, the jet diameter at faucet exit is 3.5 mm, and the faucet is 80 mm above the sink. Calculations were made to find the jet diameter at impact on the sink surface. Repeat the calculations for volumes per time that range from 1.5 10 5 m 3 /s to 6.5 10 5 m 3 /s, and graph jet diameter at as a function of the volume flow rate. -8