Solution manual for design of fluid thermal systems SI edition 4th edition by janna

Chapter Download Solution Manual for Design of Fluid Thermal Systems SI Edition 4th Edition by Janna https://getbooksolutions.com/download/solution-manual-for-design-of-fluid-thermalsystems-si-edition-4th-edition-by-janna Density, Specific Gravity, Specific Weight ◦ What is the specific gravity of 38 API oil? ◦ 141.5 38 API oil sp.gr = = 141.5 131.5 API 131.5 ◦ + 141.5 sp gr = 169.5 = 0.835 38 + ◦ The specific gravity of manometer gage oil is 0.826 What is its density and its API rating? sp gr = 0.826; ρ = 1000(0.826) = 826 kg/m ρ = 62.4(0.826) = 51.54 lbm/ft sp gr = ◦ ◦ 131.5 + API = 141.5 141.5 131.5 API 0.826 ◦ + − = API 171.3 ; 131.5 ◦ = API ◦ 39.8 API ≈ ◦ 40 API ◦ ◦ What is the difference in density between a 50 API oil and a 40 API oil? 141.5 141.5 ◦ sp gr = 131.5 API = 131.5 50 = 0.7796 for a 50 oil +◦ + 141.5 141.5 ◦ sp gr = 131.5 API = 131.5 40 = 0.826 for a 40 oil ◦ + + 0.825 − 0.7796 = 0.0455 density difference ◦ A 35 API oil has a viscosity of 0.825 N·s/m Express its viscosity in Saybolt Universal Seconds (SUS) 141.5 ◦ 35 AFI oil sp gr = 131.5 ◦API + μ = 0.825 N·s/m ν = μgc ρ 141.5 = 131.5 35 = 0.850 + −4 0.825 = 10 × 10 = 0.850(1000) Highly viscous; try ν = 0.2158 × 10 −6 10 × 10 SUS = (SUS) −4 0.2158 × 10−6 if SUS > 215 = 4633 SUS c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-1 Air is collected in a 1.2 m container and weighed using a balance as indicated in Figure P2.5 On ◦ the other end of the balance arm is 1.2 m of CO2 The air and the CO2 are at 27 C and atmospheric pressure What is the difference in weight between these two volumes? CO2 air FIGURE P2.5 ◦ Air at 27 C = 300 K has ρ = 1.177 kg/m ◦ CO2 at 27 C = 300 K has ρ = 1.797 kg/m For a volume of 1.2 m , the weight of air is 3 (1.177 kg/m )(1.2 m )(9.81 m/s ) = 13.86 N For CO2 3 (1.797 kg/m )(1.2 m )(9.81 m/s ) = 21.15 N Weight difference is 21.15 − 13.86 = 7.29N 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 N in air While submerged, the object weighs N What is the density of the liquid? Castor Oil ρ = 960 kg/m buoyant force volume = ρ mg in air − mg submerged = ρg V 9−7 551 kg/m = (0.03) 9.81 = A brass cylinder (Sp Gr = 8.5) has a diameter of 25.4 mm and a length of 101.6 mm It is submerged in a liquid of unknown density, as indicated in Figure P2.7 While submerged, the weight of the cylinder is measured as 3.56 N Determine the density of the liquid weight FIGURE P2.7 submerged object Buoyant force = mgin air − mgsubmerged = mg − 0.8 mg − 0.8 ρg V π D h buoyant force π (0.0254)2(0.1016) 5.15 10 volume = V = = = = × −5 mg = ρbV g = 8500(5.15 × 10 )(9.81) = 4.29 N mg − 0.8 4.29 − 3.56 ρ × = = 9.81(5.15 ρ = 1454 kg/m gV 5m − − 10 ) c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-2 Viscosity Actual tests on Vaseline yielded the following data: τ in N/m 200 600 000 d V /d y in 1/s 500 000 200 Determine the fluid type and the proper descriptive equation 1200 shear stress 1000 800 600 400 200 0 τ=K dV 500 1000 strain rate 1500 n dy Can be done instantly with spreadsheet; hand calculations follow for comparison purposes: dV/dy 500 1000 1200 Sum ln(dV/dy) — 6.215 6.908 7.090 20.21 ln τ — 5.298 6.397 6.908 18.60 τ 200 600 1000 ln(τ)· ln(dV/dy) · 32.93 44.19 48.98 126.1 m = Summation (ln(dV/dy)) = 136.6 b 1= 3(126.1) − 20.21(18.60) 3(136.6) 20.212 = 1.766 − 20.21 18.60 b0 = − 1.766 = −5.697 K = exp(b0) = 0.00336; dV τ = τo + K dy n n = b1 = 1.766 dV 1.766 = 0.00336 dy c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-3 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 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 200 180 stress 160 140 120 she ar 100 80 60 40 20 0 10 15 20 strain rate dV dV n n which becomes τ = τ − τo = K dy τ = τo + K dy Can be done instantly with spreadsheet; hand calculations: dV/dy 15 Sum τ τ 40 100 60 140 100 180 140 ln(dV/dy) — 1.099 1.946 2.708 5.753 ln τ — 4.094 4.605 4.942 13.64 ln(τ )· ln(dV/dy) — 4.499 8.961 13.38 26.84 m = Summation (ln(dV/dy)) = 12.33 3(26.84) − 5.753(13.64) b 1= 3(12.33) 5.7532 − 13.64 5.753 b0 = − 0.526 K = exp(b0) = 34.37; dV τ = τo + K dy = 0.526 = 3.537 n = b1 = 0.526 dV n = 0.526 40 + 34.37 dy where dV/dy is in rev/s and τ in g/cm ; these are not standard units c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-4 10 A cod-liver oil emulsion is tested with a viscometer and the following data were obtained: τ in d V /d y in rev/s 0 40 0.5 60 1.7 80 120 Graph the data and determine the fluid type Derive the descriptive equation Cod liver oil; graph excludes the first data point 140 shear stress 120 100 80 60 40 20 0 strain dV rate n τ = K dy Can be done instantly with spreadsheet; hand calculations: dV/dy 0.5 1.7 Sum ln(dV/dy) −0.6931 0.5306 1.099 1.792 2.729 τ 40 60 80 120 ln τ 3.689 4.094 4.382 4.787 16.95 ln(τ)· ln(dV/dy) −2.557 2.172 4.816 8.578 13.01 m = Summation (ln(dV/dy)) = 5.181 b 1= 4(13.01) − 2.729(16.95) 2.7292 4(5.181) = 0.4356 − 16.95 2.729 b0 = − 0.4356 = 3.537 K = exp(b0) = 51.43; dV τ = τo + K dy n = b1 = 0.4356 dV n = 51.43 0.4356 dy where dV/dy is in rev/s and τ in lbf/ft ; these are not standard units c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-5 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 0.01243 mN-m Calculate the viscosity of the fluid If the fluid density is 850 kg/m , calculate the kinematic viscosity Rotating cup viscometer R = 25.4 mm δ = 5.08 mm L = 63.5 mm ω = (10 rev/min)·(2π rad/rev)(1 /60 s) = 1.047 rad/s = T = 0.01243 × 10 −3 dV dy N-m ρ = 850 kg/m Tδ μ= μ= 2π R (R + δ)Lω −5 −3 1.243 × 10 × 5.08 × 10 −3 2π (0.0254) (0.0254 + 5.08 × 10 )(0.0635)(1.047) μ = 7.7 × 10 v −3 Pa·s −3 7.7 × 10 9.762 10 ft /s = = × − 850 9.06 10 × 6m − /s 12 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 42 mm The viscometer is used to measure the viscosity of a liquid When the outer cylinder rotates at 12 rev/min, the torque on the inner cylinder is measured −6 to be × 10 N·m Determine the kinematic viscosity of the fluid if its density is 000 kg/m R = 38/2 = 0.019 m; L = 0.08 m Routside = 42/2 = 21 mm δ = 21 − 19 = mm = 0.002 m ω = (12 rev/min)(2π/60) = 1.26 rad/s T = 3.8 × 10 μ −6 N·m ρ = 000 kg/m −6 3.8 × 10 (0.002) 2 =2π R (R + δ)(Lω) = 2π (0.019) (0.019 + 0.002)(0.08)(1.26) Tδ μ = 1.58 × 10 −3 N·s/m −3 1.58 × 10 −6 v =ρ = = 1.58 × 10 m /s 000 c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-6 13 A rotating cup viscometer has an inner cylinder diameter of 57.15 mm and an outer cylinder diameter of 62.25 mm The inner cylinder length is 76.2 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 = 28.58 mm 2(R + δ) = 62.25 mm R + δ = 31.125 ρ = 968 kg/m δ = 2.545 mm μ = 0.0421 Pa·s ω = (15 rev/ min)(2π/60) = 1.572 rad/s 2 2π (0.02858) (0.031125)(0.0762)(1.571)(0.0421) T 2π R (R + δ)(L ω)μ = δ = 0.002545 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 , vis3 cosity is 0.89 × 10 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 = 76.2 mm and z = 254 mm, determine the maximum tube size permissible zπR V ρ = 1000 kg/m Capillary tube viscometer t = ρg L 8μ μ = 0.89 × 10 −3 N·s/m z = 0.254 m L = 0.0762 m V t = Volume flow rate = AV = πR V; substituting into the equation, zπR πR V = ρg L 8μ 2 zR Rearrange and solve for V, V = ρg L 8μ The limiting value is Re < 2100; using equality, V(2R) ν = 2100; 2100μ V= R ρV(2R) = 2100 or μ zR 2ρR = ρg L 8μ Rearrange and solve for R −3 2100μ (8)(L) 2100(0.89 × 10 ) (8)(0.0762) 2 2ρ gz 2(1000) (9.81)(0.254) = = −10 R = 2.035 × 10 or −4 R = 5.88 × 10 m = 0.588 mm Any larger, flow no longer laminar c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-7 Σ F = m˙(Vout − Vin ) m˙in = m˙out frictionless flow magnitude of Vin = magnitude of Vout F [ρ AV ] = 20 mm diameter F inlet F = 2ρ AV m/s velocity − in − V ) in ρ = 000 kg/m π(0.02) A= ( V = 3.14 × 10 F = 2(1 000)(3.14 × 10 F = 5.65 N −4 −4 m V = m/s )(3) FIGURE P2.34 35 A two-dimensional, liquid jet strikes a concave semicircular object, as shown in Figure P2.35 Calculate the restraining force F Σ F = m˙(Vout − Vin ) m˙in = m˙out frictionless flow A, V F magnitude of Vin = magnitude of Vout F [ρ AV ] = inlet F = 2ρ AV ( V − in − V ) in FIGURE P2.35 36 A two-dimensional, liquid jet strikes a concave semicircular object, as shown in Figure P2.36 Calculate the restraining force F ΣF = m˙ (V gc out − V ) in m˙in = m˙out frictionless flow F magnitude of Vin = magnitude of Vout [ρ AV ]inlet A, V F= g c (−Vin − Vin ) gc = in SI units FIGURE P2.36 F= 2ρ 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 P2.37 The forces are related by F2 = 3F1 Determine the angle θ through which the liquid jet is turned c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-16 ΣF m˙ (V V ) = gc out − in m m frictionless flow ; ˙ in = ˙ out magnitude of Vin = magnitude of Vout [ρ AV ]inlet −F1 = (V outx − gc V inx ) F1 Voutx = V cos θ ; Vi n x = V F2 ρ AV [ρ AV ]inlet −F = F1 = (V cos θ − V ) = gc ρ AV Vout y = V sin θ ; ρ AV [ρ AV ]inlet F2 = (V sin θ ) = gc FIGURE P2.37 gc (cos θ − 1) [ρ AV ] inlet (V V ) outy − i ny F2 = gc (1 − cos θ ) gc θ A, V Vi ny = gc (sin θ ) F2 = 3F1; sin θ = 3(1 − cos θ ) sin θ = − cos θ T&E solution is quickest θ (1/3) sin θ − cos θ ◦ 45 0.2357 0.2929 ◦ 50 0.2553 0.3572 ◦ 40 0.2143 0.234 ◦ 35 0.1912 0.1808 ◦ 37 0.2006 0.2014 36.8 ◦ θ = 36.8 0.1997 0.1993 ◦ ◦ ◦ 38 A two-dimensional liquid jet is turned through an angle θ (0 < θ < 90 ) by a curved vane as shown in Figure P2.38 The forces are related by F1 = 2F2 Determine the angle θ through which the liquid jet is turned ΣF m˙ (V = gc V ); out − mm frictionless flow ˙ in = ˙ out in magnitude of Vin = magnitude of Vout [ρ AV ] −F1 = inlet gc (V outx − V inx θ ) F1 Voutx = −V cos θ ;Vi n x = V [ρ AV ]inlet ρ AV g g −F1 = ρ AV F1 = A, V c (−V cos θ − V ) = − c (cos θ + 1) F2 FIGURE P2.39 gc (1 + cos θ ) c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-17 [ρ AV ] inlet F2 = V out y gc (V = V sin θ ; [ρ AV ] F2 = outy − V ) i ny Vi ny = ρ AV inlet (V sin θ ) == gc gc (sin θ ) + cos θ = sin θ F1 = 2F2; T & E solution is quickest θ ◦ 45 ◦ 50 ◦ 55 ◦ 53 ◦ 54 ◦ 53.5 ◦ 53.4 ◦ 53.2 53.1 ◦ θ = 53.1 − cos θ 1.707 1.643 1.574 1.602 1.588 1.595 1.596 1.599 sin θ 1.414 1.532 1.638 1.597 1.618 1.608 1.606 1.601 1.600 1.599 ◦ Energy Equation 39 Figure P2.39 shows a water turbine located in a dam The volume flow rate through the system is 0.315 m /s The exit pipe diameter is 1.22 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.) We apply the energy equation between any two sections Section = the free surface upstream, and Section = the outlet downstream 36.6 m p2 = p1 = patm 1.22 V1 ≈ (reservoir surface velocity) m z2 = 1.83 m; z1 = 36.6 m 2 πD π (1.22) A2 = = 1.169 m = 1.83 m FIGURE P2.39 Q = 0.315 m /s Q V2 = A 0.315 = 1.169 = 0.27 m/s ρ = 1000 kg/m ρ V A = m˙ = 1000(1.169)(0.27) = 315.25 kg/s evaluated at outlet c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-18 Substituting, ∂t ρ − ∂W = p + V2 + ∂∂ t − = (0 + W − ρ gz 2− p +V2 + gzρ V A + 9.81(1 83) − (0 + + 9.81 (36.6 )) 315.25 0.27 ∂W = {(0 + 0.03645 + 17.95 − − − 359.05}315.25 ∂t ∂W + = 1.075 × 10 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 23.9 C The outlet pressure is 827 kPa For an isentropic (reversible and adiabatic) compression process, we have T p = T1 (γ −1)/γ P1 Determine the outlet temperature of the air and the power required Assume that air behaves as an ideal gas (dh = c p dT, du = cv dT, and ρ = p/ R T ) T p = T1 (γ −1)/γ P1 Determine the outlet temperature of the air and the power required Assume that air behaves as an ideal gas (dh = c p dT, du = cv dT, and ρ = p/ R T ) Solution: m˙ = 0.0438 kg Vin = Vout = unknown pin = 101.3 × 10 Pa pout = 827 × 10 Pa D A out = 0.0508 m Rair = 8.314 J/K mole T p out T in out p = out T out 2 = π D /4 = 0.00203 m γ = 1.4 c pai r = 1004 J/kg K (γ −1)/γ in T (273 out 827 × 10 23.9) = (1.4−1)/1.4 = 1.822 Tout = 296.9(1.822) 101.3 × 10 + ◦ = 540.95 k = 268 C c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-19 827 × 10 × 29 p ρ RT out = 8314(540.95) = 0.0438 m˙ V = out − ∂W ρA out = = ∂t 4.05 m/s Vout = 4.05 = 8.2 5.33(0.00203) = 2 (h + V2 + gz) ∂W Vout = (hout − hin + − ∂t 5.33 kg/m 2+ ρVA gz) in ou t − (h + V2 )ρ V AP E = (hout − hin ) = cp(Tout − Tin ) = 1004(268 − 23.9) = 2.45 × 10 J/kg (hout − hin ) = 2.45 × 10 J/kg − ∂W = (2.45 × 10 + 8.2)(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 25 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 pin = 700 kPa pout = 101.3 kPa ◦ ◦ T Tin = 25 C out = 11 C V Vin = 100 m/s out = m/s c p = 1005.7 J/(kg·K) ∂W − ∂t = h + V 2g c + V2 gz g c (h ∂ W /∂ t out hin ) − m = − + out − V m = 005.7(25 − 11) out 2 + = + + ρVA in c +g (z out − zin ) c c z gz g g in2 c ∂ W /∂ t c + 2g 23 (hout − hin ) = c p (Tout − Tin ) 2g + Vout ˙ − h z in 100 10 10 = 1.4 × −5 × ˙ − ∂ W /∂ t = × 10 J/kg ˙ m c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-20 42 A pump moving hexane is illustrated in Figure P2.42 The flow rate is 0.02 m /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 = inlet pressure gage (actually the centerline of the pipe where the pressure gage is attached), and Section = outlet pressure gage 75 mm p2 p1 p2 = 190 000 Pa z2 = 1.5 m 1.5 m p1 = −4000 Pa z1 = 1.0 m /s pump motor AV = 0.02 m 1.0 m 2 π D1 π(0.10) −3 A1 = = 7.854 × 10 m = π D2 A2 = π(0.075) = 4.42 × 10 Q 0.02 V1 = A = 7.854 10 = 2.55 m/s × − = ρ = 0.657(1 000) for hexane ∂W pV gz ∂t ρ 2g g − =+ c +c ∂W 4.52 190 000 100 mm −3 m FIGURE P2.42 Q 0.02 V2 = A = 4.42 10 = 4.52 m/s × − pV 2 − ρ2g + c + gz g ρVA c −4 000 1.5(9.81) 1.0(9.81) − ∂t = − = {289.2 + 10.22 + 14.72 + 6.088 − 3.25 − 9.81} (13.14) ∂W − ∂t ∂W 657 + + − 2.55 657 + 657(0.02) + = 4.04 × 10 N·m/s = 4.0 kW ∂t Bernoulli Equation 43 Figure 2.15 shows a venturi meter Show that the Bernoulli and continuity equations when applied combine to become 2g h Q = A2 4 − ( D2 / D1 ) Hydrostatic equation for manometer; all measurements are from the centerline p1 − ρ1 g x − ρ1 g h = p2 − ρ1 g x − ρ2 g h or p1 − p2 = −ρ1 g h c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-21 m˙1 = m˙2 ρ1 A1 V1 = ρ2 A2 V2 or A1 V1 = A2 V2 π D1 π D2 In terms of diameter, V1 = V2 = Q Bernoulli Equation p1 V1 p2 V2 ρ1 g + 2g + z1 = ρ1 g + 2g + z2 With z1 = z2, ( p1 − p2) (V2 V 2) Substitute for V in terms of Q ρ1 g = 2g ( p1 − p2) 2g ρ1 g − 2 Q (1/ A = 2ρ1 g hQ 2 − A2 1 Q 2 2 ρ g 1/ A ) c = A A2 √ 2g h = Q A − = 4 − D2 / D1 A D24 − D or finally, 2g h Q = A2 − D2 / D1 44 A jet of water issues from a kitchen faucet and falls vertically downward at a flow rate of 4.44 × −5 10 m /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 /s D1 = 0.01588 m A1 = 1.98 × 10 −4 m Q V1 = A = 0.222 m/s h = z1 = 0.3556 m D1 Bernoulli Equation p ρ1 g V1 p2 V2 h + 2g + z1 = ρ1 g + 2g + z2 p1 = p2 z1 = 0.3556 m z2 = Substituting, 0.222 + 2(9.81) V2 D2 + 9.81 0.3556= + 2(9.81) +0 which becomes c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-22 (2.512 × 10 −3 + 0.3556)(2(9.81)) = V2 or V2 = 2.65 m/s 4.44 × 10 Q A 2= V2 = π D2 1.675 = 2.65 10 1.675 = −5 D −5 × D2 = 4.62 × 10 m2 − (1.675 = π × −3 10 10 −5 ) or × m = 4.62 mm −5 45 A jet of water issues from a valve and falls vertically downward at a flow rate of × 10 m /s The valve exit is 50 mm above the ground; the jet diameter at the ground is mm Determine the diameter of the jet at the valve exit Section is the exit; section is the ground p1 V1 p V2 ρ1 g + 2g + z1 = ρ1 g D1 + 2g + z2 −5 m /s; p1 = p2 = patm ; z2 = 0; z1 = 0.05 m π(0.005) −5 D2 = mm; A2 = = 1.963 × 10 m −6 30 × 10 Q V 1.53 m/s A = −5 1.963 10 = ; = × Q = × 10 h D2 Bernoulli Equation becomes V1 V2 2g + z1 = 2g 2 V1 1.53 2(9.81) = 2(9.81) − 0.05 = 0.06931 V1 = 1.36; V1 = 1.17 m/s π D1 V A1 V1 D1 Q = = = D 4(30 × 10 = −6 π (1.17) ) π V1 10 5.7 = 4Q −3 m × D1 = 5.7 mm c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-23 46 A garden hose is used as a siphon to drain a pool, as shown in Figure P2.46 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 25 ft long 19 mm ID 1.22 m 7.6 m long FIGURE P2.46 Section is the free surface; section is the hose outlet p1 ρ1 g V1 p2 V2 + 2g + z1 = ρ1 g + 2g + z2p1 = p2 = patm ; V1 = 0; z1 = 1.22 m Substituting, V2 + + (1.22) = + √ V2 = 2(9.81) + 2(9.81)(1.22) = 4.89 m/s πD D = 19 mm A= Q = AV = 2.835 × 10 Q = 1.386 × 10 −3 −4 = 2.835 × 10 −4 m (4.89); m /s Miscellaneous Problems 47 A pump draws castor oil from a tank, as shown in Figure P2.47 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 configuration shown? If not, suggest an alternative way to measure pressure difference (All measurements are in mm.) c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-24 air air hh 559 outlet 50.8 throat inside diameter 76.2 ID 762 178 76.2 ID pump FIGURE P2.47 motor p1 − ρg x − ρair g(0.559) + ρg(0.559 + x − 0.178) = p2 ρ = 960 kg/m ρair is negligible x terms cancel; p2 − p1 = ρg(0.559 − 0.178) = 960(9.81)(0.381) = 3588 Pa Energy equation, to 2: ∂W − ∂t = V p ρ 2 ++ gz p ρ − D1 = D2 = 0.0762 m A1 V1 = A2 V2 z1 = ρ AV =ρ Q z2 = 0.178 m V 2 so ++ gzρ V A V1 = V2 The power was given as ∂W − ∂ t = 186.5 W Substituting, = ρ + = 186.5 ρQ ( p2 − p1) gz 960Q 960 + 3588 9.81(0.178) Solving for Q Q = 0.0354 m /s Now for the venturi meter, the throat diameter is Dth = 0.0508 m π Dth Ath = D = 0.0762 m −3 = 2.03 × 10 m 4 − Dth / D Q = Ath 2g h 0.0354 10 2.03 = × −3 − 2(9.81)Δh (0.0508/0.0762) c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-25 h = 12.44 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 42 mm ID pipe is used to drain a tank, as shown in Figure P2.48 Simultaneously, a 52 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 exit h z FIGURE P2.48 π (0.052) Qin = AVA = = 2.124 × 10 Qin = 2.124 × 10 −3 (1.5) = 3.19 × 10 −3 −3 m m /s Section is the free surface in the tank, and is at the exit of the pipe Apply the Bernoulli equation, to 2: p1 V1 p2 ρ1 g + 2g + z1 = ρ1 g p1 = p2 = patm ; V2 h = 2g + z2; π(0.042) A V2 + 2g + z2 V1 = 0; z1 = h; z2 = 0.04 m; the Bernoulli equation becomes At equilibrium, Qout 1.39 10 − out = V2 = 2.3 m2 ; = Qin = 3.19 × 10 Q and V × = Aout −3 m /s −3 3.19 × 10 −3 = 1.39 × 10 2.3 m/s = h = 2g + z2 = 2(9.81) + 0.04 h = 0.309 m which is independent of fluid properties, and with no friction c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-26 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 θ1 − cos θ2) Fy = (sin θ2 + sin θ2) For an angle of θ1 = 0, produce a graph of the force ratio as a function of the angle θ2 14.00 rat io 12.00 10.00 F o r c e A, V θ 8.00 θ2 6.00 y 4.00 Fx 2.00 x Fy 0.00 50 100 150 200 theta in degrees FIGURE P2.49 50 One of the examples in this chapter involved calculations made to determine the power output of a turbine in a dam (see Figure P2.50) When the flow through the turbine was 3.15 m /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 D1 36.6 m 1.22 m 280 mm 1.83 m 0D.3 cm FIGURE P2.50 FIGURE P2.51 c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-27 i m n m D i a m e t e r 1194 3.5 1044 dW/dt in Kw 895 2.5 746 448 1.5 149 0.5 597 298 0 15 30 45 h in m 0.1 0.2 0.3 −3 Volume in m x 10 51 One of the examples in this chapter dealt with a water jet issuing from a faucet The water flow rate −5 was 3.125 × 10 m /s, the jet diameter at faucet exit is 3.5 mm, and the faucet is 280 mm above the sink Calculations were made to find the jet diameter at impact on the sink surface Repeat the −5 −5 calculations for volumes per time that range from 1.25 × 10 m /s to 6.25 × 10 m /s, and graph jet diameter at as a function of the volume flow rate c 2015 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2-28 0.4 ... 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 θ1 − cos θ2) Fy = (sin θ2 + sin θ2) For an angle of. .. viscometer is used to measure the viscosity of a liquid For values of L = 40 mm, z = 250 mm, and D = 0.8 mm, determine the viscosity of the liquid The time recorded for the experiment is 12 seconds... 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,