Chapter 2 Density, Specific Gravity, Specific Weight

Transcription

1 Chapter Density, Specific Gravity, Specific Weight 1. What is the specific gravity of 38 API oil? 38 API oil sp.gr. sp. gr API The specific gravity of manometer gage oil is What is its density and its API rating? sp. gr. 0.86; ρ 1000(0.86) 86 kg/m 3 ρ 6.(0.86) 51.5 lbm/ft 3 sp. gr API API ; API API 39.8 API 0 API 3. What is the difference in density between a 50 API oil and a 0 API oil? sp. gr API for a 50 oil sp. gr API for a 0 oil density difference. 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 API μ 0.85 N s/m ν μ ρ (1000) Highly viscous; try ν (SUS) if SUS > SUS 633 SUS

2 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 kg/m 3 CO at 7 C 300 K has ρ 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 ) N For CO (1.797 kg/m 3 )(1. m 3 )(9.81 m/s ) 1.15 N Weight difference is N 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 9Nin air. While submerged, the object weighs 7N. What is the density of the liquid? Castor Oil ρ 960 kg/m 3 buoyant force volume mg in air mg submerged V ρ (0.03) kg/m ρg 7. A brass cylinder (Sp. Gr. 8.5) has a diameter of 5.mmand a length of 101.6mm. 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 volume mg 0.8 V mg ρ b Vg 8500( )(9.81).9 N ρ mg 0.8 gv ρ 15 kg/m ( ) ρg V π D h π (0.05) (0.1016) m 3 -

3 Viscosity 8. Actual tests on Vaseline yielded the following data: τ in N/m dv/dy in 1/s Determine the fluid type and the proper descriptive equation shear stress strain rate τ K ( ) dv n dy 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) Sum m 3 Summation (ln(dv/dy)) b 1 b (16.1) 0.1(18.60) 3(136.6) K exp(b 0 ) ; n b τ τ o + K ( ) dv n ( ) dv dy dy -3

4 9. A popular mayonnaise is tested with a viscometer and the following data were obtained: τ in g/cm dv/dy in rev/s Determine the fluid type and the proper descriptive equation. The topmost line is the given data, but to curve fit, we subtract 0 from all shear stress readings shear stress strain rate τ τ o + K ( ) dv n ( ) dv n which becomes τ τ τ o K dy dy Can be done instantly with spreadsheet; hand calculations: dv/dy ln(dv/dy) τ τ ln τ ln(τ ) ln(dv/dy) Sum m 3 Summation (ln(dv/dy)) 1.33 b 1 b (6.8) 5.753(13.6) 3(1.33) K exp(b 0 ) 3.37; n b τ τ o + K ( ) dv n ( ) dv dy dy where dv/dy is in rev/s and τ in g/cm ; these are not standard units. -

5 10. A cod-liver oil emulsion is tested with a viscometer and the following data were obtained: τ in dv/dy in rev/s Graph the data and determine the fluid type. Derive the descriptive equation. Cod liver oil; graph excludes the first data point. shear stress strain rate τ K ( ) dv n dy Can be done instantly with spreadsheet; hand calculations: dv/dy ln(dv/dy) τ ln τ ln(τ) ln(dv/dy) Sum m Summation (ln(dv/dy)) b 1 b (13.01).79(16.95) (5.181) K exp(b 0 ) 51.3; n b τ τ o + K ( ) dv n ( ) dv dy dy where dv/dy is in rev/s and τ in lbf/ft ; these are not standard units. -5

6 11. A rotatinup viscometer has an inner cylinder diameter of 50.8 mmand the gap between cups is 5.08 mm. The inner cylinder length is 63.5mm. 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 mn-m. Calculate the viscosity of the fluid. If the fluid density is 850 kg/m 3, calculate the kinematic viscosity. Rotatinup viscometer R 5. mm δ 5.08 mm L 63.5 mm ω (10 rev/min) (π rad/rev)(1min/60 s) 1.07 rad/s dv dy T N-m ρ 850 kg/m 3 μ Tδ πr (R + δ)lω μ π(0.05) ( )(0.0635)(1.07) μ Pa s v ft /s m /s 1. A rotatinup viscometer has an inner cylinder whose diameter is 38 mm and whose length is 80 mm. The outer cylinder has a diameter of 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 10 6 N m. Determine the kinematic viscosity of the fluid if its density is kg/m 3. R 38/ m; L 0.08 m R outside / 1 mm δ 1 19 mm 0.00 m ω (1 rev/min)(π/60) 1.6 rad/s T N m ρ kg/m 3 μ Tδ πr (R + δ)(lω) (0.00) π(0.019) ( )(0.08)(1.6) μ N s/m v ρ m /s -6

7 13. A rotatinup viscometer has an inner cylinder diameter of 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 mm R 8.58 mm (R + δ) 6.5 mm R + δ δ.55 mm ρ 968 kg/m 3 μ 0.01 Pa s ω (15 rev/ min)(π/60) 1.57 rad/s T πr (R + δ)(lω)μ δ π(0.0858) ( )(0.076)(1.571)(0.01) T N-m 1. A capillary tube viscometer is used to measure the viscosity of water (density is 1000 kg/m 3,viscosity is 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. mmand z 5 mm, determine the maximum tube size permissible. Capillary tube viscometer V t μ N s/m ρg z π R L 8μ ρ 1000 kg/m 3 z 0.5 m L m V t Volume flow rate AV πr V; substituting into the equation, πr V ρg z π R L 8μ The limiting value is R Rearrange and solve for V, V ρg z L 8μ Re < 100; using equality, V(R) ν 100; ρv(r) μ 100 or V 100μ ρr R ρg z L 8μ Rearrange and solve for R3 R 3 100μ (8)(L) ρ gz R or 100( ) (8)(0.076) (1000) (9.81)(0.5) R m mm Any larger, flow no longer laminar -7

8 15. A Saybolt viscometer is used to measure oil viscosity and the time required for m 3 of oil to pass through a standard orifice is 180 SUS. The specific gravity of the oil is found as API. Determine the absolute viscosity of the oil. For 180 SUS, ν (180) m /s API oil; sp.gr ; ρ kg/m μ ρν 806.3( ) N s/m 16. A 10 m 3 capillary tube viscometer is used to measure the viscosity of a liquid. For values of L 0 mm, z 50 mm, and D 0.8 mm, determine the viscosity of the liquid. The time recorded for the experiment is 1 seconds. ( zπr ) ( ) g 0.5π( /) (9.81) ν t 8LV 8(0.0)( (1) ) ν 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, ν (70) ν m /s 35 API oil sp. gr ρ 89.8 kg/m μ ρv 89.8( ) μ N s/m -8

9 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 1m? μ ( ) ρs ρ 1 ρg D 18V V L t L 1m D mm 0.00 m ρ s 7900 kg/m 3 ρ 1 63 μ Pa s ( ) ( ) ρs V ρ 1 ρg D μ (1 63)(9.81)( ) 18( ) V 0.015m/s Check on Re ρvd μ L t ; t 1 63(0.015 )(0.00) < 1 OK t 65.8 s 19. A mm diameter ball bearing is dropped into a viscous oil. The terminal velocity of the sphere is measured as 0.6 mm/s. What is the kinematic viscosity of the oil if its density is 800 kg/m 3? μ ( ) ρs ρ 1 ρg D 18V V L t m/s D m ρ s 7900 kg/m 3 ν μ ( ) ρ ρs gd ( ) 7900 (9.81)( ) ρ 1 18V ( ) ν m /s Check on Re VD ν ( ) < 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 Δh 17 mm and x 50.8 mm? -9

10 d open to atmosphere Acetone ρ a 787 kg/m 3 Hg ρ kg/m 3 p A + ρ a gx p atm + ρgδh Δh h p A + 787(9.81)(0.0508)(/1) (9.81)(0.17) x p A FIGURE P.0. p A 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. oil (sp gr. 0.85) air p W ρ wg ρ airg ρ Hgg 7 1 ρ cg 17 1 p atm p w 1.9(3.)(10/1) (1.9)(3.)(7/1) 0.85(1.9)(3.)(17/1) 1.7(1) p W water mercury p W 179 psf 1.1 psia FIGURE P.1.. 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 Δh on the manometer. open to atmosphere D A pressure gage pipeline Δh h 30 mm C B mercury FIGURE P.. -10

11 p D + ρ Hg gδh ρ B g(0.03) p A p D p atm (1 000)(9.81)Δh 876(9.81)(0.03) (which is a gage reading) Δh Δh Δh 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 Δh is 60 mm. Determine the density and specific gravity of the unknown fluid. air Because ρ air ρ liquid, then air hδh p A p B ρgδh; p A p B 700 N/m given; so Δh 60 mm 0.06 m, and FIGURE P.3. ρ p A p B gδh kg/m3 9.81(0.06). A U-tube manometer is used to measure the pressure difference between two air chambers, as shown in Figure P.. If the reading Δh is 15. mm, determine the pressure difference. air Because ρ air ρ liquid, then air hδh p A p B ρgδh; Δh m p A p B 1000 kg/m 3 (9.81)(0.15) p A p B 195 Pa FIGURE P.. 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 Δh is 70 mm of mercury (as shown). All dimensions are in mm. -11

12 outlet ( ) p outlet + ρg ρ Hg g(0.07) 1000 water ρg(0.0) p inlet 600 p outlet (9.81)(0.71) 13.6(1 000)(9.81)(0.07) 0 70 pump motor 1 000(9.81)(0.0) p inlet mercury inlet FIGURE P.5. p outlet p inlet p outlet p inlet 766Pa.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 castor oil water p A ρ LO g(0.076) + ρ air g(0.1016) + ρ HO g(0.17) ρ CO g(0.113) p B ρ LO 930 kg/m 3 ; ρ CO 960 kg/m 3 ρ HO 1000 kg/m 3 ρ air negligible p A p B ρ LO g(0.076) + ρ HO g(0.17) ρ CO g(0.113) p A p B 930(9.81)(0.076) 1000(9.81)(0.17) 960(9.81)(0.113) p A p B p A p B 56 Pa FIGURE P For the system of Figure P.7, determine the pressure of the air in the tank. open to atmosphere air oil ρ 86 kg/m 3 sp. gr mm 50.8 mm 15. mm p air + ρ oil g( ) ρg( ) p atm p air + 86(9.81)(0.03) 1000(9.81)(0.330) p air p air Pa FIGURE P.7. -1

13 Continuity Equation 8. Figure P.8 shows a reducing bushing. A liquid leaves the bushing at a velocity of m/s. Calculate the inlet velocity. What effect does the fluid density have? 0.5 m/s water 100 mm FIGURE P.8, P.9. 0 mm D mm 0.1 m; D 0 mm 0.0 m V m/s Density has no effect Q A 1 V 1 A V π D 1 V 1 π D V ( ) D V 1 V D V m/s 9. Figure P.9 shows a reducing bushing. Liquid enters the bushing at a velocity of 0.5m/s. Calculate the outlet velocity. D mm 0.1 m; D 0 mm 0.0 m V m/s 0.5 m/s water 100 mm FIGURE P.8, P.9. 0 mm Q A 1 V 1 A V π D 1 V 1 π D V ( ) D 1 V V 1 D V 3.13 m/s 30. Water enters the tank of Figure m 3 /s. The inlet line is 63.5 mmin 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 m 3 /s air exit Q HO in Q air out Q HO in m 3 /s P HO kg/m 3 ρ air 1.19 kg/m mm 57. mm FIGURE P.30. So Q air out AV π D V π [(0.038)] V V V air 1.66 m/s -13

14 31. An air compressor is used to pressurize a tank of volume 3m 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 + (ρ AV) out (ρ AV) in m pv RT m V RT dp dt (pav) out (pav) in Substituting, 0 V RT dp dt + p out RT out A out V out p out RT out A out V out For constant T,allRT products cancel p in RT in A in V in p in RT in A in V in V dp dt p out A out V out + p in A in V in p out p A in π(0.07) m A out Areas are equal dp 3 dt p( )(0.5) ( )() 3 dp p or dp dt dt Separating variables, p p dp p ln ( p) p dt ln ( p) ln ( (00 000)) 300( ) ln ( p) ln ( p).71 Exponentiating, p or p.3 p.3 kpa -1

15 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 kg/s. Freon-1 leaves the exchanger as a liquid (Sp. Gr ) at room temperature and pressure. Determine the exit velocity of the liquid. ṁ in ρ oυτ A out V out vapor inlet 1/ in. ID tubing FIGURE P.3. fins liquid outlet ṁ in kg/s ρ 1.915(1 000)kg/ m 3 A π D π(0.5/1) A ( ) m Substituting, ft (1 000) )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 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 0.3 kg/m 3 ). Calculate the inlet and outlet velocities of the nitrogen. Are they equal? Should they be? ṁ kg D m ρ kg/m 3 ρ 0.3 kg/m 3 A π D π(0.1016) m ṁ ρ AV V 1 ṁ ρ 1 A ( ) V m/s V 0.3( ) V 3.8 m/s Momentum Equation 3. A garden hose is used to squirt water at someone who is protecting herself with a garbage can lid. Figure P.3 shows the jet in the vicinity of the lid. Determine the restraining force F for the conditions shown. -15

16 0 mm diameter 3 m/s velocity FIGURE P.3. F Σ F ṁ(v out V in ) ṁ in ṁ out frictionless flow magnitude of V in magnitude of V out F [ρ AV] inlet ( V in V in ) F ρ AV ρ kg/m 3 A π(0.0) m V 3m/s F (1 000)( )(3) 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 F Σ F ṁ(v out V in ) ṁ in ṁ out frictionless flow magnitude of V in magnitude of V out F [ρ AV] inlet ( V in V in ) FIGURE P.35. 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 ṁ (V out V in ) ṁ in ṁ out frictionless flow F magnitude of V in magnitude of V out A, V FIGURE P.36. F [ρ AV] inlet ( V in V in ) 1 in SI units ρ AV F 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

17 Σ F ṁ (V out V in );ṁ in ṁ out frictionless flow magnitude of V in magnitude of V out F 1 [ρ AV] inlet (V outx V inx ) V outx V cos θ; V inx V A, V F F 1 θ F1 [ρ AV] inlet (V cos θ V ) ρ AV (cos θ 1) FIGURE P.37. F 1 ρ AV (1 cos θ) F [ρ AV] inlet (V outy V iny ) V outy V sin θ; V iny 0 F [ρ AV] inlet (V sin θ) ρ AV (sin θ) F 3F 1 ; sin θ 3(1 cos θ) 1 sin θ 1 cos θ T&E solution is quickest 3 θ (1/3) sin θ 1 cos θ θ 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. Σ F ṁ (V out V in ); ṁ in ṁ out frictionless flow magnitude of V in magnitude of V out θ F 1 [ρ AV] inlet (V outx V inx ) V outx V cos θ; V inx V A, V F 1 F 1 [ρ AV] inlet ( V cos θ V ) ρ AV (cos θ + 1) F F 1 ρ AV (1 + cos θ) FIGURE P

18 F [ρ AV] inlet (V outy V iny ) V outy V sin θ; V iny 0 F [ρ AV] inlet (V sin θ) ρ AV (sin θ) F 1 F ; 1 + cos θ sin θ T & E solution is quickest θ 1 cos θ sin θ 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 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.) We apply the energy equation between any two sections. Section 1 the free surface upstream, and Section the outlet downstream m p p 1 p atm 1. m V 1 0 (reservoir surface velocity) 1.83 m z 1.83 m; z m FIGURE P.39. A π D Q m 3 /s π(1.) m V Q A m/s ρ 1000 kg/m ρv A ṁ 1000(1.169)(0.7) kg/s evaluated at outlet -18

19 Substituting, W W W + W {( ) ( ) } p ρ + V p + gz ρ + V 1 + gz ρv A { ( } (1.83) ( (36.6)) {( } W 0. Air flows through a compressor at a mass flow rate of 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 101.3kPaand 3.9 C. The outlet pressure is 87 kpa. For an isentropic (reversible and adiabatic) compression process, we have T T 1 { } (γ 1)/γ p 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/rt). T T 1 { } (γ 1)/γ p 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/rt). Solution: ṁ kg V in 0 V out unknown p in Pa p out Pa D out m A out π D / m γ 1. R air 8.31 J/Kmole c pair 100 J/kg K T out T in { } (γ 1)/γ pout p in { T out } (1. 1)/1. ( ) T out 96.9(1.8) T out k 68 C -19

20 ρ out V out W p RT kg/m 3 831(50.95) ṁ ρ A out (0.0003).05 m/s V out { (h + V + gz) out (h + V + gz) in } ρv A W (h out h in + V out )ρv A ΔPE 0 (h out h in ) c p (T out T in ) 100(68 3.9) J/kg (h out h in ) J/kg W ( )(0.038) W or W 1. HP Assuming no losses 1. 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 p out kpa T in 5 C T out 11 C V in 100 m/s V out m/s c p J/(kg K) W W / ṁ {(h + V + gz ) (h + out V + gz ) } ρv A in ( ) V out (h out h in ) + + V in + g (z out z in ) 3 c (h out h in ) c p (T out T in ) z out z in W / ṁ ( ) (5 11) W/ ṁ J/kg -0

21 . A pump moving hexane is illustrated in Figure P.. The flow rate is 0.0 m 3 /s; inlet and outlet gage pressure readings are kpaand 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), p 75 mm and Section outlet pressure gage. p Pa z 1.5 m p 1 p Pa AV 0.0 m 3 /s z m 1.5 m 1.0 m pump motor A 1 π D 1 π(0.10) m 100 mm A π D π(0.075) m FIGURE P.. V 1 Q 0.0 A m/s V Q m/s A ρ 0.657(1 000) for hexane W W {( p ρ + V + gz ) { (9.81) 657 ( p ρ + V + gz ) } 1 ρv A ( (9.81) 657 )} 657(0.0) W W { } (13.1) N m/s.0 kw Bernoulli Equation 3. Figure.15 shows a venturi meter. Show that the Bernoulli and continuity equations when applied combine to become gδh Q A 1 (D /D 1 ) Hydrostatic equation for manometer; all measurements are from the centerline p 1 ρ 1 gx ρ 1 gδh p ρ 1 gx ρ gδh or p 1 p ρ 1 gδh -1

22 ṁ 1 ṁ ρ 1 A 1 V 1 ρ A V or A 1 V 1 A V In terms of diameter, π D 1 V 1 π D V Q Bernoulli Equation p 1 ρ 1 g + V 1 g + z 1 p ρ 1 g + V g + z With z 1 z, (p 1 p ) ρ 1 g 1 g (V V 1 ) Substitute for V in terms of Q (p 1 p ) g Q (1/A 1/A 1 ) ρ 1 g ρ 1 gδh ρ 1 ( ) ( ) Q 1 A Q 1 D A A 1 A D 1 A gδh Q 1 D /D 1 or finally, gδh Q A 1 D /D 1. A jet of water issues from a kitchen faucet and falls vertically downward at a flow rate of m 3 /s. At the faucet, which is mmabove the sink bottom, the jet diameter is mm. Determine the diameter of the jet where it strikes the sink. Q m 3 /s D m A m V 1 Q A 1 0. m/s h z m D 1 1 Bernoulli Equation p 1 ρ 1 g + V 1 g + z 1 p ρ 1 g + V g + z h p 1 p z m z 0 Substituting, (9.81) V (9.81) + 0 D which becomes -

23 ( )((9.81)) V or V.65 m/s A Q V m π D D π ( ) or D m.6 mm 5. A jet of water issues from a valve and falls vertically downward at a flow rate of m 3 /s.the valve exit is 50 mm above the ground; the jet diameter at the ground is 5mm. Determine the diameter of the jet at the valve exit. Section 1 is the exit; section is the ground. 1 p 1 ρ 1 g + V 1 g + z 1 p ρ 1 g + V g + z D 1 Q m 3 /s; p 1 p p atm ; z 0; z m h D 5mm;A π(0.005) m V Q m/s; A Bernoulli Equation becomes V 1 g + z 1 V g V 1 (9.81) (9.81) V ; V m/s Q A 1 V 1 π D 1 Q V 1 D 1 π V 1 ( ) D m π(1.17) D mm D -3

24 6. A garden hose is used as a siphon to drain a pool, as shown in Figure P.6. 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.6. Section 1 is the free surface; section is the hose outlet. p 1 ρ 1 g + V 1 g + z 1 p ρ 1 g + V g + z p 1 p p atm ; V 1 0; z 1 1. m Substituting, (1.) 0 + V (9.81) + 0 V (9.81)(1.).89 m/s D 19 mm A π D m Q AV (.89); Q m 3 /s Miscellaneous Problems 7. A pump draws castor oil from a tank, as shown in Figure P.7. A venturi meter with a throat diameter 50.8mmis 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 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 areinmm.) -

25 air air Δh h ID 178 outlet 50.8 throat inside diameter 76. ID pump motor FIGURE P.7. p 1 ρgx ρ air g(0.559) + ρg( x 0.178) p ρ air is negligible x terms cancel; ρ 960 kg/m 3 p p 1 ρg( ) 960(9.81)(0.381) 3588 Pa Energy equation, 1 to : W {( p ρ + V gz) ( p + ρ + V gz) } 1 + ρv A D 1 D m A 1 V 1 A V so V 1 V z 1 0 z m ρ AV ρ Q The power was given as W W Substituting, ( ) ( ) (p p 1 ) ρ Q + gz 960Q ρ (0.178) Solving for Q Q m 3 /s Now for the venturi meter, the throat diameter is D th m D m A th π D th m gδh Q A th 1 D th /D (9.81)Δh 1 (0.0508/0.076) -5

26 Δh 1. 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. 8. A mm ID pipe is used to drain a tank, as shown in Figure P.8. 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 0 mm. inlet h exit z FIGURE P.8. Q in AV A π(0.05) m Q in (1.5) m 3 /s Section 1 is the free surface in the tank, and is at the exit of the pipe. Apply the Bernoulli equation, 1to: p 1 ρ 1 g + V 1 g + z 1 p ρ 1 g + V g + z p 1 p p atm ; V 1 0; z 1 h; z 0.0 m; the Bernoulli equation becomes h V g + z ; At equilibrium, Q out Q in m 3 /s A out π(0.0) m ; and V Q m/s A out h V g + z.3 (9.81) h m which is independent of fluid properties, and with no friction -6

27 Computer Problems 9. 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 y x A, V θ 1 F y θ F x Force ratio theta in degrees FIGURE P 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. 1 D m 1. m 80 mm 1.83 m 0.3 D cm FIGURE P.50. FIGURE P

28 dw/dt in Kw h in m Diameter in mm Volume in m 3 x One of the examples in this chapter dealt with a water jet issuing from a faucet. The water flow rate was 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 m 3 /s to m 3 /s, and graph jet diameter at as a function of the volume flow rate. -8