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Appendix N Compressible Flow Tables for an Ideal Gas with k
639 APPENDIX Volume and Area Formulas: hxx xx xx r r D x xx L xx hx a b A = bh Ixx = bh3 12 Ixx = r4 4 A = 2.5981L2 Ix = 0.5127L4 A = r2 π […]
Chapter 1 Find the energy in food calories to ride for 1 hour
1.23: PROBLEM DEFINITION Apply the grid method. Situation: A cyclist is travelling along a road. P=FV. V=24mi/h,F=5lbf. Find: a) Find power in watts. b) Find the energy in food calories to ride for 1 hour. PLAN Follow the process for […]
Chapter 1 Idealize a football as a streamlined body
651 ANSWERS Solutions to Checkpoint Problems Chapter 1 1.1 e PointTM is a unit. 1.2 (e) Chapter 2 2.1 Glycerin has a higher viscosity at both temperatures. 2.2 ⫽ 2.07 ⫻ 10⫺6 m2/s 2.3 (c) 1 2 Volume […]
Chapter 1 Problem Definition Situation The List Below Identify
1.1: PROBLEM DEFINITION No solution provided because student answers will vary. 1.2: PROBLEM DEFINITION No solution provided because student answers will vary.. 1 1.3: PROBLEM DEFINITION No solution provided because student answers will vary. 1.4: PROBLEM DEFINITION Essay question. No […]
Chapter 1 Tereby affecting both aerodynamic drag and tire
4. Mass released from tank M1−M2=37.39 −26.25 M1−M2=11.14 kg 38 M2=ρ1V =6.56 kg m3×4m 3 M2=26.25 kg 1.38: PROBLEM DEFINITION Situation: Properties of air. Find: Specificweight(N/m 3). Density (kg/m3). Properties: PLAN First, apply the ideal gas law to find density. […]
Chapter 10 Homework Lq2 2g Assume 0020 Then M2054
Head loss Final calculations hp=z2−z1+hL=0.8+1.13 = 1.93 m P=γhpQ= 9693 N/m3×1.93 m ×3×10−4 P=5.61W 121 hL=µfL D+19Kb¶V2 2g=µ0.022 10 m 0.02 m +19×0.7¶(0.955 m/s)2 2×9.81 m/s2 =1.13 m 10.75: PROBLEM DEFINITION Situation: A heat exchanger is described in the problem […]
Chapter 10 Line A has a half open gate valve
10.95: PROBLEM DEFINITION Situation: Two pipes are connected in parallel. Line A has a half open gate valve, line B a fully open globe valve. Sketch: Find: Ratio of velocity in line Ato B. Assumptions: Head loss due to friction […]
Chapter 10 Power in watts needed to overcome the head loss
1. The pressure drop for a 100 ft run of pipe (∆p= 227 psf ≈1.6psi )could be 61 decreased by selecting a larger pipe diameter. 2. The power to overcome the frictional head loss is about 1/15 of a horsepower. […]
Chapter 10 Problem Definition Situation Glycerine Ows Pipe 05
2. Energy equation: 3. Head loss (laminar flow): hf=32μLV γD2 V=hfγD2 32μL =(1 m) (10000 N/m3)(0.008 m)2 32 (3.0×10−3N·s/m2)(10m) =0.667 m/s V=0.667 m/s 21 p1 γ+α1 V2 1 2g+z1+hp=p2 γ+α2 V2 2 2g+z2+ht+hL p1 γ+0+z1+0 = p2 γ+0+z2+hf 21m=20m+hf hf=1m […]
Chapter 10 Problem Definition Situation Rectangular Duct Initial State
10.85: PROBLEM DEFINITION Situation: Two reservoirs are connected by a cast-iron pipe of varying diameter. z2=110m,Q=0.3m 3/s. D1=20cm,L1=100m. D2=15cm,L2=150m. Sketch: Find: Water surface elevation in reservoir A. Properties: From Table 10.4: ks=0.26 mm. Water (10 ◦C),TableA.5:ν=1.3×10−6m2/s. SOLUTION 141 ks D20 […]
Chapter 10 Sketch Find Exit Velocity MS Discharge Ls
10.1: PROBLEM DEFINITION Situation: Kerosene flows in a pipe. Q=0.02 m3/s,D=17.7cm. Find: Is the flow laminar or turbulent? Entrance length (meters). Properties: Kerosene (20 ◦C),Table A.4,ν=2.37 ×10−6m2/s. SOLUTION Reynolds number: Entrance length: Le D=50 Le=50D=50(0.177 m) = 8.85 m 1 […]
Chapter 10 Solve the set of equations using a computer program
10.55: PROBLEM DEFINITION Situation: Water is draining out of a tank through a galvanized iron pipe. ks=0.006 in = 5 ×10−4ft,L=10ft,H=4ft. Thepipeis1–inschedule40NPS,D=1.049 in = 0.08742 ft. Find: Velocity in the pipe (ft/s). Flow rate (cfs) Assumptions: Steady flow. Component head […]
Chapter 10 The increase in velocity will increase Reynolds number
Calculate values Substitute Eq. (2) into (1) (3.28 ft) = (−2ft)+fL D V2 2g or f=5.28 µD L¶µ2g V2¶ =5.28 µ1/6ft 30 ft ¶µ2×32.2ft/s2 (5 ft/s)2¶ f=0.076 Since the resistance coefficient (f)is now known, use this value to find viscosity. […]
Chapter 10 Water exits a tank through a short galvanize
Integrate: REVIEW Possible problems with this solution: The Reynolds number is very close to the point where turbulent flow will occur and this would be an unstable condition. The flow might alternate between turbulent and laminar flow. 101 t=−42.2(1.54 −h)1/2 […]
Chapter 11 Drag force is influenced by the shape of the object
11.1: PROBLEM DEFINITION Situation: A hypothetical pressure-coefficient distribution acts on a plate. Plate length = c, Plate width (into paper) = 1.0 unit. Find:Coefficient of drag. Assumptions:Viscouseffects are negligible. SOLUTION 1. Find the force normal to the plate 2. Find […]
Chapter 11 FB is the buoyant force and FD is drag
11.50: PROBLEM DEFINITION Situation: The problem statement describes a 1.5-mm sphere moving in oil. Find: Terminal velocity of the sphere. PLAN Apply the equilibrium principle. To find the drag force, assume Stokes drag. SOLUTION Equilibrium. Since the ball moves at […]
Chapter 11 The coefficient of drag is not influenced by the car
11.33: PROBLEM DEFINITION Situation: A cartop carrier is used on an automobile. Vc=100km/h=27.8m/s, Vw=25km/h=6.94 m/s. Find: Additional power required due to the carrier. Assumptions: Density of air is ρ=1.2kg/m3. The coefficient of drag is not influenced by the car. CDis […]
Chapter 11 The coefficient of lift is specified by the data on Fig
11.64: PROBLEM DEFINITION Situation: An airplane wing has a chord of 3.5 ft. Air speed is Vo=200ft/s. Theliftis1800lbf. Theangleofattackis3 o. The coefficientofliftisspecified by the data on Fig. 11.24 in EFM10e. Find: The span of the wing. Properties: Density of air […]
Chapter 11 The rock can be idealized as a cube of size
11.19: PROBLEM DEFINITION Situation: A large rock is located on the bottom of a river. The current is strong enough so the rock moves along the bottom of the river. Find: The speed (m/s) of the current required for the […]
Chapter 12 Mass flow rate if Bernoulli equation is used
12.38: PROBLEM DEFINITION Situation: A truncated nozzle has an area of 3 cm2,the total temperature and pressure are 20oC and 300 kPa, abs and the back pressure is 90 kPa, abs. is described in the problem statement. Find:Massflow rate. Properties:FromTableA.2(EFM10e),k=1.4,R […]
Chapter 12 Nominally commercial transport aircraft fly at a Mach
12.1: PROBLEM DEFINITION Situation: The speed of sound in an ideal gas _______. (Select all that are correct): a. depends upon √Twhere Tis absolute temperature b. depends upon √Twhere Tis temperature in ◦C c. depends upon √k,wherek=cp cv,aratioofspecific heats for […]
Chapter 12 Oxygen flows from a reservoir with temperature of
Flow rate equation 21 ˙m=VρA =(268m/s)(3.16 kg/m3)(0.0060 m2) ˙m=5.08 kg/s 12.21: PROBLEM DEFINITION Situation: Oxygen flows from a reservoir with temperature of 200oCand300kPa, abs. through conduit with Mach number of 0.9. Find:(a)Velocity. (b) Pressure. (c) Temperature. Properties:FromTableA.2(EFM10e),k=1.4,R O2=260J/kg-K SOLUTION Total […]
Chapter 12 Problem Definition Situation Shock Wave Occurs Expansion
12.50: PROBLEM DEFINITION Situation: A rocket nozzle with area ratio of 4 and total pressure of 1.3 MPa exhausts to a back pressure of 30 kPa, abs. Find: The state of exit conditions. Properties: From Table A.1, (EFM 10e),k=1.4. SOLUTION […]
Chapter 13 Definition Situation Venturi Meter Described The
13.34: PROBLEM DEFINITION Situation:Waterflows (Q=0.03 m3/s) through an orifice. Pipe diameter, D=15 cm. Manometer deflection is 12 cm-Hg. Find:Orifice size: d PLAN Calculate ∆h. Then guess K and apply the orifice equation. Check the guessed value of Kby calculating a […]
Chapter 13 One mode of operation of an ultrasonic flow meter
13.54: PROBLEM DEFINITION Situation: One mode of operation of an ultrasonic flow meter involves the time for a wave to travel between two measurement stations–additional details are provided in the problem statement. Find: (a) Derive an expression for the flow […]
Chapter 13 Problem Definition Situation Water Exits Upper
13.68: PROBLEM DEFINITION Situation: Water exits an upper reservoir across a rectangular weir (L/HR=3,P/H R= 2) and then into a lower reservoir. The water exits the lower reservoir through a 60o triangular weir. Find: Ratio of head for the rectangular […]
Chapter 13 The straw will be situated far from the car body
Problem 13.1 no solution provided; answers will vary. 1 13.2: PROBLEM DEFINITION Situation: A straw (stagnation tube) and a water-filled, u-tube manometer are used to measure the speed of an automobile. Find: a. sketch the apparatus b. determine the lowest […]
Chapter 13 Would Probably Result Estimated Discharge That
13.18: PROBLEM DEFINITION Situation:Aheatedgasflows through a cylindrical stack–additional information is provided in the problem statement. Find:(a)Theratiorm/D such that the areas of the measuring segments are equal (b) The location of the probe expressed as a ratio of rc/D that corresponds […]
Chapter 14 Apply the discharge and head coefficient equations
14.19: PROBLEM DEFINITION Situation: A 14 inch diameter pump operates at 1000 rpm. Find: Plot the head-discharge curve. PLAN Apply the discharge and head coefficient equations at a series of coefficients corre- sponding to each other from Fig. 14.7 (EFM10e). […]
Chapter 14 Apply The Propeller Thrust Force Equation And
14.1: PROBLEM DEFINITION Situation:Thrustofafixed pitch propeller Find: Reason the thrust decreases with forward speed. SOLUTION The angle of attack for the propeller blade is the difference between the pitch angle and angle of flow due to forward motion, 1 α=β−tan−1µV0 […]
Chapter 14 Compare reaction turbine and centrifugal pump
14.50: PROBLEM DEFINITION Situation: A jet of water strikes the buckets of an impulse wheel and turns water by 180 degrees with bucket speed 1/2 of jet speed. Find: (a) Jet force on the bucket. (b) Resolve the discrepancy with […]
Chapter 14 Ratio of final temperature to initial temperature
Suction specific speed 41 Nss =NQ1/2(NPSH)3/4 N=690rpm Nss =690rpm ×(3,487 gpm)1/2/(36.2ft)3/4 Nss =2,760 Nss is much below 8,500; therefore, it is in a safe operating range. 14.37: PROBLEM DEFINITION Situation: A pump is needed that rotates at 1500 rpm and […]
Chapter 15 Find Determine The Ow Subcritical Supercritical Plan
15.19: PROBLEM DEFINITION How are head loss and slope related for non-uniform flow, as compared to uniform flow? Consider both rapidly and gradually varying non-uniform flow. SOLUTION •In uniform flow, velocity is constant along a streamline. Practically speaking, this requires […]
Chapter 15 Label any water surface profiles that may be classified
15.66: PROBLEM DEFINITION Situation: Rectangular channel ends with a free overfall Channel is very long width = 10 ft So=.0001 Q=120cfs One mile upstream the flow is uniform Find: Determine the classification of the water surface just before the brink […]
Chapter 15 Plan Calculate Froude Number Order
15.55: PROBLEM DEFINITION Situation: A hydraulic jump is described in the problem statement. γ=9,810 N/m2,B 3=5m. y1=40cm =0.40 m, V=10m/s. Find: Depth of flow downstream of jump (m). SOLUTION Check Fr upstream to see if the flow is really supercritical […]
Chapter 15 The Wetted Perimeter Minimum Between 063 And
15.1: PROBLEM DEFINITION Situation: Comparison of full pipes versus open channels. Find: Why is the Reynolds number for open channels different than the one for pipes? PLAN Consider the different definitions for Re for the two cases. SOLUTION For fully […]
Chapter 15 Two situations are of interest: an up step
15.38: PROBLEM DEFINITION Situation: Water flows over a broad-crested weir. Additional details are given in the problem statement. Find: The water surface elevation in the upstream reservoir (ft). PLAN Apply the Broad crested weir—Discharge equation. SOLUTION From Fig. 15.13 (EFM10e), […]
Chapter 16 Buy a container of helium and some balloons
16.1: PROBLEM DEFINITION Which of the following could be considered a model? Why? (select all that apply). a. The ideal-gas law. b. A set of instructions for using a Pitot-static tube to measure velocity. c. An airplane built from a […]
Chapter 16 Provides Method When Experimentation Analysis Not
16.13: PROBLEM DEFINITION Answer each question below re: Eq (16.72) in EFM10e. •Is the given equation in conservation form or non-conservation form? Why? •Is the given equation in invariant form or coordinate specificform?Why? SOLUTION •Non-conservation form because the equation was […]
Chapter 2 Because the viscosity of gases increases with temperature
2.55 Situation: Oxygen at 50 ◦Fand 100 ◦F. Find: Ratio of viscosities: μ100 μ50 . SOLUTION 60 Because the viscosity of gases increases with temperature μ100/μ50 >1. Correct choice is (c) . 2.56 Situation: Surface tension: (select all that apply) […]
Chapter 2 Bingham Plastic Newtonian UID Solution The Answer
2.37 Situation: A cylinder falls inside a pipe filled with oil. d=100mm,D=100.5mm. =200mm,W=15N. Find: Speed at which the cylinder slides down the pipe. Properties: SAE 20W oil (10oC) from Figure A.2: μ=0.35N·s/m2. SOLUTION 41 τ=μdV dy W πd =μVfall (D−d)/2 […]
Chapter 2 The change in velocity dV is in the direction of flow
2.20 Situation: The term dV/dy,thevelocitygradient a. has dimensions of L/T, and represents shear strain b. has dimensions of T−1, and represents the rate of shear strain SOLUTION 21 The answer is (b). See Eq. 2.14 in EFM10e, and discussion. 2.21 […]
Chapter 2 Which of these are units of density
2.1 Situation: A system is separated from its surrounding by a a. border b. boundary c. dashed line d. dividing surface SOLUTION 1 Answer is (b) boundary. See definition in §2.1. 2.2 Find: How are density and specific weight related? […]
Chapter 3 Calculate the density of helium at a gage pressure
3.1: PROBLEM DEFINITION Apply the grid method to cases a, b, c and d. a.) Situation: Pressure values need to be converted. Find: Calculate the gage pressure (kPa) corresponding to 8 in. H2O (vacuum). Solution: b.) Situation: Pressure values need […]
Chapter 3 Column Rise 120 Mm Due Pressure And
3.41: PROBLEM DEFINITION Situation: AtmosphericconditionsonMars. •Temperature at the Martian surface is T=−63 ◦C=210K The pressure at the Martian surface is p=7mbar. Find: Pressure at an elevation of 8 km. Pressure at an elevation of 30 km. Assumptions: Assume the atmosphere […]
Chapter 3 Length of chain so that gate just on verge of opening
3.110: PROBLEM DEFINITION Situation: A submerged gate is described in the problem statement. d=25cm,W=200N. y=10m,L=1m. Find: Length of chain so that gate just on verge of opening. PLAN Apply hydrostatic force equations and then sum moments about the hinge. SOLUTION […]
Chapter 3 Problem Definition Situation Regarding The Hydraulic Jack
Assumptions: Density of air is constant. Properties: Air, ρ=1.1kg/m3. Solution: Pressure at summit: psummit =pbase +∆p= 940 mbar −µ3947 Pa 1.0¶µ10−2mbar Pa ¶ psummit = 901 mbar (absolute) d.) Situation: Pressure increases with depth in a lake. ∆z=350m. Find: Pressure […]
Chapter 3 Solution The Length Gate Hydrostatic Force
Therefore, TAdoes not change with H. The correct answers are obtained by reviewing the above solution. 101 a, b, and e are valid statements. 3.70: PROBLEM DEFINITION Situation: Water exerts a load on square panel. d=1m,h=2m Find: (a) Depth of […]
Chapter 3 The Scale Would Linear Problem 363 Using
3.56: PROBLEM DEFINITION Situation: A manometer is used to measure pressure at the center of a pipe. Find: Pressure at center of pipe A (kPa). Properties: Water (10 ◦C,1atm),Table A.5, γwater =9810N/m3. PLAN Since the manometer is applied to measure […]
Chapter 3 The Vertical Component Hydrostatic Force Will The
3.82: PROBLEM DEFINITION Situation: A concrete form is described in the problem statement. y1=1.5m,θ=60◦. Find: Moment at base of form per meter of length (kN·m/m). Properties: Concrete, γ=24kN/m3. Assumptions: Assume that the form has a length of w=1meterintothepaper. PLAN Find […]
Chapter 3 To derive an equation for the load on the bolts
41 h2=4mg (S)(γwater)(πD2 1)=4(5kg)(9.81 m/s2) (0.8) (9810 N/m3)(π)(0.122m2) h2=0.553 m 3.27: PROBLEM DEFINITION Situation: An odd tank contains water, air and a liquid. . Find: Maximum gage pressure (kPa). Where will maximum pressure occur. Hydrostatic force (in kN) on top […]
Chapter 3 Which sphere has the largest buoyant force
3.95: PROBLEM DEFINITION Situation: Three spheres of the same diameter are submerged in the same body of water. One sphere is steel, one is a spherical balloon filled with water, and one is a spherical balloon filled with air. a. […]
Chapter 4 Problem Definition Situation For Path lines Streak lines And
4.1: PROBLEM DEFINITION Situation: Path of a fluid particle. Find: If a light was attached to a fluid particle and take a time exposure, would the image you photographed be a pathline or streakline? SOLUTION 1 Thepathlineisdefined as the path […]
Chapter 4 Problem Definition Situation Fuel Tank Rotated Zero gravity
4.97: PROBLEM DEFINITION Situation: Stirring a liquid in a cup. Find: Report on the contour of the surface. Provide an explanation for the observed shape. SOLUTION Stirring the cup of liquid creates a surface depressed at the center and higher […]
Chapter 4 Problem Definition Situation Velocity And Pressure Given
4.79: PROBLEM DEFINITION Situation: A two-dimensional velocity field is represented by the vector V=10xi−10yj. Find: Is the flow irrotational? SOLUTION In a two dimensional flow in the x−yplane, the flow is irrotational if (Eq. 4.34a) 81 ∂v ∂x =∂u ∂y […]
Chapter 4 Solution The Correct Answers Are And Review
4.41: PROBLEM DEFINITION Situation: Water accelerated from rest in horizontal pipe. L=80m,D=30cm,as=5m/s2. Find: Pressure at upstream end (kPa). Properties: ρ=1000kg/m3,pdownstream =90kPa. PLAN Apply Euler’s equation. SOLUTION Euler’s equation with no change in elevation 41 ∂p ∂s =−ρas =−1,000 kg/m3×5m/s2 =−5,000 […]
Chapter 4 The acceleration halfway between the entrance
4.21: PROBLEM DEFINITION Situation: In a flowing fluid, acceleration means that a fluid particle is a. changing direction b. changing speed c. changing both speed and direction d. any of the above SOLUTION 21 The correct answer is d. 4.22: […]
Chapter 4 Two Pitot tubes are connected to air-water
4.61: PROBLEM DEFINITION Situation: A Pitot tube measures the flow direction and velocity in water. Find: Explain how to design the Pitot tube. SOLUTION Three pressure taps could be located on a sphere at an equal distance from the 61 […]
Chapter 5 Definition Situation Nozzle Discharges Water Onto Plate
5.59: PROBLEM DEFINITION Situation: A sphere is falling in a cylinder filled with water. D1=8in,D2=1ft,V1=4ft/s. Find: Velocity of water at the midsection of the sphere. Sketch: PLAN Apply the continuity equation. SOLUTION As shown in the above sketch, select a […]
Chapter 5 Problem Definition Situation Air Ows Rectangular Air
5.1: PROBLEM DEFINITION Situation: Consider an automobile gas tank being filled by a nozzle. Find: (a) Discharge (gpm). (b) Time to put 50 gallons in the tank (min). (c) Cross-sectional area (ft2) of the nozzle and velocity at the exit […]
Chapter 5 Problem Definition Situation Water Ows Pipe 85
5.21: PROBLEM DEFINITION Situation: A rectangular channel has a 30oincline. u=8[exp(y)−1] m/s. y=1m,x=2m. Find: Discharge (m3/s). Mean velocity ( m/s). PLAN Apply the integral form of the flow rate equation becuse velocity is not constant over the area. SOLUTION Discharge. […]
Chapter 5 Solution Let The Pressure The Streamline Upstream
5.103: PROBLEM DEFINITION Situation: Air flows through constant area, heated pipe. D=4in,V=10m/s. p2=80kPa,p 1=100kPa. Find: Velo city at exit. Determine if the Bernoulli equation be used to relate the pressure and velocity changes. Properties: T1=20◦C,T2=50◦C. PLAN Apply the continuity equation. […]
Chapter 5 The Reynolds transport theorem is used to relate
5.39: PROBLEM DEFINITION Situation: InFig5.11in§5.2ofEFM10e, a. the CV is passing through the system. b. the system is passing through the CV. SOLUTION 41 The answer is (b), the system, which is a defined mass (think of the system as a […]
Chapter 5 Therefore The Atomizer Operate All The Pressure
5.94: PROBLEM DEFINITION Situation: Water flows in a pipe with a contraction. Q=60ft 3/s,d=2ft,D=6ft. Find: Pressure at point B. Assumptions: Water temperature is 50 ◦F. Properties: Water (50 ◦F),Table A.5: γ=62.4lbf/ft3. pA=3200psf. PLAN Apply the Bernoulli equation and the continuity […]
Chapter 5 Time to increase the density of the air in the
5.77: PROBLEM DEFINITION Situation: Atankisfilled with air from a compressor. V=10m 3,˙m=0.5ρ0 ρkg/s. Find: Time to increase the density of the air in the tank by a factor of 2. Properties: ρ0=2kg/m3 PLAN Apply the continuity equation. SOLUTION Continuity equation […]
Chapter 6 A surface force because the wing must touch
6.1: PROBLEM DEFINITION Situation: Identify the surface and body forces acting on a glider in flight. Also, sketch a free body diagram and explain how Newton’s laws of motion apply. Find: Surface and body forces acting on a glider in […]
Chapter 6 A water jet strikes a block and the block is held
6.27: PROBLEM DEFINITION Situation: A water jet strikes a block and the block is held in place by friction. v1=10m/s, ˙m=1.5 kg/s. μ=0.1,m=1kg. Find: Will the block slip? Force of the water jet on the block (N). Sketch: Assumptions: Neglect […]
Chapter 6 Assumption Given That Typical Window Withstands Force
6.70: PROBLEM DEFINITION Situation: Water flows through a 60oreducing bend–additional details are provided in the problem statement. Find: Horizontal force required to hold bend in place: Fx PLAN Apply the Bernoulli equation, then the momentum equation. SOLUTION Bernoulli equation Let […]
Chapter 6 Find Thrust The Turbofan Engine Assumptions Neglect
6.83: PROBLEM DEFINITION Situation: Lift and drag forces are being measured on an airfoil that is situated in a wind tunnel–additional details are provided in the problem statement. yp u 8m / s 2 12 m / s0.25 m 0.25 […]
Chapter 6 Problem Definition Situation Clam Shell Thrust
6.40: PROBLEM DEFINITION Situation: A clam shell thrust reverser is deployed on an aircraft engine. Find: (a) The thrust under normal operation. (b) the reverse thrust. Assumptions: Engine is stationary. Exit gas velocity unchanged at deployment. Pressure is atmospheric at […]
Chapter 6 Problem Definition Situation Gravel Ows Into Barge
2. The forces (F1and F2)are each about 40 lbf. This magnitude of force may be 21 too large for users of a toy. Or, this magnitude of force may lead to material failure (it breaks!). It is recommended that the […]
Chapter 6 Solution The Pressure Forces Acting The Inlet
6.54: PROBLEM DEFINITION Situation: An unusual nozzle creates two jets of water. d=0.5in,v 2=v3=80.2ft/s. D=3.5in.p =50psig. Find: Force required at the flange to hold the nozzle in place: F PLAN Apply the continuity equation, then the momentum equation. SOLUTION Continuity […]
Chapter 6 The Downward Load The Head Due The
where Mis the mass of the cart (mass of water moving with cart is negligible) From conservation of mass Combining terms XFx=d dt(Mvc)+ ˙m(v2x−v1) 0=Mdvc dt +ρA1(vj−vc)(vc−vj) Mdvc dt =ρA1v2 j(1 −vc vj )2 =˙mvj(1 −vc vj )2 Since the […]
Chapter 7 Apply the energy equation between the top of
7.44: PROBLEM DEFINITION Situation: Apumpfills a tank with water from a river. Dtank =5m,Dpipe =5cm. hL=10V2 2/2g,hp=20−4×104Q2. Find: Time required to fill tank to depth of 10 m. Assumptions: α=1.0. SOLUTION Energy equation (locate 1 on the surface of the […]
Chapter 7 Find electrical power using the efficiency equation
7.55: PROBLEM DEFINITION Situation: Q=500cfs, η=90%. hL=1.5V2 2g,D=7ft. z1=35ft,z2=0ft. Find: Power output from turbine. Assumptions: α=1.0. PLAN Apply the energy equation from the upstream water surface to the downstream water surface. Then apply the power equation. SOLUTION Energy equation: Power […]
Chapter 7 Find Head Loss Between Reservoir Surface And
41 p2=16.9psig 7.32: PROBLEM DEFINITION Situation: Water flows from a pressurized tank, through a valve and out a pipe. p1=100kPa,z1=8m. p2=0kPa,z 2=0m. hL=KLV2 2g,V2=10m/s. Find: The minor loss coefficient (KL). Assumptions: Steady flow. Outlet flow is turbulent so that α2=1.0. […]
Chapter 7 Find Minimum Electrical Power Watts Assumptions Neglect
Problem 7.1 Fill in the blank. Show your work. b. ____ ft ·lbf = energy to lift 10 N weight through elevation difference of 125 m. Recall that energy and work have the same dimensions. Here we are asked to […]
Chapter 7 Let the velocity in the 6 inch pipe be V6
7.83: PROBLEM DEFINITION Situation: A reservoir discharges into a pipe. Find: Draw the HGL and EGL. SOLUTION 120 37.2 m 42.6 m 2000 m 80 m 7.84: PROBLEM DEFINITION Situation: Water discharges through a turbine. Q=1000cfs, η=85%. hL=4ft,H=100ft. Find: Power […]
Chapter 7 Problem Definition Situation Water Ows Vertical Pipe
7.17: PROBLEM DEFINITION Situation: The velocity distribution in a pipe with turbulent flow is given by V Vmax =µy r0¶n Find: Derive a formula for αas a function of n. Find αfor n=1/7. SOLUTION Flow rate equation Upon integration Q=2πVmaxr2 […]
Chapter 7 Find the head loss by applying the sudden
7.70: PROBLEM DEFINITION Situation: An abrupt expansion dissipates high energy flow. D1=5ft,p1=5psig. V=25ft/s,D2=10ft. Find: (a) Horsepower lost (hp). (b) Pressure at section 2 (psig). (c) Force needed to hold expansion (lbf). Assumptions: α=1.0. Properties: Water, γ=62.4lbf/ft3. PLAN Find the head […]
Chapter 8 Find Speed Air The Wind Tunnel Match
8.51: PROBLEM DEFINITION Situation: Forces due to wind on a building are to be modeled by using a scale model in wind tunnel. 1 500 scale model. Vp=47ft/s,Vm=300ft/s. Fm=50lbf. Find: Density needed for the air in the wind tunnel ¡slug/ft3¢. […]
Chapter 8 Find The Largest Feasible Scale Ratio Solution
8.65: PROBLEM DEFINITION Situation: A model of spillway modeled in laboratory. 1 40 scale model. Vm=3.2ft/s,Qm=3.53 ft3/s. Find: Prototype velocity (ft/s) . Prototype discharge (ft/s) . PLAN UseFroudenumberscaling. SOLUTION Match Froude number Multiply both sides of Eq. (1) by Ap/Am=(Lp/Lm)2 […]
Chapter 8 From The Internet Given Ratio Pulsatile Transient
Collecting like powers gives 21 µ˙m2 D4ρ∆p¶d =µμD ˙m¶c A functional relationship is ˙m √ρ∆pD2=fµμD ˙m¶ 8.19: PROBLEM DEFINITION Situation: A torpedo-like device travels just below the water surface. Find: Identify which π−groups are significant. Justify the answer. SOLUTION •Viscous […]
Chapter 8 Qis the flow rate produced by the fan
8.1: PROBLEM DEFINITION Situation: Dimensions of density, viscosity and pressure. Find: Primary dimensions of density, viscosity and pressure. SOLUTION Density 1 [ρ]=M L3 Viscosity [μ]= M LT Pressure [p]= M LT 2 8.2: PROBLEM DEFINITION Situation: Application of the Buckingham […]
Chapter 8 Wind tunnel test done for a Beoing 787 to simulate
8.36: PROBLEM DEFINITION Situation: A large venturi meter is calibrated with a scale model using a prototype liquid. 1 10 scale model, p=400kPa. Find: The discharge ratio (Qm/Qp) Pressure difference (∆pp)expected for the prototype. SOLUTION Match Reynolds number Multiply both […]
Chapter 9 A viscosity measuring device is consists of measuring
9.18: PROBLEM DEFINITION Situation: A viscosity measuring device is consists of measuring torque on bearing. There is a 10-cm cylinder with 4-cm shaft and 4.5-cm bearing. A force of 0.6 N force on inner cylinder rotates system at 20 rpm. […]
Chapter 9 Problem Definition Situation Model Airplane With Span
9.53: PROBLEM DEFINITION Situation: Displacement thickness for a linear velocity profile. Find: Magnitude of displacement thickness. SOLUTION The equation for displacement thickness is 0 Displacement thickness δ∗=3mm 2 δ∗=1.5mm 61 δ∗=Zδ 0µ1−ρu ρ∞U∞¶dy For constant density δ∗=Zδ 0 (1 −u […]
Chapter 9 Ratio of boundary layer thickness to width
9.66: PROBLEM DEFINITION Situation: A passenger train 81 m long with a 10 m perimeter moving through air at 81.1 km/hr and 204 km/hr. Boundary layer is tripped. Find: Powerrequiredatbothspeeds. Surface resistance at both speeds. Properties: Table A.3 ν=1.41 ×10−5m2/s, […]
Chapter 9 Solution Flow Smooth Three Dierently Shaped Velocity
9.37: PROBLEM DEFINITION Situation: A liquid flows past a smooth flat plate at U0=2m/s. Find: Liquid velocity at a location x=1.0 m downstream from the leading edge and y= 0.8 mm from surface. Properties: ν=2×10−5.m2/s, μ=2×10−2N·s/m2,ρ= 1000 kg/m3 PLAN Calculate […]
Chapter 9 The shear stress is constant across the flow so
9.1: PROBLEM DEFINITION Situation: In which case is the flow caused by a pressure gradient? a. Couette flow b. Hele-Shaw flow SOLUTION 1 The correct answer is b) Hele-Shaw flow 9.2: PROBLEM DEFINITION Situation: Couette flow of liquid with temperature […]