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 control volume that is attached to the falling

sphere. Relative to the sphere, the velocity entering the control volume is V1and the

velocity exiting is V2

Continuity equation

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5.60: PROBLEM DEFINITION

Situation:

Air ﬂows in a rectangular duct.

Q=1.44 m3/s, A1=20cm×60 cm.

A2=10cm×40 cm.

Find:

Airspeedforinitialductarea,V1.

Airspeedforlatterductarea,V2.

Assumptions:

Constant air density.

PLAN

Apply the ﬂow rate equation.

SOLUTION

Flow rate equation

62

5.61: PROBLEM DEFINITION

Situation:

A pipe divides into two outlets.

D30cm =30cm,D20cm =20cm,D18cm =18cm.

V20cm =V18cm,Q=0.4m

3/s.

Find:

Dischargeineachbranch.

PLAN

Apply the ﬂow rate equation.

SOLUTION

Flow rate equation

63

5.62: PROBLEM DEFINITION

Situation:

A pipe divides into two outlets.

D30cm =30cm,D20cm =20cm,D15cm =15cm.

Q20cm =2Q15cm,Q=0.3m

3/s.

Find:

Mean velocity in each outlet branch.

SOLUTION

Continuity equation

Flow rate equation

64

5.63: PROBLEM DEFINITION

Situation:

Water ﬂows through pipe that is in series with a narrower pipe.

A pipe divides into two outlets.

D10 =10in,D6=6in.

V20cm =V15cm,Q=898 gal/min.

Find:

Mean velocity in each pipe.

PLAN

Apply the ﬂow rate equation.

SOLUTION

Flow rate equation

65

5.64: PROBLEM DEFINITION

Situation:

Water ﬂows through a tee.

DA=DB=4m,DC=2m.

VA=6m/s,VC=4m/s.

Find:

Mean velocity in outlet B.

PLAN

Apply the continuity equation.

SOLUTION

Continuity equation

66

5.65: PROBLEM DEFINITION

Situation:

Gas ﬂows in a round conduit that tapers to a smaller diameter.

D1=1.2m,D2=0.6m,V1=15m/s.

Find:

Mean velocity at section 2.

Properties:

ρ1=2.0kg/m3,ρ2=1.5kg/m3.

PLAN

Apply the continuity equation.

SOLUTION

Continuity equation

67

5.66: PROBLEM DEFINITION

Situation:

Pipes A and B are connected to an open tank.

QA=10ft

3/min,A=80ft

2,dh/dt =1in/min.

Find:

DischargeinpipeB.

If ﬂow in pipe B is entering or leaving the tank.

Sketch:

PLAN

Apply the continuity equation. Deﬁne a control volume as shown in the above sketch.

Let the c.s. move upward or downward with the water surface.

SOLUTION

Continuity equation

68

5.67: PROBLEM DEFINITION

Situation:

A tank has one inﬂow and two outﬂows.

D4in =4in=0.333 ft,V4in =10ft/s.

D6in =6in=0.5ft,V6in =7ft/s.

D3in =3in=0.25 ft,V3in =4ft/s.

Dtank =3ft.

Find:

Is the tank ﬁlling or emptying.

Rate at which the tank level is changing: dh

dt

SOLUTION

Inﬂow =(10ft/s) ³π

4´µ4

12 ft¶2

=0.8727 cfs

69

5.68: PROBLEM DEFINITION

Situation:

Atankisﬁlled with water over time.

Di=12in =1ft,Vi=1ft/s.

Do=6in=0.5ft,Vo=2ft/s.

h1=0ft,D

tank1 =1ft.

Dtank2 =2ft,h2=10ft.

Find:

At t=22s, if the the water surface will be rising or falling.

Rate at which the tank level is changing: dh

dt

Sketch:

PLAN

Apply the continuity equation. Deﬁne a control volume in which the control surface

(c.s.) is coincident with the water surface and moving with it.

SOLUTION

Continuity equation

Since Adh/dt > 0, the water level must be rising. While the water column occupies

70

the 12 in. section, the rate of rise is

Therefore, at the end of 20 sec. the water surface will be in the 2 ft. section. Then

the rise velocity will be:

71

5.69: PROBLEM DEFINITION

Situation:

A lake is fed by an inlet and has no outlet. cfs. Lake surface area is , where his

depth in feet.

Qin =1200ft

3/s,A(h)=4.5+5.5hmi2.

QEvap =13ft

3/smi

2.

Find:

Equilibrium depth of lake.

The minimum discharge to prevent the lake from drying up.

PLAN

Apply the continuity equation.

SOLUTION

Continuity equation

72

5.70: PROBLEM DEFINITION

Situation:

A nozzle discharges water onto a plate moving towards the nozzle. Plate speed

equals half the jet speed.

Qin =5ft

3/s,Vin =2Vp.

Find:

Rate at which the plate deﬂects water.

PLAN

Apply the continuity equation. Select a control volume surrounding the plate and

moving with the plate.

SOLUTION

Continuity equation

Qin =Qp

Reference velocities to the moving plate. Let Vobe the speed of the water jet relative

73

5.71: PROBLEM DEFINITION

Situation:

Atankwithadepthhhas one inﬂow ( ft3/s) and one outﬂow through a 1 ft

diameter pipe. The outﬂow velocity is .

Q=20ft

3/s,Vout =√2gh,Dout =1ft.

Find:

Equilibrium depth of liquid.

PLAN

Apply the continuity equation and the ﬂow rate equation.

SOLUTION

Continuity equation

74

5.72: PROBLEM DEFINITION

Situation:

Flows with diﬀerent speciﬁc weights enter a closed tank through ports A and B and

exit the tank through port C. Assume steady ﬂow. Details are provided on ﬁgure

with problem statement.

DA=6in,QA=3ft

3/s.

SA=0.95,DC=6in.

QB=1ft

3/s,SB=0.85,DB=4in.

Find:

Mass ﬂow rate at C.

Average velocity at C.

Speciﬁcgravityofthemixture.

Assumptions:

Steady state.

PLAN

Apply the continuity equation and the ﬂow rate equation.

SOLUTION

Continuity equation

Continuity equation, assuming incompressible ﬂow

75

5.73: PROBLEM DEFINITION

Situation:

O2and CH4are mixed in a mixer before exiting.

VO2=VCH4=5m/s.

ACH4=1cm

2,AO2=3cm

2.

Find:

Exit velocity of the gas mixture, Ve.

Properties:

From Table A.2: RO2= 260 J/kg K,RCH4=518J/kg K.

T=100◦C,ρ=1.9kg/m3,p=200kPa.

PLAN

Apply the ideal gas law to ﬁnd inlet density. Then apply the continuity equation.

SOLUTION

Ideal gas law

Continuity equation

76

5.74: PROBLEM DEFINITION

Situation:

A pipe with a series of holes is used to distribute air.

Qhole =0.67A0¡2∆p

r¢1/2.

nhole =50/m,L=10m.

Dpipe =0.5m,Dhole =2.5cm.

Find:

Velocity of air entering the pipe.

Properties:

From Table A.2: R=287J/kg K.

T=20◦C,p

pipe =100Pagage

PLAN

The total discharge out of the holes is equal to the inlet discharge.

Qin =AVin =NQhole

where Nis the number of holes and Qholeis the discharge for each hole.

SOLUTION

The total number of holes

The density in the pipe is

The ﬂow rate through the holes is

The velocity at the pipe entrance is

77

5.75: PROBLEM DEFINITION

Situation:

Water through a globe valve.

Q=10gal/min,∆r2=0.125 in.

D1=1in,D2=0.5in

Find:

Pressure drop across the valve.

Properties:

T=60◦F.

PLAN

Apply the Bernoulli equation between the 1 inch upstream pipe and the opening at

the seat of the valve

SOLUTION

The pressure drop across the globe valve is

Theareabetweenthediscandtheseatis

and the velocity in the pipe is

The speciﬁc weight of water at 60oF is 62.37 lbf/ft3.The pressure drop is

78

5.76: PROBLEM DEFINITION

Situation:

Water ﬂowthroughanoriﬁce in a pipe.

D1=2.5cm,Do=1.5cm.

0.64Ao=A2,Q=CAo³2∆p

ρ´1/2

.

∆p=10kPa,Q=1000kg/m3.

Find:

a) Derive equation for discharge.

b) Evaluate discharge across oriﬁce.

PLAN

Apply the continuity equation and the Bernoulli equation between pipe and vena

contracta. Neglect elevation change.

SOLUTION

Let point 1) be at the centerline of the upstream pipe and point 2) at the vena

contracta. The Bernoulli equation gives

The ratio of the cross-sectional area at the vena contracta to the area of the oriﬁce

is A2

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Substitute into discharge equation

For a ∆p=10kPa the discharge is

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