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

Separating variables and integrating

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5.78: PROBLEM DEFINITION

Situation:

A tire develops a slow leak.

˙m=0.68pA/√RT,V=0.5ft

3,t=3h.

Find:

Area of the leak.

Properties:

From Table A.2: R=1716ftlbf/slug R.

T=60◦F,p1=30psig, p2=25psig.

PLAN

Apply the continuity equation.

SOLUTION

Continuity equation

Let ˙mout =0.68A/√RT in the above equation

Finding area

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5.79: PROBLEM DEFINITION

Situation:

An O2bottle leaks oxygen through a small oriﬁce causing the pressure to drop.

˙m=0.68pA/√RT,V=0.1m

3.

D=0.12 mm.

Find:

Time required for the speciﬁed pressure change.

Properties:

From Table A.2: R=260J/kg K.

T=18◦C,p

0=10MPa.

p=5MPa.

PLAN

Apply the continuity equation and the ideal gas law.

SOLUTION

Continuity equation

RT

Combining previous 2 equations

Let ˙mout =0.68A/√RT in the above equation

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Finding time

84

5.80: PROBLEM DEFINITION

Situation:

A tank is draining through an oriﬁce.

h1=3m,h=0.5m.

DT=0.6m,D2=3cm

Find:

Time required for the water surface to drop the speciﬁed distance (3 to 0.5 m).

SOLUTION

From Example 5-6 the time to decrease the elevation from h1to his

85

5.81: PROBLEM DEFINITION

Situation:

A cylindrical drum of water is emptying through a pipe on the bottom.

D=2 ft,R=1 ft.

V=√2gh,L=4 ft.

d=2 ft=0.167 ft,h

0=1 ft.

Sketch:

Find:

Time to empty the drum.

PLAN

Apply the continuity equation. Let the control surface surround the water in the

tank. Let the c.s. be coincident with the moving water surface. Thus, the control

volume will decrease in volume as the tank empties. Situate the origin at the center

of the tank.

SOLUTION

Continuity equation

86

Integrate Eq. (5)

REVIEW

5.82: PROBLEM DEFINITION

Situation:

Water drains from a pressurized tank.

Ve=q2p

ρ+2gh,ho=2 m.

A=1m

2,Ae=10cm

2.

Find:

Time for the tank to empty with given supply pressure.

Time for the tank to empty if supply pressure is zero.

Properties:

p=10 kPa.

Water, Table A.5: ρ=1000kg/m3.

PLAN

Apply the continuity equation. Deﬁne a control surface coincident with the tank

walls and the top of the ﬂuid in the tank.

SOLUTION

Continuity equation

ρ+2gh

Integrating this equation gives

For zero pressure in the tank, the time to empty is

88

5.83: PROBLEM DEFINITION

Situation:

A tapered tank drains through an oriﬁce at bottom of tank.

Ve=√2gh,D=d+C1h.

h0=1m,h=20cm,d=20cm.

C1=0.3,dj=5cm.

Find:

Derive a formula for the time to drain.

Calculate the time to drain.

PLAN

Apply the continuity equation.

SOLUTION

From continuity equation

Evaluating the limits of integration gives

89

5.84: PROBLEM DEFINITION

Situation:

Water drains out of a spherical tank that begins at half full.

Ve=√2gh, R =0.5m,de=1cm.

Sketch:

Find:

Time required to empty the tank.

PLAN

Apply the continuity equation. Select a control volume that is inside of the tank and

level with the top of the liquid surface.

SOLUTION

Continuity equation

ρdV

dt =−ρAeVe

90

For R=0.5mandAe=7.85 ×10−5m2, the time to empty the tank is

91

5.85: PROBLEM DEFINITION

Situation:

A tank containing oil is draining from the bottom.

DT=2m,ho=5m.

L=6m,de=2cm.

p=(po+patm)×(L−ho)/(L−h)−patm.

dh/dt =−Ae/AT×p2gh +2p/ρ.

Find:

Predict the depth of the oil with time for a one hour period.

Properties:

ρ=880kg/m3,po=300kPa,patm =100kPa.

SOLUTION From the continuity equation

L−h¶−patm

A numerical program was developed and the numerical solution provides the following

results:

5

6

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5.86: PROBLEM DEFINITION

Situation:

Rocket Propulsion. To prepare for problems 5.87, 5.88, and 5.89 in EFM10e, use the

internet or other resources and deﬁne the following terms in the context of rocket

propulsion: (a) solid fuel, (b) grain, and (c) surface regression.

Also explain how a solid fuel rocket engine works.

SOLUTION

Answers will vary, but should include the following elements:

93

5.87: PROBLEM DEFINITION

Situation:

Propellant fuels an end-burning rocket motor.

Dc=0.1m,De=0.08 m,˙r=1.5cm/s.

Find: Gas velocity at nozzle exit plane.

Properties:

ρ=1800kg/m3,R= 415 J/kg K.

pe=10kPa,T=2200◦C.

PLAN Apply the continuity equation and the ideal gas law.

SOLUTION Ideal gas law

94

5.88: PROBLEM DEFINITION

Situation:

Propellant fuels a cylindrical-port rocket motor.

D0=0.2m,D=0.12 m.

De=0.2m,Ve=1800m/s.

pe=10kPa,˙r=1.0cm/s.

L=0.4m.

Find:

Gas density at the exit.

Properties:

ρg=2000kg/m3,R=415J/kg K.

SOLUTION Area of the grain surface (internal surface and two ends)

95

5.89: PROBLEM DEFINITION

Situation:

Propellant ﬂows through a nozzle in a rocket chamber.

˙m=0.65pcAt

√RTc

,˙r=apn

c.

n=0.3,pc=kPa.

pc=¡aρp

0.65 ¢1/(1−n)³Ag

At´1/(1−n)(RTc)1/[2(1−n)].

Find:

Derive a formula for chamber pressure.

Calculate the increase in chamber pressure if a crack increases burn area by 20%.

PLAN

Apply the ﬂow rate equation.

SOLUTION

Continuity equation. The mass ﬂux oﬀthe propellant surface equals ﬂow rate through

nozzle.

96

5.90: PROBLEM DEFINITION

Situation:

A piston moves in a cylinder and drives exhaust gas out an exhaust port in a four

cycle engine.

˙m=0.65pcAv

√RTc

,dbore =0.1m.

L=0.1m,Av=1cm

2.

V=30m/s.

Find:

Rate at which the gas density is changing in the cylinder.

Assumptions:

The gas in the cylinder is ideal and has a uniform density and pressure.

Properties:

T=600◦C,R=350J/kg K,p=300kPa.

SOLUTION

Continuity equation. Control volume is deﬁned by piston and cylinder.

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5.91: PROBLEM DEFINITION

Situation:

Gas is ﬂowing from Location 1 to 2 in a pipe expansion. The inlet density, diameter

and velocity are ρ1,D1,andV1respectively. If D2is 2D1,andV2is 1

2V1,whatis

the magnitude of ρ2?

a. ρ2=4ρ1

b. ρ2=2ρ1

c. ρ2=1

2ρ1

d. ρ2=ρ1

PLAN

Apply the continuity equation, and utilize the deﬁnition of volume ﬂowrate.

SOLUTION

ρ1

π

4D2

1V1=ρ2

π

4D2

2V2

98

5.92: PROBLEM DEFINITION

Situation:

Air is ﬂowing from a ventilation duct (cross-section 1) as shown, and is expanding

to be released into a room at cross-section 2. The area at cross-section 2, A2,is3

times A1. Assume that the density is constant. The relation between Q1and Q2is:

a. Q2=1

3Q1

b. Q2=Q1

c. Q2=3Q1

d. Q2=9Q1

PLAN

Apply the continuity equation.

SOLUTION

According to the continuity equation, Q2=Q1, because there is no storage in the

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5.93: PROBLEM DEFINITION

Situation:

Water is ﬂowing from Location 1 to 2 in this pipe expansion. D1and V1are known

at the inlet. D2and P2are known at the outlet. What equation(s) do you need to

solve for the inlet pressure P1? Neglect viscous eﬀects.

a. The continuity equation

b. The continuity equation and the ﬂow rate equation.

c. The continuity equation, the ﬂow rate equation, and the Bernoulli equation

d. There is insuﬃcient information to solve the problem

PLAN

We know that we need to apply the continuity equation, utilize the deﬁnition of

volume ﬂowrate, and use the Bernoulli equation which relates pressure, elevation,

and velocity. To verify that there is suﬃcient information to solve the problem, set

up the equations and document the knowns and unknowns.

SOLUTION

A1V1=A2V2

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