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ME 333 Fluid Mechanics Homework 1 Solutions

Nick Stelzenmuller

April 12, 2013

Problem 1

Problem Statement:

A gas may gas may be considered rarefied if it contains less than 10 12 molecuAt what pressure can air be considered rarefied at 18 C?

Approach:

Treat air as an ideal gas, use ideal gas law with consistant units, check for dimhomogeneity.

Assumptions:

Air can be treated as an ideal gas under these conditons.

Governing Equations:

P = RT Where P is pressure, is density, R is the ideal gas constant, absolute temperature.

Solution:

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From Table A.4: M = 28.97g/mol, Rair = 287 m2s2k

From the problem statement: A= 102moleculesmm3

, T = 10C= 291k

PR = 1012

molecules

mm3(1000mm)3

1m31mol

6.023 1023molecules28.97

g

mol

1kg

1000g287

m

s

PR = 1012

moleculesmm3

(1000mm)3

1m31mol

6.023 1023

molecules28.97

g

mol

1kg

1000g287

m

s

PR 4.0 kg

ms2 = 4.0 P a

Air can be considered rarefied at 18 C if the pressure is less than 4 Pa

Problem 2

Problem Statement:

Given atmospheric conditions on Mars ofT = 50 C, P= 900 P a, find :

1. The density of the Martian atmosphere, assumingRg Rco2

2. The density of air on Earth given the same conditions

3. The density ofC O2 at T = 18 C, P= 101.6kP a

Approach:

Use the ideal gas law to solve for density.

Assumptions:

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Solution:

From Table A.4: Rco2 = 189m2

s2k, Rair = 287

m2

s2k

1. mars= PRco2T

= 900 Pa189

m2

s2k223 k

= 900

kg

ms2

189m2

s2k223 k

2.16 102 kgm3

2. air = PRairT

= 900 P a

287

m2

s2k 223 k

= 900

kg

ms2

287

m2

s2

k 223 k 1.41 102 kg

m3

3. co2 = PRairT

= 101.6 kP a189

m2

s2k

223 k = 101600

kg

ms2

287m2

s2k291 k

1.85 kgm3

The density of the Martian atmospere at P= 900 P a, T = 50 C 2.16 1

The density of air at P= 900 P a, T = 50 C 1.41 102 kgm3

The density ofC O2 at P= 101.6 kP a, T = 18 C 1.85 102 kg

m3

Problem 3

Problem Statement:

A rigid tank contains helium gas at 600 kPa and 20 C. What is the change inif the temperature increases to 40 C?

Approach:

Use the ideal gas law to find the original helium gas density in the tank. A r

does not allow mass or volume change, so density must stay constant in the twe can use this fact to solve for pressure.

Assumptions:

Helium can be treated as an ideal gas at these conditions mass of gas in th

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Solution:

Constant density implies

P =RT = P1

RT1=

P2

RT2 P2 P1= P1

T2T1

1

P2

P1

= 600 kP a313 k

293 k 1 41 kP a

The pressure in the tank rises by 41 kPa when the temperature rises from 20

Problem 4

Problem Statement:

A 25 mm shaft is being pulled through a bearing filled with lubricant. The gapthe bearing and the shaft is 0.3 mm and the bearing is 0.5 m long. The lubrickinematic viscosity of= 8 104 m2/sand a density of = 910 kg/m3.

Find the force necessary to pull the shaft through the bearing at 3 m/s.

Approach:

Use the relationship between shear stress and the shear deformation of a Nfluid. = du

dy, to solve for the required force

Assumptions:

The lubricant is a Newtonian fluid, the velocity profile in the gap is linear, thcondition holds at the boundaries.

Governing Equations:

du h i th h t t th f i d i i it d

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Sketch:

Solution:

From the no-slip condition

u|bearing = 0 u|shaft= 3m/s

We assume the velocity profile is linear, so

du

dy|shaft=

u|bearing u|shafth

Using = and = FA , where A is the area of the shaft in the bearing,

=du

dy =

F

A =

u|bearing u|shaft

h

F = 2RLu|bearing u|shaft

h

F = 225mm

2 0.5m

910

kg

m3

8 104

m2

s

0 3m/s

0.3mm

F 286 N

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ME 333 Fluid Mechanics Homework 2 Solutions

Nick Stelzenmuller

April 19, 2013

Problem 1

Problem Statement:

The plunger on a hydraulic press is used to raise a piston. There is a lever on ththat multiplies the input force by a factor of 4. Given a plunger area of 0.001a piston area of 0.2 m2, what load can be raised by the piston if 1000 N is athe lever?

Approach:

Use the hydraulic force balance to find the unknown force.

Assumptions:

Nothing in the system is accelerating significantly, the hydrostatic pressure dweight of the fluid is unimportant.

Governing Equations:

P1= P2 (hydrostatic equilibrium), F =P A

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Problem 2

Problem Statement:

A cylindrical tank with hemispherical ends contains a volatile liquid and its valiquid density L=120 kPa, and atmospheric pressure PA=101 kPa. Find:

1. The gage pressure reading on the pressure gage

2. The height h of the mercury manometer

Approach:

Use the hydrostatic balance to solve for pressure, noting that we have three flusystem.

Assumptions:

Assumevapor

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Sketch:

Solution:

Pressure gage:

Leth2 be the vertical distance from the free surface of the liquid to the gage.

Pgage = Pvapor+Lgh2 PA= 120kP a+ 800kg/m3(9.81m/s2)1m

1kP a

1000P a

Pgage 26.848 kP a

Manometer:

Remembering that the pressure at a specific depth is constant, so

P2= Pvapor+Lgh2= PA+Hg gh

Pvapor PA + Lgh2 120kP a 101kP a + 800kg/m2(9 81m/s2)1m(

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Problem 3

Problem Statement:

A hinged rectangular gate is held at an angle of 70 degrees to the horizon bywith water on one side and atmosphere on the other. The gate is 1 m widelong, and the free surface meets the gate 2.5 m along its length. Find the tens

cable and the reaction force at the hinge.

Approach:

Find the resultant hydrostatic force (magnitude, direction, and location) on using the equations for hydrostatic forces on plane surfaces. Sum the momenthe hinge to find the cable tension, and sum the forces on the gate to find the

force on the hinge.

Assumptions:

System is static (in equilibrium).

Governing Equations:

Key simplifications of the hydrostatic force F =

pdAfor plane surfaces are:the hydrostatic force can be simplified to a single force vector F =hCG A, wis the depth of the centroid of the plane, applied at the center of pressure. Thepressure formula for the case when the other side of the plane is exposed to atis:yCP =

Ixxsin()hCGA

, and xCP = Ixysin()

hCGA , where I is the moment of inertia an

angle of the plane w.r.t. the horizon. xCP, and yCPare the coordinates of thepressure w.r.t. the centroid.

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Sketch:

Definition of terms on the sketch:

Fc is the tension in the cableFp is the resultant force due to hydrostatic pressureFg is the force due to the weight of the gateP1 is the centroid of the gateP2 is the centroid of the portion of the gate that is submergedP3 is the center of pressure

Solution:

Resultant hydrostatic force and center of pressure

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l3 = Ixxsin()

hCG A

The gate is rectangular, so Ixx= b(l2)3

12

l3= b(l2)

3sin()

12(l2/2)sin()bl2=

l26

Force and moment balances

The problem is now a simple statics problemto find the tension in the cabmust sum the moments about the hinge:

MA= Fp(

l22 l3) +Fg

l12cos() Fcl1sin() = 0

Fc=Fp(

l22 l3) +Fg

l12cos()

l1sin() =

l22sin()bl2(l22

l26) +Fg

l12cos()

l1sin()

Fc=l32b

6l1

+Fg

2 cot() =

9790N/m3(2.5m)3(1m)

6 3m +

5000N

2 cot(70) 9.41

The reaction force on the hinge can be found by summing the forces on the and y:

Fx =Fxp Fc+Ax= 0 Ax= Fc F

xp

Recalling that hydrostatic force operates normal to the plane, Fx

p

=Fpsin() =

Ax=l32b

6l1+

Fg2

cot() l222

bsin2()

Ax=l22b

2

l23l1

sin2()

+

Fg2

cot()

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Fyp =Fpcos() =l22

2sin()bcos()

Ay = Fg+l222

bsin()cos()

Ay = 5000N+ 9790N/m3 (2.5m)

2

2 sin(70)(1m)cos(70) 14.8 kN

The hydrostatic pressure and the weight of the gate cause a 9.41 kN force inand a reaction force on the gate at the hinge of 17.6 kN to the left and 14.8

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Problem 4

Problem Statement:

The Ballard Locks are formed by two gates as shown in the sketch. The widlocks is 24 m, the angle between the gate and perpendicular to the lock wdegrees, the water depths are 16.7 m of the Lake Union side and 8.8 m on t

Sound side. Find:1. The hydrostatic force on a single gate

2. The contact force between the two gates

3. The reaction force at the hinges

Approach:

Similarly to problem 3, this problem can be broken up into: finding a resultastatic force vector, doing the statics problem that results. The only complicatithat there is a different hydrostatic pressure on each side of the gates. We canthe resultant hydrostatic force vectors for each side and then combine them.

Assumptions:

System is in equilibrium, water on both sides is fresh water of equal densities.

Governing Equations:

Key simplifications of the hydrostatic force F =

pdAfor plane surfaces are:

the hydrostatic force can be simplified to a single force vector F =hCG A, w

is the depth of the centroid of the plane, applied at the center of pressure.

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Sketch:

Solution:

Hydrostatic forces

Solving for the resultant hydrostatic force vector for each side separately:

FL

= hL

CGA

L

Noting that the gates are vertical, we can see from the sketch that hLCG = dL2

bdL, b= w2cos()

Similarly for the resultant hydrostatic force vector on the Puget Sound side of

d

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Problem 5

Problem Statement:

A rectangular gate, hinged at the bottom, is held in place by a force FHapplcenter of the gate. The gate is 8 m high and 3 m wide, and is connected to a recpassage filled with water connected to a tank (see sketch). FH=3500 kN. Find

1. The maximum value of the depth h above the center of the gate beforebegins to open.

2. Consider how the problem would be different if the gate was hinged at t

Approach:

Find the resultant hydrostatic force (magnitude, direction, and location) using tions for hydrostatic forces on plane surfaces. Sum the moments around thefind the h. For part 2, sum the moments around the top of the gate

Assumptions:

System is static (in equilibrium).

Governing Equations:

Key simplifications of the hydrostatic force F =

pdAfor plane surfaces are:the hydrostatic force can be simplied to a single force vectorF =hCG A, whethe depth of the centroid of the plane, applied at the center of pressure. Thepressure formula for the case when the other side of the plane is exposed to atare:yCP =

Ixxsin()hCGA

, and xCP = Ixysin()

hCGA , where I is the moment of intertia an

angle of the plane w.r.t. the horizon. xCP, and yCPare the coordinates of thepressure w.r.t. the centroid.

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Sketch:

Solution:

Maximum value ofh

The hydrostatic force of the gate is FP =hCG AwherehCG is the depth of theof the gate. Here the gate is a simple vertically oriented rectangle, so hCG = the height of the gate, and b = the width of the gate, so the gate area A = bl.

FP =hbl

FPacts at the center of pressure, which for a vertically oriented rectangle is displaced downward from the centroid by a distance:

l1= Ixx

hCG A

= bl3

12hbl

= l2

12h

Summing the moments around the hinge:

Mhinge = FH

l

2

FP

l

2

l2

12h

= 0

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The case where the hinge is at the top of the gate

If the hinge were on the top of the gate the hydrostatic pressure force has a lonarm. A longer lever arm means less force is required to balance the moment atop hinge, i.e. the water depth h would be lower when the gate opens. We this reasoning by summing the moments about the top hinge:

MTop hinge

= FH l

2 F

P l

2+

l2

12h = 0

FH

l

2

=hbl

l

2+

l2

12h

FH2hb

= l

2+

l2

12h h=

FHbl

l

6

h=

3.5 106N

(9790N/m3)(3m)(8m)

8m

6

13.6m

The height of water in the tank such that the gate begins to open is approxFor the case with hinge on the top of the gate, the critical height is approxim

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Problem 6

Problem Statement:

A U-tube filled with water rotates about its axis of symmetry. The radius of this 0.1 m, and the height of the water is 0.3 m. Determine the angular velocity nfor the water to begin to vaporize in the bottom of the tube.

Approach:

Use the equation for the pressure distribution in a fluid subject to rigid bodySolve for the angular velocity s.t. the pressure at the bottom of the tube evapor pressure of water.

Assumptions:

The fluid in the tube has attained rigid body rotation. Assume standard conthe surrounding environment (T=20 C, P=101.6 kPa)

Governing Equations:

For a fluid that has attained rigid body rotation, the pressure distribution inis given by:

P =C z +1

2r22

Sketch:

"#\$ %"#\$ %

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Solution:

Let us place the origin on the axis of rotation and at the level of the bottoU-tube. Leth be the distance from the origin to the free surface, and R be tof the U-tube. At the surface P =Pa, so we can solve for the constant C:

Pa= C (h) +1

2R22 C=Pa +h

1

2R22

The water will begin to vaporize at the vapor pressure Pv, so setting the presswe can solve for :

Pv =Pa+h 1

2R2 (0) +

1

2(0)22

Pv Pa h = 1

2R22 =

Pv Pa h

(1/2)R2

The vapor pressure of water at 20 C is 2.337 kPa (White, Table A.5), so is:

=

2337P a 101600P a 9790N/m3(0.3m)

The water at the bottom of the U-tube begins to vaporizewhen the angular velocity reaches 143 rad/sec

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Problem 7

Problem Statement:

A barge has a trapezoidal shape and is 22 m long into the page (see sketch fother dimensions). If the total weight of the barge and cargo is 300 tons, whdraft, H, of the barge in seawater?

Approach:

Use Archimedes law of displacement to relate volume of displacement with bUse geometry to find an expression for volume of water displaced and solve fo

Assumptions:

The bow and stern of the barge are simply vertical. Weight is measured in Uton=8899.485 N).

Governing Equations:

Archimedes law: Buoyancy force is equal to the weight of fluid displaced.

Sketch:

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The volume of water displaced is:

V =BHL+H(Htan(30)L

W =L(BH+H2tan(30))

tan(30)H2 +BHW

L= 0

Recalling that 1 US ton is 8899.485 N, and seawater = 10050N/m3 (White, T

H=B+

B2 4tan(30)WL

2tan(30) =

8m+

(8m)2 4tan(30) 3008899.485(10050N/m3(22

2tan(30)

The barge displacement is approximately 1.37 m

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ME 333 Fluid Mechanics Homework 3 Solutions

Nick Stelzenmuller

April 26, 2013

Problem 1

Problem Statement:

Air at atmospheric conditions is drawn into a compressor at a steady rate of The compression ratio is Pout/Pin = 10, and the evolution of the gas can be as a first order approximation, to be isentropic, P /= constant, where is thspecific heats for the gas and is equal to 1.4. If the design criteria is that the vthe outlet does not exceed 70 ft/s, what is the minimum diameter for the rounthe outlet?

Approach:

Treating the compressor as a 1-D system, we can use conservation of mass tothe area of the exit pipe.

Assumptions:

System may be approximated as 1-D (i.e. flow is approximately uniform over tsection of the inlet and outlet. Flow is steady.

Governing Equations:

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Sketch:

!"#\$%

'(%#\$%

)*+ -\$..*-

Solution:

inAinVin= outAoutVout

Where:

AinVin= 15 f t3/s

Vout = 70 ft/s

We can find out using the isentropic expansion relation P/ = constantcompression ratio Pout/Pin= 10.

Pinin

=constant=Poutout

out = in

PoutPin

= 10in out= 101/in

Substituting this resulting into the conservation of mass equation and solving

inAinVin= outAoutVout= 101inAoutVout Aout=

Dout4

= inA

101/4A V

4(15f t3/ )

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Problem 2

Problem Statement:

When a 2-D liquid jet hits an inclined flat plate, it splits into two streams wspeed but uneven thickness. Assuming the shear stresses on the plate is ncalculate the resulting thicknesses,h2 and h3, as a function of the plate anglecalculate the force on the plate necessary to keep it in place.

Approach:

Conservation of mass and conservation of momentum give three scalar equatiouse to solve for the three unknowns: h2, h3, and the magnitude of the reaction

Assumptions:

The flow is steady, the velocity in each jet is approximately uniform (i.e. 1-Dmation), the fluid density is constant, there is no shear stress on the plate, theforce R is normal to the plate.

Governing Equations:

Conservation of mass for a control volume is:

d

dt

CV

dV

+

CS

(V n)dA= 0

Which can be simplified for steady, 1-D systems to:

min

= mout

Conservation of momentum for a fixed control volume is given as:

F=

d

dt

CV

V dV

+

CS

V(V n)dA= 0

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Sketch:

!

!

#\$

#%#

&

!

Solution:

We will use equations 1 and 2 to solve for the quantities h2, h3 shown in the skthe reaction force R, assumed to be normal to the plate resisting the fluid presWbe the width of the jets into the page.

From the figure:

min= hW Vmout= h2W V +h3W V

Substituting these expressions into equation 3:

min=

mout hWV =h2W V +h3W V h= h2+

Similarly for conservation of momentum:

F=

( miVi)out

( miVi)in R= hWVV + h2W VV2+

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y component: Rcos() = W h2V2

sin() +W h3V2

sin()

Now we have the three equations ([3], [4], and [5]) that we need to solve for variables of interest: R, h2, h3. Solving [3] for h2 gives h2 =h h3. Substitu[4] and [5]:

Rsin() = WhV2 W(h h3)V2cos() +W h3V

2cos()

Rcos() = W(h h3)V2sin() +W h3V

2sin()

Combining [6] and [7]:

R=WhV2 +W(h h3)V

2cos() W h3V2cos()

sin() =

W(h h3)V2sin(

cos

Simplifying:

h+ (h h3)cos() h3cos()

sin( =

(h h3)sin() +h3sin()

cos()

hcos() +hcos2() 2h3cos2() = hsin2() + 2h3sin

2()

hcos() +h(cos2() +sin2()) = 2h3(cos2() +sin2())

h3=h(1 +cos())

2

Combining [8] with [3] gives:

h2= h h3 h2 = h h(1 +cos())

2 =

h(1 cos())

2

Finally , subtituting [8] and [9] in [5] gives:

Rcos() =W h2V2sin() +W h3V

2sin()

R=

W( h(1cos())2 )V2sin() W( h(1+cos())2 )V

2sin()

cos()

Simplifying:

R=W V2sin()h

2cos() [(1 cos()) (1 +cos())]

R W V 2h i ()

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Problem 3

Problem Statement:

A 10 mm diameter jet of water is deflected by a homgeneous block (15mm x 2100 mm) that weighs 6 N. Determine the minimum flow rate needed to tip tNeglect and splashback and neglect shear forces.

Approach:

Draw a control volume around where the jet impacts the block, then use consermomentum to find the external force necessary to balance the momentum fluxforce on the block by the jet). Use statics to equate the weight moment and thecreated by the jet at the bottom right corner of the block.

Assumptions:

Flow is steady, system can be approximated as 1-D.

Governing Equations:

Conservation of momentum for a fixed control volume is given as:F=

d

dt

CV

V dV

+

CS

V(V n)dA= 0

Which can be simplified for steady, 1-D systems to:

F= ( miVi)out ( miVi)in

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Sketch:

!"# %

!"!#& %

!"!# %

!"!& %

Solution:

First define some parameters, Let:

V=jet velocity (average, since we assume 1-D jet)F=force on block by jetd=jet diameterw=block width (15 mm)h=block height (100 mm)W=weight of block

Force on the block by the jet

For this problem we can choose our control volume such that the momentum in the x-component, and the momentum out is only in the y-component. Thestatement tells us to neglect shear, so we can neglect the y-components of the mequation The x component of the momentum equation is:

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The force on the block by the jet is equal and opposite to the reaction force, s

F = R= d2

4V2

Statics

The block would overturn by pivoting about the bottom right corner; taking the

w

2 F

H

2 = 0

V =

W(w/2)

(d2/4)(h/2)

This gives the velocity of the jet s.t. the block begins to tip. The volumetric Q= V A, so

Qcritical = VcriticalA=

W(w/2)

(h/2)

d2

4 2.66 104 m3/s

The flowrate at which the block begins to tip is approximately 2.66 104 m

Problem 4

Problem Statement:

Water flows through a 90 pipe bend. The pipe cross-section is a constant 0.1flow rate is a constant 280 gpm. The pressures at the top and left side are 24 anrespectively. The total volume of the bend is 2 f t3 Calculate the net force Rto hold the pipe bend in place

Approach:

The change in mometum flux is equal to the sum of the forces acting on tholume These forces are The force required to hold the bend in place R the

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Governing Equations:

Conservation of momentum for a fixed, 1-D, steady control volume is:F=

( miVi)out

( miVi)in

Pressure force on a control volume is:

Fp=

CS(pabsolute patmospheric)(n)dA

Sketch:

Solution:

Conservation of momentum

F= mVout mVin

Resolving into x and y components:Fx= mV

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Force due to pressure

Fp=

CS

(pabsolute patmospheric)(n)dA

Resolving into x and y components:

Fxp =

inlet

(Pinlet Patm)dA= (Pinlet Patm)A

Fyp =

inlet

(Poutlet Patm)dA= (Poutlet Patm)A

Force due to weight of water in pipe bend

Fyg = Vg Where V is the volume of water in the pipe bend

Support force

Rx+Fx

p = mV Rx= (Pinlet Patm)A mV

Ry+Fyp +Fyg = mV Ry = (Pinlet Patm)A+Vg+ mV

Where:m= V= 1.937 slug

f t3280g pm

0.133681 f t3

1 US gal

1 minute

60 s

1.20839 slugs

A=0.10 f t2

V = mAV= 2f t3

Pinlet=26 psiPoutlet=24 psiPatm=14.7 psiwater = 1.937 slug/ft

3

g= 32.1740 ft/s2

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Ry = (Pinlet Patm)A+Vg+ mV = (Poutlet+Patm)A Vg+ mm

A

Ry = (24psi14.7psi)

(12in)2

1f t2

0.1f t2+1.937slug/ft3(2f t3)(32.1740f t/s2)+

Ry 264.80 lbf

The force required to hold the pipe in place is R= 168.96i+ 264.80j lbf

Problem 5

Problem Statement:

You are conducting a wind tunnel test to measure the aerodynamic drag (tance) over a cylinder. Apply the momentum principle for a control volumethe drag force per unit width, W, in terms of the upstream and downstream and pressures. The upstream velocity is a 15.6 m/s. The downstream horizonity measurements are shown in the table below. Upstream pressure is 133.5 and downstream pressure is 0.0 Pa (gage). The diameter of the cylinder is Total wind tunnel height is 240 mm. Neglect any forces on the wind tunnel whorizontal velocity profile is symmetric about the y axis.

Approach:

This problem asks us to do a force balance on the control volume shown beproblem statement says to neglect any forces on the wind tunnel walls; the foremain are: the pressure force on the inlet and outlet, and the force on the Ccylinder. The vector sum of these forces equal the change in momentum flux a

inlet and outlet of the CV. The key to this problem are choosing the correct simassumptions and being careful with the signs on the various forces.

Assumptions:

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Governing Equations:

The force balance includes pressure forces,Fp and and force on the CV by theFd. Expanding these forces for a fixed, steady-flow control volume gives:

FCV = Fp+ Fd= Change in momentum flux

FCV = CS P(n) dA + Fd= CS VV n dA

Sketch:

Solution:

Forces due to change in momentum flux

The force due to the change in momentum flux is given by:

Fm=

CS

VV ndA

Considering this equation at the inlet and outlet of our control volume gives:

Fm=

inlet

VinVin dA+

outlet

VoutVout dA

From the problem statement we know thatVinis constant, butVoutchanges wiWe have data for the outlet velocity at thirteen positions, and we need to use

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Z position (mm) Air speed (m/s) dA (m2) Mean airspeed in dA dM

-120 23.9 0.01W 23.8 6.77

-110 23.6 0.01W 23.3 6.51-100 23.0 0.01W 22.7 6.18

-90 22.4 0.01W 21.9 5.76

-80 21.4 0.01W 20.8 5.19

-70 20.2 0.01W 19.1 4.38

-60 18.0 0.01W 16.6 3.31

-50 15.2 0.01W 13.7 2.24

-40 12.1 0.01W 10.7 1.36

-30 9.2 0.01W 8.1 0.79

-20 7.0 0.01W 6.4 0.49

-10 5.8 0.01W 5.4 0.35

0 5.0 0.01W 5.4 0.35

10 5.8 0.01W 6.4 0.49

20 7.0 0.01W 8.1 0.79

30 9.2 0.01W 10.7 1.36

40 12.1 0.01W 13.7 2.2450 15.2 0.01W 16.6 3.31

60 18.0 0.01W 19.1 4.38

70 20.2 0.01W 20.8 5.19

80 21.4 0.01W 21.9 5.76

90 22.4 0.01W 22.7 6.18

100 23.0 0.01W 23.3 6.51

110 23.6 0.01W 23.8 6.77

120 23.9 Total Momentum 86.6

Table 1: The data in the two left columns were given. Area segments are definedthe measurement points, and the air speeds between adjoining measurement poaveraged for a representative air speed for each area segment. Momentum flux

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Putting this together:

F= V(HW)V+ Mout

And because we only have horizontal flow, we can put this in the scalar form Fy = (HW)V

2 +Mout

Pressure force

The force due to pressure can be similarly divided into inlet and outlet pressuCS

P(n)dA =

inlet

P(n) dA +

outlet

P(n) dA

Because we our CV is assumed to be surrounded by atmospheric pressure it is coto work in gage pressure, (P Patm). From the problem statement we know

gage pressure at the outlet is zero, so the pressure force is simply:

Fyp =Pin,gageA

Force on the CV by the cylinder

We can find the drag force on the cylinder using the force balance:

FCV = Fp+ Fd= V(HW)V+ Mout

Fd= V(HW)V+ Mout Fp

Fyd = (HW)V2in+Mout Pin,gageA

Fyd = 1.2kg/m3(0.24m)W(15.6m/s)2+86.6481W N133.5P a(0.24m)W

Fd is the force on the control volume by the cylinder, so from Newtons thir

force on the cylinder by the control volume is Fd=15.48W N to the right.

The drag on the cylinder is approximately 15.48 N per unit width in the pos

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