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CE 330 Final Exam Solutions: Hydraulics and Open Channel Flow Problems - Prof. Alfred Parr, Exams of Fluid Mechanics

The solutions to the final exam problems for ce 330, focusing on hydraulics and open channel flow. The problems include determining resultant forces, velocities, discharges, energy losses, and forces on sluice gates and vane series. Students can use this document as a study resource for understanding hydraulics concepts and solving similar problems.

Typology: Exams

Pre 2010

Uploaded on 12/15/2009

stoddard
stoddard 🇺🇸

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CE 330, Final Exam, Sp 05 Name ___________________________________________
(Problems 1-4 = 65%; problems 5-7=
35%)
Problem 1 The body is submerged in a
water tank in four different positions. Its
dimension into the tank is 10 feet and the
depth of water is 16 feet. Determine the
resultant vertical force the water exerts
on the body for each of the 4 positions.
Draw FBDs for each case on the picture below.
10
3-D View
10
10
3-D View
16’
abcd
16’
abcd
1
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff

Partial preview of the text

Download CE 330 Final Exam Solutions: Hydraulics and Open Channel Flow Problems - Prof. Alfred Parr and more Exams Fluid Mechanics in PDF only on Docsity!

CE 330, Final Exam, Sp 05 Name ___________________________________________ (Problems 1-4 = 65%; problems 5-7= 35%)

Problem 1 The body is submerged in a water tank in four different positions. Its dimension into the tank is 10 feet and the depth of water is 16 feet. Determine the

resultant vertical force the water exerts

on the body for each of the 4 positions. Draw FBDs for each case on the picture below.

3-D View

10’10’

3-D View

16’ a b (^) c

d

16’ a b (^) c

d

Problem 2 The velocity in the pipe shown below is 10 fps. Determine the discharge and the elevation of the lower tank. Then plot the EGL and HGL on the next page after filling in the values in the table for sections a-f as shown below.

Loc EGL HGL

a b c d e f

K = 1.

K = 0.

K = 0.

Z (^) a=100’

D = 6”

n = 10-5^ ft^2 /sec ks = 0.0001’

K = 1

a b c^ d^ ef

V = 10 fps

K = 1.

K = 0.

K = 0.

Z (^) a=100’

D = 6”

n = 10-5^ ft^2 /sec ks = 0.0001’

K = 1

a b c^ d^ ef

V = 10 fps

Problem 3 A 0.5-inch diameter free jet enters the vane series as shown. Determine the following (a) The magnitude of V (^) 1,abs.

30 o

45 o

u = 30 fps

20 o

V1,abs

30 o

45 o

u = 30 fps

20 o

V1,abs

20 o

V1,abs

(b) The absolute velocity vector of the outgoing jet. (c) The horsepower the jet loses going through the vane. (d) The force vector exerted by the vane series on the jet. (x and y components)

Problem 4 Water flows as a free jet from the 4-inch diameter pipe shown. Neglect friction and local losses and determine (a) The discharge. (b) The maximum height of the jet relative to the large tank water surface elevation. (c) The velocity vector just before the jet strikes the water in the 4’x4’x4’ water tank. (d) The increases in the readings of Scales A and B when the jet impinges on t water in the tank. Draw FBDs here since you will be using the momentum equation. The reading of Scale B is zero before the jet impinges.

he

(e) What is the total reading on Scale A with the impinging jet if the tank weighs 400 lb when empty?

55 o

D = 4”

1

2

3

4

4’x4’x4’ tank of water

Large tank (water)

Scale A

4’ Scale B

55 o

D = 4”

1

2

3

4

4’x4’x4’ tank of water

Large tank (water)

Scale A

4’ Scale B

Problem 6 The channel on the right has a slope of 0.001, n = .030.

yo = 9 ft

yo = 9 ft

yo = 9 ft

yo = 9 ft

(a) Determine discharge if the uniform depth of flow is 9 ft as shown. (b) Determine the Froude number for the uniform flow. (c) Determine the critical depth for the channel.

Problem 7 A long trapezoidal channel with a 15-ft base width and 1.5 to 1 side slopes flows at a uniform depth of 10 ft. If n = .020 and So = 0.0008 determine the following.

10’

y (^2)

flow

(a) The discharge and the specific energy, (b) The minimum width of an RCB required to pass the flow without creating a choke if the entrance loss coefficient is 0.5. Assume inlet control. (c) The amount by which the culvert inlet must be dropped if an 18-ft wide RCB is used and a choke is to be avoided. (d) The depth in the trapezoidal channel just upstream from the culvert if an 18-ft RCB is used with no dropped inlet. (e) For case (d), determine the distance upstream from the culvert where the depth equals 11 ft. Use the step method with 2 steps.

G C

M

G C

M

Hydrostatics ( γ )

Where G = center of gravity C = center of buoyancy M = metacenter Ioo = second area moment about plane where water surface cuts body

If GM positive, stable If GM negative, unstable

water = 62.4 lb/ft 3

dp/dz = - γ ; p/ γ + z = constant, P + γ z = constant

P absolute = P gauge + Patmospheric

Forces on an inclined plane

F = pcA = γ ybar(sin β ) A acts at (xcp, ycp) where

ycp = ybar + I/ybarA and I is the second moment about the centroidal x-axis.

Forces on a curved surface and bouyant force (hor.) Fx = Pc,x Ax; Fy = Pc,y Ay acts at center of pressure of projected area

(vert.) Fz = γ V*

acts through centroid of V*

bouyant

Fb = γ Vdisplaced

acts through centroid of Vdisplaced

y (^) cp

b/2 b/

h

b

(x cp,ycp)

y (^) cp

b/2 b/

h

b

(x cp,ycp)

3 12

1 CG where I bh V

I GM = oo^ − oo =^3

3

I bh

bh

tri

rect

I = h b h b

( )

( )

  1. (^943) ft

ρ water =

3

min

1 2

2

  1. 4

550 ( )

550 ( )

2

slug

ft

lb g

P P P

h h h

Qeff

horsepower h

Q

horsepower eff whereh

H h H h h

g

V z

P H

water water

absollute gauge atmospheric

losses friction or

t

p

p t losses

= =

= +

= +

=

=

  • = + +

= + +

γ ρ

γ

γ

γ

CONTINUITY EQUATION

For one-dimensional flows

∑ρV^ i

. A i =^ d/dt (∫ρdV)

where V (^) i and A (^) i are vectors and dV is the differential volume.

For steady flow

∑ρV^ i

. A I = 0

or for incompressible, constant density fluids

Q in = Q out

Q = VA

OPEN CHANNEL FLOW VANE PROBLEMS

Note for vanes the relative velocity vector is constant in magnitude and it is tangent to the vane everywhere.

Energy Equation H 1 = H 2 + hlosses where H = z (^) bed + y + V^2 /2g = z (^) bed+ E E = y + V^2 /2g

Mom. Eq. M 1 = M 2 + Fobs/ γ^ Friction is neglected for vanes. where M = y^2 barA + Q^ /gA where ybar is the distance Use Q* = V^ absA for a vane series from the water surface to the centroid of^ and Q* = V^ A for a single vane the cross section.

rel 2 2

Division of Flow

(Q’s are relative Q’s)

Q 2 = Q 1 (1+cosβ)/

F n = -ρ(V 1 sin β) Q 1

Q 3 = Q 1 (1-cosβ)/

Q 2

Q 1

Q 3

n

s

Division of Flow

(Q’s are relative Q’s)

Q 2 = Q 1 (1+cosβ)/

F n = -ρ(V 1 sin β) Q 1

Q 3 = Q 1 (1-cosβ)/

Q 2

Q 1

Q 3

n

s

z o oz i i z

y o o y i i y

x o ox i ix

F M V MV

F M V MV

F M V MV

or

P = Fxu = γ Q(V 1 -V 2 ) 2g

V (^) abs = Vrel +u (vector addition!!!)

a^2 = b^2 + c^2 - 2bc cos θ 1

a

b c

θ 2 θ 1

θ (^3)

sin θ 1 = sin θ 2 = sin θ 3 a b c

Momentum Eq.

ΣFx = Σ ρVx(V

.

A)

ΣFy = Σ ρVy(V.A)

ΣFz = Σ ρVz(V.A)

P = F^2

xu^ =^ γ Q(V 1 -V 2 ) (eff) (eff) 2g

Open Channel Flow Equations Definitions Energy Equation H = H + h

o c o c o c

w o f

w

MildSlope y y Critical Slopey y SteepSlopey y

A

nQ P S bed slope S frictionslope

B surfacewidth A area P wetted perimeter

gA

QB

Fr

10 / 3

(^24) / 3

3

1 2 losses^2

z E 2g

V

H z y bed

2 where = bed + + = +

2

2 2

2gA

Q

y 2g

V

E = y+ = +

γ

F

Momentum Eq. M 1 =M 2 +

where gA

Q

M y A

2 = +

Critical flow

1 gA

QB

Fr (^3) c

c

2 (^2) = =

2 c

2 c c

c c c 2gA

Q

y 2B

A

E =y + = +

Rect culverts

2

2

2

b

Q

q Vy gy

q = = E = y +

yc E c 3

g

q ; y ( gy

q Fr 1/

2 3 c

2 (^2) = = =

Uniform flow (^) 2/3 1/ w

5/ So P

A

n

Q =

Step Method

(S -S )

(1-Fr ) x x (y -y) o f,,ave

ave 2 = 1 + 2 1

2

where (^) 10/ ave

4/ 2 w,ave f,ave A

P

nQ S =(

gA

QB

F ave (^3) ave

ave

2 r^2 =

( ) 2

y y^1 y^2 ave

= ; Aave = byave + myave^2 ; Bave = b + 2 myave ; Pw , ave = b + 2 yave ( m^2 + 1 )

y

B

centroid

b m

y 1

b m

y 1

y dist fromwsto centroid

by my yA

P b y m

B b my

A by my

w

2 3

2

2

Flow from channel to culvert

(z -z ) 2g

V

E y (1 K ) 2 1

2 2 1 = 2 + + ent +

If section 2 is critical, like a culvert under inlet control, then

2g 2

V (^22) yc rect culverts =

y (^1) y 2 = y (^) c

y (^1) 2 )y (z -z^ ) y^2 = y^ c

K

i.e. E 1 =(1.5+ ent c+ 2 1

c

c B

A

2g 2

V 22

non-rect culverts =

(z -z ) 2B

A

E y (1 K ) 2 1 c

c i.e. 1 = 2 + + ent +

Flow from lake to channel

EL

Steep Channel yc > yo

y (^) c

S 2 curve

EL

Mild Channel y (^) c < yo

y (^) o

y (^) o y (^) c

CDL

CDL

NDL

NDL

EL

Steep Channel yc > yo

y (^) c

S 2 curve

EL

Mild Channel y (^) c < yo

y (^) o

y (^) o y (^) c

CDL

CDL

NDL

NDL

Steep channel

Eq. 1 ( ) lake c

ent c c E B

k A y =

Eq. 2 Mild channel

( ) lake wo

ent o o o (^) P E

A

n

k S y = ⎥

3

4

,

If slope is unknown

  1. Assume steep
  2. Calc y (^) cfrom Eq. 1
  3. Calc Qc from Fr = 1
  4. Using y (^) c and Q (^) ccalc S* from Manning’s Eq.
  5. If S*<S (^) actual, Steep slope, Q = Q c, Done
  6. If S*>Sactual , Mild slope
  7. Solve y (^) ofrom Eq. 2
  8. Calc Q from Manning’s Eq.

Alternate 4-

  1. Using y (^) ccalc Q* from Manning’s Eq.
  2. If Q*>Qc Steep slope, Q = Q (^) c, Done
  3. If Q*<Qc, Mild slope
  4. Solve y (^) ofrom Eq. 2
  5. Calc Q from Manning’s Eq.