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Solved Test 3 - The Art of Persuading Others | 010 030, Exams of Rhetoric

Material Type: Exam; Class: 010 - The Art of Persuading Others; Subject: Rhetoric; University: University of Iowa; Term: Fall 1993;

Typology: Exams

Pre 2010

Uploaded on 09/17/2009

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057:20 TEST # 3 Fall 2004
1. A laminar flow occurs between two horizontal parallel plates under a pressure
gradient dp ds ( pdecreases in the positive
s
direction). The upper plate moves
left (negative) at velocity t
u. The expression for local velocity u is given as
()
2
1
2t
dp y
uHyyu
ds H
µ
=− +
Is the magnitude of the shear stress greater at the moving plate
(
)
yH= or at the
stationary plate
()
0y=?
2. Water flows through this nozzle at a rate of 15cfs and discharges into the
atmosphere. 112 .Din= and 28.Din
=
Determine the force required at the flange
to hold the nozzle in place. Negelect gravitaional forces. Assume irrotational flow
and density of water 3
1.94
s
lug ft
ρ
=.
pf3
pf4
pf5
pf8
pf9

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  1. A laminar flow occurs between two horizontal parallel plates under a pressure

gradient dp ds^ ( p^ decreases in the positive s direction). The upper plate moves

left (negative) at velocity ut^. The expression for local velocity u^ is given as

t

dp y u Hy y u

μ ds H

Is the magnitude of the shear stress greater at the moving plate ( y^ =^ H ) or at the

stationary plate ( y^ =^0 )?

  1. Water flows through this nozzle at a rate of 15 cfs^ and discharges into the

atmosphere. D 1^ =^12 in^ .and D 2^ =^8 in .Determine the force required at the flange

to hold the nozzle in place. Negelect gravitaional forces. Assume irrotational flow

and density of water

3

ρ = 1.94 slug ft.

  1. A flat plate is oriented parallel to a 15 m s^ air flow at 20

o C (^) and atmospheric

pressure. The plate is 1.0 m long in the flow direction and 0.5 m wide. On one side

of the plate, the boundary layer is tripped at leading edge (turbulent flow on that

side). Find the total drag force on the tripped side of the plate.

  1. Water flows in the pipe shown, and the manometer deflects 80 cm^. What is f

for the pipe if V^ =^3 m s?

Prob. 1

Information and assumptions

Find

Is the magnitude of the shear stress greater at the moving plate or at the stationary plate?

Solution


Develop an equation for the shear stress du

dy

τ = μ (Eqn. = 7)


( )

dp u t H y ds H

⎝ ⎠⎝^ ⎠

Evaluate τ at y = H :

t H

dp u

H

ds H

Evaluate^ τ^ at y = 0 :

0

dp H u t

ds H

(Inter. = 2)


Observation of the velocity gradient lets one conclude that the pressure gradient dp ds is

negative. Also ut^ is negative. Therefore:

τ (^) H > τ 0 (Ans. = 1)


Prob. 2

Information and assumptions Provided in problem statement

Find

Force at flange to hold nozzle in place

Solution


Velocity calculation

( )

(^1 ) 1

Q

v ft s

A π

×

( ) ( )

2 2 2

Q

v ft s

A π

×

(Eqn. + Ans. = 2)


Bernoulli eqaution

2 2 1 1 2 2

p + ρ v = p + ρ v (Eqn.= 2)

2 2 1 2 1

p = + ρ v − v

2

= 1, 437 lbf ft (Inter.+ Ans.= 1)


x-momentum equation

Fx^ =^ m v ^2 2^ − m v ^1 1 (Eqn.= 4)

p A 1 1 (^) + F = (^) ( 1.94 × (^15) ) × 42.97 − (^) ( 1.94 × (^15) ) ×19.

( ) ( )

2 1, 437 1 1.94 15 42.97 19. 4

F

× ⎜ ⎟× + = × × −

F = − 434 lbf (^) ( act to left ) (Inter. + Ans. = 1)


Prob. 4

Information and assumptions Provided in problem statement

Find

Resistance coefficient

Solution


Manometer equation

p (^) B + γ (^) f ∆ − l γ (^) mh − γ f ⎡⎣ (^) ( ∆ − ∆ l h (^) )+ ∆ z ⎤⎦= pA

p (^) B − γ (^) mh − γ f ⎡⎣−∆ h + (^) ( z (^) AzB (^) )⎤⎦= pA

p (^) B − (^) ( γ (^) m − γ (^) f )∆ h − γ (^) f z (^) A + γ f zB = pA

( pB^ +^ γ f zB^ ) −^ ( p^ A +^ γ f z^ A ) =^ ( γ^ m −^ γ f )∆ h

p (^) zBpzA = (^) ( γ (^) m − γ f )∆ h

( ) (^1) ( 2.5 (^1) ) zB zA^ m^ f^ m

f f f

p p h h h h

− −^ ∆^ ⎛^ ⎞

h = h (^) f = 0.80 × (^) ( 2.5 − (^1) ) = 1.2 m (Eqn + Inter.+ Ans. = 3) 2

f

L V

h f D g

= (^) (Eqn = 6)

3

f

⎛ ⎞ ×

= × ⎜ ⎟× =

(Inter.+ Ans. = 1)

∆ l

∆ h

z A − zB = ∆ z

B

A

Prob. 5

Information and assumptions

Provided in problem statement From Table 10.3: K^ e =^ 0.03; Kb^ =^ 0.35; K^ E =1.

From Table A.5,

6 2 ν 10 m s − =

From Table 10.2, ks^ =0.046 mm

Find

The pump power

Solution


Energy equation from the water surface in the lower reservior to the water surface in the

upper reservior 2 2

p 1 γ+ V 1 2 g + z 1 + hp = p 2 γ + V 2 2 g + z 2 + ∑ hL

2 2 0 0 200 0 0 235 2

p e b E

V L m h m K K K f g D

⎛ ⎞

      • = + + + (^) ⎜ + + + ⎟ ⎝ ⎠ (Eqn = 7)

2

4

Q V m s A^ π

= = = ⎛ ⎞ ⎜ ⎟× ⎝ ⎠

,

2

2

V m g

=

6 6

4.44 0. Re 1.33 10 10

VD

× = = = × (^) , 0.

ks

So from Figure 10.8 f^ =0.

140 0.014 6.

L f D

= × = (^) (Inter.+ Ans. = 2)


h p = 235 − 200 + 1.01 × ( 0.03 + 0.35 + 1 + 6.53) =43.0 m

p = Q h γ p = 0.314 × 9, 790 × 43.0 = 132 kW (Inter.+ Ans. = 1)