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Material Type: Exam; Class: 010 - The Art of Persuading Others; Subject: Rhetoric; University: University of Iowa; Term: Fall 1993;
Typology: Exams
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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
Is the magnitude of the shear stress greater at the moving plate ( y^ =^ H ) or at the
stationary plate ( y^ =^0 )?
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
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.
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
( )
dp u t H y ds H
Evaluate τ at y = H :
t H
Evaluate^ τ^ at y = 0 :
0
(Inter. = 2)
Observation of the velocity gradient lets one conclude that the pressure gradient dp ds is
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
v ft s
( ) ( )
2 2 2
(Eqn. + Ans. = 2)
Bernoulli eqaution
2 2 1 1 2 2
2 2 1 2 1
2
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 = − 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 (^) m ∆ h − γ f ⎡⎣ (^) ( ∆ − ∆ l h (^) )+ ∆ z ⎤⎦= pA
p (^) B − γ (^) m ∆ h − γ f ⎡⎣−∆ h + (^) ( z (^) A − zB (^) )⎤⎦= 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 (^) zB − pzA = (^) ( 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
h f D g
= (^) (Eqn = 6)
3
f
(Inter.+ Ans. = 1)
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
2 2 0 0 200 0 0 235 2
p e b E
V L m h m K K K f g D
⎛ ⎞
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)