Download Solved Test 3 - The Art of Persuading Others | 010 030 and more Exams Rhetoric in PDF only on Docsity!
- 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 )?
- 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.
- 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.
- 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 (^) 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
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)
