MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
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MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------1. The jet pump in Figure. P1 injects water at U1 = 40 m/s through a 7.6cm pipe and entrains a
secondary flow of water U2 = 3 m/s in the annular region around the small pipe. The two flows
become fully mixed downstream, where U3 is approximately constant. For steady incompressible
flow, compute U in m/s.
Figure. P1
[solution]
Flow inlet and outlet area are
A1 = π(0.076)2/4 , A2 = π{(0.254)2-(0.076)2}/4 , A3 = π(0.254)2/4
From the mass conservation
Q1 + Q2 = Q3
Then, ρ1A1V1 + ρ2A2V2 = ρ3A3V3
And ρ1 = ρ2 = ρ2 , then mass conservation changed as
{π(0.076)2/4}(40) + [π{(0.254)2-(0.076)2}/4](3) = {π(0.254)2/4}(U3)
U3 = 6.31m/s
MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------2. A converging elbow (see Figure. P2) turns water through an angle of 135o in a vertical plane. The
flow cross section diameter is 400 mm at the elbow inlet, section (1), and 200 mm at the elbow outlet,
section (2). The elbow flow passage volume is 0.2 m3 between sections (1) and (2). The water volume
flowrate is 0.4 m3 and the elbow inlet and outlet pressures are 150 kPa and 90 kPa. The elbow mass is
12 kg. Calculate the horizontal (x direction) and vertical (z direction) anchoring forces required to hold
the elbow in place. DO NOT neglect gravitational effect.
Figure. P2
[solution]
Original linear momentum conservation derived from Reynolds transport theorem is
d
d
(mv) sys  (  v  dV )   v  (v n)dA
CS
dt
dt CV
Assume steady-state flow, and driving force is originated from net pressure and others. Then
 Pi Ai  Fi   vi  (vi ni ) Ai
So, x-direction linear momentum equation yields
-u1ρu1A1-V2cos45oρV2A2 = P1A1-FA,x+P2A2cos45o
From mass conservation
𝑚̇ = ρu1A1 = ρV2A2 = ρQ
Thus, FA,x = ρ(Q2/A1) + ρ(Q2/A2)cos45o + P1A1 + P2A2cos45o
= 25700N
z-direction linear momentum equation yields
-V2sin45oρV2A2 = P2A2sin45o - FA,z – Ww - We
Thus, FA,z = ρ(Q2/A2)sin45o + P2A2sin45o – Ww-We
= 3520N
MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------3. Two cylindrical water jets of equal size and speed strike each other as shown in Figure. P3.
Determine the speed, V, and direction, θ, of the resulting combined jet. Gravity is negligible.
Figure. P3
[solution]
Original linear momentum conservation derived from Reynolds transport theorem is
d
d
(mv) sys  (  v  dV )   v  (v n)dA
CS
dt
dt CV
Assume steady-state flow, pressure effect should be canceled out. (All pressure in inlet and outlet are
same with atmospheric pressure.) And no internal forces then,
0   vi  (vi ni ) Ai
x-direction linear momentum equation yields
-V2ρV2A2 + (Vcosθ) ρVA = 0
And y-direction linear momentum equation yields
-V1ρV1A1 + (Vsinθ) ρVA = 0
Also for mass conservation
ρV1A1 + ρV2A2 – ρVA = 0
From two linear momentum equation,
θ = cot-1(V22A2/V12A1) = 45o
Combine linear momentum equation and mass conservation. We get
-V12A1 + Vsin θ(V1A1 + V2A2) = 0
V = 2.12m/s
MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------4. When a uniform stream flows past an immersed thick cylinder, a broad low-velocity wake is
created down-stream, idealized as a V shape in Figure. P4. Pressures p1 and p2 are approximately
equal. If the flow is two-dimensional and incompressible, with width b into the paper, derive a
formula for the drag force F on the cylinder. Rewrite your result in the form of a dimensionless drag
coefficient based on body length CD=F/(ρU2bL). Neglect gravitational effect.
Figure. P4
[solution]
First, set the control volume of inlet (1) height 2H, and outlet (2) height 2L
And to be satisfy mass conservation, the integration equation existed in below will be held.
  udA    udA
1
2
H
L
0
0
2   u1dA  2  u2 dA
H
3
L
4
Then, force equilibrium is
 vi  (vi ni ) Ai  0,
L
U
y U
y
(1  )  (1  ) bdy  2 H  U 2 b   Fdrag
2
L 2
L
0
2
1
Fdrag  U 2 Lb
3
Fdrag
1
CD 

2
U Lb 3
MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------5. For the pipe-flow-reducing section of Figure. P5, D1 = 8 cm, D2 = 5 cm, and p2 = 1 atm. All fluids
are at 20°C. If V = 5 m/s and the manometer reading is h = 58 cm, estimate the total force resisted by
the flange bolts.
Figure. P5
[solution]
P1 – P2 = (γmerc – γwater)h = 71300Pa
Q1 = Q2, or (5m/s)(π/4)(0.08m)2 = V2(π/4)(0.08m)2
V2 = 12.8m/s
An atmospheric pressure act inlet (1) and outlet (2), so atmospheric pressure effect should be canceled
out. Thus only relative pressure (gauge pressure) should be valid.
∑Fx = -Fbolts + P1,gageA1 = 𝑚̇(V2 – V1)
Fbolts = (71300Pa) (π/4)(0.08m)2 – (998kg/m3) (π/4)(0.08m)2(5m/s)[12.8m/s - 5.0m/s] = 163N
MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------6. Five liters/s of water enter the rotor shown in Figure. P6 along the axis of rotation. The cross-sectional
area of each of the three nozzle exits normal to the relative velocity is 18mm2. How large is the resisting
torque required to hold the rotor stationary? How fast will the rotor spin steadily if the resisting torque
is reduced to zero and (a) θ = 0°, (b) θ = 30°, (c) θ = 60°?
Figure. P6
[solution]
Original Angular momentum conservation derived from Reynolds transport theorem is
d
 M o  dt (CV (r  v)  dV )  CS (r  v)  (v n)dA
Assume steady-state flow, then
Tshaft   (r  v )  (v n)dA
CS
Control surface is divided into 3 equal exit, then
Tshaft  3(r  v )  (Vout ) A
(r  v)  rout Vout cos , 3 Vout A  m . Then angular momentum equation yields
Tshaft = 𝑚̇routVoutcosθ
We note that
𝑚̇ = ρQ
And Vout = Q/3Anozzle
Combine all equation, we get
Tshaft = (ρQ2routcosθ)/(3Anozzle)
To determine the rotor angular velocity associated with zero shaft torque,
Tshaft = 𝑚̇rout(Woutcosθ - Uout), where Uout = routω, Wout = Q/3Anozzle
Finally we get, Tshaft = ρQrout{(Qcosθ/3Anozzle) – routω}
(a) θ = 0o, Tshaft = 231Nm and ω = 185rad/s
(b) θ = 30o, Tshaft = 200Nm and ω = 160rad/s
(c) θ = 60o, Tshaft = 116Nm and ω = 92.5rad/s
MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------7. The horizontal pump in Figure. P7 discharges 20°C water at 57 m3/h. Neglecting losses, what
power in kW is delivered to the water by the pump?
Figure. P7
[solution]
V1 = (57m3/h)/(A1) = 2.49m/s, V2 = (57m3/h)/(A2) = 22.4m/s
P1/ρg + V12/2g + hp = P2/ρg + V22/2g
120kPa/{(998kg/m3)(9.8m/s2)} + (2.49m/s)2/(2*9.8m/s2) + hp
= 400kPa/{(998kg/m3)(9.8m/s2)} + (22.4m/s)2/(2*9.8m/s2)
hp = 53.91m
P = ρgQhp = 8349W
MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------8. A venturi meter, shown in Figure. P8, is a carefully designed constriction whose pressure difference
is a measure of the flow rate in a pipe. Using Bernoulli’s equation for steady incompressible flow with
no losses, show that the flow rate Q is related to the manometer reading h by
𝑄=
𝐴2
2𝑔ℎ(𝜌𝑀 − 𝜌)
√
𝜌
√1 − (𝐷2 /𝐷1 )4
where ρM is the density of the manometer fluid.
Figure. P8
[solution]
Q = A1v1=π(D1/2)2v1 = A2v2=π(D1/2)2v2
P1+ρgh1+ρv12/2 = P2+ρgh2+ρv22/2
h1 = h 2
P1 - P2 = (ρ/2)(v22 - v12)
P1 – P2 = (ρ/2)(4Q/ π)2(1/D24 – 1/D14)
P1 – P2 = (ρM - ρ)gh
(ρM - ρ)gh = (ρ/2)(4Q/ π)2(1/D24)(1 – D24/D14)
(ρM - ρ)gh = (ρ/2)Q2(1/A22)(1 – D24/D14)
Q2 = {2(ρM - ρ)gh/ρ}A22{1/(1 – D24/D14)}
𝑄=
𝐴2
2𝑔ℎ(𝜌𝑀 − 𝜌)
√
𝜌
√1 − (𝐷2 /𝐷1 )4
MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------9. Water flows from the pipe shown in Figure. P9 as a free jet and strikes a circular flat plate. The
flow geometry shown is axisymmetrical. Determine the flowrate through the pipe and the manometer
reading, H. Hint: Use Bernoulli’s equation and DO NOT neglect gravitational effect. The fluid
velocity at the center (hole) of the circular flat plate is zero. Pressure at the pipe exit and at the
perimeter of the flat plate is atmospheric pressure.
Figure. P9
MAE221 Fluid Mechanics
Homework #3 (9 problems)
Due: 10:00pm on October 16, 2015.
-----------------------------------------------------------------------------------------------[solution]
Let’s define inlet mass flow rate point as 1, asymmetrical outlet point as 2 and manometer reading
point as 3 then Bernoulli equation is
P1/ρ + V12/2g + Z1 = P2/ρ + V22/2g + Z2, where P1 = P2 = 0, Z1 = 0 and Z2 = 0.2m
Thus, V12/2g = V22/2g + Z2
Inlet and outlet mass flow rate are same due to the mass conservation, then
A1V1 = A2V2 = Q, V1 = 1.6V2
V12/2g = (1.6V2)2/2g = V22/2g + Z2, V2 = 1.59m/s
Finally set point 2 as an origin point, then Bernoulli equation with point 2 and 3 be simplified like the
equation below.
H = V22/2g = 0.129m