You are designing a chute for grain. It is known that the average velocity, V, of the grain
traveling down the chute is a function of the chute geometry (diameter, D; height, H;
length, L), the material properties of the grain (dynamic viscosity, µ; density, ρ), as well
as gravity, g. You have constructed a model and will use dimensional analysis to relate
the properties of the model and prototype chute.
1. How many dimensionless parameters describe this system (5 points)?
a. 2
b. 3
c. 4
d. 5
e. More than 5
V=f1(D, H, L, µ, ρ, g) n=7
[V]=[L]/[T]
[ρ]=[M]/[L]3
[g]=[L]/[T]2
[µ]=[M]/[L][T]
[D]=[H]=[L]
So m=3
n-m=7-3=4
For the remainder of this problem, assume that D, g, and ρ are used as repeating
parameters.
2. Which of the following are not among the dimensionless parameters that describe this
system (8 points)?
a. V gD
b. H D
c. D (12 ρV 2 )
(
)
d. µ ρD gD
e. All of these are dimensionless parameters
П1=V/sqrt(gD)
П2=µ/[ρD*sqrt(gD)]
П3=H/L
П4=L/D
So D/[1/2ρV2] is not among the dimensionless parameters.
3. How many of the dimensionless parameters describe geometric similarity, but not
necessarily kinematic or dynamic similarity (5 points)?
a. 1
b. 2
c. 3
d. 4
e. 5
П3=H/L
П4=L/D
So, the answer is 2.
4. If the model length, Lm, is 1/5 of the prototype length Lp, what is the velocity ratio
Vm/Vp (8 points)?
a. 1/5
b. 5
c.
5
d. 1 5
e. 25
f. None of these
Lm/Lp = Dm/Dp = 1/5
Vm/sqrt(gDm) = Vp/sqrt(gDp)
Vm/Vp = sqrt(Dm/Dp)=sqrt(1/5)
5. Assume the model length, Lm, is 1/5 the prototype length, Lp. If the model length Lm
= 1 length units, and the model velocity Vm = 5 velocity unit, how long does it take
for the grain to travel from one end of the prototype chute to the other (8 points)?
a. 1/5
b. 5
c.
5
d. 1 5
e. 25
f. None of these
Tm = Lm/Vm
Lp=5Lm
Vp=sqrt(5)Vm
Tp = [5Lm] / [sqrt(5)Vp] = [(5)(1)] / [(5)sqrt(5)] = 1/sqrt(5)
6. You decide to redesign the model and use sand instead of grain. The kinematic
viscosity of sand is 1/8 that of the grain. How does this change affect the relative
diameter of the model you have to use (8 points)?
a. 4
b. 1/4
c. 8
d. 1/8
e. 1/2
f. None of these
ν=µ/ρ
µm/[ρmDm*sqrt(gDm)]= µp/[ρpDp*sqrt(gDp)]
νm / [Dm*sqrt(gDm)] = νp / [Dp*sqrt(gDp)]
νm / νp = Dm3/2 / Dp3/2
Dm / Dp = (νm / νp)2/3 = (1/8)2/3 = 1/4
7. If the ratio of inertial to gravitational forces is held constant and the diameter is
doubled, by what factor does the velocity change (8 points)?
a. 2
b.
2
c. 1/4
d. 1/ 2
e. 4
f. None of these
П1=V/sqrt(gD) = Fe = (inertial force / gravitational force)
V1/sqrt(gD1) = V2/sqrt(gD2)
V2 = V1*sqrt(D2/D1)
D2 = 2*D1
V2 = sqrt(2)*V1
A homeowner plans to pump water from a stream in their backyard to water their lawn. A schematic of the
pipe system is shown in the figure.
sprinkler
g = 9.81 m/s2
inlet pipe-to-pump
3 m coupling
1m
stream
hose-to-hose coupling
two 15.25 m lengths of 1.3
cm diameter garden hose
pump
pump-to-hose coupling
pipe inlet
10 m of 2.0 cm
diameter pipe
Details of the system are given in the following table. The design flow rate for the system is 2.5*10-4 m3/s.
Item
water density
water dynamic viscosity
inlet pipe length
inlet pipe diameter
inlet pipe roughness
hose length
hose diameter
hose roughness
pipe inlet loss coefficient
inlet pipe-to-pump coupling loss coefficient
pump-to-hose coupling loss coefficient
hose-to-hose coupling loss coefficient
pressure drop across the sprinkler
sprinkler nozzle exit diameter
hose bend losses
Value
1000 kg/m3
1*10-3 kg/(m.s)
10 m
2.0 cm
1.0 mm
two 15.25 m lengths
1.3 cm
smooth
3.0
2.0
0.2
0.1
210 kPa at 2.5*10-4 m3/s flow rate
4 mm
can be neglected
For the following questions, circle the one answer that is most correct.
(05 pts)
1.
What is the loss coefficient for the sprinkler at design conditions? Base the sprinkler loss
coefficient on the velocity just upstream of the sprinkler.
a. 1
e. 1900
b. 21
f. 118
c. 0.12
g. 0.02
d. 1600
h. 210
(05 pts)
2.
What is the friction factor for the hose at design conditions?
a. 0.0026
e. 0
b. 0.0149
f. cannot be determined from the given information
c. 0.0244
g. 0.029
d. 0.058
h. 0.011
Page 1 of 6
(05 pts)
3.
What is the velocity head, including the kinetic energy correction factor, at the sprinkler
exit?
a. 1.9 m
e. 198 m
b. 40.4 m
f. 396 m
c. 0.36 m
g. 0.01 m
d. 15.1 m
h. 20.2 m
For the next two questions, assume that the loss coefficient for the sprinkler is 1, the average velocity
in the inlet pipe is 0.80 m/s, the average velocity in the hose is 1.88 m/s, and the friction factor for the
hose is 0.01.
(10 pts)
4.
What is the total minor head loss in the system?
a. 5.4 m
e. 3.9 m
b. 0.4 m
f. 53.2 m
c. 0.2 m
g. 15.1 m
d. 5.8 m
h. 8.3 m
(10 pts)
5.
What is the total major head loss in the system?
a. 5.4 m
e. 1.2 m
b. 0.4 m
f. 53.2 m
c. 0.2 m
g. 15.1 m
d. 4.2 m
h. 8.3 m
For the next question, assume that the velocity head at the sprinkler exit, including the kinetic energy
correction factor, is 10 m and the total head loss is 15 m.
(10 pts)
6.
What is the shaft head required to operate the system?
a. 29 m
e. 13 m
b. 22 m
f. 18 m
c. 8 m
g. 275 m
d. 28 m
h. 74 m
For the next question, assume that the shaft head required to operate the system at the design flow
rate is 100 m.
(05 pts)
7.
What power must be supplied to the pump if the pump is 65% efficient?
a. 63 W
e. 27 W
b. 154 W
f. 245 W
c. 377 W
g. 159 W
d. 65 W
h. 38 W
Page 2 of 6
SOLUTION:
2
sprinkler
g = 9.81 m/s2
inlet pipe-to-pump
3 m coupling
1
1 m stream
pump
re-entrant
pipe inlet
hose-to-hose coupling
two 15.25 m lengths of 1.3
cm diameter garden hose
pump-to-hose coupling
2 m of 2.5 cm diameter
drawn tubing
1.
2.
The loss coefficient for the sprinkler may be found using the definition of a loss coefficient.
∆psprinkler
K sprinkler = 1
2
2 ρVhose
where
Q
Vhose = π 2
D
hose
4
Using the given parameters,
Q = 2.5*10-4 m3/s
Dhose = 0.013 m
∆psprinkler = 210,000 Pa
ρ = 1000 kg/m3
Vhose = 1.88 m s and Ksprinkler = 118
(1)
(2)
The friction factor for the hose may be found using the Moody plot. The Reynolds number for the flow
in the hose is:
ρV D
Re D = hose hose
(3)
µ
Using the given parameters,
Dhose = 0.013 m
ρ = 1000 kg/m3
µ = 1.0*10-3 kg/(m.s)
Vhose = 1.88 m s
the hose Reynolds number is ReD = 24500. From the Moody chart and using the smooth curve (we’re
told to consider the hose to be “smooth”), the friction factor for the hose is fhose = 0.0244.
Page 3 of 6
3.
The velocity head, including the kinetic energy correction factor, at the sprinkler’s exit is:
V2
α
2 g sprinkler
(4)
exit
where
Vsprinkler =
exit
π
4
Q
2
Dsprinkler
(5)
exit
Using the given parameters,
Q = 2.5*10-4 m3/s
Dsprinkler exit = 0.004 m
⇒ Vsprinkler = 19.9 m s
exit
and
α=1
(6)
since
ρVsprinkler Dsprinkler
Re D =
exit
exit
µ
⇒ ReD = 79600 > 2300 ⇒ Turbulent flow at the exit!
(7)
Thus,
α
4.
V2
2g
= 20.2 m
(8)
sprinkler
exit
The total minor head loss for the system includes minor losses at the pipe inlet, the pump couplings, the
hose coupling, and the sprinkler.
2

 Vpipe 
V 2
H L ,minor =  Kinlet + K inlet pipe-pump 
+  K pump-hose + K hose-hose + K sprinkler  hose
(9)
coupling
coupling

 2 g  coupling
 2g
Using the given parameters,
Kinlet = 3.0
Kinlet pipe-pump coupling = 2.0
Kpump-hose coupling = 0.2
Khose-hose coupling = 0.1
Ksprinkler = 1
Vinlet pipe = 0.80 m s
Vhose = 1.88 m s
g = 9.81 m/s2
HL,minor = 0.4 m
Page 4 of 6
5.
The total major head loss for the system includes the major losses in the inlet pipe and the hoses.
2
2
 L V
 L V
H L ,major = f  
+f 
 D  2 g pipe
 D  2 g hoses
(10)
Using the given parameters,
Lpipe = 10 m
Dpipe = 0.02 m
Vpipe = 0.80 m/s
Lhoses = 30.5 m
Dhoses = 0.013 m
Vhoses = 1.88 m/s
fhoses = 0.01
fpipe = 0.073 (from the Moody chart)
ρVpipe Dpipe
ε
1*10−3 m
where Re D =
=
= 0.05
= 15900 and
D pipe 2*10−2 m
µ
(11)
HL,major = 5.4 m
6.
The required shaft head may be found by applying the Extended Bernoulli Equation from the free
surface of the stream to the outlet of the sprinkler.
 p
  p

V2
V2
+
α
+
z
=
+
α
+ z  − HL + HS
(12)

 
2g
2g
 ρg
2  ρ g
1
V2 
 p − p1   V22
HS =  2
− α 1  + ( z2 − z1 ) + H L
 + α
2g 
 ρ g   2g
where
p1 = p2 = patm
V1 ≈ 0
V2 = Vsprinkler
(13)
(14)
(15)
(16)
exit
HL = HL,minor + HL,major
Using the given parameters,
HL = 15 m
α
V2
2g
(17)
= 10 m
sprinkler
exit
z2 – z1 = 3 m
HS = 28 m
7.
The power that must be supplied to the pump with the given efficiency is:
Winto
S ρ QgH S
mgH
water
W
=
=
=
into
pump
η
η
η
Using the given parameters,
HS = 100 m
η = 0.65
ρ = 1000 kg/m3
Q = 2.5*10-4 m3/s
g = 9.81 m/s2
Page 5 of 6
(18)
Winto = 377 W
pump
Page 6 of 6