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PHYS207 Assignment 4 Real Fluids
Q1. A 1:100 model submarine is 1.25 m long and is towed underwater at a velocity of 4.4 m/s. At
what velocity will the full scale submarine operate for dynamically similar forces? (Note: this is not a
design target but a laboratory examination of similitude eg similar levels of turbulence).
This is a submerged situation, so the key dimensionless number is Reynolds. Using the model data,
the Reynolds number is:
Re = VD / = 4.4(1.25)1000/0.001 = 5.5x106
(We can use any characteristic length, so long as we use the same length in model and prototype. In
this case we have used the hull length. The fluid data states only that it is water, so we can assume
rough figures of 1000 kg/m3 and 0.001 Pa.s for the density and viscosity.)
The prototype will have a similar Reynolds number for similar forces:
Re = VD / = V(125)1000/0.001 = 5.5x106
The remaining unknown is the velocity:
V = 0.001(5.5x106
)/(125(1000)) = 0.044 m/s
This agrees with the point that submerged models at high speed are dynamically similar to submerged
prototypes at low speeds. But is it realistic to assume that the full scale submarine will behave this
way? 0.044 m/s is a fairly low speed…
Q2. The towing force on the model submarine of the previous question was measured as 7.4 N. What
kinematically similar force would be applied to the prototype to attain the calculated speed?
For similitude of forces we need to use the Euler number, which in Reynolds cases leads to a force
ratio of
Fr = r (p. 121 study guide)
The fluid density is the same for the model and the prototype, so Fr = r = 1
This means that the forces on the prototype are the same as on the model for kinematically similar
speeds.
Fp = 7.4 N
No wonder the prototype velocity is so slow! This motive force is tiny compared to the actual power
needed for even a small submarine. The point is that dynamic similitude, while providing a useful
point of comparison between model and prototype, is not realistic as a description of actual
performance.

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Q3. The submarine of the previous example is also tested in a high float surface propulsion mode. For
a towing force of 6.2 N a slightly higher velocity of 4.8 m/s is obtained. What are the corresponding
prototype velocity and propulsion force?
Surface propulsion is a Froude number situation: We can work out the Froude number for the model
and use it to work out the prototype velocity:
gy
V
Fr
For the model, Fr = 4.8/√(9.8(1.25)) = 1.37
(Again we can use any characteristic length, so long as we use the same length in model and
prototype)
For the prototype, Fr = V/√(9.8(125)) = 1.37
This leaves only V: V = 1.37√(9.8(125)) = 48 m/s
This seems more realistic but in fact it only represents a design point for similar dynamic effects
compared to the model.
The propulsion force ratio is similarly determined from the Euler and Froude numbers:
Fr = rLr
3
( study guide)
= Lr
3
(same density for model and prototype)
That is Fm/Fp = (1/100)3
And Fp = Fm(1003
) = 6.2(1003
) = 6.2 MN thrust.
At 48 m/s this would require a power of about 300MW, a very unrealistic design target but not
unacceptable for establishing wave friction loads for the experimental design process.
Q4. A reservoir’s water surface is at an elevation of 80 m. The reservoir drains via a cast iron pipe
which is 250 m long and 400 mm in diameter into a pre-treatment pond. If the pond and outfall are at
an elevation of 40 m, what flowrate can the pipeline deliver?
It helps to draw a diagram:
80 m
40 m250 m
HGL
Entry losses
losses

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The hydraulic gradeline loses a small incremental amount due to entry turbulence and then a larger
gradual amount due to pipe friction.
The entry losses are he = ke V2
/2g where ke is about 0.5
The pipe losses are hf = fLV2
/2gD where f is an as yet unknown friction factor.
The two losses add up to the total head difference between the two water surfaces:
he + hf = 40 m
Let’s assume a value of f = 0.02 and check that later:
V2
(ke/2g + fL/2gD) = V2
(0.5/(2(9.8)) + 0.02(250)/(2(9.8)0.4) = V2
(0.663) = 40 m
So V = √(40/0.663) = 7.7 m/s
This gives a Reynolds number of Re = 7.7(0.4)1000/0.001 = 3.1x106
Using the moody diagram for cast iron this gives a friction factor of 0.012. Repeat the calculation with
this value:
V2
(0.5/(2(9.8)) + 0.012(250)/(2(9.8)0.4) = V2
(0.408) = 40 m
V = √(40/0.408) = 9.9 m/s
Re = 9.9(0.4)1000/0.001 = 4.0x106
The value of f does not change for this small difference in Re.
Hence V = 9.9 m/s and Q = VA = V( D2
/4) = 9.9( 0.42
/4) = 1.24 m3
/s
Q5. A pipeline needs to be designed to carry water from the pre-treatment pond of the previous
example to the treatment works 200 m away. The treatment works are at an elevation of 30 m and
have an internal working head of 15 m. A pump of design head 25 m is specified to supply 300 l/s
from the pond to the treatment works. What diameter pipe is required to carry this amount?
Let’s do a diagram again:
The total headloss between the pump and the pressurised works is (40 + 25) – (30 + 15) = 20 m. This
is shared between the pipe losses and an exit loss into the works tank:
P40 m
30 m
45 m
+ 15 m
+ 25 m
200 m

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ht = he + hf = keV2
/2g + fLV2
/2gD = 20 m
ke is an exit coefficient, ke ~ 1.0 and f is the friction factor, again guessed initially as 0.02.
We don’t know the velocity or the diameter of the pipe, but we do know the flowrate
Q = VA = V D2
/4
Q2
= V2 2
D4
/16 = 0.32
= 0.09
So V2
= 0.09(16)/( 2
D4
) = 0.146/D4
Putting in the other numerical values we can express the headloss as a function of D:
ht = he + hf = 0.0074/D4
+ 0.03/D5
= 20 m
We can ignore the first (small) term and evaluate D = 5
√(0.03/20) = 0.272 m
V is then V = √0.146/D4
= 5.1 m/s
We need to check if our assumed f is correct: Calculate a Reynolds number:
Re = 5.1(0.272)1000/0.001 = 1.374x106
Assuming cast iron pipe, this is f = 0.013
The solution amounts to correcting this factor:
D = 5
√(0.03x(0.013/0.02)/20) = 0.250 m
V = √0.146/D4
= 6.1 m/s
Re = 6.1(0.250)1000/0.001 = 1.525x106
The friction factor does not change much for this small change in Reynolds number:
D = 250 mm or next standard size up
Q6. An airfoil of 37 m2
area (1 wing only) has an angle of attack of 6o
and is travelling at 25 m/s. The
aircraft has a total profile area of 4.2 m2
in the direction of travel. If the coefficient of drag varies
linearly from 0.04 at 4o
AoA to 0.12 at 14o
AoA, what power is required to maintain this velocity at
4o
C temperature and 0.9x105
Pa absolute pressure? If the lift coefficient is inversely proportional to
the square root of the drag coefficient, what is the lift on the wing at this speed? Corrections added
via Moodle discussion.
Start by looking at the drag equation:
pD A
U
CDrag
2
2
Total Drag Force
We can work out the drag coefficient CD by interpolating the data given
We can work out the density using the absolute temperature and pressure

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The slope of the CD profile is k = (0.12 – 0.04) / (14 – 4) = 0.008
The value of CD at 6o
is thus CD(6o
) = 0.04 + 0.008(6 – 4) = 0.056
The density of air at 4o
C (277 K) and 0.9x105
Pa is
= P/RT = 0.9x105/(287(277)) = 1.132 kg/m3
The total profile area must increase as the angle of attack increases. In the absence of blueprints, let’s
assume it is approximately the base profile area + the wing area x the sine of the angle of attack:
Ap = 4.2 + 2(37)sin(6) = 11.94 m2
.
So we can work out the drag force on the wing:
pD A
U
CDrag
2
2
= 0.056(1.132)252
(11.94)/2 = 237 N
The power required to maintain that velocity is
Power = F.V = 237(25) = 5.9 kW
The lift coefficient is inversely proportional to the square root of the drag coefficient:
CL = CD
-1/2
= 0.056-1/2
= 4.2
And the lift force is
WL A
U
CLift
2
2
= 4.2(1.132)252
(37)/2 = 55 kN
AoA
14o
4o
6o
0.04
0.12