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QUESTION 3 2006
(a) The head loss in a pipe can be expressed in the form hf = KQ^2. Two pipes having constants K 1 and K 2 are to be considered as a single equivalent pipe. Determine the value K 3 of this single pipe when the two are laid:
i. in series ii. in parallel.
SOLUTION PART A
i. In series the flow is the same and total head loss is the sum of the two.
hf1 = k 1 Q^2 hf2 = k 2 Q^2 hf1 + hf2 = k 3 Q^2 = k 1 Q^2 + k 2 Q^2 Hence k 3 = k 1 + k 2
ii. In parallel the friction heads are the same and the flows different. hf = k 1 Q 12 Q 1 = (hf /k 1 )1/ hf = k 2 Q 22 Q 2 = (hf /k 2 )1/ hf = k 3 (Q 1 + Q 2 ) 2
(^1212)
3
(^1212)
3
1 2
f 2
f 1
f 3
2
2
f 1
f f 3
kk
k
k
k
kk
k
k
1 k
kk
2h k
h k
h k k
h k
h h k
(b) When the flow rates are expressed in litres per second and the head losses in metres, K values for the pipe systems shown are as given in the table. Under a particular set of inputs and demands the network experienced the flow rates indicated.
The head loss in the system was considered to be excessive and a second pipe was alongside pipe 3 so that they carried flow in parallel. The equivalent single pipe for these two pipes has k = 0. ms^2 /litre^2. When the pipe had been installed the pipe flows shown changed but the inputs and demands on the system remained the same.
Use the flows shown as initially assumed flows and apply an iterative method of network analysis to determine the changed flows in the pipes. Make only two rounds of corrections to the initial flows.
Pipe 1 2 4 5 K ms^2 /l^2 0.000570 0.012118 0.001698 0. (Pipe 3 has K = 0.000818 in question)
SOLUTION PART B
The problem must be solved as two loops with a common pipe 3. Start with loop 1 with the flows shown. Data is shown for initial guess. Note clockwise flow is positive.
Starting data First iteration loop 1 (pipes 1, 2 and 3) PIPE K Q hf hf/Q 1 0.000570 204 23.7212 0. 2 0.012118 104 131.068 1. 3 0.000818 -123 -12.376 0. 142.4 1.
- 2 2 x1.
2 h/Q
h δQ f
= f^ = =
∑ Correct all flows in loop 1 by subtracting 48.
First correction shown above First Iteration loop 2 (pipes 3,5 and 4) PIPE K Q hf hf/Q 3 0.000570 171.2 23.97 0. 5 0.006946 -123 -105.08 -0. 4 0.001698 -173 -50.82 -0. -131.93 1.
- 208 2 x1.
2 h/Q
h δQ f
= f^ = =−
Correct all flows in loop 2 by subtracting -51.
Second correction
D204 Q5 2004
(a) Compare and contrast the following two iterative calculation methods for complex networks of pipes. (i) the head balance method (also known as the Hardy Cross or loop method). (ii) the flow balance method (also known as the quantity balance or nodal method.
Explain briefly how and in what situation each of the methods may be used and state which of the correction methods shown at the end of this question is used in which method.
SOLUTION PART (a) The nodal balance method is used for solving problems involving many pipes with a common junction where the total flow into the junction must be zero. The correction factor used for iteration is
Q/h f
2 Q
H
The flow balance method is used for problems with multiple loops where the total head loss around a given loop is zero. The correction factor to be used is
∆ = −^ ∑
2 h/Q
h Q f
f
(b) Water is supplied from a large reservoir at A to a pipe network BCDE as shown, in the diagram. The frictional resistances of the various pipes are given by the K value in the table which may be used with the formula hf = KQ^2 to relate the magnitude of head loss hf in the pipeline to the volumetric flow rate Q. Water is drawn at constant flow rates from the network at nodes C and D. The static heads (elevation + pressure head) at nodes B, C and D are 100m, 65m and 61m respectively above the local datum. Calculate the discharges at C and D and the water level in reservoir A. (The data has been added to diagram to aid the solution) Use no more than 3 iterations and 3 significant figures
TABLE Pipeline AB BC CD DE CE BE K s^2 m^5 4 40 110 25 25
SOLUTION PART (b) The problem must be solved as two loops with a common pipe EC. First calculate the flows in known pipes.
BC hf = 35 m Q = (hf /K)1/2^ = (35/40) 1/2^ = 0.935 m^3 /s CD hf = 4 m Q = (hf /K)1/2^ = (4/110) 1/2^ = 0.191 m^3 /s
The solution evolves around doing a flow balance at node E.
1st ITERATION Guess hE = 80 PIPE K hf Q = (hf/K)1/2^ Q/hf BE 35 20 0.756 0.0378 (into junction) EC 25 -15 -0.775 0.0516 (out of junction) ED 25 -19 -0.872 0.0349 (out of junction) Totals -0.89 0.
- 16
2 x(-0.89) Q/h
2 Q
h f
f = =−
2nd ITERATION Guess hE = 80 -13.16 = 66. PIPE K hf Q = (hf/K)1/2^ Q/hf BE 35 20 0.0294 0. EC 25 -15 -0.148 0. ED 25 -19 -0.083 0. Totals 0.219 0.
2 x(0.219) Q/h
2 Q
h f
f = =
3rd ITERATION Guess hE = 66.84 + 1.69 = = 68. PIPE K hf Q = (hf/K)1/2^ Q/hf BE 35 20 0.948 0. EC 25 -15 -0.376 0. ED 25 -19 -0.549 0. Totals 0.0241 0.
Further iterations will show only minor corrections giving flows of 0.945, -0.388 and -0.557. If these figures are used you get the answers given by the examiner.
We can now calculate Qc and Qd 0.935 + 0.376 – 0.191+ Qc = 0 Qc = -1.12 m^3 /s
0.549 + 0.191 + Qd = 0 Qd = -0.74 m^3 /s
Total flow from the reservoir is 1.12 + 0.74 = 1.86 m^3 /s Head loss pipe AB is hf = kQ^2 = 4 x 1.86^2 = 13.8 m Head at entrance to pipe is 113.8 m
(c) The diagram shows two loops of a horizontal network with inflows and outflows in m^3 /s. The K values of the seven pipes are given in the table. The pressure head at node A is 25 m. Calculate the flow rate through each pipe and the pressure head at each node. No more than two rounds of iteration are required, and final values of pressure heads may be rounded to the nearest metre.
Pipe AB BC CD DE BE EF AF K (s^2 /m^5 ) 2 2 20 20 10 10 10
SOLUTION part (c)
The problem must be solved as two loops with a common pipe BE. First make a guess at the flow rates. Bear in mind that the net flow is zero at all nodes.
Data shown for initial guess
Start with loop ABEFA
2 x12.
2 h /Q
h Q f
∆ =− f^ =− =−
PIPE K Q hf hf/Q ∑
AB 2 1.9 7.22 3.
BE 10 0.2 0.4 2
EF 10 -0.6 -3.6 6
FA 10 -0.1 -0.1 1
Totals 3.92 12.
Correct all flows in this loop by adding -0.
First loop correction
Now do loop BCDEB
PIPE K Q hf hf/Q BC 2 0.7 0.98 1. CD 20 0.2 0.8 4 DE 20 -0.3 -1.8 6 BE 10 0.04688 0.021973 0. Totals 0.001973 11.
2 x11.
2 h/Q
h Q f
∆ =− f^ =− =−
Correct loop 2 The initial guess was so good that the correction is minor
Second iteration of loop 1 is:
2 x14.
2 h/Q
h Q f
∆ =− f^ =− =−
PIPE K Q hf hf/Q ∑
AB 2 1.74688 6.10314 3.
BE 10 -0.0468 -0.02189 0.
EF 10 -0.7531 -5.67197 7.
FA 10 -0.2531 -0.64072 2.
Final solution is
Pipe AB BC CD DE BE EF AF Q 1.75 0.7 0.2 -0.3 ±0.05 -0.75 -0.
Head at A is 25 m and rounding off the hf values Head B is 25 - 6 = 19 m Head at C is 19 – 1 = 18 m Head at D = 18 - 1 = 17 m Head at E = 17 + 2 = 19 m or 19 - 0 = 19 m Head at F = 19 + 6 = 25 m or 25 -1 = 24 m error due to rounding off values.
FLUID MECHANICS D209 Q10 1996
(a) (i) Distinguish between impulse and reaction water turbines.
(ii) Describe three different types of reaction turbine and specify appropriate conditions under which each type of machine would be used.
(b) A turbine is required to work under a total head of water of 28 m and to operate at 7. rev/s. A one-quarter scale model of the proposed turbine is to be tested under a total head of water of 10.8 m.
(i) Determine the speed at which the model should be operated in order to predict the performance of the full scale turbine.
(ii) At the speed described in (i), the model develops 100 kW of power at a discharge of 1.085 m^3 /s. Calculate the corresponding power developed by the full-scale turbine.
(iii) Calculate the specific speed, stating the units used, of the full-scale turbine and specify the type of machine.
IMPULSE – All the pressure is converted into Kinetic Energy in the nozzles and the KE is converted into mechanical power by the rotor.
REACTION – All the pressure is used in the rotor to accelerate the fluid over the vanes and the reaction force to this produces a torque and mechanical power.
In practice turbines like the Francis Wheel are partly impulse and partly reaction.
(b) The diagram illustrates the different types of turbines. On the left we have high pressure and on the right low pressure. The optimal efficiency for each occurs at a particular specific speed as indicated.
The Francis Wheel used Ns ≈ 75 and is used with fairly high heads and flow rates.
The mixed design uses lower heads and larger flow rates with Ns ≈ 200
The Axial flow (Kaplan) turbine uses the lowest head and is almost a free wheel propeller driven by the stream. Ns ≈ 600
The units for Head are m, speed is rev/min and flow is m^3 /s
(b) H1 = 28 m N 1 = 7.14 rev/s
Model ¼ scale H 2 = 10.8 m
The dimensionless equation for turbines is (^) ⎟ ⎠
3 53 N 2 D 2
g∆H ND
Q
φ ρN D
P
The head coefficient must be the same for both.
2 2 2 1
2 2 ND
g∆H N D
g ∆H ⎟ ⎠
2 2 2 1
2 2 ND
∆H
N D
∆ H
2 1
(^22) 1
2 N D / 4
7. 14 D
2 N 2
10.8x 16
- 14
hence N = 17.73 rev/s for the model.
The flow coefficients must also be the same
2
3 1
3 ND
Q
ND
Q
27.82m /
- 73
1.085x7.184x 4 ND
ND
Q Q^3
3 3 2 2
3 1 1 1 = 2 = =
The Power Coefficients must be the same.
2
3 5 1
(^3 5) ρND
P
ρND
P
6687.5kW
7.14 x 4 100 x ρND
ρND P P 3
3 5 5 2
3 2
5 1
3 1 1 = 2 = =
7.14x 60 x27. H
NQ
Ns (^) 3/
1/ 3/
1/ (^1 1) = = ⎟
This would indicate the a mixed flow turbine (The official examiners answer is a Kaplan)
FLUID MECHANICS D209 Q11 1996
It is required to pump water at a rate of 0.0160 m 3 /s against a total head of 30.5 m. Four geometrically similar pumps, whose sizes are 100 mm, 125 mm, 225 mm and 300 mm, are available.
The characteristics of the 100 mm size pump, tested at 150 rad/s, are tabulated below.
Discharge 0 0.0076 0.0151 0.0226 0.0302 m 3 /s Head 43.9 46.1 43.9 34.2 14.6 m Efficiency 0 48 66 66 45 %
(a) Determine which pump should be used, and the speed at which it should be driven, so that maximum possible efficiency is obtained.
(b) If, temporarily, only the 125 mm pump is available, determine the speed of operation and the input power from the motor, necessary to satisfy the head and discharge requirements.
By plotting the data for the 100 mm pump we can determine that the optimal point (for max efficiency) is when Q = 0.0188 m^3 /s and H = 40 m. The peak efficiency is 68%
For the 100 mm pump H = 40 m Q = 0.0188 m^3 /s N = 150 rad/s
1.293 rad/s 40
150 x0. H
NQ
Ns (^) 3/
1/ 3/
1/ (^1 1) = = ⎟
= (12.34 rev/min)
For the required condition
3/
1/
- 5
Nx0.
- 293 = Hence N = 131 rad/s (1251 rev/min)
For the optimal size, remember that condition (1) is the optimal condition of the pump and condition (2) is the actual operating conditions.
Equating Flow Coefficients we get 1 0.0188x 131
0.016x 150 QN
Q N
D
D
1 / (^31) / 3
1 2
2 1 1
Equating head coefficients we get we get 1 40
H
H
N
N
D
D
1
2 2
1 1
The 100 mm seems to be the best.
(b) 125 mm pump at the same speed The larger pump must slower to obtain the same flow. First calculate the corresponding flow and head for the 100 mm pump.
For the same Flow coefficient 1.953Q 1.953xQ 100
Q
D
D
Q 0. 016 Q 1 1
3 1
3
1
2 2 1 ⎟ = = ⎠
For the same Head coefficient (^1)
2 1
2
1
2 2 1 100 1.^562 H
H
D
D
H 40 H ⎟ =
Q 1 0 0.0076 0.0151 0.0226 0.0302 m 3 /s H 1 43.9 46.1 43.9 34.2 14.6 m Efficiency 0 48 66 66 45 % Q 2 0 0.0148 0.0295 0.0441 0. H 2 68.6 72 68.6 53.4 22.
Plotting H 2 and Q 2 gives the curve shown. It is assumed that the efficiency is unchanged.
As can be seen we cannot obtain the required operating point at 150 rad/s. For the same flow coefficient between at two different speeds
3 A A
A 3 B B
B N D
Q
N D
Q
A
B B AN
N
Q =Q
For the same Head Coefficient at two different speeds
2 A
2 A
A 2 A
2 A
A N D
gH N D
g H = (^2) A
2 B 2 A A
2 B B A N
N
H
N
N
H =H =
Substitute A
B A
B Q
Q
N
N
= to eliminate the speed
2
A
B B A Q
Q
H H ⎟⎟
2
B
A A (^) Q
Q
H ⎟⎟
= H (^) B Where A and B correspond to different speeds.
For the case in hand let B be the values at the new speed and A the values at 150 rad/s
2 A
2 A A (^) 0.016 119141Q
Q
H 30. 5 ⎟ =
Calculate the flows at the new speed for the 125 mm pump.
Efficiency 0 48 66 66 45 % Q (^) A 0 0.0148 0.0295 0.0441 0. H (^) B 0 26 104
Plotting H (^) B we get the result shown. We require the speed to produce operating point B for the same size (125 mm).
Q)// /7^ 7c'
,r/ ,
AssA^ e
i> =^ / 7a ,*n"
/ € , - * ' /Q (^) €* /tzu (^) 11a,.-^
a>, e /$ "4-t
v,/.) a^
0 z = n > - / z - -^7 t ' t 2 v^ / a s. > / L :!?.
/ F z = c ; / A t r - ' c - t S t^ r.^ 1 2 '^ o ' < " f \
: --- .- VY?----+
/ 2. q e A
. , i. , d -. '. -
= (^) /. , a 9 "..!
V < t. 2 o a
Q A V u t z - -^ / 2.?. " 6 -^ 2 ' a o^ A l^ i o '^ -
v r "?. 7 L^. n z. 3 a u^ ;^ / p. P /
V,.!6 ?tc^ |$e = V.'1 "^ fle'?/ ZS^ :^ 4.^ 6 2 4^ * ' ' = / t 4 2 * '. t
,. c t ,?. 2? - - /. 3 5 9^ L : - -
7s.+a
'./t
ti2. " (^) 7< 3 .*'it
li&"*Lt (^) RG <c, ./&'.'€- et , 3 9 ''/^ (^) r + .6 2^ E
l&*n 1z:s4 : 3 <a'>
/ r ' : A - ' p a n a C' a. , -^ A l a. + A =^ t t t^ / * t^ - t , t S a , > 4 , 2 /. 3.^ J =^ / 2 , 2 3 * z. t
A k ( a c * - ; ) = t t 2 t s - 3 =^ f^ 2 3 -
q ' > i l r. 2 3 ; 7 9 , 3 , -
D t =^ l / r uz :^ / € '^ / 2 ,^ f u 4 ' " 1 ' j^ /. P -^ k * J
t , t. E : / n 3 4 L^ :^ r s -
/n ut
t, s1s,/t.r"^ ?:
FLUID MECHANICS D203 Q11 1998
The water surface in a reservoir is 18 m above the water surface level of a river. The reservoir is to be supplied with a steady flow rate of 1100 1itres/min of water from the river using a centrifugal pump. The suction and delivery pipes will have a diameter of 100 mm and total equivalent length of 120 m. The friction factor f for the pipes may be assumed to be 0.020. Three geometrically similar pumps with impeller diameters of 165 mm, 182 mm and 214 mm respectively are available and test results for the 182 mm diameter impeller pump running at 3000 rev/min are given in the table.
(a) Determine which pump is the most appropriate to use for this application and give reasons for your choice.
(b) Calculate the pump speed which will match the supply requirements and determine the power required to drive the pump under these conditions. The water density is 1000 kg/m^3.
Table for 182 mm at 3000 rev/min discharge q (litres/min) 0 500 1000 1500 2000 2500 head H (m) 43.8 42.5 38.8 33.0 25.2 16. overall efficiency η(%) 0 38 61 71 71 54
Bernoulli ∆H is the head added by the pump
h 1 + z 1 + u 12 /2g + ∆H= h 2 + z 2 + u 22 /2g + hf + exit loss velocity = 0 at free surface 0 + 0 + 0 + ∆H = 0 + 18 + 0 + hf + exit loss ∆H = 18 + hf + exit loss
2g
u 2gx0.
0.02x 120 xu ∆H 18
2 2 = + + 127.32Q πx0.
Q
u = 2 =
∆H = 18 +20656.7Q^2 Given Q = 1.1/60 = 0.01833 m^3 /s ∆H = 18 + 20656.7(0.01833)^2 = 24.94 m
Plot the pump characteristic for 182 mm and 3000 rev/min
The optimal point is at 1750 litres/min (0.0292 m^3 /s) with H = 30 m and η = 72% approx
The required Ns is
40 30
3000 x0. H
NQ
Ns (^) 3/
1/ 3/
1/ = = =
To achieve this, the speed must be changed to produce the required head and flow.
3/
1/ 3/
1/
- 94
Nx0. H
NQ
Ns = 40 = =
N = 3296.6 rev/min
A higher speed means a smaller pump is required. Choose the 165 mm pump. We need to determine the operating characteristics of this pump when running at the same speed (3000 rev/min).
© D.J.Dunn freestudy.co.uk
FLUID MECHANICS D203 Q1 1998
1 (a) State the conditions under which the Bernoulli equation is applicable.
(b) Water of density 1000 kg/m^3 is flowing into inlet (1) of the pipe-junction shown in the diagram. at a steady flow rate of 0.22 m^3 /s. The volume of water in the junction is 0.016 m^3. The centre of the outlet (3) is situated 600 mm vertically above the main horizontal pipe running between (1) and (2). The water pressure at (1) is 230 kN/m^2 and at (2) is 200 kN/m^2 ; energy losses in the flow are considered negligible. Determine.
(i) the water pressure at (3).
(ii) the flows leaving the junction at (2) and (3).
(iii) the magnitude and direction of the force acting on the junction as a result of the flow.
D 1 = 0.25m D 2 = 0.15m D 3 = 0.1m Q1= 0.22 m^3 /s p 1 =230 x 10^3 N/m^2 p 2 =200 x 10^3 N/m^2
- 482 m/s A
Q
u
0.007854m 4
D
0.00177m A 4
D
0.0491m A 4
D
A
1
1 1
2
2 3 3
2
2 2 2
2
2 1 1
π π π
CONSERVATION OF ENERGY OF A STREAMLINE FROM (1) TO (2)
- 847 m/s A
Q
CONSERVATIONOFMASS Q Q Q 0.22-0.158 0.616m/s u
FLOWRATE Q A u 0.158m /s
u 8. 95 m/s
u 200 x 10 1000 2
230 x 10 1000
u p 2
u p
3
3 3
3 3 1 2
3 2 2 2
2
2 3 2
2 3
2 2 2
2 1 1
+ρ = +ρ
CONSERVATION OF ENERGY OF A STREAMLINE FROM (1) TO (3)
3 2 3
2 3
2 3
3
2 3 3
2 1 1
p 203. 4 x 10 N/m
1000 x9.81x0. 2
p 1000 2
230 x 10 1000
ρgz 2
u p 2
u p
+ρ = +ρ +
© D.J.Dunn freestudy.co.uk
FORCES
Force at (1) F 1 = m 1 u 1 + A 1 p 1 → F 1 = 11390 N
Force at (2) F 2 = m 2 u 2 + A 2 p 2 ← F 2 = 4960 N
Force at (3) F 3 = m 3 u 3 + A 3 p 3 at 45o F3 = 2080 N
Horizontal component = 2080 cos 45o^ = 1470 N← Vertical component = 2080 sin 45o^ = 1470 N ↓
Total horizontal force = 11390 – 4960 – 1470 = 4960 N → Total vertical force = 1470 N ↓
RESULTANT FORCE = {49600^2 + 1470^2 }1/2^ = 51731 N Angle = tan-1^ 1470/4960 = 16.5o
© D.J.Dunn freestudy.co.uk
