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Pump Example.mcd
1
4/17/00
A pump in the system shown draws water from a sump and delivers it to an open tank through 1250 ft of new 4” nominal diameter, schedule 40 steel pipe. The vertical suction pipe is 5 ft long and includes a foot valve with hinged disk and a 90 deg standard elbow. The discharge line includes a foot valve with hinged disk and a lift check valve, and a fully open gate valve. The design flow rate is 200 gpm. 1. Find the head head losses losses in the the suction suction and discha discharge rge lines. lines. 2. Ca Calc lcul ulat atee th thee NP NPSH SHA. A. 3. Select a pump suitable suitable for for this application given the pump selection selection charts charts in Figures Figures D.1 D.1 and D.2. D.2. 4. Fin Find d the appropri appropriate ate rotatio rotation n speed for for the pump pump to meet meet design design conditio conditions. ns. 5. Comm Comment ent on the the expected expected efficie efficiency, ncy, estima estimate te it and the the power power required required by the the pump in hp. hp. 6. Draw the the pump characteris characteristic tic curve (Head vs. vs. Volume Flowrate) and the system curve curve on one graph
Properties of Water
w( T )
w( T )
0.001792
100 0
kg m sec
kg 3
exp
1.94
kg
0.0178
m
4.8
C
1.7
3
(
T
273. 27 3.15 15 K T
273.15 K
4 C )
1.7
m
Given Pipe: 4" nominal SC40 steel Discharge Side: Total discharge pipe length: Two 90 deg standard elbows One angle lift check valve One gate valve (fully open)
L Dp
1250 12 50 ft
6.74
273. 27 3.15 15 K T
273.15 K
2
C
K
Pump Example.mcd
2
4/17/00
Suction Side: Total suction pipe length:
L Sp
5 ft
One foot valve with hinged disk One 90 deg standard elbow Design Flowrate:
Qd
200
gal min
Solution From Table E2.1 Pipe Diameter:
D p
4.026 in D
Pipe Area:
A p
2
p
4
From Figure 8.15 Roughness:
e
0.00015 ft
e D p
4.471 10
4
Find loss coefficients for all hydraulic components From Table 8.3: Gate Valve (fully open) Equivalent length:
L eGV
8 D p
L eFV
75 D p
L eCV
55 D p
Foot Valve with hinged disk: Equivalent length: Check Valve, angle lift Equivalent length: Standard Elbow, 90 deg Equivalent length:
L eE
30 D p
From Table 8.1: Entrance from Reservoir to Pipe Loss Factor: From Figure 8.17:
Exit from Pipe to Reservoir Loss Factor:
K e
0.78
K ex
1
Reservoirs are large so velocities in the reservoirs are negligible Lower Reservoir Surface State 0:P o
0 psi
(gage)
Vo
0
Upper Reservoir Surface State 1:P 1
0 psi
(gage)
V1
0
Suction of Pump State S:
zS
28.62 ft
m sec m sec
zo
24 ft
z1
289 ft
Pump Example.mcd
3
Temperature:
4/17/00
25 C
Tw
Atmospheric Press ure:
P atm
14.7psi
Water Viscosity:
w Tw
9.075 10
Water Density:
w Tw
996.851
Velocity in Pipe:
V Qd
A p
D
Reynolds Number in Pipe
m sec
kg 3
m
Q
V( Q )
kg
4
Re p ( Q )
1.536
m sec
p V ( Q )
Re p Q d
5 1.726 10
Larger than 2300 so flow is Turbulent Find Friction Factor: f
0.25 log
e D p 3.7
1
Given
2 log
0.5
f
2
5.74 Re p Q d
0.9
e
2.51
D p 3.7
f( Q )
0.5
Re p ( Q ) f
f Qd
Find( f )
0.018824
Head Losses:Sum of Major (friction in pipe) and Minor (hydraulic components) head losses Discharge Side: Major
Minor
Total
h DM( Q )
h Dm ( Q )
f( Q )
h DL( Q )
f( Q )
2
L Dp V ( Q ) 2 D p
h DM Q d
2 g
L eE
L eCV
L eGV
D p
D p
D p
h DM( Q )
h Dm( Q )
K ex
27.691 ft
V( Q ) 2 g
h DL Q d
2
h Dm Q d
29 ft
1.309 ft
Pump Example.mcd
4
4/17/00
Suction Side: Major
h SM( Q )
Minor
h Sm( Q )
Total
f( Q )
h SL( Q )
L Sp V ( Q ) 2
f( Q )
h SM Q d
2 g
D p L eE
L eFV
D p
D p
h SM( Q )
K e
V( Q )
0.111 ft
2
h Sm Q d
2 g
h Sm( Q )
h SL Q d
1.088 ft
1.199 ft
To find the NPSHA we need to find the stagnation head (sum of the static and dynamic heads) at the suction point of the pump. This can be found by applying the energy equation for steady, incompressible flow between States 0 and S: A simple though mechanistic way of writing th e energy equation for steady, incompress ible flow between two points is by considering the head balance: Total Head In (begining state) + Head Gained (along the fluid path) = Total Head Out (final state) + Head Lost (along the fluid path) where
Total Head=Static+Dynamic+Hydrostatic
Thus, for our purposes: Total head at State 0:
Total head at State S:
h to
2
Po
Vo
g
2 g
zo
h to
24 ft
htS =hSt +zS
So applying the balance between states 0 (begining state) and S (final state), and solving for the stagnation head at Suction we get: h St( Q )
h to
zS
h SL( Q )
Total Pressure at Suction:
P St( Q )
gh
St( Q )
Vapor Pressure for Water at T=25C from Steam Tables:
P atm
5.819 ft
P St Q d
12.185 psi
6
Pv Pv
h St Q d
0.003169 10 Pa
0.46 psi
Pump Example.mcd
5
4/17/00
Net Positive Suction Head NPSHA( Q )
P St( Q )
Pv
NPSHA Q d
g
27.132 ft
The Pump Head can be calculated by applying the energy equation for steady, incompressible flow between States 0 and 1: Considering that: Total head at State 0:
h to
Total head at State 1:
h t1
2
Po
Vo
g
2 g
P1
V1
g
2 g
zo
h to
24 ft
z1
h t1
289 ft
2
writing the head balance, and solving for the required head of the pump we get
h P( Q )
h t1
h to
h SL( Q )
h DL( Q )
hP Q d
295.199 ft
So the operating point of the required pump must be Head
hP Q d
295.199 ft at a Flow Rate of Q d 200
gal min
The Power that will be transmitted to the fluid is Pw( Q )
g h
P( Q ) Q
Pw Q d
11.099 kW
Looking at Pump Selection Charts such as those of Figures D.1 and D.2 we see that a pump type 4AE12 (4" Suction) could do the job at a rotation speed between 3550 and 1750 RPM Let us look at the pump characteristics for this type 4AE12. They are given for 3550 and 1750 RPM in Figures D.5 and D.4 respectively. We see that we will not be able to meet our requirement at these speeds. So we need to determine the speed at which we need to run a 4AE12 type pump to do the job. Let us pick an impeller size of:
Di
11 in
Pump Example.mcd
6
4/17/00
We can obtain the head vs. flowrate characteristic for this speed knowing that the dependence of the head on the flowrate is quadratic: H p (Q p ) H p0
BQ p2
RPM
1 60
Hz
So we read the head at zero flow off Figure D.4 for the chosen impeller size at 1750 RPM: H p0
125 ft
1750 RPM
we also read the Hp, Qp pair for one more point on the characteristic, say the maximum efficiency point. gal H pe 105 ft Q pe 460 min Then we can calculate B:
H p0
B
H pe 2
B
9.452 10
Q pe H p Q p
H p0
B Q p
5
ft min
2
2
gal
2
From similarity considerations the coefficient B will be independent of pump rotation speed. If we change the speed of rotation only H p0 will change. The requirement we have to meet is dictated by our system's operating point Head h P Q d
295.199 ftat a Flow Rate of Q d 200
Therefore we can calculate the required Hp0.
H p0r
hP Q d
2
B Q d
H p0r 298.98 ft
To find out the required pump speed we use the similarity law H p1 2 2 1 D1
H p2
2 2 2D2
Because we are not changing impeller size we have:
r
H p0r H p0
3 r 2.706 10
RPM
gal min
Pump Example.mcd
7
4/17/00
Speed must be a fraction of 3550 RPM, so
r
49 64
3550 RPM
3 r 2.718 10
RPM
Close enough
So an 4AE12 pump at 2718 RPM will do the job but the efficiency is going to be very poor. It is evident from inspection of the pump characteristics (Figures D.4 and D5) that the operating point is very far from the maximum efficiency region for this class of pumps. There will also be no problem with the NPSH. Let us estimate the efficiency to be:
p
0.45
The power we have to supply the pump with to do the job is
