ME 364
A hydroelectric facility extracts power from a reservoir at an elevation of 800 m discharging into a tailrace at an elevation of 200 m. A penstock...
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Problem：A hydroelectric facility extracts power from a reservoir at an elevation of 800 m

discharging into a tailrace at an elevation of 200 m. A penstock 1500 m long and 0.8 m in diameter connects the reservoir to the tailrace. Minor losses are K = 3 and C = 1500. Determine the power extracted by the turbine if the flow rate is 2 m3/s. If a Francis turbine is to be used, specify an appropriate speed.

I find a solution from Coursehero. My answer in V and Re is different from this answer, I don't know why it is. Could you help me with it?

Thank you very much.     6. A hydroelectric facility extracts power from a reservoir at an elevation of 800 m
discharging into a tailrace at an elevation of 200 m. A penstock 1500 m long and 0.8 m
in diameter connects the reservoir to the tailrace. Minor losses are K = 3 and C = 1500.
Determine the power extracted by the turbine if the flow rate is 2 m /s. If a Francis
turbine is to be used, specify an appropriate speed.
ORIGIN = 1
Set origin for counters to 1 from the default value of 0.
Input the pipe geometry:
wa
o.com
Diameter in mm
Length in m
Roughness in mm:
e
D:=
(0.8
0.8
. m
L:=
750
750
- m
=: 3
0.046
0.046
mm (commercial pipe)
Input the system boundary (initial and end) conditions:
Pressures in Pa
Elevations in m:
resour
seHer
Po J
(o)
- Pa
800
200
Input the loss coefficients:
K factor
Equivalent length
Number of pipes
(1500
Input the fluid properties:
stu
N:= length(D)
Density in kg/m3
Kinematic Viscosity in m2/s
P := 1000
v := 1.14 . 10-6
m
sec
Input the flow rate in cms:
Initial guess on turbine change in head:
Q:= 2.0.
sec
re
Ws := 100 - newton .
KE
Define constants and adjust units for consistency:
E:= 9.806
Ec:= 1 .
m . kg
newton . sec
Define the functions for Reynolds number, fully-rough friction factor, and friction factor.
Re(q, d) := -
4.9
it . d - v
fy (d, =) :=
0.3086
E
1.1172
(3.7- d) f( q, d, z)=
0.3086
if Re(q, d) &gt; 2300
1.11
log
6.9
Re(q, d)
3.7 - d
64
Re(q, d)
otherwise
The generalized energy equation is:
Given
Po - Pa
N
Ws
8
P - E
- Ect Zo- Za+ S
i= 1
2
(D.)
E
Q. D ,, ).
D.
1 + K + C . f/D, ;)
W, = Find( W.)
Ws = -5 559 x 10' m. 32
Additional output of useful quantities:
1 = 1 ..N
V(q, D) :=
4 - 9
vas
T . D2
D. =
Va, D. =
Re q, D.) =
f q, D., &amp; =
0.8 m
6.299
0.8
6.299
m . s
4.42-106
0.011
0.011
4.42.106
0.011
rseHero.con
0.011
Powerextracted := p . Q - Ws
Powerextracted = -1.112 x 10* . kw
The change in head across the device is
Ah =.
-WS
Ah = 566.856 m
eso
A reasonable efficiency is 0.90. Powerour := 0.90 - Powerextracted
Powerof = -1.001 x 10* . KW Francis turbine
NSp = 2.0
WJW /P
N SP = 2. 0. 17 = 0.9
(9H ) 3/4
= 2.0
N SP = WJW
= 2.0
Diameter : D = 0-8m
Length: L= 1500 m. Roughness 9= 0.246
Input the system boundary
29 ! =
250
x x /JW
K = 3 .
C = 1500.
0. 8
p = 1000 kg / m ?
V : = 1. 14 x 10 me / S
Q = 2 m3/s
9 = 9. 806 m / 2 2
gc = 1 m. kq
Wt JC
=
Pa - Pb
N . S 2
P. 9 - gc + Za - Zb -
Q 2
D 4 . 9
( f IQ. D. E ). D
fT ID . 9 ) =
0- 3086
Log 1 1 3.7- 17 1 10 &quot;
- = 0.0/ (Re12, 1) ) =
4 0
+ 1 + C. fr ( D. E ))
TEDS = 2.79 x 106
f (Q. D. E ) =
0 -3086
log
6-9
Re l QD )
E
2
if Re ( Q. D ) &gt; 230
64
Re (Q. 17 )
otherwise
f 1 Q , D . 2 ) =
0 096 41:1172
- 2.19 4 106 + ( 3. 7 x08)
= - J
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