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# Hydraulics Elec & Mech Fall 2022 - Quizz 1 with solution (3)

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```UNIVERSITE SAINT-JOSEPH
FACULTE D’INGENIERIE
E.S.I.B
Le jeudi 3 Novembre, 2022
Cursus Ing&eacute;nieur G&eacute;nie M&eacute;canique
Cursus Ing&eacute;nieur G&eacute;nie Electrique
Semestre 3
QUIZZ 1 D’HYDRAULIQUE
Dur&eacute;e 01H – Documents et manuscrits permis, Calculatrices non-programmables autoris&eacute;es
Nom et Pr&eacute;nom :
Q1
Q2
Total
Bar&egrave;me
40
60
100
Note
Lire attentivement :
• Cette &eacute;preuve comporte 9 pages, 2 probl&egrave;mes.
• Les donn&eacute;es des probl&egrave;mes sont exprim&eacute;es en anglais et fran&ccedil;ais (en italique).
• La clart&eacute; de votre &eacute;criture sera not&eacute;e
• On adoptera pour la tension de vapeur saturante &laquo; 𝑡𝑣 &raquo;, la masse volumique &laquo; 𝜌 &raquo;, la
pression atmosph&eacute;rique &laquo; 𝑝𝑎𝑡𝑚 &raquo;, l’acc&eacute;l&eacute;ration de la pesanteur &laquo; 𝑔 &raquo; les valeurs
suivantes :
𝑡𝑣 = 𝑝𝑣 = 0.2𝑚, 𝜌 = 1000 𝑘𝑔⁄𝑚3, 𝑝𝑎𝑡𝑚 = 105 𝑁⁄𝑚2 ou 10m d’eau, 𝑔 = .81 𝑚⁄𝑠2.
𝛾
Page 1 of 7
RES/Fall 2022/Hydraulics Quiz 1
Problem 1
A pipe 1000 m long with a diameter of D=0.5 m and a friction factor f=0.02 delivers water from a
reservoir situated 500 m above sea level to a nozzle situated 300 m above sea level. The jet from the
nozzle is used to drive a small impulse turbine. Assuming the minor losses in the nozzle of the form 0.04
V2 /(2g), find the jet diameter Dj, that will result in maximum power in the jet. Inlet minor losses from
the reservoir to the pipe are neglected. Use g = 10 m/s2 and ρ = 1000 kg/m3.
Use the following equation to calculate the power:
𝑃𝑗 = 𝛾𝑄
𝑉𝑗2
2𝑔
Hint:
One method to solve this problem is to use the energy equation and the continuity equation and find Dj
by trial and error (start with Dj=0.1 m to a maximum Dj of 0.5m) to obtain the highest power.
Probl&egrave;me 1
Un tuyau de 1000 m de long avec un diam&egrave;tre de D = 0,5 m et un facteur de frottement f = 0,02 am&egrave;ne
l'eau d'un r&eacute;servoir situ&eacute; &agrave; 500 m au-dessus du niveau de la mer &agrave; une buse situ&eacute;e &agrave; 300 m au-dessus du
niveau de la mer. Le jet de la tuy&egrave;re est utilis&eacute; pour entra&icirc;ner une petite turbine &agrave; impulsion. En
supposant des pertes mineures dans la tuy&egrave;re de la forme 0,04 V2 /(2g), trouver le diam&egrave;tre du jet Dj,
qui se traduira par une puissance maximale dans le jet. Les pertes mineures &agrave; l'entr&eacute;e du r&eacute;servoir vers
la conduite sont n&eacute;glig&eacute;es. Utilisez g = 10 m/s2 et ρ = 1000 kg/m3.
Utilisez l'&eacute;quation suivante pour calculer la puissance :
𝑉𝑗2
𝑃𝑗 = 𝛾𝑄
2𝑔
Indice :
Une m&eacute;thode pour r&eacute;soudre ce probl&egrave;me consiste &agrave; utiliser l'&eacute;quation d'&eacute;nergie et l'&eacute;quation de
continuit&eacute; et &agrave; trouver Dj par essais et erreurs (commencez avec Dj = 0,1 m jusqu'&agrave; un Dj maximum de
0,5 m) pour obtenir la puissance la plus &eacute;lev&eacute;e.
Solution
Energy equation:
500 -0.02*(1000/0.5) *(Vp2/2g) – 0.04 Vj2/2g = 300 + Vj2/2g
Using the continuity equation:
ApVp = AjVj
Dp2Vp = Dj2Vj
0.25Vp = Dj2Vj
Vp = 4 Dj2Vj
Page 2 of 7
RES/Fall 2022/Hydraulics Quiz 1
Substituting this expression for Vp in the energy equation gives:
200 = (Vj2/2g) *(1.04+640Dj4)
Assuming different values for Dj, we can compute corresponding values of Vj and Q and then compute
the jet power as follows:
Dj (m)
V (m/s)
Aj (m2)
Qj=AjVj (m3/s)
0.10
60.2
0.0079
0.20
44.0
0.0314
0.30
25.4
0.0707
0.40
15.2
0.1257
0.50
9.9
0.1963
A 0.2 m diameter is the optimum, it will give the maximum power.
Page 3 of 7
RES/Fall 2022/Hydraulics Quiz 1
0.47
1.38
1.79
1.90
1.94
Pj (hp)
856,438.7
1,340,125
575,821.7
218,548.6
94,466.72
Problem 2:
A centrifugal pump draws water from a reservoir to an elevated tank. The difference in water levels
between the reservoir and the tank is 10 m. The pipe between them has length 𝐿=150 m, diameter
𝐷=150 mm and friction factor f=0.02. Minor losses can be combined into an overall minor loss
coefficient, 𝐾, which is unknown. The characteristics of the pump at the operational speed are given in
the table below.
a)
b)
c)
d)
Determine the head loss due to friction as a function of discharge.
The flow is 46 L/sec. Find the pump head and power consumption at the flow.
Determine the overall minor loss coefficient, 𝐾.
After a rearrangement of facilities, the elevated tank is raised by 15 m and the pipe lengthened
by 70 m. Through careful engineering, minor losses have been significantly reduced and can be
assumed to be negligible (𝐾≈0). If the same discharge is to be maintained, find the new rotation
speed of the pump.
Given data:
ρ= 1000 kg/m3, g = 9.81 m/s2
Probl&egrave;me 2
Une pompe centrifuge aspire l'eau d'un r&eacute;servoir vers un r&eacute;servoir sur&eacute;lev&eacute;. La diff&eacute;rence de niveaux
d'eau entre les 2 r&eacute;servoirs est de 10 m. Le tuyau entre eux a une longueur 𝐿=150 m, un diam&egrave;tre
𝐷=150 mm et un facteur de frottement f=0,02. Les pertes mineures peuvent &ecirc;tre combin&eacute;es en un
coefficient global de perte mineure, 𝐾, qui est inconnu. Les caract&eacute;ristiques de la pompe &agrave; la vitesse de
fonctionnement sont donn&eacute;es dans le tableau ci-dessous.
a)
b)
c)
d)
D&eacute;terminer la perte de charge due au frottement en fonction du d&eacute;bit.
Le d&eacute;bit est de 46 L/sec. Trouvez &laquo; the pump head &raquo; et la consommation d'&eacute;nergie au d&eacute;bit.
D&eacute;terminez le coefficient global de perte mineure, 𝐾.
Apr&egrave;s un r&eacute;am&eacute;nagement des installations, le r&eacute;servoir sur&eacute;lev&eacute; est sur&eacute;lev&eacute; de 15 m et la
canalisation allong&eacute;e de 70 m. Gr&acirc;ce &agrave; une ing&eacute;nierie soign&eacute;e, les pertes mineures ont &eacute;t&eacute;
consid&eacute;rablement r&eacute;duites et peuvent &ecirc;tre consid&eacute;r&eacute;es comme n&eacute;gligeables (𝐾≈0). Si le m&ecirc;me
d&eacute;bit doit &ecirc;tre maintenu, trouver &laquo; the system head &raquo;.
Given data:
ρ= 1000 kg/m3, g = 9.81 m/s2
Page 4 of 7
RES/Fall 2022/Hydraulics Quiz 1
In the above equation λ is considered the friction factor “f”
b) The discharge 𝑄 = 46 L/s (= 0.046 m3/sec). Draw graphs of 𝐻 vs 𝑄 and 𝜂 vs 𝑄 to determine pump
head and efficiency at the required flow.
At the required flow, 𝐻 = 23.8 m 𝜂 = 62.0% (= 0.62)
Page 5 of 7
RES/Fall 2022/Hydraulics Quiz 1
c) At the required flow, pump head equals the system requirement (static lift + frictional losses + minor
losses):
Page 6 of 7
RES/Fall 2022/Hydraulics Quiz 1
Page 7 of 7
RES/Fall 2022/Hydraulics Quiz 1
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