Class notes
1_Hydraulics Practice Quiz with Answer and Solution
- Course
- Institution
This note contains practice quizzes for civil engineering hydraulics. It includes practice problems with answers and solutions for the study guide.
[Show more]
Seller
Follow
DocuPlex
Content preview
Refresher - HYDRAULICS Quiz 1PROBLEM 1-3:
A jet of water 25 mm in diameter and having a velocity of 7.5 m/s strikes against a
plate at right angles.
➀ Determine the force on the plate if the plate is fixed.
➁ Determine the force on the plate if the plate is moving in the same direction
as the jet with a uniform velocity of 3 m/s.
V’=0
Solution:
➀ Force on the plate if the plate is fixed: V1=7.5 m/s
Q = AV
π
Q = (0.025)2 (7.5) = 0.00368 m3 / s
4
QW 0.00368(9810)
F= (V1 - V2 ) = (7.5 - 0)
g 9.81
F = 27.6 N
➁ Force on the plate if the plate is moving in the same direction as the jet with a
uniform velocity of 3 m/s.
U = V1 - V1
V =3.0
’
U = 7.5 - 3 = 4.5 m/s
Q = AU
V1=7.5 m/s
π
Q' = (0.025)2 (4.5) = 0.0022 m3 / s
4
QW 0.0022(9810)
F= (V1 - V1 ) = (7.5 - 3)
g 9.81
F = 9.94 N
, Refresher - HYDRAULICS Quiz 1
PROBLEM 3-5:
A water tank has a sloping inclined at 45˚ with the horizontal. The total depth of water in the tank is 6.7 m.
A water jet issues from an orifice located on the inclined side of the tank under a hydrostatic head of 4 m.
or that orifice is located 2.7 m. vertically above the bottom of the tank. Coefficient of velocity is 1.0.
Negcting air resistance on the jet.
➀ Determine the maximum height of the issuing jet arises above the level of the center of orifice in
meters.
➁ Determine the time it takes for this particle of the jet to sit the ground that is 1.20 m. below the
bottom of the tank in seconds.
③ Determine the horizontal distance on the ground traveled by the jet from the center of the orifice in
meters.
Solution:
➀ Max. height the jet arises above the center of orifice:
V = Cv 2gH
w.s.
V1
V = 1.0 2(9.81)(4) V 2 =0
V
4m V
V = 8.86 m/s h 45˚
45˚
V1 = 8.86 Sin45˚= 6.26 2.7m V3
45˚ 45˚
V2 2 = V1 2 – 2gh
0 = (6.26)2 – 2(9.81)h 1.2m
h = 2 m.
➁ Time for the particle to sit the ground: w.s.
V2 = V1 – gt1 V1 V 2 =0
V
0 = 6.26 – 9.81t1 4m
B
t1 = 0.64 sec. A
45˚
Time to travel from A to B is t1 = 0.64 sec. 2.7m 45˚
V3
5.9
Time to travel from B to C: 1.2m
5.9 = ½ gt2 2 C
5.9 = ½ (9.81)t2 2
t2 = 1.1 sec.
Total time for the particle to sit the ground = 0.64 + 1.1 = 1.74 sec.
③ Horizontal distance:
V3 = 8.86 Cos 45˚ = 6.26 m/s
D = V3 t
D = 6.26(1.74) = 10.89 m.
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller DocuPlex. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $7.99. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
72042 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now