Download Hydraulic Engineering First Test, 2007 (100/106) Name and ID: ___

Hydraulic Engineering
First Test, 2007 (100/106)
Name and ID: ___Key________________________
Rules: one page cheat sheet one sided allowed; Walton text notes and textbook. No: cell
phones, non-ABET calculators, or other materials. Fill in the answers in the space provided
with units; use the white space to show your work. You MUST show work to receive credit.
1. (30 pts) Water flows through 100 feet of four inch galvanized iron pipe at 2 ft3/s. The entrance
and exit pipes are sharp, R/D for the bend is 1, the 50% closed gate valve has a K of 0.7.
B
.
a) (4) The minor loss coefficients are:
Kentrance = 0.5, Kexit = 1.0, Kelbow = 0.35, Kvalve = 0.7
b) (4) Solving the energy equation for the elevation of the top of the second tank gives (don’t give the
numbers but be sure to zero out appropriate terms):
z2 = z1 - hloss
c) (3) The Reynolds number is:
V = Q/A = 2/(pi (2/12)^2) = 22.9 ft/s
Re = V D/v = 22.9*0.3333/(1.08 10^-5) = 707000
d) (4) The friction factor is:
e = 0.0005 ft from Table 3.1 e/D = 0.0005/0.33333 = 0.0015
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f = 0.022 (from equations) allow anything from 0.0205 to 0.0235 to account for chart reading
errors
e) (4) The head loss is:
hl = V2/2g (fL/D + sum of K’s) = 22.9^2/(2 *32.2)*(0.022 100/0.3333 + 0.5 + 1 + 0.35 + 0.7) = 74.5 ft
Note: there will be some slop in the answer from the range of f, allow +/- 10%
f) (2) The elevation of the top of the second tank is:
335 – 74.5 = 260.5 ft (same range of uncertainty allowed, about +/- 10%
g) (4) If the pressure at the bottom of the second tank is 311.9 lbf/ft2, how deep is the second tank?
P/(rho g) = z = 311.9/62.4) = 5 ft
h) (5) If point B is 5 feet inside the pipe and the upper tank is 3 ft deep, solve the energy equation for
the pressure at point B (put in all known numbers into the derived equation but DO NOT solve it for
a number unless you’re bored)
PB = rho g (z1-zB-Vb^2/2g – Vb^2/2g(fL/D+K)) = 62.4 (8 – 22.9^2/2*32.2 - 22.9^2/2*32.2(0.022 5/0.33
+ 0.5) = -432 lbf/ft^2 (don’t have to solve for final number
Put your work below here ***********************************************************
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2. (20) When water is flowing at a depth of 1.49 feet in the triangular channel shown with slope of
0.006 and n of 0.012.
a) (5) Find the hydraulic radius:
A = y2
P = 2 Sqrt(2) y R = A/P = y2 /2 Sqrt(2) y = y/(2 Sqrt(2)) = 0.527
b) (5) Find the discharge:
1.49/0.012 1.49^2 0.527^0.67 0.006^0.5 = 13.9 cfs
Assuming that the discharge is 20 cfs (note: this incorrect assumption allows the second part of the
problem to be independent of getting the first part right) find:
c (5) the critical depth
Q2/g = D A2 when Froude number is 1; D = A/T where Q is discharge, T is top width = 2y, A is area
= y2, D is hydraulic depth = A/T,
Q2/g = A2/T plugging in for A = y2 and T = 2y gives ycritical = (Q2/g)0.2 = (202/32.2)0.2 = 1.66 ft
d) (5) and the Froude number
Fr = V/Sqrt(g A/T) = (Q/A)/ Sqrt(g A/T) = 20/(1.492 Sqrt(32.2 (1.49/2)) = 1.84
Put work below here **************************************************************
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4.(9)
(3) The surface profile classification just upstream of the jump is:
(3) The surface profile classification just downstream of the jump is:
(3) If baffle blocks are placed at A, the jump will:
a) Move downstream
b) Move upstream
c) Stay in the same location
Answer: H3 before the jump, H2 after the jump, The baffle blocks move the jump upstream (b)
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3. (4) A pipe holder is placed at the bend. Show the force the holder puts on the pipe to keep it in
place using arrows. The arrow shown is the direction of flow.
In the bend the water velocity is increased in the downward direction and decreased in the X direction
requiring the reactive forces shown
4. (4) The following equation gives energy in units of (circle one): a) energy per unit mass, b)
energy per unit weight, c) energy per unit volume, d) total energy, e) energy per unit of velocity.
In the SI system the units are: __Pascals_______
5. (7) Sketch the EGL and HGL on the figure below. Mark the point of greatest velocity.
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6. (8) Label on the graph: system curve, pump performance curve, pump efficiency curve, power
used curves.
Top curve is pump efficiency
7. (3) Mario is standing on a: __broad crested weir_________________________________.
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8. (9) Water is present at a depth of 1.6 meters above a sluice gate. On the figure put Point A
above the sluice gate, Point B just below the gate and Point C just after the hydraulic jump.
Label the portions for subcritical, supercritical, and critical flow.
9. (4) Label the lines for turbulent and laminar flow:
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10. (8) If the velocity 3.13 m/s, and the figure is drawn to scale, find the (approximate) pressure
head in the pipe at P, S, Q, and R.
The velocity head is the difference between the two lines V2/2g = 0.5 m. P and Q are right on the HGL
so pressure is zero, S is about 1.5 m above the HGL so pressure is about -1.5 m, R is about 1.0 m below
HGL so pressure head is +1.0 m
Note in grading that I asked for pressure when I meant pressure head so give some slack
Put work below here ***********************************************************
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