Physics 201: Fluid Mechanics Assignment - Pressure and Flow

This physics assignment solution provides detailed answers to a series of fluid mechanics problems. The assignment includes matching questions on pressure and flow rate concepts, as well as calculations involving standard atmospheric pressure, pressure at a specific depth, and flow rates in pipes with varying areas. It also covers the Hazen-Williams method for calculating flow rates in pressurized water pipes, and the determination of flow rates in sewer pipes under different conditions. Additionally, the solution includes the calculation of flow rate through a V-notch weir. The workings for each problem are clearly presented, including formulas, unit conversions, and step-by-step calculations, making it a comprehensive resource for understanding and solving fluid mechanics problems. The document covers a range of topics including pressure, flow rate, cross-sectional area, and the application of the Hazen-Williams formula.

Physics Assignment
Student Name:
Instructor Name:
Course Number:
24 April 2019

Assessment
Q U E S T I O N 1
Match each item to the appropriate answer.
Pressure
Standard atmospheric pressure
What is the pressure at 15 ft under
water?
10 ft3/s
What is the flow rate of water in a pipe
flowing full with an area of 0.3 m2 and
velocity of 2.5 m/s?
What is a pipe’s inside cross-sectional
area, if the inside diameter is 11.9
inches?
Referring to Figure 2-12 on page 31 in
the textbook, if QA=3 ft3/s and QC=1.7
ft3/s, what is QB?
If a 400-mm diameter pipe with a pipe
roughness coefficient of 100 flows full
of pressurized water with a head loss
of 0.4 ft per 1,000 ft of pipeline, what is
the flow rate? Use the Hazen-Williams
method.
If a 1-m diameter sewer pipe is flowing
at a depth of 0.4 m and has a flow rate
of 0.15 m3/s, what will be the flow rate
when the pipe flows full?
What is the flow rate through a V-notch
weir having a head of 0.42 ft?
A
. 4488 gal/min
B
. 0.29 ft3/s
C
. Flow rate times slope
D
. 14.7 psi
E. 0.77 ft2
F. 0.037 m3/s (37 liter/s)
G
. Force divided by area
H
. 6.45 psi
I. 0.75 m3/s
J. 0.44 m3/s
K. 1.3 ft3/s
WORKINGS
 Pressure is defined as the force acting normally( perpendicularly) per unit area.
In other words it is Force divided by area.
 Standard atmospheric pressure
It supports 760 mm (29.92 inches) 0f mercury at sea level.
This is equivalent to 101 325 pascals.
1 Psi ( pound per square inch)= 6894.76 Pa.
G.
D.
H.
A.
I.
E.
K.
F.
A.
B.

101325 Pa= 101325
6894.76 psi
=14.6959 psi
=14.7 psi
 Pressure at 15 ft under water.
Pressure in liquids is given by P=height of liquid ×density of liquid ×pull of gravity
P= 15× 30.48
100 m×1000kg/m3 × 9.8N/kg= 44805.6 pascals1
Psi ( pound per square inch)= 6894.76 Pa.
44805.6 pa= 44805.6
6894.76
= 6.498 psi
=6.45 psi is approximately equal to 6.498 psi
 10 ft3/s
1ft3/s=(30.48cm)3/s=28316.64659 cm3/s
4488gal/min= 4488
60 gal/s
=74.8 gal/s
But 1 gal/s=3785.411784 cm3/s
74.8 gal/s= 74.8×3587.411784 cm3/s
=283148.8014 cm3/s
= 283148.8014
28316.64659 =9.9999 ft3/s
=10 ft3/s
Thus 10ft3/s = 4488 gal/min
 Flow rate =area × velocity
=0.3 m2× 2.5 m/s
=0.75 m3/s
 D=11.9 inches
=11.9× 2.54 cm=30.226 cm
1 foot= 30.48 cm
30.226 cm= 30.226
30.48 ft
=0.9917 ft
Area=πr2
A= 22
7 × 0.9917
2 × 0.9917
2
=0.7727 ft2

=0.77 ft2
 QA=QB +QC……………………………….(i)
QA= 3 ft3/s ,QC=1.7 ft3/s
Substituting for QA= 3 ft3/s and QC=1.7 ft3/s in (i) we have
3 ft3/s = QB + 1.7 ft3/s
QB=3 ft3/s - 1.7 ft3/s =1.3 ft3/s
=1.3 ft3/s
 D=400 mm=40 cm , r=20cm
C=100
L=1000 ft
Hf=0.4 ft
Applying Hazen-Williams Formula i.e
Hf =L
[ V
1.5873
√R × 1
KC ]1.852
Where Hf=Head loss due to friction
V= Velocity
K= Unit conversion (1.318)
L-Length
R=Hydraulic radius= D
4 = 40
4 =10cm= 10
30.48 ft=0.3281 ft
C=roughness coefficient
Substituting the known values into the Hazen-Williams Formula, we can determine the velocity
V
0.4=1000[ V
1.5873
√0.3281 × 1
1.318 ×100 ]^1.852
0.4
1000=[ V
1.5873
√0.3281 × 1
1.318 ×100 ]^1.852
0.0004= [ V
131.8 ×0.4955 ]^1.852

0.0004= [ V
65.3069 ]^1.852
0.00040.54=[ V
65.3069 ]
V=0.00040.54×65.31
V=0.9552 ft/s
Flow rate Q=Area A×Velocity V
Q= AV
A=πr2
r =20cm = 20
30.48 ft=0.6562 ft
A=πr2= 22
7 ×(0.6562ft)2=1.353 ft2
Q=AV=1.353 ft2×0.9552 ft/s
=1.2924 ft3/s
= 1.3 ft3/s= 1.3× (0.3048m)3/s
=0.0368119 m3/s
=0.037 m3/s (37 liters/s)
 Flow rate Q=Area × velocity
Assume that we take a pipe with two ends A and B both with varying cross sectional
areas;then
Q=A1V1=A2V2= constant.
Therefore flow rate at end A=flow rate at end B
Flow rate at narrow end=Flow rate at wide end
Flow rate at 0.4m diameter=Flow rate at 1m diameter
Flow rate at 1m diameter(full pipe)=0.15m3/s
=0.15m3/s= 0.2 m3/s (1d.p)
=0.2m3/s = 0.2
0.02832ft3/s=7.062 ft3/s
= 10 ft3/s (nearest tens)
=4488 gal/min
 Flow rate for through a V-notch weir is given by
Q= 2.49H2.48

Where Q=Discharge in cfs.
H=Head over the weir in ft
Q= 2.49H2.48=2.49(0.422.48)
=0.2896 cfs
=0.2896 ft3/s
=0.29 ft3/s