Physics Assignment

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