Sample Calculation of Sieve Analysis
Wt . of Bowl ( g )=120 g
Wt . of Bowl+Sand ( g )=1620 g
Wt . of Sand ( g )= (Wt . of Bowl+ Sand )(Wt . of Bowl)
( 1620 )( 120 )
1500 g
Sieve 4
Sieve Wt . ( g )=467.8 g
Sieve Wt .+ Retained Wt . ( g )=468 g
Wt . Retai

Sample Calculations
Part A
Temp=70.5
N
=9810 3
m
g=981
cm
sec 2
W =30 lbs
Q=
W 4.48
c m3
106
=
t
sec
V=
Q 479.03
cm
=
=90.23
A 5.309
sec
2
(
30lbs4.48
c m3
c m3
10 6
=479.03
sec
sec
N
9810 3 ( 28.6 s )
m
)
2
Velocity Head=
V
90.23
=
=4.15 cm
2 g 2 ( 981

1.
A 36 sewer pipe is flowing half full with a flowrate of 6000 gpm. Find the average velocity in the
pipe. (3.77 ft/s)
2.
A building foundation is excavated to a depth of 20. The building footprint is 100 x 150. The site
has filled in with 10 of water. W

Sample Calculations
Temp= 70.5
(d/ cm 3 )= 978.9
g (m/ s 2 ) = 9.81
b = width of the plane = 7.5 cm,
a = depth from beam to top of plane = 10 cm,
d = height of the plane = 10 cm
LW = length of lever arm = 27.5 cm
Fully Submerged
Height of water above th

1.
In a typical toilet a floating ball is used to close off the water filling the cistern when it reaches a
certain level see schematic below. If 10 lbs of force are required to shut the valve what size does the
ball have to be in inches? (ans = 9)
2.
A h

1.
The fluid in the pipe is water. The density of the manometer fluid is 560 lb/f 3. Find the
differential pressure in psi. Which point on the pipe has the higher pressure?
2.
There is liquid in the pipe with a density of 30 lb/f3. The manometer liquid ha

Fluid Properties Problems
1.
At what temp in oF and oC is water at its most dense.
2.
At what temp in oF and oC is water at its least dense.
3.
Fifty thousand gallons of groundwater at 50 oF is pumped into an elevated storage tank
where the temperature ri

A Glossary of Literary Terms
A few general terms
Term
Plot
Definition
Examples
The order of arrangement of ideas and/or incidents that make up a Complex, simple
story in a play or a novel. A plot can be Complex or Simple.
Form: Structural design and patte

NYU Tandon School of Engineering
Chemical and Biomolecular Engineering(CBE)
Course Outline: CM-UY 1004 General Chemistry for Engineers
Fall 2016
Professor: Dr. Robinson-Surry
D-Lecture (15703), Mon. & Wed. 11:00 AM 12:20 PM, Pfizer Auditorium
To contact p

Group 4 Lab Report-Viscosity of Seawater-Siddhant Bafna
Teacher Prompt
To devise an experiment with the help of umbrella topic water and group sub-topic seawater
Aim
To measure and evaluate the viscosity of seawater us

Sample Calculation
Tank Orifice (Tecquipment)
Run 1:
Flow:
Q=
(
(
w4.448
c m3
106
t
sec
)
30 lbs4.448
c m3
106
N
sec
9802 2 62.3 sec
m
)
c m3
218.52
sec
Difference in R:
|R RR L|
|14.4 threads4.24 threads|
10.16
threads.127 cm
1 thread
1.29 cm
Coefficien

=9810
N
3
M
=.998
gm
c m3
W= 15 lbs
t=22.60 s
Flow:
Q=
4.448 W
c m3
10 6
t
sec
3
4.448 (15)
cm
106
N
s
(9810 3 )(22.60 s)
M
c m3
300.9
s
Force Actual:
F jet=.2 y
1
.2
.2lbs
.2
lb444840 dynes
=88968 dynes
1lbs
Sample Calculation
(Flat Plate)
=9810
N
M

Equation Derivation for Illustration 10
Initial equil . no water jet :
M o =0
Eq (1 ) :+ X F s Z ( Tare Wt . ) 6 ( 1.2lb )=0
Equil . witha water jet :
M o =0
Eq ( 2 ) :+ X F s Z ( Tare Wt . )+6 F Jet ( 6+ y ) (1.2 ) =0
Subtract Eq ( 1 ) Eq (2) :
Eq ( 2

W/C = 6.0
% Fine aggregate = 35%
% Coarse aggregate = 65%
Specific Gravity = 2.60
Sample Calc.
For 1 Sac of Cement
gal
lb
( H 2 O ) =6.0 sac 8.345 gal 1 sac =50.07lb ( H 2 O )
( Agg )=5.5
lb
l bcement
l bcement
1 sac =517lb ( Agg )
sac
94
Fine aggregate=5

Calibration of the Universal Testing Machine using the Proving Ring:
[Original Loading + 1F] Cycle Data
Div0T = Zero dial load reading
DivT= Loaded micrometer dial reading
DivT0 = Equivalent divisions at T =78 oF
T= Temperature in oF
C= Calibration Factor

Calibration of the Universal Testing Machine using the Proving Ring:
[Original Loading + 200 LBS] Cycle Data
Div0T = Zero dial load reading
DivT= Loaded micrometer dial reading
DivT0 = Equivalent divisions at T =78 oF
T= Temperature in oF
C= Calibration F

Calibration of the Universal Testing Machine using the Proving Ring: Original Loading Cycle
Data
Div0T = Zero dial load reading
DivT= Loaded micrometer dial reading
DivT0 = Equivalent divisions at T =78 oF
T= Temperature in oF
C= Calibration Factor
PActua

Sample #1:
Length=2
Thickness=2
Width=4
Area ( A )=8 i n2
Modulus of Elasticity=1,600,00 psi
Moment of Inertia:
3
I=
bd
12
4
2
3
2.66 in
4
Failure Exp. Load:
P=28,000 lbs
Failure Exp. Stress:
=
P
A
28,000 lbs
8 in 2
3,500 psi
Radius of Gyration:
R=
I
A

Data for Oak in Compression:
Length(Lo )=5.875
Width=2
Thickness=2
2
Area ( A)=4 i n
Load ( P )=3500 lbs
Divisions:
1=.0010
.001
=1.0.
Stress:
P
A
=
(3500 lbs)
( 4 in 2)
875 psi
Strain:
=
Lo
.0010
5.875
.0001702
Modulus of Elasticity:
E=
(875 psi)
( .

Sample Calculation
(1-inch diameter pipe)
Flow 1 (Venturi):
Head loss:
h1=6
6
h2=9
8
h=|h2h 1|
6
9 6
8
15.75
15.752.54
cm
=40.005 cm
Flow:
Q=92.4 h
92.4 40.005 cm
584.43
cm
s
3
Flow 1 (Pipes):
Area:
1 2.54
A=
D
4
cm
=2.54 cm
2
2.54 cm
2
2
5.06 c m
L

Name_Section_
Experiment 2
Chromatography
CM1004
Fall 2016
1
Chromatography
In your text (Chang and Goldsby 7th Ed) : 1.3 Classifications of Matter
1.4 Physical and Chemical Properties of Matter
Extra reading: 12.2 Intermolecular Forces
Purpose : Two type