College of Engineering & Petroleum
Mechanical Engineering Department
ME 259: Introduction to Design
CANS
By
Zainab Alsaffar
ID: 2201115949
May 12, 2022
Abstract:
Cans manufacturing is being processed by factories to produce billions of cans with different
dimensions and costs. The main purpose of this experiment is to define the actual cost of the
produced cans by knowing the volume and the height with respect to the diameter ratio. The lowest
and highest cost is being detected by filling a table with a provided different values using Microsoft
Excel. In addition, three cans with different dimensions is being compared with the calculated
values, in order to obtain the supermarket cost with respect to the manufacturing cost. The results
shows that the lowest cost has 1 H/D and 0.25 L. According to the calculations and actual results,
the cans in the supermarket were sold with an excessive cost in comparison to the calculated
values. As a result, we can detect that there is a huge difference between the manufacturing price
and the market price. A supervisor must be employed above these factories to maintain the cost
efficiency of the produced products, in order to serve the customer needs.
I
Table of Contents
Abstract: ........................................................................................................................................................ I
List of Figures and Tables: ........................................................................................................................ III
Nomenclature: ............................................................................................................................................ IV
Introduction: ................................................................................................................................................ 1
Background: ................................................................................................................................................. 1
Results and Discussion: ............................................................................................................................... 4
Conclusion: ................................................................................................................................................... 5
Acknowledgments ........................................................................................................................................ 6
Appendix: ..................................................................................................................................................... 7
II
List of Figures and Tables:
Figure 1: values of volume and H/D ratio. .................................................................................................... 1
Table 1: Cans results...................................................................................................................................... 3
Figure 2: cost vs H/D chart. ........................................................................................................................... 3
Figure 3: Three different cans........................................................................................................................ 4
Table 2: Real cans measurements. ................................................................................................................. 4
Figure 4: Actual cans details.......................................................................................................................... 4
III
Nomenclature:
At
Total Area
Ac Cross-sectional Area
D
Diameter
H
Height
X
H/D ratio
V
Volume
KD Kuwaiti Dinar
IV
Introduction:
Billions of cans are being produced by factories each year for many different uses, depending
on its size, shape, and material. In this report, the optimized can will be determined by considering
that cans are made from thin aluminum sheet which costs about 0.300 KD (300 fils) per square
meter. The main target of this experiment is to define the minimum cost with the best height to
diameter ratio (H/D) and respect to the volume of the can by using Microsoft Excel.
Background:
The costs will be measured according to the volume and (H/D) ratio. The volume in liter will be
in the x-axis where (H/D) will be in the y-axis with many different values that starts by the value
0.25 and is increasing by 0.25 (+ 0.25), as shown in figure 1. A specific equations will be used to
fill these blanks, in order to get the minimum cost with the best ratio by comparing all the values
with each other.
Figure 1: values of volume and H/D ratio.
Since the cost is given per square meter as a unit (KD/m2), so we need to convert this unit into
KD only, to calculate the cost of each blank in the given table.
1
The total area of the can equals 2 multiply by the area of the circle which represents the top and
the bottom + the area of the triangle which represents the circumference of the can.
𝜋
At = 2 × (4 ×D2) + (πD×H)
𝜋
At = (2 ×D2) + (πD×H)
The volume of the cylindrical can is the product between the cross-sectional area and the height of
the can.
V = Ac × H
𝜋
V= (4 ×D2) × H
𝐇
Suppose that 𝑥 = 𝐃
So, by cross-multiplication, we will get H = 𝑥×D
By substituting the value of H in the volume equation to get a new equation that connects between
the diameter, volume, and X.
D=(
𝟒𝐕 1/3
)
𝜋𝑥
By substituting the value of X and D in the total area equation, a new equation will be formed, that
relate the total area with the volume and H/D ratio. (Detailed in the appendix)
𝟒𝐕
𝟏
𝐇
At = π ( 𝛑 (𝐇/𝐃) )2/3 + ( 2 + 𝐃 )
Data has been calculated by adding this function in Microsoft Excel, and Liter has been converted
to m3 by dividing its value with 1000:
F(x) = (PI()*POWER(4*C2/(PI()*B3*1000),2/3)*(0.5+B3))*0.3
(As shown in table 1 and figure 2, detailed derivation in the appendix)
2
Table 1: Cans results.
Volume in Liter
Height / Diameter
0.25
0.0083037
0.0069747
0.0066534
0.0065907
0.0066263
0.0067062
0.0068076
0.0069198
0.0070369
0.0071559
0.0072750
0.0073931
0.0075095
0.0076241
0.0077364
0.0078465
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
0.50
0.0131813
0.0110716
0.0105615
0.0104620
0.0105186
0.0106454
0.0108064
0.0109844
0.0111704
0.0113593
0.0115483
0.0117358
0.0119207
0.0121024
0.0122808
0.0124556
0.75
0.0172724
0.0145079
0.0138395
0.0137091
0.0137832
0.0139494
0.0141604
0.0143937
0.0146374
0.0148849
0.0151326
0.0153782
0.0156205
0.0158587
0.0160924
0.0163214
1.00
0.0209241
0.0175751
0.0167654
0.0166074
0.0166972
0.0168985
0.0171541
0.0174367
0.0177319
0.0180318
0.0183318
0.0186294
0.0189229
0.0192114
0.0194945
0.0197720
1.25
0.0242802
0.0203941
0.0194545
0.0192712
0.0193753
0.0196089
0.0199056
0.0202335
0.0205761
0.0209241
0.0212722
0.0216175
0.0219580
0.0222929
0.0226214
0.0229434
1.50
0.0274183
0.0230299
0.0219689
0.0217619
0.0218795
0.0221432
0.0224783
0.0228486
0.0232354
0.0236283
0.0240215
0.0244114
0.0247960
0.0251741
0.0255451
0.0259086
2.50
3.00
3.50
V=1.25L
V=1.50L
1.75
0.0303858
0.0255225
0.0243466
0.0241173
0.0242476
0.0245399
0.0249111
0.0253215
0.0257502
0.0261857
0.0266214
0.0270535
0.0274797
0.0278987
0.0283099
0.0287128
2.00
0.0332149
0.0278987
0.0266134
0.0263627
0.0265051
0.0268246
0.0272305
0.0276791
0.0281477
0.0286237
0.0291000
0.0295723
0.0300382
0.0304962
0.0309456
0.0313861
Cost vs H/D
0.035
0.030
Cost (KD)
0.025
0.020
0.015
0.010
0.005
0.000
0.00
V=0.25L
0.50
V=0.5L
1.00
1.50
V=0.75L
2.00
V=1.00L
H/D
V=1.75L
4.00
4.50
V=2.00L
Figure 2: cost vs H/D chart.
As shown in the chart above, the best H/D ratio that gives the minimum cost occurs when H/D=1
and V= 0.25L.
3
Three cans were brought from the supermarket with different H/D ratios and volume, in order to
compare the real results with the calculated results, the cans are ordered from the right to the left,
as shown in figure 3.
Figure 3: Three different cans.
Results and Discussion:
As shown in table 2, Can 3 has the lowest cost which equals 0.100 KD. Its volume equals
0.0001668 L and H/D ratio equals 0.01700, as shown in table 2. In table 1, the lowest cost was
0.0065907 KD with 1 H/D and 0.25 L. A detailed cans is shown in figure 4, to compare between
the three cans easily. According to the calculations and actual results, the cans were sold with an
excessive cost in comparison to the calculated values.
Table 2: Real cans measurements.
Voulme (L)
Can1 - Pepsi 0.0002676
Can2 - Lipton 0.0002519
Can3 - 7 up
0.0001668
H/D
0.02568
0.02037
0.01700
Cost (KD)
0.150
0.350
0.100
actual cans details
100%
90%
80%
70%
60%
50%
40%
30%
20%
10%
0%
Voulme (L)
Can1 - Pepsi
H/D
Can2 - Lipton
Cost (KD)
Can3 - 7 up
Figure 4: Actual cans details.
4
Conclusion:
In conclusion, it is very important to measure the dimensions and determine the cost of the cans
before entering the manufacturing process, in order to save money and provide the best results to
the consumers. From the previous calculations, we can detect that there is a huge difference
between the manufacturing price and the market price. Professional factories must avoid deceiving
the customer because of their urgent need and they should really care about the customer service
in all the fields. In addition, table 1 shows that the minimum cost equals 0.0065907 KD and the
highest cost equals 0.0332149 KD.
5
Acknowledgments
I would like to thank Dr. Abdullah Alajmi for extending the deadline of the report, for his
patience and kindness with us, and for helping me and answering my questions on time.
6
Appendix:
Appendix 1: Derivation equations and detailed calculations
𝜋
(1)
At = ( ×D2) + (πD×H)
(2)
V = Ac × H
(3)
𝜋
(4)
At = 2 × (4 ×D2) + (πD×H)
𝜋
2
V= ( ×D2) × H
4
𝐇
Let, 𝑥 = 𝐃
(5)
So, H = 𝑥×D
(6)
𝜋
(7)
𝜋
(8)
V= 4 + D2 (𝑥D)
V= 4 + D3 𝑥
D = √𝜋𝑥
4𝑉
(9)
D = ( 𝜋𝑥 )1/3
𝟒𝐕
(10)
𝜋
(11)
𝜋
(12)
𝜋
(13)
3
At = 2 D2 + πDH
At = 2 D2 + πD (𝑥D)
At = D2 + πD2 (𝑥)
2
1
At = πD2 ( 2 + 𝑥 )
𝟒𝐕
(14)
𝟏
At = π ( 𝛑 (𝑥) )2/3 + ( 2 + 𝑥 )
𝟒𝐕
𝟏
𝐇
At = π ( 𝛑 (𝐇/𝐃) )2/3 + ( 2 + 𝐃 )
(15)
(16)
7