Progressive Delivery Required in Mechanical Engineering
Module 1 Activity 1 – Failure Distribution Curve
- Introduction
- Purpose
This report proposes depicts the testing and examination of paper clasps to decide the weakness disappointment of various materials, for example, metals or plastics.
- Problem
The investigation of the quality of materials is critical in our everyday lives. This is on the grounds that the information on this can spare us from perilous outcomes in an occasion a few materials should break because of weakness. Consequently, this analysis will fill in as a decent reference for both plastic and metal paper cuts.
1.3 Scope
This examination would not cover all the metal, clay or polymer materials. It is constrained to the investigation of metal and plastic paper cuts.[unique_solution]
- Background
2.1 Theory
Researcher investigates materials, for example, metals, pottery or polymers and designers to uncover their mechanical properties to figure out what utilize the materials may have. A case of a mechanical property that is tried for is the measure of pressure a material can deal with before it breaks. The pressure a material can take before it breaks quantifies the quality of the material. Additionally, as a material gets more established, it can deal with less pressure, which can make it come up short at a lot of lower stresses.
2.2 Research
Weariness is an exceptionally regular method of disappointment for materials and has been read for quite a long time. Weariness happens each day in objects that you know about. A model, plane wings exhaustion a huge number of cycles on each flight and extensions weakness each time a vehicle rolls over them. In any case, in light of the fact that a material is experiencing weakness doesn’t imply that it will consistently break. Specialists run cautious examinations so they can be certain that things won’t break because of weakness while you are utilizing them.
- Tests and Evaluation
- Apparatus
As gave in the task sheet, the contraption that was are:
- Small metal paper cuts
- Small plastic covered paper cuts
- Metric ruler Protractor
Figure 11. Paper Clip
- Procedure
Additionally gave in the task sheet, the technique is as per the following:
- Using your hands, open up the internal circle of the paperclip so it makes 180-degree edge with the external circle.
- Bend the paperclip back to its unique position. This considers one stacking cycle.
- Repeat stages 1 and 2 until the paperclip breaks. Record the quantity of stacking cycles that pass. During the genuine examination, record all perceptions (i.e., changes in surface completion, shading, and so on.).
- After finishing the investigation for 180-degree point, make a speculation underneath:
- Repeat stages 1 through 3 for 45, 90, and 270-degree points.
- Record the accompanying for every paperclip in Table A.
- Look at the break surface and record any perceptions in the Table B.
- Plot your outcomes on the diagrams demonstrated as follows.
- Repeat whole investigation for a paperclip of an alternate size or material, (for example, plastic).
The tables below show the numerical results of the experiment.
Table A
Type of Paper Clip | Angle of Rotation | Cycles to Failure |
Plastic Clip | 270 | 4 |
Plastic Clip | 180 | 9 |
Plastic Clip | 90 | 20 |
Plastic Clip | 45 | 89 |
Metal Clip | 270 | 2 |
Metal Clip | 180 | 5 |
Metal Clip | 90 | 22 |
Metal Clip | 45 | 38 |
Table B
Type of Paper Clip | Description of Failure |
Metal Clip | Appeared twisted before it broke |
Plastic Clip | Turned white then broke |
4.2 Interpretation
From the information gave over, the plastic paper cut had more quality than the metal paper cut since it took more pivots before it broke. A potential clarification for this could be because of the way that metals shouldn’t be bowed; their principle capacity will be to convey more weight. Instead of the plastic paper cut that is increasingly adaptable.
5. Conclusion
5.1 Assessment
There was a direct connection between the point of pivot and the cycles to disappointment in the two materials of paper cuts. As the edge of revolution diminished, the quantity of cycles expanded for the two materials. It was likewise observed that metal paper clasps could not take an excessive number of cycles before disappointment because of its unbending nature.
5.2 Recommendation
More materials ought to likewise be considered. This will realize greater assorted variety in the experimentation. Likewise, it is astute to have more points of revolution to give progressively precise outcomes.
Module 1 Activity 2 – Identifying Failure Sources
Failure Distribution Curve
The underlying method
- Utilize your thumb and pointer finger of each hand and draw the two end circles of the clasp separated quite far so the middle curve fixes up.
- Curve the clasp to and fro from completely open through to contacting the fingers.
- While checking the quantity of cycles rehash the open-close cycle until the wire breaks. Record the cycle check.
Break 20 paper clasps and diagram the outcomes on a dissipate plot.
Second strategy for improving the unwavering quality of the paper cuts.
- To Open the Paper Clip, one hand will hold the clasp by the more drawn out, external circle.
- With the thumb and pointer of the other hand, handle the littler, inward circle.
- Force the littler, inward circle out and down 90 degrees with the goal that a correct edge is shaped
Drive the littler inward circle up and in 90 degrees so the littler circle is come back to the first upstanding situation in accordance with the bigger, external circle this finishes one cycle.
Rehash until the paper cut breaks and Count and record the cycles-to-disappointment for each clasp.
Rehash the above method utilizing the other 19 paper clasps and record the cycles to disappointment for each clasp and graph the outcomes on a disperse plot.
(C ) Third procedure for improving the reliability of the paper clips.
- To Open the Paper Clip, one hand will hold the clasp by the more drawn out, external circle.
- With the thumb and pointer of the other hand, handle the littler, inward circle.
- Force the littler, inward circle out and down 45 degrees with the goal that a correct edge is shaped.
- Drive the littler internal circle up and in 90 degrees so the littler circle is come back to the first upstanding situation in accordance with the bigger, external circle this finishes one cycle.
- Rehash until the paper cut breaks and Count and record the cycles-to-disappointment for each clasp.
- Rehash the above strategy utilizing the other 19 paper clasps and record the cycles to disappointment for each clasp and outline the outcomes on a disperse plot.
Module 1 Activity 3 – Cumulative Failure and Reliability Plots
DATA OBTAINED.
Cycles-to-failure from the first procedure. 4,4, 5,5, 5, 5.5,5.5,5,5.5, 6, 6,6.5,6,6.5,6.5,6,4,4.5,6.5,5. cycles.
Cycles-to-failure when the paper clips are placed at 90°: 16,17 17,16, 18, 21, 22, 23, 16,23,22,21,18, 16,17,19,23,21,22,18. cycles.
Cycles-to-failure when the paper clips are placed at 45°: 58, 59,61,63, 65,67,69, 72,73, 78,81,83,85,85, 86 ,63,65,59,72,65. cycles.
Observations and analysis.
It is seen that the when the paper cuts is at 180 degrees the quantity of cycles to disappointment are least. Decreasing of edge degrees is an increasing speed factor for the disappointment of the paper cuts.
Module 1 Activity 5 – Separate Failure Mode Distribution Plot
Module 1 Activity 4 – Failure Distribution and Weibull Plot
Module 1 Activity 6 – Pump Reliability Improvement Business Case
Objective:
The target of this venture was to plan a framework to cool tomatoes to save freshness preceding being put in refrigeration trucks for delivery. The plan will comprise of a hydro cooler that will successfully cool water that is being recycled to cool tomatoes on a transport line. The framework portrayed in the report will successfully cool the tomatoes to the right temperature, ration water, and keep up a spotless cooling condition.
Hypothesis:
Hydro cooling is a procedure of easing back the aging of produces in the wake of gathering by submerging in chilly water. For cooling the water, a hydro cooler comprises of a fume pressure refrigeration framework. The framework comprise of four parts: a blower, condenser, development valve, and evaporator. The blower begins the refrigeration cycle; it takes in low-pressure, low-temperature refrigerant in gas structure and packs it into a high-pressure, high-temperature gas. The condenser takes in the gas from the blower and water streams over the curls to expel heat from the refrigerant. As the refrigerant loses heat, it will start to gather until the entirety of the gas has consolidated into a fluid. The development esteem takes in fluid from the condenser and limits the progression of the refrigerant. At the point when the high-pressure fluid experiences the development valve it enters the evaporator. The evaporator makes the refrigerant begin to dissipate once again into gas which makes the refrigerant become freezing and assimilate a great deal of warmth. This is the place water collaborates to basically chill itself. The warmth from the water is moved to the refrigerant and the procedure begins once again. The length of the procedure relies upon the size of the produce, and the time can run from minutes to hours. With the utilization of a hydrocooler, it is basic to keep up clean water to anticipate sullying so the produce doesn’t get acquainted with microscopic organisms.
For deciding hydrocooling rates for some foods grown from the ground, numerous investigations have been led so as to ascertain to what extent it would take to cool certain produce to an ideal temperature. For this to be resolved, a Decimal Temperature Difference (DTD) and estimated size of the produce is required. To decide DTD, the accompanying condition is utilized:
Eq. 1
where T is the objective temperature, W is temperature of the water, and P is the beginning temperature of the produce. For deciding the surmised size of the produce, the accompanying table is utilized and is isolated into seven classifications relying upon size:
Table 1: Size chart for cooling vegetables/fruits
A – Greens |
B – Beans, peas, asparagus |
C – Small cucumbers, radishes, beets (<1.5” dia.) |
D – Small apples and peaches, slicing cucumbers |
E – Sweet corn, apples, and peaches |
F – Large apples and peaches (>3” dia.) |
G – Cantaloupes, large eggplants |
With the DTD and size category determined, the following graph is used to determine the active minutes needed to effectively cool the produce:
Figure 1: Graph for determining cooling time.
Strength (HP) is a unit of estimation of intensity, or the rate at which work is finished. For some engines, the premise of yield is drive. For this structure, the power yield is to be communicated in kilo-watts (kW) and the change from HP to kW is communicated as the accompanying condition:
Eq. 2
Essentially, to figure the measure of intensity in kW while given working Voltage (V) and working Amperage (An), Eq. 3 is utilized:
Eq. 3
With control in kW, it is conceivable to compute cost based off the activity time. Accepting the expense of activity is $0.10 per kilo-watt-hour, the all-out working expense can be communicated as pursues:
Eq. 4
where T is the allocated time of activity in hours (hr).
Design:
Figure 2: The schematic of the tomato cooling system.
The framework comprises of a transport line that will convey tomatoes driven by an engine. As the tomatoes are gradually moved, they are splashed with cool water. This water gathers in a catch container situated under the produce and transport line. Since the water is presently tainted, it is sent through a siphon which at that point experiences a sand channel to expel earth and disinfector to guarantee drinking quality. The perfect water is then gathered in a spotless water supply. From the spotless water store, the water is siphoned to the chiller, which cools the water to the necessary temperature, and is then sent to the sprayers to chill the tomatoes. The procedure rehashes as important.
Siphons:
In a building based framework at times a siphon is found in its schematic. Siphons are a key factor in frameworks where fluid or gas substances are being moved starting with one area then onto the next to enter the following stage in the framework. A siphon is basically the engine that produces pressure in hoses or channels to drive water or another substance to the following area in a framework. Notwithstanding the earlier explanation, siphons are required between each huge part in a framework. For this particular tomato cooling framework two siphons would be required.
One of the siphons would push water from the catch skillet under the transport to the repository. This siphon would be moderately little since it would not be moving a lot of water. It would just make a most extreme 55 pounds for each square inch (psi) of weight and move close to 3 gallons of water for every moment (GPM). The particular siphon that was wanted for utilize was found to draw 12 Volts (V) and 7.5 Amperes (A). Utilizing these qualities for the siphon and Equation 3 a worth was determined for the power required to run the siphon to be 0.09kW.
A subsequent siphon would need to be actualized to move the water back to the transport once it endured the store. However, this siphon would have the option to deliver a weight somewhere in the range of 500 and 1000 psi to create the fog at the spouts that is looked for after. It would likewise should be able to do going at any rate 5 GPM. This siphon is enlisted at 6.0 HP, so changing over to kW yields an estimation of 4.4742kW required to run this siphon.
Transport line:
In industry transport lines are exceptionally valuable and effective. They are utilized to move an article from direct A toward point B, or for this situation used to gradually ship tomatoes from one side of the spout filled encased passage to the following. More excellent transport frameworks will have movable belt speed settings.
A transport line explicit to a tomato cooling framework would need to move gradually enough for the tomatoes to produce the full results of the cooling water. Utilizing a 15 foot long transport that is 4 foot wide and separating the length by the necessary minutes of dynamic cooling to drop the temperature of the tomato from 75° F to 40° F yields a necessary speed of 0.32 feet every moment (ft/min) for the belt. A particular transport engine was found to utilize 1HP or 0.7457kW. This engine likewise determined working at 3,450 cycles for each moment (RPM). What’s more, the utilization of a gearbox would help with accomplishing the necessary precise speed of 0.32ft/min.
Spouts:
Cooling tomatoes utilizing new cool water is perfect when one considers keeping tomatoes new. To achieve this said freshness the tomato needs to suffer showering of cold water fog by means of hoses and spouts. Indeed, for this particular structure 5 hoses for each foot of width that run the full 15 foot length of the belt and 8 spouts each hose each 2 feet of length. A particular spout was found to yield 0.028GMP. Since 5 hoses would be utilized with 8 spouts on every one, increasing the yield estimation of 0.028 by a complete 40 spouts yields an all-out stream pace of 1.12GPM.
Chiller:
Another key factor to a tomato cooling framework is the cooling framework alone. For water to be sufficiently cold to cool the naturally picked warm tomatoes a water-cooler chiller would be actualized between the filtration framework and the repository to guarantee all water heading out to the supply is at the necessary temperature. This chiller is another fundamental machine that is controlled by power. In particular, a chiller was seen as
Module 1 Activity 7 – Conveyor Failures Timeline
Time Interval (h)
| Number of Observed Failures
|
0–2 | 6 |
2–6 | 12 |
6–10 | 7 |
10–15 | 6 |
15–25 | 7 |
25–100 | 2 |
λ = n / T = 40/400 = 0.1/hr
P(0<t<2) = e-λt1 – e-λt2 = e-λ*0 – e-λ*2 = 1 – 0.08187 = 0.1813
Calculating Expected Failures freuquency :
Fexp(0<t<2) = N * P(0<t<2) = 40 * 0.1813 = 7.252
Similarly doing the calculation to all the values
After that χ2 statistic = (Observed failure – Expected failure)2/ Expected failue
Time Interval (h)
| Number of Observed Failures | P(t1 < t < t2) | Expected Failure, Fexp | χ2 Stat |
0–2 | 6 | 0.1813 | 7.252 | 0.2162 |
2–6 | 12 | 0.2699 | 10.796 | 0.1342 |
6–10 | 7 | 0.1809 | 7.237 | 0.0078 |
10–15 | 6 | 0.1447 | 5.7899 | 0.0076 |
15–25 | 7 | 0.1410 | 5.641 | 0.3270 |
25–100 | 2 | 0.08203 | 3.2815 | 0.5 |
Total χ2 Stat, W = 1.197
Rejection region, R = χ(1-α) 2 (k-1-m)
Where k = number of data intervals = 6
m = number of parameters of null hypothesis distribution estimated from the data
α = level of significance.
R = χ(0.95)2 (6 – 1 – 1) = 2.3719 * 4 = 9.4877
As W < R –
do not reject the null hypothesis at 95% confidence level implying the above problem follows exponential distribution.
The following time-to-failure data with the ranked value of ti. Test the hypothesis that the data fit a normal distribution. (Use the Kolmogorov test for this purpose.)
Event | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Time to Failure (h) | 10.3 | 12.4 | 13.7 | 13.9 | 14.1 | 14.2 | 14.4 | 15.0 | 15.9 | 16.1 |
Mean of the sample data,
= / 10
= 140/10
= 14
Calculating Variance of the sample data,
S2 =
Calculating Z =
S – sample standard deviation.
Sn(ti) = i/n
Where n =10
Event (i) | Time to failure (ti) | Sn(ti)
| Sn(ti-1) | Z = | F(ti) From table | |F(ti) -Sn(ti)|
| |F(ti) -Sn(ti-1)|
|
1 | 10.3 | 0.1 | 0.0 | -2.194 | 0.0142 | 0.08574 | 0.0142 |
2 | 12.4 | 0.2 | 0.1 | -0.9491 | 0.1713 | 0.0287 | 0.0713 |
3 | 13.7 | 0.3 | 0.2 | -0.1779 | 0.4294 | 0.1294 | 0.2294 |
4 | 13.9 | 0.4 | 0.3 | -0.0593 | 0.4764 | 0.0764 | 0.1764 |
5 | 14.1 | 0.5 | 0.4 | 0.0593 | 0.5236 | 0.0236 | 0.1236 |
6 | 14.2 | 0.6 | 0.5 | 0.1186 | 0.5472 | 0.0528 | 0.0472 |
7 | 14.4 | 0.7 | 0.6 | 0.2373 | 0.5938 | 0.1062 | 0.0062 |
8 | 15.0 | 0.8 | 0.7 | 0.5932 | 0.7235 | 0.0765 | 0.0235 |
9 | 15.9 | 0.9 | 0.8 | 1.1270 | 0.8701 | 0.0299 | 0.0701 |
10 | 16.1 | 1.0 | 0.9 | 1.2472 | 0.8936 | 0.1064 | 0.0064 |
Module 1 Activity 8 – Conveyor Reliability
From the above table we have observed the max K-S statistic from the columns
[|F(ti) -Sn(ti)|and |F(ti) -Sn(ti-1)|]
The max K-S statistic is 0.2294.
Now we must find the rejection region R at 5% significance level.
Dn(α) = 0.409 from tables of Critical value for the Kolmogonov-Smirnov statistic
Rejection region, R ≥ Dn(α) is defined.
From the values : K-S statistic value < Dn(α)
Therefore we accept the null hypothesis i.e, the above model follows normal distribution at 95% confidence level.
- Mean of time to failure : = 235
Standard deviation of the time to failure =
Z = (X- )/ σ
P(T<200) = 0.05
Z = (200-235)/σ
But P(Z) = 0.05 => Z = -1.64
There σ = 35/1.64
= 21.28
Module 1 Activity 9 – Risk Ranking
- P(L<t) = 0.01
- For 0.01 the value of Z = -2.326
Using Z = (X- )/ σ
-2.326 = (L0.01 – 235) / 21.28
L0.01 = 185.5hrs
- P(T>250 | T>200) = =
- As this model follows normal distribution
- P(T<250) = P(Z<(250-235)/21.28) = P(Z<0.704) = 0.75804
- P(T<200) = P(Z<(200-235)/21.28) = P(Z<-1.644) = 0.505
- P(T>250 | T>200) =
µy = 12000, σy = 2000
µt = ln = 9.379
σt = ln = 0.1655
P(B>12000)
First we have find Z values for which from the tables we get P(B<12000).
Z = = = 0.0827
P(Z=0.0827) = 0.532
P(B>12000) = 1 – P(B<12000) = 1 – 0.532 = 0.468
P(L=12000 | L>10000)
=
P(L > 10000)
Following above procedure
Z = = = -1.0187
P(Z) = 0.1538
P(L>10000) = 1 – P(L<10000) = 1 – 0.1538 = 0.846
P(L=12000 | L>10000) = 0.468 * 1/ 0.846 = 0.55324