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Engineering
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Topic:

Compute And Discuss 5 Engineering Tasks And Show My Workings (Coursework Sample)

Instructions:

I Was Required To Compute And Discuss 5 Engineering Tasks And Show My Workings.

source..
Content:

Course Work - Problem Based Tasks
Name
Professor
Course
5/6/18
Task 1
* Write-up for the Liquid Pressure Experiment
Experiment: Investigating how Water Pressure Relates with Depth, ("Investigating the pressure of a water column")
Aim
The aim of this experiment is to determine the relationship that pressure in liquid has with depth. The experiment will help in demonstrating how the pressure of water behave in varied depths, ("Investigating the pressure of a water column").
Apparatus
1 A large plastic jar with 1L capacity
2 A hammer
3 A nail
4 A 30cm ruler
5 3 dot rubber stoppers
Procedure
1 Using the ruler, measure a distance of 1In from the bottom of the container and punch a hole using the nail and the hammer.
2 Add to 2 more holes in side, but the same side of the jar with an interval of an inch.
3 Using the dot rubber stickers, close the holes punched in side of the jar.
4 Fill the jar with water then remove the rubber stoppers to allow water to flow.
5 Using a ruler, take measurements of the distance of water that squirts from each of the three holes.
6 Using the data, plot a graph of distance of water level (depth) against the distance of water that squirts from each hole of the jar.
Results
It is observed that the hole that is near the bottom of the jar squirts water to a far distance followed by the second hole from the bottom and lastly the top most hole squirts water to the shortest distance from the jar, ("Investigating the pressure of a water column").
* Reading Tabulation
Depth (in meters)

Distance of Water Squirt from the Can (in meters)

Pressure

0.01m

0.01m

0.098Pa

0.02m

0.02m

0.196Pa

0.03m

0.03m

0.294Pa

p=gxh
Where p= pressure of water
g= the acceleration of water due to gravity
h=the depth of water
P1 = (9.8m/s2x0.01m)=0.098Pa
P2 = (9.8m/s2x0.02m)=0.196Pa
P3 = (9.8m/s2x0.03m)=0.294Pa
* Hand Drawn Graph
Task 2
* Determining speed of water coming out of a tap
If water flow, the principle of mass conversion applies. Therefore, if a water flows from a tap, there is a steady mass of water that comes out of the tap, so it is important to note that the volume or mass of water that flowed into the tank is equal to the same volume of water that has flowed out of the tank through a tap, ("Continuity Equation").
Source: ("Continuity Equation")
Therefore, the speed with which water flowed out of the tap is
Rate= (pAV)
* The aim of this experiment is to verify the continuity equation using through Bernoulli’s effect, ("Bernoulli Experiment Lab Report," 2016). In the experiment, a glass tube having different cross sectional areas is used and it is fitted with vertical tubes. By filling the glass tube with water, I fitted the vertical tubes on the glass tube and filled them with water to the same level. After opening an outlet at one end, I fixed an inlet that over the same amount of water as the amount that was flowing out.
In the experiment, I was able to identify the level of water in the tubes and recorded every slight changed. The level of P1 and P3 were the same while that of P2 was lower since the pressure high due to low velocity. Of P1 and P3, the velocity was low that is why the pressure was high, ("Bernoulli Experiment Lab Report," 2016).
Task 3
* Electric Appliances and Operating Voltage
Electric Appliance

Operating Voltage

Electric Power Consumption

* Blender

240V

350W

* Microwave Oven

230V

700W

* Electric Iron Box

220V

1450W

* Electric Lamp

220V

18W

* Calculate Resistance
In Oman, the electricity supply voltage is 240V while the standard frequency is 50Hz, ("Travel Adaptor for Oman").
According to Ohm’s Law,
Current= Power/Voltage
But Resistance= Voltage/Current
Electric Appliance

Current=(power/voltage)

* Blender

(350/240)= 1.4583A

* Microwave Oven

(700/230)= 3.0435A

* Electric Iron Box

(1450/220)= 6.5909A

* Electric Lamp

(18/220)= 0.0818A

Therefore, Resistance=Voltage/Current
Electric Appliance

Resistance=(Voltage/Current)

* Blender

(240/1.4583)= 164.5752Ω

* Microwave Oven

(240/3.0435)= 78.8566Ω

* Electric Iron Box

(240/6.5909)= 36.4138Ω

* Electric Lamp

(240/0.0818)= 2933.9853Ω

* Calculate the total resistance of the appliance and current passing through the network


Resistance

Appliance 1:

Blender

164.5752Ω

Appliance 2:

Microwave Oven

78.8566Ω

Appliance 3:

Electric Iron Box

36.4138Ω

Appliance 4:

Electric Lamp

2933.9853Ω

For a parallel resistance
1/Rt=1/R1+1/R2+…
For a Series Resistance
RT=R1+R2+…
Therefore, we are going to compute resistance for appliance 1 and 2, then calculate that of appliance 3 and 4. After getting the resistance we will use the formula for parallel resistance.
So,
RT1= 164.5752Ω + 78.8566Ω = 243.4318Ω
RT2=2933.9853 Ω +36.4138 Ω = 2970.3991Ω
Therefore, computing the Parallel resistance
1/R= (1/243.4318) + (1/2970.3991)= 0.007973
But now,
R=0.00444-1
Resistance=224.993Ω
* Critical Comment on the Need for Parallel Domestic Wiring System
Parallel domestic circuit system is the standard domestic wiring system that should be used at home due to its advantages and convenience, ("GCSE Physics: Parallel Circuits"). In this case, the parallel circuits has it components connected in a parallel composition to form a circuit that distributes electric power equally to every appliance. This system has four advantages compared with series. In parallel system, there is advantage of independent components of the circuit, there is consistent voltage throughout the circuit, it is easy to add other appliances or rather additional components to the circuit, and it is simple, safe and reliable circuit system to use at home, ("GCSE Physics: Parallel Circuits").
The reliability of the parallel system is that the electric contractor will easily fix the system in your home and the system will not only be safe, it will be simple to control. Therefore, proper electric writing for the parallel system will make it easier for you to add any component compatible with the system without changing the voltage. This is because in parallel system, additional components does not lead to increase in resistance since the more components are added to the system, the less the resistance. Using the parallel circuit system is safe especially while using with component s such as air conditioners, ("GCSE Physics: Parallel Circuits"). Most of the domestic appliance need approximately 110 Volts of electricity to operate. Therefore, in a parallel circuit system, all appliances get the same operating voltage as the source according to standard operating voltage of each component. In this case, it is possible for each componen...
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