Course - Higher Nationals in Engineering
Unit 21 in Higher Nationals Engineering dives into the world of electric machines, the workhorses of our technological age. From generators that create electricity to motors that power our appliances, you'll explore how these devices convert energy in various ways. The course delves into transformers, motors, generators, along with transducers and actuators, analyzing their construction, operation, and applications in different fields. By understanding their characteristics and operating parameters, you'll gain a solid foundation in the machines that shape our electrical world.
Table of Contents
- Question 1
- Question 1 i) Transformers
- Step-up transformers
- Step-down transformers
- Isolation 3
- Question 1 ii) Losses and efficiency of transformer
- Question 1 iii a) Calculating efficiency
- Transformer A: Single PHASE
- Transformer B: Single PHASE
- Transformer C: THREE PHASE
- Transformer D: THREE PHASE
- Question 1 iii b) Selection of the transformer
- Question 2
- Question 2 i) Selecting best motor design for lift
- Question 2 ii) Methods of starting induction and synchronous motors
- Question 2 iii) Evaluating efficiency
- Question 3
- Question 3 a) Types of generators
- DC generators
- Induction Generators
- Synchronous Generators
- Question 3 b) Identifying a generator for an application
- Question 3 c) Accessing generators available
- References
Unit 21 Electric Machines
Question 1
LO1 Assess the constructional features and applications of transformers
i) Examine shell and core type transformer options with the customer. Discuss the application for each type of transformer.
Transformers
Step-up transformers
Physical Construction
Step-up transformers are used in raising the voltage at the output. The primary voltage is raised by a certain value depending on the transformer ratio to a higher secondary voltage output. The construction is based on increasing the number of secondary windings on the secondary core but ensuring that the primary winding has a lower ratio to the secondary winding.
This increases the voltage but at the same time lowers the current. The transformers are used by the transmission grid, this is when the power needs to be transmitted over long distances (Abbasi, 2018). We need to ensure that the voltage is maintained and losses do not affect the voltage reaching the anticipated regions.
Application
Step-up transformers are mainly used in the distribution lines. They increase the voltage in the transmission line to help in balancing the losses that might be incurred during the transmission.
Starting machines. Some machines that needs to be started requires some external starting device or a transformer to help in powering the system. To power up the system transformers rise the voltage hence the machines gets started.
Step-down transformers
Construction
Step-up transformers are design by ensuring that the primary windings are more compared to the secondary winding on the core. This allows the current to get raised easily hence there will be a drop in voltage. The ratio of primary to the secondary can be in a ratio of 2: 1 which implies that there will be a loss of almost halfway depending on the core and copper losses.
Application
Step down transformers are used in homesteads in charging systems like the phone chargers to prevent excess voltage from destroying the device.
Other applications include in the welding machines. The welding machines requires to have high current hence we reduce the voltage to increase the current. In that way the welding machines are able to heat a surface.
Isolation
Physical Construction
The construction is based on ensuring high insulation on the windings on both the primary and the secondary core. Once the current is passed over these transformers it is filtered and only the AC is allowed to pass. Capacitive coupling is used in ensuring that the filter is effective.
Application
They are used as pulse transformers for filtering the AC and DC currents by ensuring that only the AC current is collected as the output.
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Question 1 ii) Illustrate to the customer the various windings, connections. Show on an equivalent circuit the iron and copper losses.
ii) Losses and efficiency of transformer
Losses on a transformer can be as a result of many things like the type of wire used in the windings, the core type and other things. The material used is under the copper losses on the transformers. In terms of the core, there are many ways that the core can be designed like the circular core and the standard rectangular core. The core losses might vary depending on the type used.
Eddy currents mainly are the currents which alter the magnetic field during the generation. As the current increases, there could be the currents that alter the direction and the magnetic file. These are referred to the eddy currents.
Question 1 iii a) Calculate the full load efficiency of all four transformers, A, B, C and D both single and three-phase
a) Calculating efficiency
Transformer A: Single PHASE
The transformer output power will = transformer rating x power factor
50000 X0.85=42500 .
Efficiency at full load = output/(input+ open circuit+ short circuit)
=42500/((42500+300+600) )
0.9793*100
97.93%
Transformer B: Single PHASE
The transformer output power will = transformer rating x power factor
80000 X 0.95= 76000
Efficiency at full load = output/(input+ open circuit+ short circuit)
=76000/(76000+500+250)
=0.9902
=99.02%
Transformer C: THREE PHASE
The transformer output power will = transformer rating x power factor
=350 000 X 0.78=273000 .
Efficiency at full load = output/(input+ open circuit+ short circuit)
273000/(273000+75+50)
0.99954
=99.95%
Transformer D: THREE PHASE
The transformer output power will = transformer rating x power factor
=450 000 X 0.87=391 500 .
Efficiency at full load = output/(input+ open circuit+ short circuit)
391500/(391500+400+300)
0.9982
=99.82%
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Question 1 iii b) Assess and select one single phase and one three phase transformer to be used in the distribution system for the hotel. Select the most appropriate transformer for this application.
b) Selection of the transformer
In The section of the transformer we consider the loading power and the efficiency of the transformer. In the description we have been given that the anticipated load will be 300 KVA. From here we will select a value of transformer output which is near or corresponds to the anticipated load but maintain the efficiency.
For the single phase transformer we make a selection of TRANSFORMER B- Reason for the selection is the high efficiency which is 99%.
For the Three phase transformer we select the TRANSFORMER C- Reason is due to the high power efficiency of 99.95.
Question 2
LO2 Analyse the starting methods and applications of the three-phase induction motors and synchronous machines
Question 2 i) Select the best type of motor design to operate the service lift and justify your answer.
i) Selecting best motor design for lift
Lift motors selection is mainly based on the horse power of the motor. Most motors are designed with specifications of the load they can lift and the height they can support the weight at.
Since we have been given the height, it implies that the force that will be needed to hold the mass or load at that height will be Load * acceleration due to gravity which is 9.8.. Hence the force will be Load * 9.8
The force = Load * 9.8
Then to get the power required to hold the mass over the specified height will be
Load* 9.8 * height
= load * 9.8 * 2200m
Synchronous motor are powerful and have the ability to hold a mass at a constant speed. Despite adding the load at the input of the motor, the motor will continuously work despite the change in mass.
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Question 2 ii) Analyse the different methods of starting induction motors and synchronous machines and suggest the best type of starter for the service lift motor.
ii) Methods of starting induction and synchronous motors
Using DC machines coupled: Small dc machines can be coupled to the synchronous motors in order to induce the voltage and start the motor. This is the common way of starting synchronous machines
Dumper windings: The short-circuiting a motor is used normally as an additional method in starting the motors. Once there is a delay with the DC machines we can perform short circuiting which is way more effective method of starting the motor. Here additional windings on the normal windings are added to short-circuit the system until it starts.
Question 2 Critically evaluate the efficiency of the motors that your company has in stock and make a recommendation for a new motor.
iii) Evaluating efficiency
Comparing the original motor to the others, I would recommend the selection of the first option motor. This is due to a number of reasons like the additional protection based on thermal. The original motor does not have thermal protection and these could be an additional layer to the motor to protect the individuals and the motor itself from the thermal changes on the motor.
The other factor is due to the high Horse-power, considering that we are running a high altitude, more power is required to raise the load to that height.
Question 3
LO3 Investigate the types of generators available in the industry by assessing their practical application
Question 3 Explain to them the types of generators (DC, induction and synchronous) and their construction.
a) Types of generators
DC generators
Just like any other generator, DC generators are used in converting the mechanical energy in electrical energy. It is mainly composed of the Rotor, Stator, commutator and the Armaturewindings (Zhang, 2017).
It operates in consideration to the faraday's law where we understand that when a current carrying conductor is kept on a magnetic field electromotive force is induced on the conductor. This emf is the voltage.
They are mainly used in starting other motors like the synchronous motors.
Induction Generators
This generator produces the power when the rotor is run faster compared to the synchronous speed. If these happens electrical power is produced in turn and can be supplied. The synchronous generator speed runs at a constant speed and the scenario comes when the speed is in excess in that way the electrical power is supplied (Hu, 2018).
Synchronous Generators
The synchronous generator runs at a constant speed called the synchronous speed and is used on converting the mechanical energy in electrical energy. The major components include the rotor, stator and the armature windings.
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Question 3 Show your staff how to identify a generator for a specific application based on the available generators.
b) Identifying a generator for an application
The provided information is that the company is running 2 welders and the six drillers. This implies that the power consumed is high and if all run at the same time we will have the power consumption in the company being very high hence a powerful generator will be required to run the company. The total power that can be consumed running at the same time would be 3.2+ 0.6+ 1.8+ 0.6 = 6.2KW. The calculation considers a total run time the company which is 3+6+5+8 = 22hours. This is a very high power. To ensure we have a generator supplying sufficient power to the company we have to consider the rating on the generator.
For such a case scenario we should consider a powerful generator and the one that can run at a constant speed. This help in ensuring continuity in power generation by the generator. The best match for such information would be the synchronous generator with a rating of 7KW.
Question 3 c) Access the generators that the company has available as seen in table 4, to make the best possible choice of recommendation for the customer, taking generator efficiency into consideration.
c) Accessing generators available
For the customer's specification of a total of 6.2KW generator to supply the company, we can get the best selection from the stock. The time required to main the power generation is 22 hours which is next to 24/7 day and night. In the selection e also consider the maximum device consuming the power. From the specifications provided by the customer, we have the arc welders consuming the most power which is at 3.2 Kw.
All the generators in the stock uses gasoline which implies that all can fit the customer's requirements in terms of fuel.
To get the output power for the first generator will be 3.2*0.85 = 2.72Kw.
The output for the second option generator is 4* 0.8 = 3.2Kw
The output for the third generator is 2.6* 0.78 =2.028 Kw
In relation to the one with the highest output we would select the second option since it gives the output of 3.2Kw which matches the highest power consumption device.
FAQ: Transformers - Construction and Applications
- 1. What are the main parts of a transformer?
- 2. What are the different types of transformer construction?
- 3. How do transformers work?
- 4. What are the applications of transformers?
- 6. What are the main types of generators based on output current?
- 7. How are generators categorized based on fuel source?
- 8. Practical Applications of Generator Types?
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