Group1_Module5_Design_Concepts
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Apr 3, 2024
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Group 1 – Module 5 Assignment
Background
:
Since our overall problem statement is to design a power system that ensures 6-sigma power reliability (0.999999 uptime), we need to use a simple single-family home that may have an average power usage for Florida. Here is a quick reference to start: "However, in 2018, a survey done by the U.S. Energy Information Administration showed that the
average home in the United States used 10,972 kWh for the entire year, which is 914 kWh each month. In Florida, the survey showed that the average home used 1,110 kWh." https://www.floridarealtymarketplace.com/blog/how
-
much
-
power
-
does
-
a
-
house
-
use.html
We would base out system on 1,100 kWh and need to design a system that will compensate for all normal appliances in the house: "If you divide 10,715 kWh by 365 (days in a year), you’ll get the average number of kilowatthours used per day, which is 29.36 kWh. If you multiply that by 1,000, you can find the energy consumption in watts that occur in 24 hours, or 29,360 watts. If you then divide that by 24,
you’ll find that the average household requires 1,223 watts of power." The immediate needs will have to overcome starting and normal operations: "To determine how many watts your generator requires to run all your home appliances, you’ll need to add up the watts they use to run continuously. Then, you’ll add the highest number of starting watts your appliances use. For example, consider that you have the following: An air conditioner that requires 1,100 running watts and 1,700 starting watts A refrigerator that requires 800 running watts and 2,400 starting watts A television that requires 500 running watts and 0 starting watts A tea kettle that requires 600 running watts and 0 starting watts You’ll add your running watts to get 1,100 + 800 + 500 + 600 = 3,000. You’ll then add the highest starting wattage of all your appliances to that total. In this case, the highest starting wattage is 2,400 from the refrigerator. So, you’ll need a generator with at least 5,400 starting watts and 3,000 running watts." https://www.forbes.com/home
-
improvement/home/how
-
many
-
watts
-
run
-
house/
Key
Acceptance Criteria
: Effectively provide emergency power to the home by: •
Switching over from grid power to backup power with minimal* delay. •
Providing equivalent power capacity as existing grid power supply. •
Providing days* of continuous power to the home. * Note: Needs to be more specific; remove ambiguity. Design Concepts
:
1.
Grid power (normal); Solar Panels; Large Battery backup for immediate transfer and to power
the house at night and during low power creation times (cloudy, rain, etc.).
2.
Grid power (normal); Solar Panels; Small wall mounted battery pack (15 mins or less for
emergency transfers); Wind Turbine for consistent power during low power creation times
(cloudy, rain, etc.). 3.
Grid power (normal); Solar Panels; Small wall mounted battery pack (15 mins or less for
emergency transfers); LP/Gas/Propane generator for consistent power during low power creation
times (cloudy, rain, etc.). Pugh Matrix
: 6-Sigma Power System for Residential Homes
Critical Quality
Weight (1 being least important, 10 being most important)
Grid Power; Solar Panels; Large battery bank
Grid Power; Solar Panels; Wall mounted battery pack; Wind turbine
Grid Power; Solar Panels; Wall mounted battery pack; Gas/LP generator
Concept 4
Concept 5
Concept 6
Cost (single family home)
10
-1
1
0
Installation
5
1
-1
-1
Reliability
8
1
0
0
Maintenance
4
0
1
0
ROI
6
-1
-1
1
Warranty
4
-1
0
1
Chosen Design
: There are some very distinct disadvantages to 1 and 2 (batteries are corrosive, heavy loads on foundation, Wind Turbine large, issues with neighbors, noisy, wind mitigation, etc.). Based on the above Pugh Matrix, the chosen design is concept number 3. Total "1s"
2
2
2
0
0
0
Total "0s"
1
2
3
0
0
0
Total "-1s"
3
2
1
0
0
0
Total -7
3
5
0
0
0
Summary Table
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Related Questions
Questions:
1.
b.
As the power plant engineer, you are asked by the power plant manager to design a new steam-based power
plant with a net power output target of 200 MW with a thermal efficiency of 40%. Thermodynamics calculation
should be conducted for the known condition illustrated in Figure 1.
Boiler
P₁ =
8.0 MPa
Saturated
vapor
Condenser
Pump
Turbine
of Vapor and Gas]
Cooling
water
8.0 MPa
0.008 MPa
tw.
Saturated
liquid at 0.008 MPa
Figure 1. Illustration of steam turbine system and its T-S diagram
a. Calculate the back work ratio, the system's mass flow rate, and the cooling water's mass flow rate.
What kind of efforts should be conducted if the plant manager asks you to increase the system's efficiency
to 50% ? Explain and provide proof in the form of calculation!
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These sketches are done on attached engineering or typing paper, unless
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nd
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Natural gas BTU: 1,028 Btu per cubic foot
Oil BTU: 138,590 Btu per gallon
Natural Gas BTU= 80698000 Btu
Oil BTU = 71512440 Btu
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Mobile power plants, such as the one illustrated below, are used in remote locations.
Flowing water turns a turbine, which then moves a coil of wire through a magnetic
field.
Mobile Power Plant
Moving water
Stream
flowing downhill
Location I
Water turning turbine
Location II
-Spinning turbine
-Permanent magnet
Rotating wire coil
Split-ring
commutator
Metal brush
10. The components that connect to the spinning turbine in the illustration on the
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O ammeter
O generator
O motor
1.
O transformer
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MacBook Pro
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Problem Statement
A heat engine draws 14 kW of heat from a reservoir at 900 K. The work it produces drives
a refrigerator which produces a refrigeration capacity of 12 kW. Both devices reject heat to
a reservoir at 300 K. What is the lowest possible temperature of the space cooled by the
refrigerator?
Answer Table
Correct
Stage
Description
Your Answer
Answer
*
1 Lowest temperature of cold space (K)
Due Date
Grade
(%)
Part
Weight Attempt Action/Message Type
Nov 7, 2024 11:59 pm 0.0
1
1/5
Submit
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Natural gas BTU: 1,028 Btu per cubic foot
Oil BTU: 138,590 Btu per gallon
Natural Gas BTU= 80,698,000
Oil BTU = 71,512,440
The home that used a natural gas furnace used more energy for heating.
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Item
Power requirement per smelting processing line 0.6 MW
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- Questions: 1. b. As the power plant engineer, you are asked by the power plant manager to design a new steam-based power plant with a net power output target of 200 MW with a thermal efficiency of 40%. Thermodynamics calculation should be conducted for the known condition illustrated in Figure 1. Boiler P₁ = 8.0 MPa Saturated vapor Condenser Pump Turbine of Vapor and Gas] Cooling water 8.0 MPa 0.008 MPa tw. Saturated liquid at 0.008 MPa Figure 1. Illustration of steam turbine system and its T-S diagram a. Calculate the back work ratio, the system's mass flow rate, and the cooling water's mass flow rate. What kind of efforts should be conducted if the plant manager asks you to increase the system's efficiency to 50% ? Explain and provide proof in the form of calculation! VIOLarrow_forwardSubmit correct and complete solutions. give propriate Explanation. Provide step-by-step detailed explanations.arrow_forwardPart 1: Do the Analysis of the Ocean Thermal Gradient Power Plant shown below. Your Analysis will be easier to do in EES but it is up to you. Your EES program must be well documented and documentation in your code should reference system sketches. (The cycle and individual components) These sketches are done on attached engineering or typing paper, unless you are able to draw them in EES. You must validate your results with hand calculations on engineering paper that invoke the 1 and 2 Law from the perspective of the entire cycle, not the individual components. Of course, a system sketch is required. nd 1. An ocean thermal gradient power plant using a simple non-ideal Rankine Cycle operates with a peak boiler temperature of 70 °F and a condenser temperature of 40 °F. The warm surface water of the ocean is supplying the thermal energy to the boiler. Assume a high source temperature of 80 °F. The cooler water deeper in the ocean is the sink for heat rejection at the condenser. Assume a…arrow_forward
- Need help with this question asap waited over 2 hours.arrow_forwardAnswer question 2 based on the results of question 1. You have a natural gas furnace in your home that used 78,500 cubic feet of natural gas for heating last winter. Your neighbor has a furnace that burns heating oil, and used 516 gallons of heating oil last winter. You can convert the natural gas and heating oil consumption data into Btu to determine which home used more energy for heating. Natural gas BTU: 1,028 Btu per cubic foot Oil BTU: 138,590 Btu per gallon Natural Gas BTU= 80698000 Btu Oil BTU = 71512440 Btu The home that used a natural gas furnace used more energy for heating. 2. You need a new furnace for your home, and you are comparing systems that use natural gas and heating oil. One factor to consider is the cost of fuel. You can compare the price of the fuels on an equal basis by dividing the price per unit of the fuels by the Btu content per unit of the fuels to get a price per million Btu. Assume Natural gas price = $10.50 per thousand cubic…arrow_forwardAn electric hot water heater consumes 3.1 kilowatts of electricity and converts it to heat. How long will it take the water heater to heat a 67 gallon tank of water from 10 degrees Celsius to 50 degrees Celsius? (1 kilogram of water is 0.37 gallons, 1 Calorie = 4200 J). It may be helpful to refer back to the weekly handout for guidance on this problem. Your final answer should be in minutes (rounded to the nearest 10 minutes).arrow_forward
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