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Sphinxa [80]
3 years ago
13

A food-services company with a 480 V, three-phase service entrance has the following set of loads:  A 7 ton walk-in refrigerati

on unit:3 COP of 1.2 and 0.78 lagging power factor  A conveyor belt system consisting of 10 continuous-duty induction motors: each 5 HP, 81% efficient, PF = 0.73 lagging  A 20 kVA T8 fluorescent lighting system with electronic ballasting (PF = 0.93 leading)  A 75,000 BTU/hr food dehydrator4 Consider the case when all systems are fully loaded. Calculate the reactive power required to compensate this customer’s load to a 0.95 lagging power factor.
Engineering
1 answer:
Mila [183]3 years ago
4 0

Answer:

A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

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An AI operated automatic garbage collection system

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3 years ago
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A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni
arlik [135]

Answer:

The population size would be p' = 5000

The yield would be    MaxYield = 2082 \ fishes \ per \ year

Explanation:

So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing

 

    So the current  yield which is mathematically represented as

                               \frac{dN}{dt} =   \frac{2000}{1 \ year }

 Where dN is the change in the number of fish

            and dt is the change in time

So in order to obtain the solution we need to obtain the  rate of growth

    For this we would be making use of the growth rate equation which is

                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

  Where N is the population of the fish which is given as 4,000 fishes

          and  K is the carrying capacity which is given as 10,000 fishes

             r is the growth rate

        Substituting these values into the equation

                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

                      Max Yield  = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by p' would be

                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

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3 years ago
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System Integration summary
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Answer:

System integration can be defined as the progressive linking and testing of system components to merge their functional and technical characteristics into a comprehensive interoperable system.

Explanation:

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Air enters a compressor operating at steady state at 1 bar, 290 K, with a mass flow rate of 0.1 kg/s and exits at 980 K, 10 bar.
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Answer:

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Explanation:

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