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yuradex [85]
2 years ago
8

Consider a building whose annual air-conditioning load is estimated to be 40,000 kWh in an area where the unit cost of electrici

ty is $0.10/kWh. Two air conditioners are considered for the building. Air conditioner A has a seasonal average COP of 2.3 and costs $5500 to purchase and install. Air conditioner B has a seasonal average COP of 3.6 and costs $7000 to purchase and install. All else being equal, determine which air conditioner is a better buy.
Engineering
1 answer:
sdas [7]2 years ago
7 0

Answer:

Air Conditioner B is a better buy for the consumer.

Explanation:

We are given

Annual load = 40,000kWh

Cost = $0.1/kWh

COPa = 2.3

COPb = 3.6

Purchase and Installation charge of A = $5500

Purchase and Installation charge of B = $7000  

We need to make our decision between a cheaper but inefficient and a costlier but efficient Air conditioner for the building.

We assume that all other factors for both the air conditioner remains same and are comparable in all other aspects except cost and efficiency.

The air conditioner which will cost less during its life time will be a better buy. The total cost of the unit during its lifetime like its (servicing, maintenance, etc) can be determined by performing a life cycle analysis, but a simpler method to determine it is by calculating the simple payback period. The energy and cost saving of more efficient Air conditioner is

Energy saving = (Annual energy usage of A) - (Annual energy usage of B)

                        = Annual Load( \frac{1}{COPa }  - \frac{1}{COPb}

                        = 40,000kWh( \frac{1}{2.3} - \frac{1}{3.6})

                        = 40000(0.157)

                        = 6280kWh/year

Cost saving = (6280kWh/year)*($0.1)  

                    = $628.02/year

The installation cost difference of the Air conditioner s

    Cost difference =  $7000-$5500

                               = $1500

Therefore a more efficient Air conditioner B will pay for $1500 cost differential for the first year.

A cost conscious buyer will go for a costly but efficient Air conditioner B which is a better buy in this case also the Air conditioner will last at least for 15 years. But it would be difficult if the cost of the electricity in the area of the consumer is less than $0.1/kWh or if the annual load is less than 40,000kWh.

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<h3>How to solve algebra word problem?</h3>

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The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.
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b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

                           i (t) = 6*e^(-2*t)

                           v (t) = 10*di / dt

Find:

( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

-  The amount of charge Q delivered can be determined by:                      

                                       dQ = i(t) . dt

                  Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt

- Integrate and evaluate the on the interval:

                   = 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C

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                 v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

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                 P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

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- Integrate and evaluate the on the interval:

                  W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ

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