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yuradex [85]
3 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]3 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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the highest rate of heat transfer allowed is 0.9306 kW

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v_f = 0.001057 m³/kg

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h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

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Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

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