A plant uses an air compressor 24 hours a day, 360 days per year. The compressor

Question

A plant uses an air compressor 24 hours a day, 360 days per year. The compressor provides 500 cubic feet per minute (cfm) at a specific efficiency of 0.27 kW per cfm. The facility is considering purchasing a new compressor for $18,638, which has an specific efficiency of 0.22 kW per cfm. The salvage value of the current compressor is $3,635. The facility pays $0.09 per kWh for electricity. Assuming a 4 year lifetime on the new compressor, and an interest rate of 5% what is the Present Worth of the project?

Explanation / Answer

purchase of new compressor = $18638

salvage value of old compressor = $ 3635

effective price = $ 15003

amount of electricity consumption saved in a year =

old machine comsumes 500cfm at an efficiency of 0.27 KW per cfm .. so in an hour electricity consumes -

60mins X 500cfm X 0.27 KW =8100 KWh

for 24 hours it consumes - 8100kWh X 24 hrs = 194400 kWh

cost incurred for a DAY on electricity = 194400kWh X $0.09 per kWh = $17496

new machinery -

new machine comsumes 500cfm at an efficiency of 0.22 KW per cfm .. so in an hour electricity consumes -

60mins X 500cfm X 0.22 KW =6600 KWh

for 24 hours it consumes - 6600kWh X 24 hrs = 158400 kWh

cost incurred for a DAY on electricity = 158400kWh X $0.09 per kWh = $14256

total cost of electricity saved in a day of electricity = $(17496-14256) = $3240

Total cost saved in a year =$ 3240 X 360 = $1166400

now calculating the PV of all the cost saved over 4 years @5% = $ 4135997

cost of the project = inflow - outflow

$4135997- $15003 = $4120994.( PV of the project )

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