To determine which pumping system is more cost-effective, we will calculate the annual equivalent cost (AEC) for each system and then compare them.
Let's start with System A:
1. Calculate the annual electricity cost for System A:
Power consumption = Pump Power / Efficiency of Motor
Power consumption = 12 hp / 0.5 = 24 kW
Electricity consumption per hour = Power consumption x Operating hours per year
Electricity consumption per hour = 24 kW x hours = 24 kWh x hours
Electricity cost per year = Electricity consumption per hour x Cost of electricity
Electricity cost per year = 24 kWh/h x $0.05/h = $1.20/h
Total annual electricity cost = Electricity cost per year x Operating hours per year
Total annual electricity cost = $1.20/h x hours = $1.20 x hours
2. Calculate the annual equivalent cost (AEC) for System A:
AEC = Initial Cost + (Annual electricity cost x Present worth factor for MARR, 10 years) - Salvage Value
AEC = $2,500 + ($1.20 x hours x 6.417) - $100
Now, let's calculate the AEC for System B:
1. Calculate the annual electricity cost for System B:
Power consumption = 18 hp / 0.85 = 21.18 kW
Electricity consumption per hour = 21.18 kW x hours
Electricity cost per year = 21.18 kWh/h x $0.05/h = $1.059/h
Total annual electricity cost = $1.059 x hours
2. Calculate the AEC for System B:
AEC = $5,800 + ($1.059 x hours x 6.417) - $220
a. To find the number of operating hours per year at which you would be indifferent between the two pumping systems, set the AEC of System A equal to the AEC of System B and solve for "hours."
AEC of System A = AEC of System B
$2,500 + ($1.20 x hours x 6.417) - $100 = $5,800 + ($1.059 x hours x 6.417) - $220
After solving for "hours," you will know the point at which you would be indifferent between the two systems.
b. If the systems are expected to run for 3,000 hours per year, you can compare the AEC of System A and System B at this operating time. Choose the system with the lower AEC as the recommended option.
An agricultural field needs to be irrigated using water from a nearby lake. Two different pumping systems are being considered.
System A:
*Pump Power: 12 hp
*Efficiency of Motor: 0.5
*Initial Cost: $2,500
*Salvage Value: $100
*Useful Life: 10 years
System B:
*Pump Power: 18 hp
*Efficiency of Motor: 0.85
*Initial Cost: $5,800
*Salvage Value: $220
*Useful Life: 10 years
The Minimum Attractive Rate of Return (MARR) is 8% per year, and the cost of electricity is $0.05 per kWh. Assume that 1 horsepower (hp) equals 0.746 kilowatts.
a. At how many operating hours per year would you be indifferent between the two pumping systems?
b. If the systems are expected to run for 3,000 hours per year, which system should be recommended?
1 answer