Question

As energy costs continue to rise, power efficiency is increasingly important. Acme Chemical is evaluating 2 different electric motors to drive a mixing motor and needs to perform a present economy study. The motor will produce 75 hp and will be operated 8 hours per day, 365 days for one year (maintenance will be performed on second shift—assume no down time during operation), after which time the motor will have no value. Select the most economical motor. Assume Acme’ s electric power costs $0.16 per kWh. (1 hp = 0.746 kW).
Motor A Motor B
Purchase price $3,200 $5,900
Annual maintenance cost $250 $450
Efficiency 75% 85%

Answer 1

Answer:

The answer is below

Explanation:

For motor A, efficiency = 75%= 0.75 hence:

$Operating\ cost\ of\ motor\ A=\frac{75\ hp}{0.75} *\frac{0.746\ kW}{hp}*\frac{\ 0.16}{kWh} *\frac{8\ hr}{day}*\frac{365\ days}{year} \\\\Operating\ cost\ of\ motor\ A=\ 34853$

Total cost for motor A = operating cost + purchase cost = $34853 + $3200

Total cost for motor A = $38053

For motor B, efficiency = 85% = 0.85

$Operating\ cost\ of\ motor\ B=\frac{75\ hp}{0.85} *\frac{0.746\ kW}{hp}*\frac{\ 0.16}{kWh} *\frac{8\ hr}{day}*\frac{365\ days}{year} \\\\Operating\ cost\ of\ motor\ B=\ 30753$

Total cost for motor B = operating cost + purchase cost = $30753 + $5900

Total cost for motor B = $36653

Therefore motor B is more economical since it has a lesser total cost