Engineering Economics 1) A new building is going to require 40,000 conventional

Question

Engineering Economics

1) A new building is going to require 40,000 conventional watts of lightning. It is expected that the equivalent full wattage hours of operation per year is 1500. With conventional lights, the light bulb cost will be $.05 per watt. With high efficient fluorescent bulbs, an 18 watt bulb is equivalent to a 40 watt conventional bulb (same light with lower heat). However, the HiEf bulb costs $57/watt. What is the payback period for the HiEf option. The price of electricity is $.12/kWh.

2) A typical natural gas water heater has an efficiency of 87% and costs $250 annually to operate if the cost of natural gas is $1.50/therm. A typical heat-pump hot water heater has a COP of 2.5 but costs $1050 more than the NGWH. What is the payback return on the HP water heater if the extra money was borrowed at 4% but the loan was paid off with the savings? (How long will it take to pay off the loan?)

Explanation / Answer

1. let the payback period be 'x' years

First consider conventional lighting.

Watt required = 40000

cost per watt = $ 0.05

Total initial cost = 40000 * 0.05 = $2000

Electricity bill cost per year = 0.12 * ( No. of KWh per year)

No. of KWh per year = 1500 * 40000 / 1000 = 60000 KWh

Electricity bill cost for x years = x* 0.12 * 60000 = 7200* x

Total cost for x years = 2000 + 7200 * x

Now consider high efficient fluorescent bulbs,

Total watt required = 18/40 * 40000 W = 18000 W ( 18 w is sufficient for conventional 40 W)

initial cost = 18000 * 57 / 18 = $ 57000

Electricity bill cost of x years = x* (18000/1000 * 1500) * 0.12 = 3240 * x

Total cost for x years = 57000 + 3240 * x

Now, equating both total costs,

57000 + 3240 * x = 2000 + 7200 * x

x = 13.9

Hence the payback period for Hi Ef lighting is approx 14 years

2. Extra cost of NGWH per year = $ 250 - ($ 250* 0.87 /2.5) = $ 250 - $ 87 = $ 163

Initial extra cost for heat-pump hot water heater = $ 1050

Now we have to pay $ 1050 at 4% annual rate in yealr installments of $163

let x be the number of years

Now 1050 ( 1.04) ^ x = 163 * ( (1.04^x) - 1) / (1.04 -1)

Solve for x.

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