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1. An industrial facility that needs a continuous supply of process heat is cons

ID: 1716649 • Letter: 1

Question

1. An industrial facility that needs a continuous supply of process heat is considering a 65 kW microturbine to help fill that demand. Waste heat recovery will offset fuel needed by its existing 80% efficient boiler. The microturbine has a 30% electrical efficiency and it recovers 48% of the fuel energy as usable heat.

a)     Find the Energy-Chargeable-to-Power (ECP) for this microturbine.

b)     Natural gas for the microturbine and existing boiler costs $4/MBtu. Find the operating cost chargeable to power (CCP)

c)     Suppose the 65kW microturbine costs $100,000 and has an annual O&M cost of $40/kw per year. The capacity factor of this microturbine is 95% and the owner uses a fixed charge rate of 12%/yr and 1.42 levelizing factor for O&M cost. What is the cost of electricity from the microturbine?

d)     If the facility currently pays 10.0¢/kWh for energy, plus demand charges of $7/kW-mo, what would be the annual monetary savings of the microturbine?

Explanation / Answer

a) Let 1 KW = 3412.142 Btu/hr

Finding the thermal input to the 30 percent efficient, 65 kW microturbine

Micorturbine input = 65/.30 x 3412.142

= 739297.43 Btu/hr

Since 48 percent of that is delivered as usable heat

microturbine usable heat=0.48x739297.43

=354862.768 Btu/hr

The displaced fuel for the 75 percent efficient boiler will be

displaced boiler heat= 354862.786 / 0.75

=473150.357 Btu/hr

ECP = (microturbine input - displaced boiler heat)/ 65KW

= 4094.57 Btu/KWh

b) CCP= ECPx $4/MBtu

= $0.010236/ KWh

c) the amortized cost of microturbine = $ 100000 x 0.12/yr

= $12000/yr

Annual O&M cost = $ 40/kw x 65 KW = $ 2600

Let mircoturbine operates = 8000 hr / yr

Annual fuel cost for electricity = 65 KW x CCP x 8000 hr/yr

= $5322.72 / yr

Electricity cost = ($12000+$2600+$5322.72)/65 x 8000 hr/yr = $0.03831 / KWh

d) The value of the energy saved would be

energy saving = 65 KW x 8000 hr/yr x ($ 0.1- $0.03831)

= $ 32078.8 / yr

Assuming the peak demand is reduced by the full 65 kW saves

demand saving = 65 KW x $7 /mo-KW x 12 mo/ yr

= $ 5460 + $ 32078.8

Total annual saving = $37538.8 / yr