Question 11.AE.3: A pumped-storage, hydro-power plant (Figure E11.3.1) is to b......
A pumped-storage, hydro-power plant (Figure E11.3.1) is to be designed [18, 19, 146– 148] for a rated power P_{rated}=250 MW and a rated energy capacity 1500 MWh per day. It consists of an upper and a lower reservoir with a water capacity C_{H2O} each. In addition there must be an emergency reserve for 625 MWh. Water evaporation must be taken into account and is 10% per year for each reservoir, and the precipitation per year is 20 inches. The maximum and minimum elevations of the upper reservoir are 1050 ft and 1000 ft, respectively. The maximum and minimum elevations for the lower reservoir are 200 ft and 150 ft, respectively. The water turbine is of the Francis type and it is coupled with a salient-pole synchronous machine with p=24 poles, which can be used as a generator for generating electricity by releasing the water from the upper reservoir to the lower one, and as a motor for pumping the water from the lower reservoir to the upper one. A capacity factor of 100% can be assumed: in a real application the capacity factor may vary between 70-90%.
a) If the power efficiencies of the water turbine, the synchronous generator, and the Δ-Y transformer are η_{turbine}=0.8, η_{synchronous\ machine}=η_{Y-Δ\ transformer} =0.95, respectively, compute the required turbine input power P_{\text {turbine }}^{\text {requared }}.
b) Provided the head of the water is H=850 ft, the frictional losses between water and pipe amount to 15%, and the water flow measured in cubic feet per second is Q=6000 cfs, compute the mechanical power available at the turbine input P_{kW} [146–148].
c) How does \mathrm{P}_{\text {turbine }}^{\text {required }} required compare with P_{kW}?
d) Compute the specific speed N_q [146–148]. Is the selection of the Francis turbine justified?
e) What other types of water turbines exist?
f) What is the amount of water the upper or lower reservoirs must hold to generate E= (1500 MWh+625 MWh)=2125 MWh per day during an 11.3 hour period?
g) Is the given precipitation per year sufficient to replace the evaporated water? If not, what is the required “rain-catch” area to replace the yearly water loss through evaporation?
h) The pumped-storage plant delivers the energy E=(1500 MWh+625 MWh)= 2125 MWh per day for which customers pay $0.20/kWh due to peak-power generation. What is the payback period of this pumped storage plant if the construction price is $3000 per installed power capacity of 1 kW, the cost for pumping is $0.03/ kWh, and the interest rate is 3%?
a) Computation of the turbine input power \mathrm{P}_{\text {turbine }}^{\text {required }}:
P_{turbine\_imput }^{\text {required }}=\frac{P_{\text {rated }}}{\eta_{\text {turbine }} \cdot \eta_{synchronous\_machine } \cdot \eta_{\Delta-Y_{-} \text {transformer }}}=346.25 \mathrm{MW}b) Computation of the mechanical power available at the turbine input P_{kW}:
P_{\mathrm{kW}}=\frac{H \cdot Q \cdot W \cdot 0.746}{550}=\frac{722.5 \cdot 6000 \cdot 62.4 \cdot 0.746}{550}=366.9 \mathrm{~kW},
where H is measured in ft, Q is measured in cubic feet per second (cfs), W is the weight of 1 cubic foot of water of 62.4 (lb-force)/(ft)³ , 0.746 kW=1 hp, and 550 is a constant converting ft(lb-force)/s to horse power (hp).
c) How does P_{\text {turbine }}^{\text {Fequired }} compare with P_{kW}?
P_{turtine\_input }^{\text {required }} \leq P_{\mathrm{k} W} \text {. }d) Computation of the specific speed. Is the selection of the Francis turbine justified?
Synchronous speed of generator n_{\mathrm{s}}=\frac{120 \cdot f}{p}=300 \mathrm{rpm}, \quad Q=6000 \cdot 0.0283= 169.8 \mathrm{~m}^3 / \mathrm{s}, H_{\text {eff }}=722.5 \cdot 0.3048=220.22 \mathrm{~m}. Specific speed of a water turbine is now N_{\mathrm{q}}=n_{\mathrm{s}} \frac{Q^{0.5}}{H_{\mathrm{eff}}^{0.75}}=68.38, which means a Francis turbine is satisfactory.
e) What other types of water turbines exist?
Pelton wheel, Kaplan turbine, bulb-type turbine, mixed-flow turbine, propeller turbine, and Gorlov turbine.
f) The amount of water the upper or lower reservoirs must hold:
Q=6000 cfs, therefore, the water required during one day is Capacity_{H2O} = 6000\cdot 0\cdot 0283\cdot 3600\cdot 11\cdot 3=6.908 \cdot 10^6 m³ , this requires two (an upper and lower) reservoirs of 20 m depth, 588 m long and 588 m wide, each. As is well known the water inlets at the upper and lower reservoirs will be in the middle of the reservoir depths in order to prevent that sand and rocks enter the turbine and the pump. This is to say the 20 m depths are effective depths and the actual depths might be more than twice as large, say 50 m.
g) Required “rain-catch” area:
Capacity_{H2Oevaporation}=6.908\cdot 10^5 m³, Capacity_{H2O} +Capacity_{H2Oevaporation} =6.908.10^6 +6.908.10^5 =7.599.10^6 m³ . For 20 inches of precipitation one gets per year Capacity_{H2Oprecipitation}=20\cdot 0\cdot 0254 (588)² =1.756.10^5 m³ . Note, Capacity_{H2Oevaporation} exceeds Capacity_{H2Oprecipitation} from this follows that a rain-catch area is required. At the time of commissioning the plant the additional rain-catch area is required to fill the upper reservoir; some run-off will be required after the commissioning of the plant. The additional water to be supplied initially by the rain-catch area is Capacity_{H2O\ additional}=Capacity_{H2O} +Capacity_{H2Oevaporation} – Capacity_{H2O\ precipitation}=7.599\cdot 10^6 m³ -1.756.10^5 m³ =7.42106 m³ requiring a rain-catch area of (area) rain-catch=7.42106 /(200.0254)=14.61.10^6 m² =14.61 km² .
h) Payback period of pumped storage plant:
Earnings = $2125.10³ 0\cdot 20\cdot 365y = 155.125.10^6 y, expenses = $750.10^6+ (2125\cdot10³ 0\cdot03\cdot 365y)/(0\cdot 95\cdot0\cdot95\cdot0\cdot80\cdot0\cdot85) =$750.10^6 +$37.92y.10^6 .
Payback period in y years without interest payments: y=750.10^6 /(155.125.10^6 – 37.92.10^6 )=6.4 years
Payback period with interest payments: 155.125y=750(1.03)^y +37.92y resulting in about y=8 years.