Power and Energy Potential study

Data and approach

Gross head $= 2250-1650 = 600$ m. Only friction loss is considered. Overall efficiency $\eta = 0.95\times0.93\times0.99\times0.99 = 0.8659$ (hydraulic × turbine × generator × transformer). Energy is computed month by month (730 h per month), with the friction loss recalculated for the flow used.

Head loss at design flow $Q_d = 18$ m³/s (Darcy-Weisbach)

Tunnel: $A=\frac{\pi}{4}(3)^2 = 7.069$ m², $V = 2.546$ m/s

$$h_{f,t} = f\frac{L}{D}\frac{V^2}{2g} = 0.014\times\frac{4000}{3}\times\frac{2.546^2}{2\times9.81} = 6.17\ \text{m}$$

Penstock: $A=\frac{\pi}{4}(2.2)^2 = 3.801$ m², $V = 4.735$ m/s

$$h_{f,p} = 0.012\times\frac{1000}{2.2}\times\frac{4.735^2}{2\times9.81} = 6.23\ \text{m}$$

Net head at $Q_d$: $H_n = 600-6.17-6.23 = 587.60$ m. (Loss varies as $Q^2$ for smaller flows.)

(i) Installed capacity, energy and plant factor

Environmental release $=10\%\times$ minimum monthly flow $=0.1\times3.4=0.34$ m³/s, so the flow available for power is $Q-0.34$.

$$P = 9.81\,\eta\,H_n\,Q_d = 9.81\times0.8659\times587.60\times18 = 89,846\ \text{kW} = 89.85\ \text{MW}$$

• Total annual energy $= 447.73$ GWh
• Primary (firm) energy, from the minimum available flow $3.06$ m³/s (available in every month): $136.54$ GWh
• Secondary energy $=447.73-136.54 = 311.19$ GWh
• Plant factor $=\dfrac{447.73\times10^6}{89.85\times10^3\times8760} = 0.569$ (56.9%)

(ii) Reservoir for 4-hour daily peaking

In the driest month the available inflow is $Q_{in} = 3.06$ m³/s. The plant is to run for 4 hours a day at the design flow of 18 m³/s and store water in the other 20 hours. The daily inflow is 264,384 m³ and the peak use is $18\times4\times3600 = 259,200$ m³, which is slightly smaller, so 4-hour peaking at 18 m³/s is possible.

During the 4 peak hours the draw-down from the reservoir is

$$V = (Q_d - Q_{in})\times t = (18-3.06)\times4\times3600 = 215,136\ \text{m}^3 \approx 2.15\times10^5\ \text{m}^3$$

The same volume must be stored in the 20 off-peak hours ($3.06\times20\times3600 = 220,320$ m³ available), so it fills in time.

Answer: P = 89.8 MW; total energy = 447.7 GWh; primary energy = 136.5 GWh; secondary energy = 311.2 GWh; plant factor = 0.57; pondage volume ≈ 215,136 m³ (≈ 2.15 × 10⁵ m³).

The generator efficiency is missing in the question; it is assumed to be 99%, the same as the transformer (all other data as given).

Data and approach

Gross head $= 2250-1650 = 600$ m. Only friction loss is considered. Overall efficiency $\eta = 0.95\times0.93\times0.99\times0.99 = 0.8659$ (hydraulic × turbine × generator × transformer). Energy is computed month by month (730 h per month), with the friction loss recalculated for the flow used.

Head loss at design flow $Q_d = 18$ m³/s (Darcy-Weisbach)

Tunnel: $A=\frac{\pi}{4}(3)^2 = 7.069$ m², $V = 2.546$ m/s

$$h_{f,t} = f\frac{L}{D}\frac{V^2}{2g} = 0.014\times\frac{4000}{3}\times\frac{2.546^2}{2\times9.81} = 6.17\ \text{m}$$

Penstock: $A=\frac{\pi}{4}(2.2)^2 = 3.801$ m², $V = 4.735$ m/s

$$h_{f,p} = 0.012\times\frac{1000}{2.2}\times\frac{4.735^2}{2\times9.81} = 6.23\ \text{m}$$

Net head at $Q_d$: $H_n = 600-6.17-6.23 = 587.60$ m. (Loss varies as $Q^2$ for smaller flows.)

(i) Installed capacity, energy and plant factor

Environmental release $=10\%\times$ minimum monthly flow $=0.1\times3.4=0.34$ m³/s, so the flow available for power is $Q-0.34$.

$$P = 9.81\,\eta\,H_n\,Q_d = 9.81\times0.8659\times587.60\times18 = 89,846\ \text{kW} = 89.85\ \text{MW}$$

• Total annual energy $= 452.92$ GWh
• Primary (firm) energy, from the minimum available flow $3.06$ m³/s (available in every month): $136.54$ GWh
• Secondary energy $=452.92-136.54 = 316.38$ GWh
• Plant factor $=\dfrac{452.92\times10^6}{89.85\times10^3\times8760} = 0.575$ (57.5%)

(ii) Reservoir for 4-hour daily peaking

In the driest month the available inflow is $Q_{in} = 3.06$ m³/s. The plant is to run for 4 hours a day at the design flow of 18 m³/s and store water in the other 20 hours. The daily inflow is 264,384 m³ and the peak use is $18\times4\times3600 = 259,200$ m³, which is slightly smaller, so 4-hour peaking at 18 m³/s is possible.

During the 4 peak hours the draw-down from the reservoir is

$$V = (Q_d - Q_{in})\times t = (18-3.06)\times4\times3600 = 215,136\ \text{m}^3 \approx 2.15\times10^5\ \text{m}^3$$

The same volume must be stored in the 20 off-peak hours ($3.06\times20\times3600 = 220,320$ m³ available), so it fills in time.

Answer: P = 89.8 MW; total energy = 452.9 GWh; primary energy = 136.5 GWh; secondary energy = 316.4 GWh; plant factor = 0.58; pondage volume ≈ 215,136 m³ (≈ 2.15 × 10⁵ m³).

Gross potential is the theoretical power of the river between two sections, using all the flow and the full head difference with 100% efficiency. Net potential is the power that can actually be developed after deducting head loss, efficiency and the part of the flow that cannot be used.

Gross potential

Divide the reach into small sub-reaches where the flow is nearly constant (the flow increases downstream with tributaries and catchment area). For each sub-reach:

$$P_{gross} = \sum \rho g Q_i \Delta H_i = \sum 9.81\, Q_i\, \Delta H_i \ \ (\text{kW})$$

where $Q_i$ is the mean flow (m³/s) and $\Delta H_i$ is the drop in elevation (m) of the sub-reach. Annual energy = $P \times 8760$ kWh. Flow comes from the flow series at the two sections (area-ratio method for ungauged sites), and elevations from topographic maps or survey.

Net (technical/economic) potential

• Choose the design discharge $Q_d$ from the flow duration curve (e.g. $Q_{40}$ for RoR), after leaving the environmental flow.
• Net head $H_n = H_{gross} - h_f$, where $h_f$ includes the intake, desander, waterway and penstock losses (about 5-10% of gross head).
• Overall efficiency $\eta$ of the turbine, generator and transformer (about 80-90%).

$$P_{net} = 9.81\, \eta\, Q_d\, H_n \ \ (\text{kW})$$

• Annual energy is found from the monthly flows up to $Q_d$. Economic potential is the part for which benefit-cost ratio is greater than one.

The installed capacity of a RoR plant is optimised by comparing the extra cost and the extra benefit of each increase in capacity. Since flow varies throughout the year, a bigger plant uses the large flow only for a short time, so there is an optimum point.

Procedure

1. Prepare the flow duration curve (FDC) from the long-term flow series (after the environmental release).
2. Select a number of trial design discharges $Q_d$ (e.g. $Q_{30}$, $Q_{40}$, $Q_{50}$ ...). For each, compute the power $P = 9.81\,\eta\,H_n\,Q_d$ and the energy:

$$E = \sum 9.81\,\eta\,H_n\,\min(Q_i,Q_d)\,t_i$$

3. Compute annual benefit = energy × price (firm and secondary energy may have different prices).
4. Compute annual cost = (fixed cost + variable cost × P) × capital recovery factor + O&M. Fixed cost includes the headworks, civil structures, access and so on; variable cost is proportional to the capacity (turbine, generator, penstock).
5. Plot the net benefit (benefit − cost), the benefit/cost ratio and the marginal values against the capacity.

Criterion

• Maximum net benefit (NPV), or
• Marginal benefit = marginal cost: increase capacity until the extra benefit from the last kW (= price × energy produced by it during the time it is used) equals its extra annual cost. Beyond this point the machines run only a few weeks per year and do not pay back.

Other considerations: the cost per kW, the plant factor (usually 50-70% for RoR), the grid demand, the risk and the possibility of staged development.

Optimum capacity of a RoR plant is where the marginal benefit of the last unit of capacity equals its marginal cost (equivalently the net benefit is maximum). The fixed cost does not change with capacity, so it does not affect the optimum.

Data and constants

• Fixed cost = NRs 10,00,00,00,000 (= NRs 10,000 million); variable cost = NRs 1,50,000/kW (present worth); price = NRs 10/kWh
• Turbine 95%, generator 97%, so $\eta = 0.95\times0.97 = 0.9215$; no O&M cost is given
• "Power factor" in the question is read as the plant factor.
• Net head $H = 150$ m; overall efficiency $\eta = 0.9215$
• Capital recovery factor $CRF = \dfrac{i(1+i)^n}{(1+i)^n-1} = 0.10139$ for $i=10\%$, $n=45$ years
• Energy per month: $E_m = 9.81\,\eta\,H\,Q\,(730\ \text{h})$ (8760/12 h per month)
• Annual cost $=(\text{fixed}+\text{variable}\times P)\,CRF$; annual benefit = energy × price

Flow available for power

available flows (m³/s) in descending order: 1200, 1000, 800, 500, 450, 200, 150, 100, 95, 90, 85, 80.

Marginal check

Each extra 1 kW of capacity above the flow level is used only in the months in which the flow is greater than the capacity flow. Benefit of 1 kW used for $k$ months $= 0.73\,k$ MWh × price. Benefit per kW for each month it is used = 730 kWh × NRs 10 = NRs 7,300; so a kW must be used for at least 2.08 months, i.e. at least 3 months in a year, to pay for itself. The 3rd highest flow is 800 m³/s, so capacity beyond 800 m³/s (used only 2 months) is not justified.

Trial design discharges

The net benefit is greatest at $Q_d = 800$ m³/s.

Results

• Installed capacity: $P = 9.81\times0.9215\times150\times800 = 1,084,790\ \text{kW} = 1,084.79$ MW
• Firm (primary) energy, from the minimum flow 80 m³/s all year: $9.81\times0.9215\times150\times80\times8760 = 950,275,865\ \text{kWh} = 950.28$ GWh
• Total energy: $4,107.96$ GWh
• Secondary energy $= 4,107.96-950.28 = 3,157.69$ GWh
• Plant factor $= \dfrac{E}{P\times8760} = \dfrac{4,107.96\times10^6}{1,084,790\times8760} = 0.432$ (43.2%)

Answer: best installed capacity = 1,084.8 MW (design flow 800 m³/s); firm energy = 950.3 GWh; secondary energy = 3,157.7 GWh; total energy = 4,108.0 GWh; plant factor = 0.43.

Optimum capacity of a RoR plant is where the marginal benefit of the last unit of capacity equals its marginal cost (equivalently the net benefit is maximum). The fixed cost does not change with capacity, so it does not affect the optimum.

Data and constants

• Fixed cost = USD 2795 × 10⁶; variable cost = USD 645/kW; O&M = 2.5% of the variable cost per year; price = USD 38/MWh = USD 0.038/kWh for primary and secondary energy
• Overall efficiency $\eta = 0.89$ and head = 48 m (constant)
• Net head $H = 48$ m; overall efficiency $\eta = 0.8900$
• Capital recovery factor $CRF = \dfrac{i(1+i)^n}{(1+i)^n-1} = 0.11101$ for $i=11\%$, $n=45$ years
• Energy per month: $E_m = 9.81\,\eta\,H\,Q\,(730\ \text{h})$ (8760/12 h per month)
• Annual cost $=(\text{fixed}+\text{variable}\times P)\,CRF + \text{O\&M}$; annual benefit = energy × price

Flow available for power

available flows (m³/s) in descending order: 93, 73, 68, 53, 48, 43, 38, 36, 33, 28, 23, 18.

Marginal check

Each extra 1 kW of capacity above the flow level is used only in the months in which the flow is greater than the capacity flow. Benefit of 1 kW used for $k$ months $= 0.73\,k$ MWh × price. Benefit per kW for each month it is used $=0.73\times38 = 27.74$ USD/yr; the annual cost of 1 kW is USD 87.73, so a kW must be used for at least 3.16 months, i.e. at least 4 months. The 4th highest flow is 53 m³/s, so the optimum design flow is 53 m³/s.

Trial design discharges

The net benefit is greatest at $Q_d = 53$ m³/s.

Results

• Installed capacity: $P = 9.81\times0.8900\times48\times53 = 22,211\ \text{kW} = 22.21$ MW
• Firm (primary) energy, from the minimum flow 18 m³/s all year: $9.81\times0.8900\times48\times18\times8760 = 66,081,039\ \text{kWh} = 66.08$ GWh
• Total energy: $146.54$ GWh
• Secondary energy $= 146.54-66.08 = 80.46$ GWh
• Plant factor $= \dfrac{E}{P\times8760} = \dfrac{146.54\times10^6}{22,211\times8760} = 0.753$ (75.3%)

Note: with the fixed cost taken exactly as stated (USD 2795 million), the net benefit is negative for every capacity (the project does not pay for itself). The optimum capacity is the same because the fixed cost is independent of capacity; the fixed cost is possibly a misprint in the question.

Answer: best installed capacity = 22.2 MW (design flow 53 m³/s); firm energy = 66.1 GWh; secondary energy = 80.5 GWh; total energy = 146.5 GWh; plant factor = 0.75.

Optimum capacity of a RoR plant is where the marginal benefit of the last unit of capacity equals its marginal cost (equivalently the net benefit is maximum). The fixed cost does not change with capacity, so it does not affect the optimum.

Data and constants

• Net head $= 229-12 = 217$ m (gross head minus average transition loss)
• Efficiency $\eta = 0.94\times0.92\times0.99 = 0.8562$ (turbine × gearbox × generator)
• Fixed cost US$ 1200 × 10⁶; variable cost US$ 500/kW (present worth, annualised with CRF); price US$ 48/MWh = 0.048 US$/kWh
• 10 m³/s is left in the river, so available flow $=Q-10$
• Net head $H = 217$ m; overall efficiency $\eta = 0.8562$
• Capital recovery factor $CRF = \dfrac{i(1+i)^n}{(1+i)^n-1} = 0.12042$ for $i=12\%$, $n=50$ years
• Energy per month: $E_m = 9.81\,\eta\,H\,Q\,(730\ \text{h})$ (8760/12 h per month)
• Annual cost $=(\text{fixed}+\text{variable}\times P)\,CRF$; annual benefit = energy × price

Flow available for power

Environmental release of 10 m³/s is deducted: available flows (m³/s) in descending order: 2532, 2004, 1440, 1001, 820, 583, 199, 186, 108, 105, 83, 79.

Marginal check

Each extra 1 kW of capacity above the flow level is used only in the months in which the flow is greater than the capacity flow. Benefit of 1 kW used for $k$ months $= 0.73\,k$ MWh × price. Benefit per kW per month of use $=0.73\times48 = 35.04$ USD/yr; annual cost of 1 kW $= 500\times0.12042 = 60.21$ USD, so a kW must be used at least 1.72 months, i.e. at least 2 months. The 2nd highest flow is 2014 m³/s (2004 m³/s after the 10 m³/s release); beyond that the capacity is used for only 1 month.

Trial design discharges

The net benefit is greatest at $Q_d = 2004$ m³/s (available flow after release; river flow 2014 m³/s).

Results

• Installed capacity: $P = 9.81\times0.8562\times217\times2004 = 3,652,392\ \text{kW} = 3,652.39$ MW
• Firm (primary) energy, from the minimum flow 79 m³/s all year: $9.81\times0.8562\times217\times79\times8760 = 1,261,277,982\ \text{kWh} = 1,261.28$ GWh
• Total energy: $11,457.94$ GWh
• Secondary energy $= 11,457.94-1,261.28 = 10,196.66$ GWh
• Plant factor $= \dfrac{E}{P\times8760} = \dfrac{11,457.94\times10^6}{3,652,392\times8760} = 0.358$ (35.8%)

Answer: best installed capacity = 3,652.4 MW (design flow 2004 m³/s); firm energy = 1,261.3 GWh; secondary energy = 10,196.7 GWh; total energy = 11,457.9 GWh; plant factor = 0.36.

Capacity should be increased as long as the marginal benefit (MB) of the added capacity is at least its marginal cost (MC). The fixed cost ($60 × 10⁶) does not depend on the capacity, so it does not enter the marginal comparison.

Data

• Interest $i=12\%$, life $n=35$ years: $CRF=\dfrac{i(1+i)^n}{(1+i)^n-1}=0.12232$
• Annual cost of 1 kW of electro-mechanical equipment $= 650\,(CRF+0.03) = 650\times(0.12232+0.03) = 99.01$ USD per kW per year
• Energy price = $0.08/kWh

Method

The power duration table gives the percentage of time each power level is equalled or exceeded. Raising the capacity from the next lower power level to the level $P_k$ adds $\Delta P$ kW, which can be used for only the percentage of time $t_k$ for which that power is available:

$$\Delta E = \Delta P \times \frac{t_k}{100} \times 8760,\quad MB = 0.08\,\Delta E,\quad MC = c\,\Delta P$$

Decision

The ratio MB/MC is below 1 for the step 604 → 805 kW (MB = $11,734 < MC = $19,900), so this extra capacity is not worth building. The step 564 → 604 kW has MB/MC = 1.18 (more than 1), so it is justified, and all lower steps are also justified.

Answer: the optimum installed capacity is 604 kW (the capacity exceeded 16.67% of the time).

(If the curve is interpolated linearly between the 8.33% and 16.67% points, the break-even occurs at about 14.3% of the time, which is roughly 660 kW; with the tabulated power levels, 604 kW is the answer.)

a) Flow duration curve

Arrange the 12 monthly flows in descending order and assign the exceedance probability by the Weibull formula $p = \dfrac{m}{N+1}\times100$ with $N = 12$.

Plot Q (vertical) against percentage of time (horizontal) and join the points with a smooth curve:

 Q (m³/s)
 150 |*
 125 | *
 100 |   *
  75 |      * *
  50 |           *
  25 |               * *
  10 |                    *
     +--------------------------
      0  20  40  60  80  100 %
         time flow is exceeded

b) Power at the mean flow

Mean flow $\bar Q = \dfrac{\sum Q}{12} = \dfrac{870}{12} = 72.5$ m³/s

$$P = \gamma Q H \eta = 9.81\times72.5\times100\times0.85 = 60,454.12\ \text{kW} = 60.45\ \text{MW}$$

The mean flow of 72.5 m³/s is exceeded about 46% of the time (six of the twelve months have flows above it).

Answer: mean flow = 72.5 m³/s; power at mean flow ≈ 60.45 MW (the FDC is tabulated above).

Flow duration curve

Rank the monthly flows in descending order and take $p=\dfrac{m}{N+1}\times100$ with $N=12$.

 Q (m³/s)
  72 |*
  56 | *
  45 |-+--*-*-----  design flow
  38 |        * *
  30 |           * *
  22 |               *
  16 |                  *
     +----------------------
      0  20  40  60  80 100 %

The FDC is flat at 45 m³/s for about 31% of the time (months with flow above the design flow).

Energy

Constants: $H=60$ m, $\eta=0.82$, $Q_d=45$ m³/s, 730 h per month.

• Installed capacity $P=9.81\times0.82\times60\times45 = 21,719$ kW $=21.72$ MW
• Firm energy is based on the flow available all the time (minimum flow 16 m³/s):

$$P_{firm} = 9.81\times0.82\times60\times16 = 7,722\ \text{kW},\quad E_{firm} = 7,722\times8760 = 67.65\ \text{GWh}$$

• Total energy with $Q_{used}=\min(Q,45)$: sum of used flows = 420 m³/s-months, so mean used flow = 35 m³/s

$$E_{total} = 9.81\times0.82\times60\times35\times8760 = 147.98\ \text{GWh}$$

• Secondary energy $=E_{total}-E_{firm} = 147.98-67.65 = 80.33$ GWh

Answer: firm energy ≈ 67.6 GWh; secondary energy ≈ 80.3 GWh; total ≈ 148.0 GWh per year.

Data and assumptions

Net head $H=250$ m. Dry months (Nov-May, 7 months) use stored water for 4 h/day of peaking. Wet months (Jun-Oct, 5 months) the plant runs as RoR for 24 h. One month $=730$ h $=2.628\times10^6$ s. Overall efficiency is not given, so power is first found for $\eta=1$ and then for a typical $\eta=0.85$; the energy ratio does not depend on $\eta$.

Design flow $Q_{25}$

Arrange the 12 monthly flows in descending order: 1200, 1100, 800, 500, 350, 200, 200, 120, 105, 100, 80, 80. The flow equalled or exceeded 25% of the time (3 months out of 12) is the 3rd value:

$$Q_{25} = 800\ \text{m}^3/\text{s}$$

Is the stored water enough for peaking at $Q_{25}$?

• Dry-season inflow volume $=\sum Q\times 2.628\times10^6 = 885\times2.628\times10^6 = 2,326\times10^6$ m³
• Plus live storage $=1100\times10^6$ m³, so water available $=3,426\times10^6$ m³
• Peaking hours in 7 months $=7\times730\times\frac{4}{24} = 851.7$ h
• Water needed to run at $Q_{25}$ for these hours $=800\times851.7\times3600 = 2,453\times10^6$ m³ (less than available)

So the turbines can run at $Q_{25}=800$ m³/s during the 4 peak hours every day, and the plant capacity is controlled by the design flow (the water available would allow up to 1,117 m³/s).

Maximum power

$$P_{max} = 9.81\,\eta\,Q\,H = 9.81\times\eta\times800\times250 = 1,962\,\eta\ \text{MW}$$

• With $\eta = 1$: $P_{max} = 1,962$ MW
• With $\eta = 0.85$: $P_{max} = 1,668$ MW

Wet-season to dry-season energy ratio

Wet season flows used: $\min(Q,800)$ = 500, 800, 800, 800, 350, total $=3250$ m³/s-months.

$$E_{wet} \propto 3250\times730 = 2,372,500,\qquad E_{dry} \propto 800\times851.7 = 681,333$$ $$\frac{E_{wet}}{E_{dry}} = \frac{2,372,500}{681,333} = 3.48$$

Answer: maximum power = 1,962 MW for $\eta=1$ (1,668 MW for $\eta=0.85$); wet season energy / dry season energy = 3.48.

Maximum generating capacity

The minimum flow gives the average power available throughout the day:

$$P_{avg} = \gamma Q H \eta = 9.81\times25\times30\times0.60 = 4,414.5\ \text{kW}$$

With pondage the plant can follow the daily load, and the load factor is

$$\text{Load factor} = \frac{P_{avg}}{P_{max}} = 0.65 \Rightarrow P_{max} = \frac{4,414.5}{0.65} = 6,791.5\ \text{kW} \approx 6.79\ \text{MW}$$

Volume of pondage (20 h average load + 4 h peak load)

Peak load: $P_{max}=6,791.5$ kW for 4 h. The average over the day must equal $P_{avg}$, so the off-peak load for 20 h is

$$P_{off} = \frac{24\times4,414.5 - 4\times6,791.5}{20} = 3,939.1\ \text{kW}$$

Flow needed at peak:

$$Q_{peak} = \frac{P_{max}}{9.81\times30\times0.6} = \frac{6,791.5}{176.58} = 38.46\ \text{m}^3/\text{s}$$

The river supplies 25 m³/s, so the deficit of $(38.46-25)$ m³/s for 4 hours must come from the pond:

$$V = (38.46-25)\times4\times3600 = 193,846\ \text{m}^3 \approx 1.94\times10^5\ \text{m}^3$$

(The pond fills during the 20 off-peak hours by $(25-22.31)\times20\times3600 = 193,846$ m³, the same volume.)

Answer: generator capacity = 6.79 MW; pondage volume ≈ 193,846 m³ (≈ 1.94 × 10⁵ m³).

Data

Inflow $Q=40$ m³/s; daily inflow volume $=40\times86400 = 3,456,000$ m³. The plant works 6 days (Sunday-Friday) and on Saturday the whole flow is stored. $H_{full}=25$ m, maximum fluctuation $=1$ m, $\eta_t=0.80$, $\eta_g=0.90$.

(i) Surface area of the pond

1. Weekly pondage (Saturday flow): all of Saturday's inflow is stored for use on the other days.

$$V_1 = 40\times86400 = 3,456,000\ \text{m}^3$$

2. Daily pondage: two requirements are given.

• To take care of the daily flow fluctuation: $0.30\times3,456,000 = 1,036,800$ m³
• For the 50% daily load factor: mean flow to the turbines on working days $=40\times\frac{7}{6}=46.67$ m³/s, so the pondage $=0.2\times46.67\times86400 = 806,400$ m³

The pond must satisfy both, so the larger value is taken: $V_2=1,036,800$ m³.

3. Total live storage:

$$V = V_1+V_2 = 3,456,000+1,036,800 = 4,492,800\ \text{m}^3$$

The allowed draw-down is 1 m, so

$$A = \frac{V}{\Delta h} = \frac{4,492,800}{1} = 4,492,800\ \text{m}^2 \approx 4.49\ \text{km}^2$$

(ii) Weekly output at the switchboard

The whole weekly inflow is used: $40\times7\times24 = 6,720$ m³/s·h. The pond level goes down by 1 m, so the average head is $25-0.5 = 24.5$ m.

$$E = 9.81\,\eta_t\eta_g\,H\,\sum Q\,t = 9.81\times0.80\times0.90\times24.5\times6,720 = 1,162,885\ \text{kWh}$$

(With the full-pond head of 25 m, the figure would be 1,186,618 kWh.)

Answer: pond surface area ≈ 4.49 km² (4,492,800 m²); weekly output ≈ 1.16 GWh (1,162,885 kWh).

(i) Pure RoR plant (no pondage)

Only the minimum flow can be used continuously:

$$P_1 = 9.81\,\eta\,Q\,H = 9.81\times0.85\times30\times70 = 17,511\ \text{kW} = 17.51\ \text{MW}$$

(ii) Peaking plant, 6 h peak, two units

Let $Q_u$ be the discharge of one unit at full load. Off-peak (18 h) only one unit runs, using $Q_u$. In peak hours (6 h) both units run, using $2Q_u$.

Water stored in the 18 off-peak hours (inflow 30 m³/s) after 5% loss:

$$V_s = 0.95\,(30-Q_u)\times18\times3600$$

During the 6 peak hours the extra flow required is $(2Q_u-30)$, supplied from storage:

$$(2Q_u-30)\times6\times3600 = 0.95\,(30-Q_u)\times18\times3600$$ $$12Q_u - 180 = 513 - 17.1\,Q_u \Rightarrow Q_u = \frac{693}{29.1} = 23.814\ \text{m}^3/\text{s}$$

Peak discharge $=2Q_u=47.629$ m³/s.

$$P_2 = 9.81\times0.85\times47.629\times70 = 27,801\ \text{kW} = 27.80\ \text{MW}$$

Check: pondage volume $=400,825$ m³ (about 4.01 × 10⁵ m³). Off-peak one unit uses 23.81 m³/s (less than 30), so the surplus goes to the pond.

Answer: (i) installed capacity = 17.51 MW (pure RoR); (ii) installed capacity = 27.80 MW with 6-hour peaking (2 units of 13.90 MW each), a gain of 1.59 times.

i) Pure run-of-river plant

Minimum flow is 74 m³/s (February).

$$P_1 = 9.81\times0.85\times74\times200 = 123,410\ \text{kW} = 123.41\ \text{MW}$$

ii) Peaking plant with three units, 4 h peaking

One unit (discharge $Q_u$) runs at full load for the 20 off-peak hours; all three units ($3Q_u$) run for the 4 peak hours. Daily water balance on the minimum flow of 74 m³/s:

$$20\,Q_u + 4\,(3Q_u) = 24\times74 \Rightarrow 32\,Q_u = 1776 \Rightarrow Q_u = 55.50\ \text{m}^3/\text{s}$$

Peak discharge $=3Q_u=166.50$ m³/s.

$$P_2 = 9.81\times0.85\times166.50\times200 = 277,672\ \text{kW} = 277.67\ \text{MW}$$

(Each unit is 92.56 MW. Check storage: off-peak stored $=(74-55.5)\times20 = 370$ m³/s·h; peak draw $=(166.5-74)\times4=370$ m³/s·h, equal.)

iii) Increase in benefit during the minimum flow month

Daily energy is the same in both cases ($24\times123.41 = 2,962$ MWh).

• RoR (constant output): 4 peak hours at Rs 12 and 20 off-peak hours at Rs 6 per kWh:

$$B_1 = 123,410\times(4\times12+20\times6) = \text{Rs }20,732,846\text{ per day}$$

• Peaking plant: $B_2 = 92,557\times20\times6 + 277,672\times4\times12 = \text{Rs }24,435,140$ per day

Increase $=3,702,294$ Rs/day $\approx$ Rs 3.70 million per day, or about Rs 111.1 million for a 30-day month.

Answer: (i) 123.4 MW; (ii) 277.7 MW; (iii) extra benefit ≈ Rs 3.70 million per day (≈ Rs 111 million per month).

Data

$H=100$ m, $\eta=0.85$, so $9.81\,\eta\,H = 833.85$ kW per m³/s. One month $=730$ h $=2.628\times10^6$ s.

i) RoR designed for $Q_{40}$

Flows in descending order: 2500, 2100, 2000, 1800, 900, 500, 320, 300, 300, 140, 120, 100. By the Weibull formula $p=\frac{m}{N+1}\times100$ with $N=12$, $p=40\%$ gives $m=0.4\times13=5.2$, which lies between 900 (m = 5) and 500 (m = 6):

$$Q_{40} = 900 - 0.2\,(900-500) = 820\ \text{m}^3/\text{s}$$ $$P = 833.85\times820 = 683,757\ \text{kW} = 683.8\ \text{MW}$$

• Spill (flow above $Q_{40}$) by month (m³/s): Jun 980, Jul 1180, Aug 1680, Sep 1280, Oct 80, total $=5,200$ m³/s-months.

$$E_{spill} = 833.85\times5,200\times730 = 3,165\ \text{GWh}$$

• Firm energy (minimum flow 100 m³/s available all year):

$$E_{firm} = 833.85\times100\times8760 = 730.5\ \text{GWh}$$

• (Total energy of the RoR plant $= 833.85\times5,880\times730 = 3,579$ GWh, so the non-firm energy is 2,849 GWh.)

ii) Storage with full regulation

The design discharge equals the mean flow:

$$\bar Q = \frac{11,080}{12} = 923.33\ \text{m}^3/\text{s}$$

Deficits ($\bar Q-Q$) in the months with flow below the mean, from Oct to May (m³/s-months): Oct 23.33, Nov 423.33, Dec 623.33, Jan 823.33, Feb 803.33, Mar 783.33, Apr 623.33, May 603.33. Sum $=4,706.67$ (this equals the surplus of Jun-Sep, which refills the reservoir).

$$V_s = 4,706.67\times2.628\times10^6 = 12,369\times10^6\ \text{m}^3 \approx 12.37\ \text{billion m}^3$$

iii) Installed capacity and energy of the storage project

$$P = 833.85\times923.33 = 769,922\ \text{kW} = 769.9\ \text{MW}$$ $$E = P\times8760 = 6,745\ \text{GWh/year}$$

(The head is taken as constant at 100 m.)

Answer: (i) P = 683.8 MW at $Q_{40}=820$ m³/s; spill energy ≈ 3,165 GWh; firm energy = 730.5 GWh. (ii) storage = 12,369 Mm³. (iii) P = 769.9 MW; E = 6,745 GWh.

Monthly flows from $Q_t = 5.589t^2 - 51.275t + 139.94$

$H=150$ m, $\eta=0.85$: $9.81\,\eta\,H = 1250.77$ kW per m³/s.

Design discharge $Q_{40}$

Descending flows: 329.46, 252.18, 186.09, 131.17, 94.25, 87.44, 59.75, 54.88, 36.42, 33.49, 24.26, 23.29. Weibull position for 40%: $m=0.4\times13=5.2$ (between the 5th flow 94.25 and the 6th flow 87.44):

$$Q_{40} = 94.25 - 0.2\,(94.25-87.44) = 92.89\ \text{m}^3/\text{s}$$

a) RoR project

$$P = 1250.77\times92.89 = 116,185\ \text{kW} = 116.18\ \text{MW}$$

The minimum monthly flow is 23.29 m³/s (February), the flow available in all months, so

$$E_{firm} = 1250.77\times23.29\times8760 = 255,183,616\ \text{kWh} = 255.2\ \text{GWh}$$

(Total energy with flows limited to $Q_{40}$ is 715.8 GWh; the part above firm energy is secondary energy.)

b) Peaking run-of-river (6 h daily peaking)

Daily water volume at $Q_{40}$: $Q_{40}\times24\times3600$. Half of it (50%) is used during the 18 off-peak hours and the other half is stored (pond volume $=0.5\times92.89\times86400=4,012,865$ m³) and used in the 6 peak hours:

$$Q_{peak} = \frac{0.5\,Q_{40}\times24}{6} = 2\,Q_{40} = 185.78\ \text{m}^3/\text{s}$$ $$P_{PRoR} = 1250.77\times185.78 = 232,370\ \text{kW} = 232.37\ \text{MW}$$

Off-peak the plant uses $0.5\,Q_{40}\times24/18 = 61.93$ m³/s. The installed capacity is therefore twice that of the RoR plant.

Answer: (a) RoR: installed capacity = 116.2 MW, firm energy = 255.2 GWh; (b) PRoR: installed capacity = 232.4 MW.

Data

$H=150$ m, $\eta=\eta_t\eta_g = 0.90\times0.95=0.855$. Environmental flow = 10% of each month's flow, so the flow available for power is $0.9\,Q$. One month $=2.628\times10^6$ s.

Sum of flows $=900$ m³/s, mean $=75$ m³/s. Mean available flow $=0.9\times75 = 67.5$ m³/s. With full regulation the design discharge is this mean flow, so the plant runs at constant $Q_d=67.5$ m³/s.

Reservoir capacity (mass-curve method)

The reservoir is full at the start of December, when inflow drops below the draft. Cumulative of (0.9Q − 67.5):

The cumulative reaches its lowest value of $-166.5$ m³/s-months in June; from July the surplus refills the reservoir.

$$V = 166.5\times2.628\times10^6 = 0.0\times10^6\ \text{m}^3 \approx 0\ \text{Mm}^3$$

Installed capacity

$$P = 9.81\,\eta\,H\,Q_d = 9.81\times0.855\times150\times67.5 = 84,924\ \text{kW} = 84.92\ \text{MW}$$

Annual energy

$$E = P\times8760 = 743.9\ \text{GWh}$$

(The head is taken as constant at 150 m; in practice the head varies with the reservoir level.)

Answer: reservoir capacity ≈ 0 Mm³; installed capacity ≈ 84.9 MW; annual energy ≈ 744 GWh.

Water needed by the turbine

Take 1 HP = 0.7457 kW. Output $=1000\times0.7457 = 745.7$ kW, net head 30 m, overall efficiency 0.80:

$$Q = \frac{P}{9.81\,\eta\,H} = \frac{745.7}{9.81\times0.80\times30} = 3.167\ \text{m}^3/\text{s}$$

Monthly demand volume $D = 3.167\times2.628\times10^6 = 8.32\times10^6$ m³ $=8.32$ Mm³ (constant draft).

As the data are given (1000 HP)

The demand (8.32 Mm³/month) is far below the smallest monthly inflow (70 Mm³), so the inflow is greater than the demand in every month and the reservoir never goes short.

• (i) Minimum capacity required $=0$ Mm³ (no storage is needed for this load).
• (ii) Water wasted $=\sum(\text{inflow}) - 12D = 1067 - 99.88 = 967.1$ Mm³ per year.

If the turbine is 10,000 HP (probable intended load)

The inflow data (mean 88.9 Mm³/month) match a turbine about ten times larger, so the working for 10,000 HP is also shown. $P=7,457$ kW, $Q=31.67$ m³/s, monthly demand $D=83.24$ Mm³.

The reservoir (full at the start of the year) falls to empty in May, so the minimum capacity is the deficit accumulated from February to May: 29.9 Mm³. The inflow after June refills it, and the water that cannot be stored (spill) adds to 76.4 Mm³ per year.

Answer: for the 1000 HP load stated, minimum capacity = 0 and wasted water = 967 Mm³. For a 10,000 HP turbine, minimum capacity ≈ 29.9 Mm³ and wasted water ≈ 76.4 Mm³.

Units are m³/s-months; one m³/s-month $=2.628\times10^6$ m³.

i) Mass curve

Mean inflow $=\dfrac{\sum Q}{12} = \dfrac{179}{12} = 14.917$ m³/s, taken as the uniform draft.

 Cum.
 volume
  179 |                          ____*   inflow mass curve
      |                   ___---
      |              _--*      draft line (slope = mean)
      |        __.-*      .-'
      |   _.-*      _.--'
      | *     _.--'
      +---------------------------
       Oct                    Sep

The mass curve of inflow lies above the draft line from October to February (surplus) and below it from March to September (deficit, the reservoir is drawn down).

ii) Storage capacity and critical period

The largest vertical gap between the draft line (tangent to the mass curve at the end of February) and the mass curve is at the end of September. This is the sum of the deficits from March to September:

$$S = \sum(\bar Q - Q)_{Mar-Sep} = (4.917+6.917+7.917+8.917+10.917+11.917+11.917) = 63.42\ \text{m}^3/\text{s-months}$$ $$S = 63.42\times2.628\times10^6 = 166.7\times10^6\ \text{m}^3$$

Critical period $=7$ months (March to September), when the reservoir goes from full to empty. It refills by about February of the next year.

iii) Draft = 80% of mean inflow

$0.8\times14.917 = 11.933$ m³/s. Deficit months are those with inflow below this value (Mar to Sep: 10, 8, 7, 6, 4, 3, 3):

$$S_{80} = (1.933+3.933+4.933+5.933+7.933+8.933+8.933) = 42.53\ \text{m}^3/\text{s-months} = 111.8\times10^6\ \text{m}^3$$

The critical period is still March to September (7 months). The storage required is reduced from 166.7 Mm³ to 111.8 Mm³, a saving of 33%, but the water supplied also falls by 20%.

Answer: storage for the mean-flow draft = 166.7 Mm³ (63.42 m³/s-months), critical period = 7 months; storage for 80% draft = 111.8 Mm³ (42.53 m³/s-months).

The reservoir must supply the mean flow as a constant draft, so the storage is the largest cumulative deficit.

Mean monthly volume $=\dfrac{6,671}{12} = 555.9\times10^6$ m³.

Months 4 to 8 have a surplus and fill the reservoir. Months 9 to 12 and 1 to 3 (the following year) are deficit months. Cumulative deficit from month 9 (reservoir full at the start of month 9):

The maximum cumulative deficit is 1802.4 × 10⁶ m³.

Answer: required storage capacity ≈ 1,802 × 10⁶ m³ (about 1,802 million m³).

Volumes are in 10⁶ m³ per month.

(i) Maximum uniform draw-off, reservoir capacity, minimum initial storage

Total inflow of the driest year $=28.75$. The maximum uniform draw-off possible in the year is the mean monthly inflow:

$$x = \frac{28.75}{12} = 2.396\times10^6\ \text{m}^3/\text{month}$$

Mass-curve table (cumulative of inflow − draw-off, starting from January):

• The cumulative reaches its highest value (+4.062) at the end of March and its lowest (-6.458) at the end of October.
• Minimum initial storage (January 1st) needed so that the demand is never short: $6.458\times10^6$ m³.
• Reservoir capacity = peak − trough $=4.062-(-6.458) = 10.521\times10^6$ m³.

(ii) Capacity 8 × 10⁶ m³, initial storage 4 × 10⁶ m³

The reservoir is smaller than 10.52, so some water is spilled when it is full and the draw-off must be reduced so that the storage never becomes negative. Trial routing gives the largest draw-off for which the reservoir just reaches zero (in October):

(a) Maximum uniform draw-off $=2.036\times10^6$ m³/month ($\approx0.77$ m³/s).

(b) Spillage $=1.143\times10^6$ m³ in the year (in March, when the reservoir is full).

Answer: (i) draw-off = 2.396 Mm³/month, capacity = 10.52 Mm³, minimum initial storage = 6.46 Mm³. (ii) (a) draw-off = 2.036 Mm³/month; (b) spillage = 1.14 Mm³.