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Assessment of Resource Potential of Submersible Water Turbine

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Presentation on theme: "Assessment of Resource Potential of Submersible Water Turbine"— Presentation transcript:

1 Assessment of Resource Potential of Submersible Water Turbine
On 8th Feb, 2018 at 10:00 HRS PST Author: Dr. Muhammad Nawaz Akhtar

2 Assessment of Resource Potential
Following SAARC Member States are blessed with a good irrigation canal system ; India Pakistan Moreover, following SAARC Member States have plenty of rivers carrying hydrokinetic electrical power potential; Afghanistan Bhutan Nepal Bangladesh Sri Lanka Principle of Operation : As that of wind turbines Similar design concepts may be employed. The only difference; Density of water is 850 times greater than the density of air. Energy of water stream = 850 times greater than a wind turbine for the same air flow. Dissemination Webinar

3 Types of resource potential
Theoretical Potential Technical Potential Economical potential Dissemination Webinar

4 Calculation of resource potential of selected site
1. Approach to calculate resource potential 2. Calculation of theoretical resource potential 3. Calculation of Technical resource potential 4. Calculation of Economic resource potential Dissemination Webinar

5 1. Approach to Calculate Resource Potential
A survey was conducted to measure the following parameters to assess the potential of electrical power from the upper Jehlum canal. Flow of water/discharge (m3/s) Velocity of water (m/s) Depth of water (m) Availability of water in days over the year Dissemination Webinar

6 Theoretical Potential
The theoretical resource potential is the annual average amount of physical energy that is hypothetically available.  It can be calculated by following considerations; Average flow rate of water (m3/s) Change in hydraulic head between the beginning and end of canal (m) Dissemination Webinar

7 2. Calculation of theoretical resource potential of Selected Site
Following are the general parameters of the Upper Jehlum Canal. Velocity of water in the main stream = m/s. Velocity of water in the channels under the bridge = m/s No. of channels under the bridge= 17 A raised siphon concrete x-section of length= 73.2 m (near to Jaggu where 114 small turbines can be installed). Dissemination Webinar

8 Existing Parameters of Upper Jehlum canal
Two sections of Upper Jehlum canal First section up to 74.4 km 2nd Section from km Discharge, Q (m3/s) 116.09 Bed, width, B (m) 69.54 65.5 Full Supply Depth, (m) 3.2 2 Side Slope (Rise /Run) 1.5 1 Longitudinal Slope (1/Slope) Head/Length 6667 Dissemination Webinar

9 Dissemination Webinar
History of daily discharge report of Upper Jehlum canal (February –March 2017). Dissemination Webinar

10 Theoretical Potential It was calculated based on the data available in the Table presented at Slide#8, by utilizing the following expression; Pth = γ Q ΔH ……… Eq-1 Where, γ = Specific weight of water [9789 N/m3)] at 20 oC….. Q = Flow rate of water (m3/s) ΔH= Change in hydraulic head between the beginning and end of canal (m). Length of Canal= km (two sections) Dissemination Webinar

11 Continued from Previous Slide
Both Sections with different geometries and water flow rates. Hence, we have to calculate the theoretical power potential for the two sections separately. First section length =74.4 km 2nd Section length =74.4 to km Dissemination Webinar

12 Calculation of Theoretical Resource Potential (Fist section) (P1th)
Using the Equation Pth = γ Q ΔH ……… γ = 9789 N/m3 Q= m3/s ΔH= m………… [Slope (1/6667) x length (74.4 Km)] We get P1th = 9789 x x 11.16 = w = MW Dissemination Webinar

13 Calculation of Theoretical Resource Potential (2nd Section) (P2th)
By using the Equation Pth = γ Q ΔH ……… γ = 9789 N/m3 Q= m3/s ΔH= 7.95 m………… [Slope (1/6667) x length (53 Km)] We get P2th = 9789 x x 7.95 = w = 9.5 MW Total Theoretical Resource Potential of this canal, Pth = P1th + P2th = MW MW Pth = MW Dissemination Webinar

14 Technical Potential The technical resource potential is the portion of a theoretical resource that can be captured and converted into electricity using a specific technology. It can be calculated by following considerations; Velocity of flowing water stream Width of canal or river Depth of water Rotor diameter of turbine Inter Turbine spacing in the arrays Distance between two arrays Availability of resource in days System and topographic constraints Land use constraints System performance, etc. Temperature of water Dissemination Webinar

15 5KW floating type submersible turbine developed by Smart-Hydro Germany
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16 3. Calculation of Technical Resource Potential (Ptech)
Due to parametric variations the canal has been divided into following 4 sections; Water Level Crossing Bridges = 4 Siphons = 18 1st section of canal 2nd section of canal Dissemination Webinar

17 Installation of two turbines in one channel under the bridge
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18 Technical Potential under Bridges (PBtech) 1/2
General Parameters; No. of channels under the bridge = 17 Channel width = 6.1 m Water depth in the channel = 1 m Velocity of water in channel slope = 3.6 m/s Rotor diameter of the turbine = 1m Proposed minimum distance between two turbines = 2.5 m Number of turbines in one channel = 2 Total number of turbines under the bridge = 17 x 2 = 34 Number of similar level crossings (n) = 4 Total turbines on the 4 level crossings = 34 x 4 = 136 Dissemination Webinar

19 Technical Potential under Bridges (PBtech) 2/2
Technical Potential for 136 turbines =? (Characteristic curve on next slide#21). Power produced under the bridges = 136 x 5 Kw = 680 Kw Losses in cascading and = 68 Kw (PBtech)= Kw = MW Dissemination Webinar

20 Operating curve of the 5 kW submersible turbine used as a model for calculation
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21 2. Calculation of Technical Resource Potential on the Siphons (Pstech)
No of Siphons = 18 Velocity of flowing water on Siphon= 1.08 m/s Width of the siphon = m Width of one turbine = 1m Proposed minimum distance between two turbines = 2.5 m Number of turbines proposed in one row = / (2.5+1) As proposed on Slide#21 = 19 The length of the Siphon = 91.5 m Proposed distance between two rows = 15D (15x 1m) = 15 m Total rows of turbines = 91.5/15 = 6 rows Number of turbines on one siphon = 19 x 6 = 114 Total turbines = 114 x 18 = 2052 Dissemination Webinar

22 Proposed schematic arrays of the turbines on one of the siphon head
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23 Calculation of Technical Resource Potential on the Siphons (Pstech)
2/2 By using the proposed model turbine with; Rotor diameter = 1 m Max. Power = 5 Kw Reference, characteristic curve of this turbine on Slide #20 At a water velocity= 1.08 m/s The expected power from this turbine = 200 W Total power from the siphon = 200 x W = 410 KW Losses in cascading and = = 369 KW (Pstech) = MW Dissemination Webinar

24 Technical Potential in the 1st Section Main canal (PM1tech)
1/3 Technical Potential in Canal length of km = (PM1tech)H Velocity of flowing water = 0.91 m/s Width of canal = 68.62 Depth of water in canal = 2.86 m Number of turbines proposed in one row = / (2.5+1) Number of turbines in one row = 19 Proposed distance between two rows = 30D (30x 1m) Number of rows in m length of canal = 74400/30 = 2480 No of rows missed by roads & =248 No of rows accounted for in 6*9 = 54 No of rows accounted for under Bridges = 4 Net No. of rows =2480-( ) = 2174 Number of turbines in 1st section of main canal= 2174 x 19 = 41306 Dissemination Webinar

25 Technical Potential in the 1st Section Main canal (PM1tech) 2/3
Proposed schematic arrays of the turbines that may be installed in the main canal. Dissemination Webinar

26 Technical Potential in the 1st Section Main canal (PM1tech) 3/3
By using the proposed model turbine with; Rotor diameter = 1 m Max. Power = 5 Kw Reference, characteristic curve of this turbine on Slide #20 At a water velocity = 0.91 m/s The expected power from this turbine = 200 w Total power from 1st section = 200 x W = 8261 KW = 8.26 MW Losses in cascading and = MW Net power from 1st Section = 7.43 MW The proposed schematic arrays of the turbines in the main canal have been shown in the slide# 23. Technical Potential in the 1st Section Main canal = 7.43 MW Dissemination Webinar

27 Technical Potential in the 2nd Section (PM2tech)T. 1/2. (from 74
Technical Potential in the 2nd Section (PM2tech)T / (from km) Length of 2nd section = 53 km Velocity of flowing water = 0.91 m/s Width of canal = m Depth of water in canal = 2 m Number of turbines proposed in one row = / (2.5+1) Number of turbines in one row = 18 Proposed distance between two rows = 30D (30 x 1m) Number of rows in m length of canal = 53000/30 = 1766 No. of rows missed by roads & = 176 No of rows accounted for on 6 x = 54 Net No. of rows = 1766-( ) = 1536 Number of turbines in 2nd section of main canal= 1536 x 18 = 27648 Dissemination Webinar

28 Technical Potential in the 2nd Section (PM2tech)T 2/2
By using the proposed model turbine with; Rotor diameter = 1 m Max. Power = 5 Kw Reference, characteristic curve of this turbine on Slide #18 At a water velocity = 0.91 m/s The expected power from this turbine = 200 W Total power from 2nd section = 200 x W = 5529 KW = 5.53 MW Losses in cascading and = MW Net Technical Potential in the 2nd Section = 5.53 – 0.55 MW = 4.98 MW Total power from main canal = (PM1tech)H + (PM2tech)T (PMtech) = MW = MW Dissemination Webinar

29 Calculation of Technical Resource Potential (Ptech) (PBtech) + (PStech) + (PMtech)
= MW Net resource potential of this canal = MW This much power can power; 2 KW= or Cottage KW = 669 or Irrigation tube KW = 669 Dissemination Webinar

30 Considerations for economical Potential; Projected cost of the project
Projected power from the project Availability of water resource in days Availability of plant factor Upfront tariff Annual O&M cost Annual revenue Annual profit Dissemination Webinar

31 Calculation of Economical Resource Potential
Projected Power from the Upper Jehlum Canal = KW = x 24 Considering, Canal and Turbines availability for 300 days annually = x 24 x = KWh As upfront tariff for’ Micro hydro power plants by NEPRA (Pakistan) = Rs.12.7 / KWh Hence, Annual revenue from the upper Jehlum canal = PKRs x 12.7 =PK Rs Million PKRs. 110 =1US$ = /110 = US$ Million Dissemination Webinar

32 Calculation of Economical Resource Potential 2/2
The cost of the project For hydrokinetic power plants ; The average equipment cost/ MW = U.S $ 3.0 – 4.0 Million (M) The project cost/MW @50% of Equipment cost = US$4.0 M x = Hence, installed cost of the project = U.S $ x 4.0 x 1.5 M = U.S $ 82.5 M Annual O&M cost of the project (2.4% of the project cost) = U.S $ 82.5 x 2.4/100 = U.S $ M Annual net profit = U.S $ (11.12 – 1.98 M = U.S $ 9.14 M = PKRs M Economic Resource Potential of the canal = PKRs M annually Payback period = Total cost of the project/Annual net profit = U.S $ 82.5 M / U.S $ 9.14 M = 9 years Which is a reasonable period over the project life cycle, in this regard. Dissemination Webinar

33 Governing equation for the Theoretical Resource Potential (Pth) is ;
Generic model for the calculation of hydrokinetic resource potential of any site Governing equation for the Theoretical Resource Potential (Pth) is ; Pth = γ Q ΔH ………….. Eq-1 Where, γ = The specific weight of water. Q = Flow rate of water (M3/s). ΔH = Elevation difference of the starting and ending points of the canal or river (m). Dissemination Webinar

34 Dissemination Webinar

35 Governing equation for Economic Resource Potential is;
Economic Potential = Power Produced (Kw) * Hours/day (H)* No. of days turbine produced power (D) x Tariff Dissemination Webinar

36 Dissemination Webinar
QUESTIONS & ANSWERS ? ? ? ? ? ? ? ! ! ! ! ! ! ! ! ! Dissemination Webinar


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