bhandariroshan / simulating-power-generation-in-reservoir-based-hydropower

Simulating-Power-Generation-in-Reservoir-Based-Hydropower

This algorithm helps simulate power generated for different time during pick hour and off hour in a reservoir using height of the water available in reservoid for each day for 12 months

NOTE: To run the code, you need to have python 2.7 installed Running the code:

1. python power.py

2. Input following:

   a. Number of days in the month

   b. Input in the month

   c. Number of "I" in per peak shift

   d. Number of hrs in non peak shift

Equation of height of water in reservoir

f(h)= 13557.8282850575x2 - 23743654.1706026000x + 10398688014.4791000000

Equation of derivative of height

f'(h)=2* 13557.8282850575-23743654.1706026000

Algorithm:

1. Ask for intial height of water "h", power to be generated "p" ,month count=1
2. Main loop starts here; month count = monthcount + 1
3. If month count > 12, exit all the loops go to 4 and type year finished 3.1 day=1

3.1.1 Ask for number of days in the month and input in the month, number of "I" in per peak shift, number of hrs in non peak shift

3.1.1.1 day = day+1

3.1.1.2 if day>no. of days in month, go to step 2

3.1.1.3 First shift loop starts here; i=1

3.1.1.3.1 dh=f(h)*(p/9810/h-input)*3600*10/f'(h)

3.1.1.3.2 h=h-dh

3.1.1.3.3 i=i+1

3.1.1.3.4 if i> number of "I" per peak shift, go to 3.1.1.4 else go to 3.1.1.3.1

3.1.1.4 new volume=f(h)+inputnumber of hours in non peak shift3600

3.1.1.5 new height h= solution of eqn(13557.8282850575x2 - 23743654.1706026000x + 10398688014.4791000000 = new volume from step 3.1.1.4) precision upto 0.001 needed here

3.1.1.6 second shift loop starts here; i=1

3.1.1.6.1  dh=f(h)*(p/9810/h-input)*3600*10/f'(h)

3.1.1.6.2  h=h-dh

3.1.1.6.3  i=i+1

3.1.1.6.4  if i> number of "I" per peak shift, go to step 3.1.1.7 else, go to 3.1.1.6.1

3.1.1.7 new volume=f(h)+input*number of hours in non peak shift*3600

3.1.1.8 new height h= solution of eqn(13557.8282850575x2 - 23743654.1706026000x + 10398688014.4791000000=new volume from step 3.1.1.7) *precision upto  0.001 needed here*

3.1.1.9 print the end of the day height "h" and go to 3.1.1.1

4. Print f(h) at the end of year

5. End