I recently received a question regarding the placement of a hydropower plant on a diverted outlet. Since I had not done that before, I developed this example to demonstrate this in HEC-ResSim.
In the figure below, there are two reservoirs on the main stem of the river. There is also a tributary entering the main stem at the junction named "confluence".
There is a diverted outlet coming from the upstream reservoir and re-entering the main stem at the junction named "hydro return".
Here are some of the pertinent parameters of the model:
1.) There is 1,200 cfs entering the system at both the upstream end of the main stem and the upstream end of the tributary.
2.) The upstream reservoir has a controlled outlet at the dam (capacity 1,500 cfs) and the diverted outlet from the reservoir. On the diverted outlet, there is a power plant with a hydraulic capacity of 1,500 cfs, a generating capacity of 4 MW, and an efficiency of 85%. The station use and hydraulic losses are set to zero in our example.
3.) The downstream reservoir has a controlled outlet with a capacity of 1,500 cfs.
4.) The tailwater associated with the power plant is a constant 50 ft.
5.) The top of conservation for both reservoirs is set to 75 ft and this elevation is held throughout the entire simulation by passing inflow.
Note that on the upstream reservoir, two outlet groups are modeled. One outlet group is at the dam and one is the diverted outlet from the reservoir. Also note that there is a tailwater associated with the power plant. This is needed for HEC-ResSim to compute the generation.
The operations set for the upstream reservoir contains no rules. Since our simulation will start at the top of conservation, inflow will be passed to hold the top of conservation.
Since we have two outlet groups at the upstream reservoir, I set the release allocation to "Balanced" to have half the inflow go through the dam and half the inflow go to the diverted outlet.
The downstream reservoir will also pass inflow.
Below are the lookback conditions. Both reservoirs start at the top of conservation of 75 ft. The release through the dam of the upstream reservoir is 600 cfs (half the inflow). The other half of the inflow goes to the diverted outlet. The downstream reservoir will be passing 600 cfs of inflow since the 600 cfs going through the diverted outlet re-enters the system below the downstream reservoir.
In the reach below the upstream reservoir, there is 600 cfs for both the lookback and simulation period. This is what we should expect to see.
At the junction named "confluence", there is 1,800 cfs. Of this total, 600 cfs comes from the main stem while 1,200 cfs comes from the tributary. The 1,200 cfs is shown as the local contribution in the figure below.
At the junction named "hydro return", the total flow is 2,400 cfs. This is due to an additional 600 cfs coming in from the diverted outlet.
The computation of power is shown below.
Power computed in MW is equal to (Q*w*h*e) / 737,560
Q in cfs
w in lb/ft^2
h is head differential in feet (75 ft - 50 ft for this example)
e is efficiency
MW = (600*62.4*25*.85)/737,560 = 1.08 MW
The line at the top of the plot shows the potential generation assuming the entire hydraulic capacity is used:
MW = (1500*62.4*25*.85)/737,560 = 2.70 MW
In the figure below, there are two reservoirs on the main stem of the river. There is also a tributary entering the main stem at the junction named "confluence".
There is a diverted outlet coming from the upstream reservoir and re-entering the main stem at the junction named "hydro return".
Here are some of the pertinent parameters of the model:
1.) There is 1,200 cfs entering the system at both the upstream end of the main stem and the upstream end of the tributary.
2.) The upstream reservoir has a controlled outlet at the dam (capacity 1,500 cfs) and the diverted outlet from the reservoir. On the diverted outlet, there is a power plant with a hydraulic capacity of 1,500 cfs, a generating capacity of 4 MW, and an efficiency of 85%. The station use and hydraulic losses are set to zero in our example.
3.) The downstream reservoir has a controlled outlet with a capacity of 1,500 cfs.
4.) The tailwater associated with the power plant is a constant 50 ft.
5.) The top of conservation for both reservoirs is set to 75 ft and this elevation is held throughout the entire simulation by passing inflow.
Note that on the upstream reservoir, two outlet groups are modeled. One outlet group is at the dam and one is the diverted outlet from the reservoir. Also note that there is a tailwater associated with the power plant. This is needed for HEC-ResSim to compute the generation.
The operations set for the upstream reservoir contains no rules. Since our simulation will start at the top of conservation, inflow will be passed to hold the top of conservation.
Since we have two outlet groups at the upstream reservoir, I set the release allocation to "Balanced" to have half the inflow go through the dam and half the inflow go to the diverted outlet.
The downstream reservoir will also pass inflow.
Below are the lookback conditions. Both reservoirs start at the top of conservation of 75 ft. The release through the dam of the upstream reservoir is 600 cfs (half the inflow). The other half of the inflow goes to the diverted outlet. The downstream reservoir will be passing 600 cfs of inflow since the 600 cfs going through the diverted outlet re-enters the system below the downstream reservoir.
In the reach below the upstream reservoir, there is 600 cfs for both the lookback and simulation period. This is what we should expect to see.
At the junction named "confluence", there is 1,800 cfs. Of this total, 600 cfs comes from the main stem while 1,200 cfs comes from the tributary. The 1,200 cfs is shown as the local contribution in the figure below.
At the junction named "hydro return", the total flow is 2,400 cfs. This is due to an additional 600 cfs coming in from the diverted outlet.
The computation of power is shown below.
Power computed in MW is equal to (Q*w*h*e) / 737,560
Q in cfs
w in lb/ft^2
h is head differential in feet (75 ft - 50 ft for this example)
e is efficiency
MW = (600*62.4*25*.85)/737,560 = 1.08 MW
The line at the top of the plot shows the potential generation assuming the entire hydraulic capacity is used:
MW = (1500*62.4*25*.85)/737,560 = 2.70 MW
Dear Kevin,
ReplyDeleteI have 2 reservoirs (A,B) and a diversion canal from B to A. I need a simple canal release rule based on elevation (modeled values) in both reservoirs i.e. if elevB > elevA then canal diverts water (flow > 0)) else if elevB<=elevA the diversion closed (flow = 0).
I have been trying multiple (if else), setups but anyway I am getting flow in the diversion canal when elevB<= elevA.
In addition to the diversion canal, I have ecological outlet in conservation zone, and spillway in the flood zone of reservoir B.
Any hints, thank you very much,
Best regards,
Greg