0
\$\begingroup\$

Here is the generator I intend to use on my project which I obtained from this site.

generator
(source: seeedstudio.com)

Could you help me to get a tip/direction on how to start the calculation for this. I search some (might be wrong), I would need to calculate/measure the head and flow. but I don't have the flow. also the device's efficiency isn't also listed on its parameters.

This is the description of this generator given by the seller:

  1. Unique hydropower separation technology, the unique dual clutch, magnetic circuit coupling volume is only about 4/5 of the normal micro hydroelectric generator, the output power to ordinary hydroelectric generator 2-3 times.

  2. High efficiency, long life of waterway design and low flow loss, low water pressure (0.5 KG), load short circuiting, also can start smoothly

The specifications supplied by the seller are shown below but these do not match those given for apparently similar units sold by other sources. I bought the unit from a local seller due to difficulties with importing them in my situation.

Maximum working pressure: 1.75MPa
Working pressure: 0-1.75Mpa
Test pressure: 1.2MPa
Temperature range:-0-120 degrees Celsius
Flow range: 1.5-20L/min (press 0.05-0.2MPa)
Maximum output voltage: 5v VDC
Maximum water flow rate: 10L/min
Shell material: nylon
Weight:200g

Sold by

Thank you so much for the help.

\$\endgroup\$
6
  • 3
    \$\begingroup\$ If you post all of the information that you have been able to find, someone may be willing to explain it to you. A question that asks for volunteers to research something for you is likely to be closed by moderators. \$\endgroup\$
    – user80875
    Commented Dec 6, 2015 at 15:43
  • 1
    \$\begingroup\$ efficiency is the work done over the work expected to be done. you need an experiment to calculate the efficiency. \$\endgroup\$
    – AHB
    Commented Dec 6, 2015 at 15:44
  • 3
    \$\begingroup\$ Best tip : get a proper datasheet for it. The old rule : no datasheet, no sale is a good one here. \$\endgroup\$
    – user16324
    Commented Dec 6, 2015 at 15:59
  • 2
    \$\begingroup\$ Leon Heller, PeterJ, Daniel Grillo, nidhin, Respawned Fluff -> the question is clear and Charles' answer and my answer address it OK. You can take him out the back and rough him up u bit and give him a blood nose or whatever else makes you feel good when dealing with beginners BUT putting the question on hold is unproductive, at best. \$\endgroup\$
    – Russell McMahon ♦
    Commented Dec 8, 2015 at 4:14
  • 1
    \$\begingroup\$ @RussellMcMahon None of the down/close votes were mine in the first place. But given the changes I'm happy to vote for a re-open. I hear you on poor treatment of newbies, which is why I pointed out what's missing and waited. \$\endgroup\$
    – user16324
    Commented Dec 9, 2015 at 12:28

2 Answers 2

Reset to default
4
\$\begingroup\$

Test this by allowing water to flow through it from a container with a measured volume placed a measured distance above the generator.
Record the time required for a measured volume of water to flow.

  • The rate of flow is volume divided by time.

  • Input power is rate of flow X head X density X units of measure constant.

  • Output power is voltage X current.

  • Efficiency is output power / input power.

\$\endgroup\$
4
  • \$\begingroup\$ Good answer. Easier measurement is to place a container of known volume under outlet and measure time for it to fill (or reach calibrated mark). This way you can have large source tank and move container into flow and use stopwatch to determine fill time. (I use this method for various things with good success). Change in head of source tank may be an issue of course - but more so usually with a small header tank. \$\endgroup\$
    – Russell McMahon ♦
    Commented Dec 6, 2015 at 19:40
  • 1
    \$\begingroup\$ Thank you sir! I really appreciate this. While I'm embarrassed to have lack of knowledge for this and had to ask further questions, How could I measure the head and flow? I've search how to do this, but I only found how to measure head and flow of a river, or a dam, or the like. How could I do this on a faucet sir? Thank you! \$\endgroup\$
    – Nico Dioso
    Commented Dec 9, 2015 at 6:22
  • \$\begingroup\$ @NicoDioso Flow rate is explained in my comment above. Measure the time taken to output a measured volume of water from the turbine. eg iof it outputs 1 litre in 15 seconds the flow rate is 4 litres per minute or 4000cc/60 =~ 67 cc/second. \$\endgroup\$
    – Russell McMahon ♦
    Commented Dec 9, 2015 at 8:34
  • \$\begingroup\$ @NicoDioso Head is the total vertical height between water source and water outlet. Power in water = head in metres x flow in kg/second x 9.8 m/s/s (g) close enough to 10 x h x m . So if you had say 3 metres of head x 66cc/second h=3m m = 0.066 kg/s and power in ~= 3 x 0.066 x 10 =~ 2 Watts. \$\endgroup\$
    – Russell McMahon ♦
    Commented Dec 9, 2015 at 8:38
3
\$\begingroup\$

Very good information (which may be correct) is given on the devices Wiki page which is here

Also see Seed studio specification page

Including:

enter image description here

and

enter image description here

The information provided appears not quite complete and you may have to make a few assumptions - eg from the graphs you can get voltage versus flow rate and pressure but you may need to infer the current it will provide at a given flow rate (compared to the maximum stated current) but the results should give a good indication of what can be expected. I'd expect well under 50% overall efficincy but I may be pleasantly surprised.

Practical measurement of your system will provide the definitive answer. Using what Charles has said:

Efficiency = Volts_out x Current_out / (Water_head x Water_flow_rate x g)

g = 9.8 m/s/s = gravitational acceleration

If you use "head" in metres and flow rate in litres/second then the bottom line becomes h x f x g = metres x litres/second x 9.8 (for water with sg =1).

Related:

Seed Studio data

Working Pressure: < 2KGF
Pressure Drop: 0.4 Bar (3.0 L / Min )
Peak Current: 1.4 A
Stand-by Current: Max. 1.4A
Voltage Output: 3.6VDC --- 700mAh
Power Output: 125mW (4LPM)
Battery Capacity: 300mAh
Max. Flow: 20 L / Min ( 2KGF)
Water-resistance: IP4
High limited voltage: 4.3VDC
Dimensions: L 84.5 mm - W 64.5 mm - H 81 mm
Wire Length: 84 mm
Jack Type: JWT C2521 Terminal Plug
Connect Thread Gauge: G 1/2"
Weight: 165 ± 5g
Burst Pressure: 20KGF
Normal Operation Temp. 4 °C~80 °C
Max. Intermittent Operating Temp. 110 °C max. 30 min
Generating Model: NACuM Core (PAT NO.DE202006004800)
Battery Type: Li-polymer Battery with PCM

See Wiki page for more specifications.

\$\endgroup\$
6
  • \$\begingroup\$ I guess when the OP is so lazy as to not even post a link or product name... reverse image search is how you answer this :-) \$\endgroup\$
    – got trolled too much this week
    Commented Dec 8, 2015 at 3:26
  • \$\begingroup\$ @RespawnedFluff Eaxch to their own, mate. Apparently closing the question and making it hard (or harder) for a person new to the list to deal with people is another way to do it. (Or, perhaps, the people listed above really can't understand what the OP is asking :-) ). RM \$\endgroup\$
    – Russell McMahon ♦
    Commented Dec 8, 2015 at 5:51
  • \$\begingroup\$ @RespawnedFluff More slept now :-) - The OP is from the Philippines. English may or may not be a familiar language. Ability to ask questions well may not have been taught (yet) even though it may seem it should be obvious. Laziness may well be part of the equation but also may be less or much less than seems plausible. Giving a newcomer the benefit of the doubt to help them get going sometimes pays dividends and seldom (but sometimes) causes vast pain. Putting questions on hold can have adverse affects on some vastly beyond what many may realise - that's based on my discussions with a ... \$\endgroup\$
    – Russell McMahon ♦
    Commented Dec 8, 2015 at 10:09
  • \$\begingroup\$ ... number of people who have had that happen to them. Some continue on at an unacceptable level while others benefit from the initial tutelage. Some benefit from the school of hard knocks but it's hard to say who and those who don't may be repelled. I've seen our most famous bovver boy drive off an international expert (in the use of cryo cooled CCDs for astronomy) because the BB did not realise that the expert's terminology was biased by his primary language being French. (Demand -> demande -> ask being an example I recall :-) ). I've seen a person who has designed satellite electronics ... \$\endgroup\$
    – Russell McMahon ♦
    Commented Dec 8, 2015 at 10:15
  • 1
    \$\begingroup\$ Thanks for the responses sir. I'm sorry for the lack of details in my question. I have just used that image, but the specifications is different from what I've also searched on the web. These were really helpful thanks! \$\endgroup\$
    – Nico Dioso
    Commented Dec 9, 2015 at 6:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.