Mae cymuned fach ger Aberystwyth yn dymuno gosod tyrbin trydan dŵr micro yn yr afon leol. Eich tasg chi yw canfod a all yr afon gynhyrchu digon o drydan drwy gydol o flwyddyn ac a yw'n cyd-fynd ag anghenion ynni tymhorol y pentref.
Byddwn yn darparu'r hafaliadau peirianneg angenrheidiol, data llif yr afon a gofynion trydan cartrefi i'ch helpu i wneud penderfyniad.
Cliciwch y botwm 'nesaf' i ddechrau arni.
Mae gan y rhan o'r afon a ddewiswyd ar gyfer y gosodiad hwn gwymp uchder (neu ben) yn mesur 3.2 metr.
Byddai'r tyrbin a ddewiswyd yn rhedeg gyda gwerth effeithlonrwydd o 0.6 - sy'n golygu am bob joule o ynni potensial sy'n dod i mewn i'r system, caiff 0.6 joule o drydan ei gynhyrchu.
Mae rhai cysonion sydd angen i ni eu cadw mewn cof hefyd:
Dwysedd dŵr = 1000kg/m3
Disgyrchiant = 9.8m/s2

Dyma'r hafaliad y mae peirianwyr yn ei ddefnyddio i gyfrifo cynhyrchu pŵer o dyrbin trydan dŵr:
P = ρghQη
Gall hafaliadau deimlo'n frawychus weithiau, yn enwedig os ydyn nhw'n cynnwys llythrennau o wyddorau eraill (megis rho (ρ) ac eta (η) Groeg yn yr achos hwn).
Fodd bynnag, does dim symbolau mathemategol 'ynfyd' yn yr hafaliad hwn, dim ond pum newidyn a gynrychiolir gan wahanol lythrennau/symbolau a gaiff eu lluosi gyda'i gilydd i roi cyfanswm y pŵer (P) - a fesurir mewn watiau (jouleau fesul eiliad) - a gynhyrchir.
Gadewch i ni edrych ar beth sydd angen cymryd lle pob llythyren/symbol:
ρ = dwysedd dŵr (kg/m3)
g = disgyrchiant (m/s2)
h = uchder y cwymp (m)
Q = cyfradd llif (m3/s)
η = cyfradd effeithlonrwydd y tyrbin
Rydyn ni eisoes wedi darparu'r gwerthoedd ar gyfer popeth ar wahân i gyfradd llif yn y manylion uchod. Trafodir y gwerth coll, Q, yn yr adran nesaf gan fod cyfradd llif afon yn dymhorol ac felly'n amrywio drwy gydol y flwyddyn.
Mae ymchwil a gynhaliwyd ar yr afon yn darparu'r cyfartaleddau cyfradd llif hyn ar gyfer pob tymor:
| Tymor | Cyfradd llif (m³/s) | Nodiadau |
|---|---|---|
| Gaeaf | 1.8 | Glawiad trwm, llif cryf |
| Gwanwyn | 1.2 | Llif cymedrol |
| Haf | 0.6 | Llif isel, perygl o sychder |
| Hydref | 1.4 | Glawiad cynyddol |
Gan ddefnyddio'r ffigurau cyfradd llif hyn dylech nawr allu cwblhau'r hafaliad cynhyrchu pŵer ar gyfer pob tymor.
Yr hafaliad yw P = ρghQη
Mae hyn yn golygu yn y gaeaf:
P = 1000 x 9.8 x 3.2 x 1.8 x 0.6
felly, P = 33,868.8 Wat neu 33.87kW (i 2 bwynt degol)
Ar gyfer y tymhorau eraill, ailadroddwch y broses uchod gan newid dim ond gwerth Q (1.8 yn yr uchod) am y gyfradd llif ar gyfer y tymor hwnnw fel y'i nodir yn y tabl uchod.
Nawr mae gennym ni syniad faint o ynni y gellid ei gynhyrchu gan y gwaith trydan dŵr micro - ond faint o bŵer sydd ei angen ar y pentref?
Mae arolygon o'r pentref yn dangos defnydd ynni tymhorol fesul aelwyd ar gyfartaledd. Mae'r canlyniadau i'w gweld yn y tabl canlynol.
| Tymor | Cyfartaledd fesul Aelwyd (kW) | Nodiadau |
|---|---|---|
| Gaeaf | 1.3 | Y galw mwyaf: gwresogi, golau, gweithgaredd dan do |
| Gwanwyn | 1.0 | Tymor trosiannol, defnydd cymedrol |
| Haf | 0.7 | Y galw isaf: dyddiau hir o olau dydd, ychydig o wresogi |
| Hydref | 1.1 | Nosweithiau'n byrhau, gwresogi'n dechrau eto |
Mae 20 o aelwydydd yn y gymuned.
Bellach mae gennych chi'r holl wybodaeth ynghylch cynhyrchu ynni tymhorol o'r gwaith trydan dŵr arfaethedig, ac anghenion ynni'r gymuned fach. Gyda hynny, ceisiwch ateb y cwestiynau canlynol:
A all y tyrbin ddiwallu anghenion pŵer y gymuned drwy gydol y flwyddyn?
Os na all, a fyddai storfa batri yn cynnig ateb? Hynny yw, a oes digon o bŵer yn cael ei gynhyrchu drwy'r flwyddyn?
Allwch chi feddwl am ddewis 'gwyrdd' arall ar gyfer unrhyw brinder tymhorol?
A small community near Aberystwyth wants to install a micro-hydro turbine in their local river. Your job is to work out whether the river can generate enough electricity throughout the year and whether it matches the village's seasonal energy needs.
We shall provide you with the necessary engineering equations, river flow data, and household electricity demands to help you make a decision.
Click on the 'next' button below to get started.
The chosen section of river for this installation has a height drop (or head) measuring 3.2 metres.
The chosen turbine would run with an efficiency value of 0.6 - meaning that for every joule of potential energy entering the system, 0.6 joules of electricity is produced.
There are some constants we need to remind ourselves of too:
Water density = 1000kg/m3
Gravity = 9.8m/s2

Here is the equation engineers use to calcualte power production from a hydroelectric turbine:
P = ρghQη
Equations can sometimes seem daunting, especially with the inclusion of letters from other alphabets (such as the Greek rho (ρ) and eta (η) in this case).
However, there are no 'crazy' mathematical symbols in this equation, just five variables represented by different letters/symbols which are all multiplied together to give us the amount of power (P) - measured in watts (joules per second) - is produced.
So, let us look at what each letter/symbol needs replacing with:
ρ = water density (kg/m3)
g = gravity (m/s2)
h = drop height (m)
Q = flow rate (m3/s)
η = efficiency ratio of the turbine
We have already provided the values for everything except flow rate in the key details above. The missing Q value is covered in the next section as the flow rate of a river is seasonal and therefore varies throughout the year.
Research conducted on the river provides these flow rate averages for each season:
| Season | Flow Rate (m³/s) | Notes |
|---|---|---|
| Winter | 1.8 | Heavy rainfall, strong flow |
| Spring | 1.2 | Moderate flow |
| Summer | 0.6 | Low flow, drought risk |
| Autumn | 1.4 | Increased rainfall |
Using these flow rate figures you should now be able to complete the power generation equation for each season.
The equation is P = ρghQη
This means that in Winter:
P = 1000 x 9.8 x 3.2 x 1.8 x 0.6
so, P = 33,868.8 Watts or 33.87kW (to 2 decimal places)
For the other seasons repeat the above process while only replacing the Q value (1.8 in the above) with the flow rate for that season as given in the table above.
We now have an idea of how much energy could be generated by our micro-hydroelectric plant - but how much power does the village need?
Surveys of the village show the following average seasonal usage of energy per household. The results are shown in the following table.
| Season | Average per Household (kW) | Notes |
|---|---|---|
| Winter | 1.3 | Highest demand: heating, lighting, indoor activity |
| Spring | 1.0 | Transitional season, moderate use |
| Summer | 0.7 | Lowest demand: long daylight, little heating |
| Autumn | 1.1 | Nights drawing in, heating begins again |
There are 20 households in this community.
You now have all the information regarding seasonal energy production from the proposed hydroelectric plant, and the energy needs of this small community. With this, please attempt the following questions:
Can the turbine support the community's power needs throughout the year?
If not, would battery storage provide a solution? In other words, is enough power being produced throughout the year?
Can you think of another 'green' alternative to any seasonal shortfalls?