Hydro power Flashcards

Question

What is the main factor contributing to the specific work of a reaction turbine?

Answer 1

It is the absolute total pressure difference over the turbine that contributes to the specific work that can be extracted. In a reaction turbine we need a device that extracts work by reducing the static pressure while keeping the velocuty before and after the device similar

Answer 2

The draft tube, shaped as a diffuser, should reduce the flow velocity leaving the runner, so it can reach (almost) zero at the tailwater surface with as little losses as possible. This minimizes p2 and therefore maximizes Wt The area expansion can't be too fast, since the flow will separate and increase losses The head losses between draft tube and tailwater can be reduced by increasing the length of the draft tube, but that would increase the ead between the turbine outlet to the draft tube as well as the construction costs

Answer 3

Wt = c2^2/2-c3^2/2-g*deltaHt Specific work is change in velocity (except losses). Maximum work is given for c3 = 0. We need a device that removes the jet velocity

Answer 4

deltaW = g(H-deltaHtot)

Answer 5

eta = (1-deltaHtot/H)

Answer 6

hw-1. Intake losses (including channel flow, trash racks and inlet design), penstock friction losses, losses in spiral casing and wicket gate Ht. Poor turbine design, flow shear and inappropriate flow at off-design H2-3: Poor draft tube design, draft tube friction losses H3-tw: Losses due to sudden expansion

Answer 7

hw-1: Intake losses (including channel flow, trash racks and inlet design), penstock friction losses Hn: Losses in nozzle Ht: Losses in turbine z3: The distance that the water falls down is also a loss, but is comparatively small for impulse turbines that are used at very high head

Answer 8

Hf = f*(L/dh)*(c^2)/(2*g) where f is calculated as: f = 64/Re for laminar flow (Re<2300) and using Haaland formula for turbulent flow. Also possible to obtain from graphs

Answer 9

Hm = K*c^2/(2*g) K is a dimensionless loss coefficient

Answer 10

* The extended penstock just adds losses and reduces overall efficiency!

Answer 11

If P = 0, we get H = deltaHtot This means that the head is balanced by the losses. All losses depend on the flow rate, i.e. we will get a flow rate that gives the balance If we have no losses, meaning that H = 0, the upper reservoir will instantly empty into the lower reservoir until H = 0, i.e. Q = inf if H <> 0

Answer 12

deltaW = U1*c1-U2*c2 wher U = omega*r and c is the tangential flow component For a perfectly axial flow (U1=U2) the specific work is extracted by changing the tangential flow velocity

Answer 13

* Before and after the turbine, the flow ideally has only a ameridional component, given by the volume flow rate (which is preserved at all cross-sections) * The flow is first accelerated in the tangential direction (doesn't contribute to the flowrate) in a steady part of the machine, at (almost) constant z. Thus, the static pressure will decrease. * The flow is then decelerated in the tangential direction, but now in a rotating part of the machine. The deceleration in the inertial coordinate system correspond to a counteracting acceleration in the rotating coordinate system. Thus, the static pressure will further decrease * The deceleration/acceleration of the tangential velocity component in the rotating part of the machine is extracted as work rate (power)

Answer 14

* The flow is accelerated in nozzles to jets in atmospheric pressure, tangential to the turbine shaft. Thusm the static pressure will decrease completely to the atmospheric pressure * The flow is then decelerated to zero in the tangential direction, but in a rotating part of the machine. The deceleration in the inertial coordinate system correspond to a counteracting acceleration in the rotating coordinate system * The water has close to zero velocity directly after the turbine, and simply falls down to the floor where it exits to the tailwater * The deceleration/acceleration of the tangential velocity component in the rotating part of the machine is extracted as work rate (power)

Answer 15

Ratio of the static pressure drop in the rotating part of the device to the total static pressure drop over the entire device A device with a degree of reaction of zero is an impulse device (no pressure drop in the rotating part of the device)

Answer 16

eta_o = power available at the output shaft/maximum power available in the fluid flow * Mechanical losses occur dur to friction in bearings and seald, etc. These losses are hard to estimate. For small machines it could be 5% or more but for large machines it can reduce to 1% * The hydraulic efficiency is given by: eta_h = power supplied from the flow to the rotor/maximum power available in the fluid flow eta_o = eta_h*eta_m

Answer 17

* Pelton turbines are usually directly coupled to an electric generator, which means that it must run at a certain rotational speed * The flow, Q, is regulated using "spears/needles" in the nozzles * For sudden changes in load, deflector plates remove the jets from the buckets * The flow rate must be changed gradually to prevent "water hammer" * The rotor consists of a circular disk with a number of buckets * Ine or more nozzles direct their jet in the tangential direction in atmospheric pressure (in air) * A sharp splitter splits the jet into two equal parts * The bucket redirects the flow into two opposite directions and prevents the flow from hittint the vack of the next bucket to avoid reducing the efficiency * The absolute flow velocity after the buckets should be close to zero

Answer 18

* The spiral casing distributes the water evenly at the guide vane entrance (and gives the flow an initial swirl) * The stay vanes hold the structure together (high pressure!) * The pivotable guide vanes control the flow rate and runner inlet swirl. Static pressure is reduced * The swirl from the guide vanes is removed by the runner and converted into shaft work rate. Static pressure further reduced * I.e. only part of the pressure drop occurs before the runner, and the rest occurs in the runner itself * The flow leaves the turbine axially through a diffuser (draft tube), with almost no remaining swirl * The water completely fills the turbine passages

Answer 19

* Considered as an important integral (passive) part of reaction turbines * Shaped as a diffuser to reduce the kinetic energy of the flow and recover the static pressure (reducing it after the runner) * Tailrace outlet must be under water so that the draft tube is completely filled with water * Turbine may be placed above tailwater surface, but cavitation may then be an issue * Remaining swirl after the runner may cause pressure fluctuations

Answer 20

* Variation of the flow of a river over a year * Percentage of time that the flow in a river is equal to or exceeds a particular flow rate * On x-axis percentage and on y-axis flow rate

Answer 21

Assume that all the flow in the river goes through the power plant, but can be stored in the dam for later use

Answer 22

Assume that all the flow in the river goes through the power plant, at the same rate as in the river, i.e. not possible to control the power output in the same way

Answer 23

Relative discharge (flow rate) is the ratio between the current discharge and the designated flow rate A turbine that operates at the designed flow rate will have a relative discharge of 100%