We have an hydrogenerator with followings technical data; 16000 kVA, 6.3 kV, I stator=1466 A, n=500rpm., m=3 phase, R stator=0.010 ohm, GD2=85 tm2 and we want use electrical braking system by the short circuit method. When I calculated the time for braking the generator I found t=870 sec.. In my opinion this is not acceptable. Probably the calculation is not good. If someone know more about this problem please give me more details. I have calculated the moment produced at shortcircuit with this formula: mxRxI2x60/(2x3.1415xn) [Nm]. If this is not good please correct me. Thank you.
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Electrical braking at hydrogenerators
- Thread starter nelvex
- Start date
LionelHutz
Electrical
My first guess without even looking at the data. If you want to remove energy then you need to give the energy some place to go. Connecting resistors instead of a short circuit gives you some place to dump the energy.
Yes LionelHutz is correct. A resistor bank is connected in to the stator by a CB. How the excitation system works in your case is not explained. Often a separately supplied static exciter would be used and set for a certain field current at the start of the braking sequence. As the unit slows down the braking effect reduces as the generator voltage drops as a function of speed. This would be a function of current squared in the resistor.
At about 30 percent speed the mechanical brakes would bring the unit to a full stop as in a normal hydro unit.
The braking excitation would be tripped off at or before this point.
The excitation can not remain on until near zero speed as the slip rings would be damaged. Cooling of the unit at reduced speed has to be considered.
Other methods would raise the excitation current at the lower speeds to obtain a faster stop.
Why not consult the OEM for a solution.
regards, rasevskii
At about 30 percent speed the mechanical brakes would bring the unit to a full stop as in a normal hydro unit.
The braking excitation would be tripped off at or before this point.
The excitation can not remain on until near zero speed as the slip rings would be damaged. Cooling of the unit at reduced speed has to be considered.
Other methods would raise the excitation current at the lower speeds to obtain a faster stop.
Why not consult the OEM for a solution.
regards, rasevskii
- Thread starter
- #4
Thanks to the intervention of you, I would be glad if someone could possibly indicate a book or a site where the electrical braking for Hydro-generators is better presented. I must solve this problem and I must know more details about this (formulas, calculation, etc.). I know generally information about electrical braking but this is not sufficient for me.
Thank you.
Thank you.
LionelHutz
Electrical
If you go back to basics then it's not that difficult a problem. It's not simple but it's really not that hard either.
You can calculate the starting rotational energy or angular kinetic energy in the machine.
So, pick a resistor. As an arbritary value, size it for twice rated current at rated voltage. With synchronous motors, we size for around 2.5 times to 3 times current so I suspect a similar current level could be used for this machine. Maybe someone else with experience in this can comment on the resistor sizing.
Then, you can calculate how long it takes to dissipate the energy from the machine into the resistor. Easiest might be to solve the problem programmatically. You will likely have to take a small time step and determine how much energy is transferred in that time step. Then, you can calculate the end conditions (the new speed, voltage, current and power transfer) of that time step to use during the next small time step. This will likely require a small program or a mathcad loop to calculate.
You will also have to determine what happens with the generator output voltage as the machine slows down. If the excitation remains applied then the voltage should decay linearly with speed. However, if you lose excitation as the voltage drops then you might have the problem of no braking as you slow down.
As a general note, the rotational energy goes down by the square of the speed and the power being transferred goes down by the square of the output voltage. So, the speed should drop fairly linearly.
You can calculate the starting rotational energy or angular kinetic energy in the machine.
So, pick a resistor. As an arbritary value, size it for twice rated current at rated voltage. With synchronous motors, we size for around 2.5 times to 3 times current so I suspect a similar current level could be used for this machine. Maybe someone else with experience in this can comment on the resistor sizing.
Then, you can calculate how long it takes to dissipate the energy from the machine into the resistor. Easiest might be to solve the problem programmatically. You will likely have to take a small time step and determine how much energy is transferred in that time step. Then, you can calculate the end conditions (the new speed, voltage, current and power transfer) of that time step to use during the next small time step. This will likely require a small program or a mathcad loop to calculate.
You will also have to determine what happens with the generator output voltage as the machine slows down. If the excitation remains applied then the voltage should decay linearly with speed. However, if you lose excitation as the voltage drops then you might have the problem of no braking as you slow down.
As a general note, the rotational energy goes down by the square of the speed and the power being transferred goes down by the square of the output voltage. So, the speed should drop fairly linearly.
electricpete
Electrical
fwiw, I tried to solve the problem outlined by David (his assumptions sound reasonable to me):
Rotational equivalent of F=M*A = M * d/dt(velocity):
T = J * dw/dt [eq1]
Equation for Torque associated with I^2*R losses
T = - 3*I^2*R / w [eq2]
where factor of 3 accounts for 3 phases and (-) sign indicates deceleraiton.
Substitute T from 2 into 1:
-3*I^2*R / w = J * dw/dt [eq3]
* Assume:
V = k*w [eq4] (neglect iron nonlinearities etc and assume induced stator voltage proportional to w... seems reasonable to me since airgap controls the flux more than iron, and we assume excitation is held constant.)
** Assume:
I = V/R [eq5] (assumes inductive/synchronous reactance is negligible. That seems reasonable to me. If we were running the machine at full load into a resistor R100%, that R100% would be higher than synch reactance... and further we presume you will use a load bank quite a bit larger than R100%. And finally of course inductive reactance will decrease further as speed frequency decreases.
Substitute 5 into 4:
I = k*w/R [eq6]
Substitute I from 6 into 3
-3*(k*w/R)^2 *R / w = J * dw/dt
Cancel out w and R on LHS:
-3*k^2*w/R = J * dw/dt
Solve for dt:
dt = -J*R / (3*w*k^2) dw
Use dummy variable t' and w' instead of t and w
dt' = -J*R / (3*w*k^2) dw'
Integrate (t', w') = (0,w0) to (t,w):
t = -J*R/[3*k^2]* ln (w) + -J*R/[3*k^2] * ln (w0)
t = -J*R/[3*k^2] *ln (w/w0)
w/w0 = exp(-t*3*k^2 / [J*R])
w = w0* exp(-t*3*k^2 / [J*R])
Symbols:
R = total per-phase resistance in stator (internal plus external)
w = rotating speed in radians/ time
k = V/w where V = line to neutral voltage at full speed, w = full speedin radians/time
I = stator current
V = voltage induced in stator loop on per-phase basis
T = total torque (all three phases)
J = rotating inertia for example kg*m^2. Need some work to convert inertia into this form. I am familiar with alternate expression of inertia (H) in seconds required to accelerate machine to full speed at full torque. I am not familiar with your units GD2=85 tm2... what is tm2?
=====================================
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Rotational equivalent of F=M*A = M * d/dt(velocity):
T = J * dw/dt [eq1]
Equation for Torque associated with I^2*R losses
T = - 3*I^2*R / w [eq2]
where factor of 3 accounts for 3 phases and (-) sign indicates deceleraiton.
Substitute T from 2 into 1:
-3*I^2*R / w = J * dw/dt [eq3]
* Assume:
V = k*w [eq4] (neglect iron nonlinearities etc and assume induced stator voltage proportional to w... seems reasonable to me since airgap controls the flux more than iron, and we assume excitation is held constant.)
** Assume:
I = V/R [eq5] (assumes inductive/synchronous reactance is negligible. That seems reasonable to me. If we were running the machine at full load into a resistor R100%, that R100% would be higher than synch reactance... and further we presume you will use a load bank quite a bit larger than R100%. And finally of course inductive reactance will decrease further as speed frequency decreases.
Substitute 5 into 4:
I = k*w/R [eq6]
Substitute I from 6 into 3
-3*(k*w/R)^2 *R / w = J * dw/dt
Cancel out w and R on LHS:
-3*k^2*w/R = J * dw/dt
Solve for dt:
dt = -J*R / (3*w*k^2) dw
Use dummy variable t' and w' instead of t and w
dt' = -J*R / (3*w*k^2) dw'
Integrate (t', w') = (0,w0) to (t,w):
t = -J*R/[3*k^2]* ln (w) + -J*R/[3*k^2] * ln (w0)
t = -J*R/[3*k^2] *ln (w/w0)
w/w0 = exp(-t*3*k^2 / [J*R])
w = w0* exp(-t*3*k^2 / [J*R])
Symbols:
R = total per-phase resistance in stator (internal plus external)
w = rotating speed in radians/ time
k = V/w where V = line to neutral voltage at full speed, w = full speedin radians/time
I = stator current
V = voltage induced in stator loop on per-phase basis
T = total torque (all three phases)
J = rotating inertia for example kg*m^2. Need some work to convert inertia into this form. I am familiar with alternate expression of inertia (H) in seconds required to accelerate machine to full speed at full torque. I am not familiar with your units GD2=85 tm2... what is tm2?
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electricpete
Electrical
Minor correction (doesn't affect result):
t = -J*R/[3*k^2]* ln (w) + -J*R/[3*k^2] * ln (w0)
should've been
t = -J*R/[3*k^2]* ln (w) - -J*R/[3*k^2] * ln (w0)
=====================================
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t = -J*R/[3*k^2]* ln (w) + -J*R/[3*k^2] * ln (w0)
should've been
t = -J*R/[3*k^2]* ln (w) - -J*R/[3*k^2] * ln (w0)
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
nelvex:
Why do you want to decelerate your hydro generator by electrical braking? This type of action makes more sense for a pumped-storage motor/generator connected to a pumpturbine. Such units have two directions of rotation and the rotor has to be brought from generator mode to zero speed as quickly as possible before the pump mode can be initiated.
There is no formula for calculating braking times for hydro generators mainly because it is unknown how high the hydraulic losses generated by the turbine runner rotating in water are. Other generator loss components when the generator is disconnected from the grid and coasting de-excited are windage and bearing friction losses. These are more or less negligible as they decrease considerably with falling speed.
It is not true that the electrical braking effect reduces with speed decrease. Its the other way around. The generator voltage reduces with speed as well as the synchronous reactance Xd. Because E/Xd (this is the braking current) remains constant with speed decrease, the braking losses in the armature winding remain constant, too. Therefore: The braking torque increases with speed decrease.
In practice a braking current of 120% rated current is selected. Cooling of the armature winding is reduced at low speeds but temperature increase can be considered as negligible (low single digit degrees/Kelvin). Mechanical brakes can assist the electrical braking at low speed and the braking excitation will be cut off shortly before the unit comes to rest.
Braking resistors can be very effective but the extra investment may not warrant such expenses.
Best regards
Wolf
Why do you want to decelerate your hydro generator by electrical braking? This type of action makes more sense for a pumped-storage motor/generator connected to a pumpturbine. Such units have two directions of rotation and the rotor has to be brought from generator mode to zero speed as quickly as possible before the pump mode can be initiated.
There is no formula for calculating braking times for hydro generators mainly because it is unknown how high the hydraulic losses generated by the turbine runner rotating in water are. Other generator loss components when the generator is disconnected from the grid and coasting de-excited are windage and bearing friction losses. These are more or less negligible as they decrease considerably with falling speed.
It is not true that the electrical braking effect reduces with speed decrease. Its the other way around. The generator voltage reduces with speed as well as the synchronous reactance Xd. Because E/Xd (this is the braking current) remains constant with speed decrease, the braking losses in the armature winding remain constant, too. Therefore: The braking torque increases with speed decrease.
In practice a braking current of 120% rated current is selected. Cooling of the armature winding is reduced at low speeds but temperature increase can be considered as negligible (low single digit degrees/Kelvin). Mechanical brakes can assist the electrical braking at low speed and the braking excitation will be cut off shortly before the unit comes to rest.
Braking resistors can be very effective but the extra investment may not warrant such expenses.
Best regards
Wolf
Hi Wolf
Have you come across any hydro generators being electrically braked by short circuit method ?
Muthu
Have you come across any hydro generators being electrically braked by short circuit method ?
Muthu
electricpete
Electrical
Just to mention - the behavior would depend on assumptions about presence and size of braking resistor. I stated my assumptions. What is more common I don't know.
Good point. That was not accounted for in my discussion.
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With the generator data available the ohmic loss portion of the armature winding can be calculated to about 80 kW (at 75 degrees C). The stray losses are estimated to be 40 kW. At rated current the total braking losses therefore are in the region of 120 kW, without application of a braking resistor. It can be assumed that a braking resistor bank has practically no stray losses. Therefore, with a braking resistor bank of 3 x 0.010 ohm, for instance (equivalent to the armature resistance), we have a total braking loss of 200 kW at rated current and of almost 300 kW at 120% rated current.
Electrical braking is quite common for hydro generators. The Sanxia (Three Gorges) generators, for instance, have a rating of 840 MVA and an electrical braking system is installed. The main advantage for electrical braking is that the mechanical brakes step in at a speed of 2 or 3 percent only, thus saving brake pad wear and reducing brake dust pollution.
Pumped-storage units equipped with pumpturbines have two directions of rotation. Electrical braking is a must here for switching the unit from generator to motor mode quickly. However, most pumpturbine driven units have a starting converter to run-up the motor/generator to rated speed for pump mode operation. In turn this converter can be used as a braking device when the unit has to switch from generator to motor service. The kinetic energy stored in the rotating parts then can be fed back into the grid.
Regards
Wolf
Electrical braking is quite common for hydro generators. The Sanxia (Three Gorges) generators, for instance, have a rating of 840 MVA and an electrical braking system is installed. The main advantage for electrical braking is that the mechanical brakes step in at a speed of 2 or 3 percent only, thus saving brake pad wear and reducing brake dust pollution.
Pumped-storage units equipped with pumpturbines have two directions of rotation. Electrical braking is a must here for switching the unit from generator to motor mode quickly. However, most pumpturbine driven units have a starting converter to run-up the motor/generator to rated speed for pump mode operation. In turn this converter can be used as a braking device when the unit has to switch from generator to motor service. The kinetic energy stored in the rotating parts then can be fed back into the grid.
Regards
Wolf
I stand corrected. In my original post I had assumed that the external braking resistor was of such an ohmic value that the Xd of the generator could be ignored. Not so. The current is determined by the voltage divided by the vectorial sum of the Xd ohms + Rt ohms. Where Rt is the stator DC ohms plus the external braking resistor ohms. Correct me if I am still wrong...
Since Xd and voltage decrease with speed linearly /??/ therefore the current would remain nearly constant if Rt is small compared to Xd...is this correct? Assuming that the excitation current remains constant, that is.
Now, do the Three Gorges units have external braking resistors or not? That would be interesting for such large units. What is the normal RPM and what is the stopping time?
regards, rasevskii
Since Xd and voltage decrease with speed linearly /??/ therefore the current would remain nearly constant if Rt is small compared to Xd...is this correct? Assuming that the excitation current remains constant, that is.
Now, do the Three Gorges units have external braking resistors or not? That would be interesting for such large units. What is the normal RPM and what is the stopping time?
regards, rasevskii
- Thread starter
- #13
Thank you all for your answers. The intention is short circuiting the phases without using the phase resistances.
electricpete:
"GD2 = 85 tm2 ... what is tm2?"
GD ^ 2 is Flywheel effect (tm2 .... tones x m ^ 2)
GD ^ 2 = 4 x g x J
wolf39
Armature Winding loss = 68 kW;
stray losses = 29 kW.
mechanical loss at rated power (ventilation and bearing loss) approx. 120 kW.
I do not know how much is the hydraulic losses generated by the turbine runner rotating in water. The thrust bearing is self-lubricated and is designed for normal starting and stopping using high pressure oil injection.
As I said, I generally known the phenomenon, but at question "how is time to brake?" I do not know yet to respond. Perhaps it is not necessary to give an exact amount but at least the order of magnitude, ie, 70 seconds or 700 seconds I must to give. I realize that this calculation can not be done exactly. As I said in my first post in my calculation I have obtained 870 seconds and I think that is not correct.
I would ask how much was the time for brake the Sanxia generators or other hydrogenerators.
Thank you all for your intervention and apologize for my English.
electricpete:
"GD2 = 85 tm2 ... what is tm2?"
GD ^ 2 is Flywheel effect (tm2 .... tones x m ^ 2)
GD ^ 2 = 4 x g x J
wolf39
Armature Winding loss = 68 kW;
stray losses = 29 kW.
mechanical loss at rated power (ventilation and bearing loss) approx. 120 kW.
I do not know how much is the hydraulic losses generated by the turbine runner rotating in water. The thrust bearing is self-lubricated and is designed for normal starting and stopping using high pressure oil injection.
As I said, I generally known the phenomenon, but at question "how is time to brake?" I do not know yet to respond. Perhaps it is not necessary to give an exact amount but at least the order of magnitude, ie, 70 seconds or 700 seconds I must to give. I realize that this calculation can not be done exactly. As I said in my first post in my calculation I have obtained 870 seconds and I think that is not correct.
I would ask how much was the time for brake the Sanxia generators or other hydrogenerators.
Thank you all for your intervention and apologize for my English.
rasevskii:
You are correct. The braking current is determined by the voltage devided by the vectorial sum of Xd and R (the latter is the sum of armature winding resistance and braking resistor). Voltage and Xd decrease linearly with speed and the short circuit braking current remains nearly constant until the Xd drops to figures where the R portion of the vectorial sum is not negligible any more. The excitation current is always being kept constant.
All hydro generator electrical braking equipment that I know of did rely on the resistance of the armature winding only (no external resistors). For the 840 MVA Sanxia generators it would be quite impractical to install a braking resistor bank as this device had to be designed for a rated current of 24,250 amps. I'll try to find some information on electrical braking times in my archive but I vaguely remember a figure of about 5 minutes for the Sanxia units. Rated speed of the Sanxia generators is 75 rpm.
Regards
Wolf
You are correct. The braking current is determined by the voltage devided by the vectorial sum of Xd and R (the latter is the sum of armature winding resistance and braking resistor). Voltage and Xd decrease linearly with speed and the short circuit braking current remains nearly constant until the Xd drops to figures where the R portion of the vectorial sum is not negligible any more. The excitation current is always being kept constant.
All hydro generator electrical braking equipment that I know of did rely on the resistance of the armature winding only (no external resistors). For the 840 MVA Sanxia generators it would be quite impractical to install a braking resistor bank as this device had to be designed for a rated current of 24,250 amps. I'll try to find some information on electrical braking times in my archive but I vaguely remember a figure of about 5 minutes for the Sanxia units. Rated speed of the Sanxia generators is 75 rpm.
Regards
Wolf
nelvex:
When short-circuiting the generator terminals, you use the generator phase resistances for electric braking. The braking can be amplified by connecting braking resistors in series with the stator winding terminals. I've never seen such external braking resistors used for hydro generators but for synchronous motors this may make sense.
With stator winding resistances of 3 x 0.010 ohms (I assumed that these figures were based on the standard temperature of 20 degrees C), I arrive at an ohmic loss figure of 78 kW at 75 degrees C, not 68 kW. However, it may be that the above resistances were based on 75 degrees C already. This important info was missing.
One more thing is important: What type of hydroturbine is in use, Francis or Pelton? Pelton runners rotate in air with relatively low friction losses compared with Francis runners, which are rotating in water. In your case I assume a Francis turbine.
Under normal circumstances one should let the hydro unit coast down to approximately 50% rated speed before the electrical braking steps in. Reason: Deceleration is at its maximum at high speeds because turbine friction losses are very high in this region, compared with the electrical braking losses. It therefore makes no sense to commence electrical braking at speeds of above 50% rated speed.
I've found some examples in my archive for hydro generators coupled to Francis runners. Deceleration times with the turbine runner rotating in water from 100% to 50% rated speed were between 60 seconds and 100 seconds, depending on the moment of inertia of the rotating parts (generator rotor plus turbine runner). When electrical braking with 120% rated generator current steps in at 50% rated speed, one has on average to add about 200 seconds to the above figures. A total braking time of about 5 minutes would therefore be a fair figure to deal with.
Regards
Wolf
When short-circuiting the generator terminals, you use the generator phase resistances for electric braking. The braking can be amplified by connecting braking resistors in series with the stator winding terminals. I've never seen such external braking resistors used for hydro generators but for synchronous motors this may make sense.
With stator winding resistances of 3 x 0.010 ohms (I assumed that these figures were based on the standard temperature of 20 degrees C), I arrive at an ohmic loss figure of 78 kW at 75 degrees C, not 68 kW. However, it may be that the above resistances were based on 75 degrees C already. This important info was missing.
One more thing is important: What type of hydroturbine is in use, Francis or Pelton? Pelton runners rotate in air with relatively low friction losses compared with Francis runners, which are rotating in water. In your case I assume a Francis turbine.
Under normal circumstances one should let the hydro unit coast down to approximately 50% rated speed before the electrical braking steps in. Reason: Deceleration is at its maximum at high speeds because turbine friction losses are very high in this region, compared with the electrical braking losses. It therefore makes no sense to commence electrical braking at speeds of above 50% rated speed.
I've found some examples in my archive for hydro generators coupled to Francis runners. Deceleration times with the turbine runner rotating in water from 100% to 50% rated speed were between 60 seconds and 100 seconds, depending on the moment of inertia of the rotating parts (generator rotor plus turbine runner). When electrical braking with 120% rated generator current steps in at 50% rated speed, one has on average to add about 200 seconds to the above figures. A total braking time of about 5 minutes would therefore be a fair figure to deal with.
Regards
Wolf
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