A Functionality of draft Tube in Turbine

[Mikealvarado100]

Hi
You know draft Tube after a Turbine decreases velocity and increases head of flow (it changes kinematic energy of flow to pressure energy). It is used at outlet of turbine to increase pressure head from sub-atmospheric pressure to zero.
Now this is the question: Why is pressure at the outlet of turbine, negative? Flow has a high pressure at inlet of turbine. it rotates turbine and then moves out of turbine. What happens here which results in negative pressure?

 

[benny_91]

Hi
You know draft Tube after a Turbine decreases velocity and increases head of flow (it changes kinematic energy of flow to pressure energy). It is used at outlet of turbine to increase pressure head from sub-atmospheric pressure to zero.
Now this is the question: Why is pressure at the outlet of turbine, negative? Flow has a high pressure at inlet of turbine. it rotates turbine and then moves out of turbine. What happens here which results in negative pressure?

Draft tube itself is the reason for negative pressure just at the exit of the reaction turbine. Without the draft tube the pressure at the exit of the turbine would be atmospheric pressure if the area of cross section of the pipe carrying water from the turbine exit to the tailrace is kept constant.
I can give a mathematical explanation to your question.
First of all, the head produced by a turbine can be obtained by using Bernoulli's equation:
P₁/ρg + v₁²/2g + z₁ = P₂/ρg + v₂²/2g + z₂ + H_turbine
where subscripts 1 and 2 denote the inlet and exit states of the control volume respectively.
Rearranging the above equation we get,
H_turbine = (P₁-P₂)/ρg - (v₂²-v₁²)/2g + z₁-z₂
Now by selecting our control volume appropriately we can obtain the required equation. Let the inlet of the control volume be the inlet to the turbine and the exit of the control volume be the tailrace where the pressure of water exiting through the pipe is atmospheric pressure (since the tailrace surface is at atmospheric pressure)
Let the subscript 't' denote the conditions at the tailrace. When a draft tube is used the value of v_t is considerably reduced. Hence from the below equation we can see that the head produced by the turbine increases considerably due to this change in the exit velocity. For simplicity let us assume that z₁-z₂ is negligible. Therefore we have
H_turbine = (P₁-P_t)/ρg - (v_t²-v₁²)/2g ...(1)
Here we can see as v_t decreases H_turbine increases. Hence a draft tube helps in increasing the net power obtained from a turbine.
Now let us see how the pressure at the exit of the turbine decreases below the atmospheric pressure when a draft tube is used. For this let us select the control volume as follows
Let the inlet of the control volume be the inlet of the turbine and the exit of the control volume be the exit of the turbine. Neglect changes in elevation. Using Bernoulli's equation we have:
H_turbine = (P₁-P₂)/ρg - (v₂²-v₁²)/2g ...(2)
where the subscripts 1 and 2 denote the state of the fluid at the inlet and exit of the turbine respectively.
As the turbine head is the same equations (1) and (2) can be equated. Therefore,
(P₁-P_t)/ρg - (v_t²-v₁²)/2g = (P₁-P₂)/ρg - (v₂²-v₁²)/2g
This can be further simplified as
P_t/ρ + v_t²/2 = P₂/ρ + v₂²/2
Now P_t or the pressure at the tailrace is atmospheric pressure which can be given the value 0. Due to the draft tube v₂ or velocity at the exit of the turbine is considerably greater than the velocity on reaching the tailrace. Substituting P_t=0 in the above equation we get
P₂ = -ρ{(v₂²-v_t²)/2}
Therefore in reaction turbines pressure at the exit of the turbine is below atmospheric pressure.

 

[Mikealvarado100]

Dear Benny-91
Thank you very much for your explanation in detail. I've got it mathematically.
But I am confused yet about concept of this negative pressure (At outlet of turbine). How does turbine make a negative pressure at it's outlet from a positive head at it's inlet, with regard to this point that the turbine does not work on fluid, but fluid does work on turbine. How can flow make itself negative while transferring through turbine?
Thank you again.

 

[Mikealvarado100]

Regarding to previous thread:
My problem is the misunderstanding of this subject CONCEPTUALLY, not mathematically and explanation of this concept is needed.

 

[benny_91]

Regarding to previous thread:
My problem is the misunderstanding of this subject CONCEPTUALLY, not mathematically and explanation of this concept is needed.

Well first of all the concept of negative pressure is not a big thing to worry about. This is with reference to the local atmospheric pressure. Forget negative pressure for a while. Do you have any problem in accepting the fact that pressure of the fluid at the exit of the turbine will be lesser than the pressure at the inlet. If that is acceptable then we can proceed further. You may very well know that: Gauge pressure = Absolute pressure - Local atmospheric pressure. It is the gauge pressure that becomes negative in the above analysis. The absolute pressure can never be negative. When the absolute pressure at the exit goes below the atmospheric pressure we say that the gauge pressure of the fluid is negative.
Suppose there is a turbine which changes the pressure of the fluid flowing through it from 150kPa to 100kPa. Let the local atmospheric pressure be 110kPa. So you see, the gauge pressure of the fluid at the exit of the turbine is 100-110= -10kPa which is of course a negative value. Now suppose the same turbine setup is taken to a very high mountain where the local atmospheric pressure is 90kPa. The turbine will perform in the same manner reducing the pressure of the working fluid from 150kPa to 100kPa. But in this case the gauge pressure is positive at the exit. So whether the gauge pressure at the exit of the turbine is positive or negative does not depend upon the turbine. It depends on the local atmospheric pressure.
Now tell me, can the fluid from the exit of the turbine be discharged to the tail race (which is atmospheric pressure) without a change in its pressure. The answer is NO. In the first case we will have to increase the area of cross section of the pipe from turbine exit to tail race so that the pressure increases from 90kPa to 100kPa or in terms of gauge pressure: from -10kPa to 0kPa. Similarly in the second case a reduction in pipe cross section would be necessary to decrease the pressure from 110kPa to 100kPa.
Now you might ask what if we provide a constant diameter pipe from turbine exit to tail race. In that case the turbine exit pressure of the fluid will never go above or below the atmospheric pressure. Hence the pressure at turbine exit will be equal to the pressure at tail race.

 

[Mikealvarado100]

benny_91
Thank you very much. I've got it.
At both situations of your example, if there is not any draft tube between turbine and tail race, then what happened? Is functionality and operation of the turbine changed (to produce zero pressure at the exit)? Or pressure at the exit of the turbine will be negative (for first case) and will be positive (for second case)? What happens if draft tube of a turbine is removed?
Thanx

 

[benny_91]

benny_91
Thank you very much. I've got it.
At both situations of your example, if there is not any draft tube between turbine and tail race, then what happened? Is functionality and operation of the turbine changed (to produce zero pressure at the exit)? Or pressure at the exit of the turbine will be negative (for first case) and will be positive (for second case)? What happens if draft tube of a turbine is removed?
Thanx

If a constant diameter pipe is provided between the turbine exit and the tail race neither of the above two cases will exist. In both cases the fluid pressure at the exit of the turbine will be atmospheric pressure and will remain at the same pressure throughout its flow from turbine exit to tail race through that constant diameter pipe. That is in first case the turbine exit pressure will be 110kPa and in the second case it will be 90kPa.

As for what a draft tube actually does let us take an example in which we do not provide a draft tube in the beginning. We check all the parameters. Then we provide a draft tube to the same turbine and then check the same parameters to see if there is any change.
So now we have a turbine which does not have a draft tube (which means a constant diameter pipe connects the turbine exit to the tail race). Hence the fluid pressure at the exit of the turbine is same as that at the tail race which is atmospheric pressure. Now in this turbine we also find that the velocity of the fluid exiting at the tail race is quite high and we are not very happy about it as we are aware that a lot of kinetic energy is wasted in this way. Now we want to decrease the fluid velocity at the tail race without any energy loss. So what is the best and easiest thing to do? Yep! Increase the area of cross section of the pipe at tail race (this increase in area should be gradual and not sudden since sudden increase in cross sectional area can convert the kinetic energy of the fluid into heat which will be dissipated out of the system and we do not want that). That means we have actually provided a draft tube at the tail race. The exit diameter of the draft tube is such that now the velocity of the fluid exiting from the draft tube is very low compared to the previous case without the draft tube. Now what is the pressure of the fluid at the exit of the draft tube? Of course it will be atmospheric pressure as the fluid itself is exposed to atmosphere there. But the area of cross section at the turbine exit is now lower than the area of cross section at the draft tube exit. This means the pressure of the fluid at turbine exit must be lower than the pressure of the fluid at draft tube exit. (this result that can be obtained by using continuity equation and Bernoulli's equation). But the pressure of the fluid at draft tube exit which is nothing but atmospheric pressure has been taken as 0. Hence the pressure at turbine exit must be negative.
Hope this answers all your questions!

 

[Mikealvarado100]

Dear benny
I really really thank you for your explanation completely.
By using draft tube, what happens for that kinetic energy we saved? Let me explain my interpret:
Suppose a turbine which changes fluid pressure from 150 kpa to 100 kpa and atmospheric pressure is 90 kpa (such as your example). Without draft tube, the turbine uses 60 kpa (difference 150 and 90) with (e.g.) 20 m/s velocity at tail race.
With draft tube, it uses more than 60 kpa (e.g. 80 kpa which is difference between 150 and 70 kpa. 70 kpa is negative pressure [70 < 90] because of using draft tube),with (e.g.) 4 m/s velocity at the tail race.
Now the energy saved is 16*16/(2*9.81). This energy has resulted that 20 kpa (not consider units).
Is my interpret correct or need to be correct?
Thanx.

 

[benny_91]

Dear benny
I really really thank you for your explanation completely.
By using draft tube, what happens for that kinetic energy we saved? Let me explain my interpret:
Suppose a turbine which changes fluid pressure from 150 kpa to 100 kpa and atmospheric pressure is 90 kpa (such as your example). Without draft tube, the turbine uses 60 kpa (difference 150 and 90) with (e.g.) 20 m/s velocity at tail race.
With draft tube, it uses more than 60 kpa (e.g. 80 kpa which is difference between 150 and 70 kpa. 70 kpa is negative pressure [70 < 90] because of using draft tube),with (e.g.) 4 m/s velocity at the tail race.
Now the energy saved is 16*16/(2*9.81). This energy has resulted that 20 kpa (not consider units).
Is my interpret correct or need to be correct?
Thanx.

Well the example which I gave regarding a hypothetical t

Dear benny
I really really thank you for your explanation completely.
By using draft tube, what happens for that kinetic energy we saved? Let me explain my interpret:
Suppose a turbine which changes fluid pressure from 150 kpa to 100 kpa and atmospheric pressure is 90 kpa (such as your example). Without draft tube, the turbine uses 60 kpa (difference 150 and 90) with (e.g.) 20 m/s velocity at tail race.
With draft tube, it uses more than 60 kpa (e.g. 80 kpa which is difference between 150 and 70 kpa. 70 kpa is negative pressure [70 < 90] because of using draft tube),with (e.g.) 4 m/s velocity at the tail race.
Now the energy saved is 16*16/(2*9.81). This energy has resulted that 20 kpa (not consider units).
Is my interpret correct or need to be correct?
Thanx.

First of all don't thank me. I understood this stuff only when I tried to explain it to you. This same problem used to bother me a lot during the 3rd year of my engineering. So I have gained from this. Proceeding to your question: Well the example which I gave you was not completely right because without a draft tube or any tube of varying cross section between the turbine exit and the tail race, the fluid pressure at the turbine exit will always be the same as that at the tail race that is the local atmospheric pressure. So in a region with local atmosphere as 90kPa the fluid will always exit the turbine at 90kPa if a draft tube is not used.
So I shall rewrite your own previous statement with necessary corrections. Almost all of it is correct though! So here goes:
"Suppose a turbine which changes fluid pressure from 150 kpa to 90 kpa and atmospheric pressure is 90 kpa (such as your example). Without draft tube, the turbine uses 60 kpa (difference 150 and 90) with (e.g.) 20 m/s velocity at tail race.
With draft tube, it uses more than 60 kpa (e.g. 80 kpa which is difference between 150 and 70 kpa. 70 kpa is negative pressure [70 < 90] because of using draft tube),with (e.g.) 4 m/s velocity at the tail race.
Now the energy saved is the difference of kinetic energy head in the two cases that is (20²-4²)/2*9.81. This energy has resulted that 20 kpa (not consider units)."

 

[Mikealvarado100]

Hi
Have a look at P.4 of below video:
http://www.slideshare.net/swargpatel283/draft-tube
"If the pressure at the exit of the turbine is lower than the pressure of fluid in the tailrace, a back flow of liquid into the turbine can result in significant damage."
How does reverse flow can occur in this case (from tailrace to turbine)? "If the pressure at the exit of the turbine is lower than the pressure of fluid in the tailrace" means that the pressure at exit of the turbine is NEGATIVE.
Let me say that I agree with you and think pressure at exit of turbine must be zero not negative. But P. 4 of the video has to explained.

 

[benny_91]

Hi
Have a look at P.4 of below video:
http://www.slideshare.net/swargpatel283/draft-tube
"If the pressure at the exit of the turbine is lower than the pressure of fluid in the tailrace, a back flow of liquid into the turbine can result in significant damage."
How does reverse flow can occur in this case (from tailrace to turbine)? "If the pressure at the exit of the turbine is lower than the pressure of fluid in the tailrace" means that the pressure at exit of the turbine is NEGATIVE.
Let me say that I agree with you and think pressure at exit of turbine must be zero not negative. But P. 4 of the video has to explained.

If a draft tube is used gauge pressure at the exit of the turbine will be negative. If no draft tube is present gauge pressure at the exit of the turbine will be 0.
It is not always true that flow takes place from a high pressure region to a low pressure region. For example, consider the throat and the diverging part at the throat exit of a venturimeter as our control volume. The pressure at the throat is lower than the pressure at the diverging section but even then flow takes place from the throat to the diverging section. Same is the case of flow from the turbine exit to the tailrace through the draft tube. For any flow to take place in a particular direction, the flow should satisfy two equations:
1. The energy conservation equation (also known as Bernoulli's equation)
2. The entropy equation (The entropy of the universe must increase)
Both these equations are satisfied in the above case!

 

[Mikealvarado100]

benny
I've got it. This was one of best discussions I had. THANK you very much, in spite of your mis-tendency.
Let's have discussion about your last point: How does 'entropy equation' can be explained for this case?
Please describe a little about it.