Mixing
Physic: Calculated Waterflow based on Power generation of turbine
$begingroup$
I have a turbine which produces 50MW, the water falls down from three meters height.
How much water does flow per second?
Please do not answer this question, but rather give an explanation how I can calculate it myself.
physics
asked Jan 9 at 21:38
TimTim
31
$endgroup$
|
show 6 more comments
$begingroup$
I have a turbine which produces 50MW, the water falls down from three meters height.
How much water does flow per second?
Please do not answer this question, but rather give an explanation how I can calculate it myself.
physics
asked Jan 9 at 21:38
TimTim
31
$endgroup$
$begingroup$
How is this about mathematics?
$endgroup$
– copper.hat
Jan 9 at 21:41
$begingroup$
@copper.hat I can't quite follow what your question is. Do you mean why I posted it on Mathematics? As I didn't see a Physics SE. And physic belongs to math. The math is calculating the waterflow.
$endgroup$
– Tim
Jan 9 at 21:44
$begingroup$
And what is the difficulty? It is not about math. It is a physics homework question.
$endgroup$
– copper.hat
Jan 9 at 21:45
$begingroup$
I don't know how I can get the water flow out of generated Power and fall height.
$endgroup$
– Tim
Jan 9 at 21:46
1
$begingroup$
Mhm. 1 m^3 = 1000 Liters = 1000kg so W=1000*3=3000 So if I understand right, 50'000'000 / 3000=16666.6.. Does this mean a waterflow of 16'666L/s or 16qm of water per seconde?
$endgroup$
– Tim
Jan 9 at 21:51
|
show 6 more comments
$begingroup$
I have a turbine which produces 50MW, the water falls down from three meters height.
How much water does flow per second?
Please do not answer this question, but rather give an explanation how I can calculate it myself.
physics
asked Jan 9 at 21:38
TimTim
31
$endgroup$
I have a turbine which produces 50MW, the water falls down from three meters height.
How much water does flow per second?
Please do not answer this question, but rather give an explanation how I can calculate it myself.
physics
physics
asked Jan 9 at 21:38
TimTim
31
asked Jan 9 at 21:38
TimTim
31
asked Jan 9 at 21:38
TimTim
31
asked Jan 9 at 21:38
TimTim
31
asked Jan 9 at 21:38
TimTim
31
31
$begingroup$
How is this about mathematics?
$endgroup$
– copper.hat
Jan 9 at 21:41
$begingroup$
@copper.hat I can't quite follow what your question is. Do you mean why I posted it on Mathematics? As I didn't see a Physics SE. And physic belongs to math. The math is calculating the waterflow.
$endgroup$
– Tim
Jan 9 at 21:44
$begingroup$
And what is the difficulty? It is not about math. It is a physics homework question.
$endgroup$
– copper.hat
Jan 9 at 21:45
$begingroup$
I don't know how I can get the water flow out of generated Power and fall height.
$endgroup$
– Tim
Jan 9 at 21:46
1
$begingroup$
Mhm. 1 m^3 = 1000 Liters = 1000kg so W=1000*3=3000 So if I understand right, 50'000'000 / 3000=16666.6.. Does this mean a waterflow of 16'666L/s or 16qm of water per seconde?
$endgroup$
– Tim
Jan 9 at 21:51
|
show 6 more comments
$begingroup$
How is this about mathematics?
$endgroup$
– copper.hat
Jan 9 at 21:41
$begingroup$
@copper.hat I can't quite follow what your question is. Do you mean why I posted it on Mathematics? As I didn't see a Physics SE. And physic belongs to math. The math is calculating the waterflow.
$endgroup$
– Tim
Jan 9 at 21:44
$begingroup$
And what is the difficulty? It is not about math. It is a physics homework question.
$endgroup$
– copper.hat
Jan 9 at 21:45
$begingroup$
I don't know how I can get the water flow out of generated Power and fall height.
$endgroup$
– Tim
Jan 9 at 21:46
1
$begingroup$
Mhm. 1 m^3 = 1000 Liters = 1000kg so W=1000*3=3000 So if I understand right, 50'000'000 / 3000=16666.6.. Does this mean a waterflow of 16'666L/s or 16qm of water per seconde?
$endgroup$
– Tim
Jan 9 at 21:51
$begingroup$
How is this about mathematics?
$endgroup$
– copper.hat
Jan 9 at 21:41
$begingroup$
How is this about mathematics?
$endgroup$
– copper.hat
Jan 9 at 21:41
$begingroup$
@copper.hat I can't quite follow what your question is. Do you mean why I posted it on Mathematics? As I didn't see a Physics SE. And physic belongs to math. The math is calculating the waterflow.
$endgroup$
– Tim
Jan 9 at 21:44
$begingroup$
@copper.hat I can't quite follow what your question is. Do you mean why I posted it on Mathematics? As I didn't see a Physics SE. And physic belongs to math. The math is calculating the waterflow.
$endgroup$
– Tim
Jan 9 at 21:44
$begingroup$
And what is the difficulty? It is not about math. It is a physics homework question.
$endgroup$
– copper.hat
Jan 9 at 21:45
$begingroup$
And what is the difficulty? It is not about math. It is a physics homework question.
$endgroup$
– copper.hat
Jan 9 at 21:45
$begingroup$
I don't know how I can get the water flow out of generated Power and fall height.
$endgroup$
– Tim
Jan 9 at 21:46
$begingroup$
I don't know how I can get the water flow out of generated Power and fall height.
$endgroup$
– Tim
Jan 9 at 21:46
1
1
$begingroup$
Mhm. 1 m^3 = 1000 Liters = 1000kg so W=1000*3=3000 So if I understand right, 50'000'000 / 3000=16666.6.. Does this mean a waterflow of 16'666L/s or 16qm of water per seconde?
$endgroup$
– Tim
Jan 9 at 21:51
$begingroup$
Mhm. 1 m^3 = 1000 Liters = 1000kg so W=1000*3=3000 So if I understand right, 50'000'000 / 3000=16666.6.. Does this mean a waterflow of 16'666L/s or 16qm of water per seconde?
$endgroup$
– Tim
Jan 9 at 21:51
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This is physics, not math.
If a mass $m$ of water falls through a height $h$ then the potential energy change is
$mgh$.
If $dot{m}$ of water passes through a height $h$ then the potential energy change per unit time is $dot{m}gh$.
If water has an average density of $rho$, and the volume flow rate is $dot{V}$ then the
mass flow rate is $dot{m} = rho dot{V}$.
If all the potential energy is converted to usable work $P$ in the turbine then we
have $P = rho dot{V} gh$.
You have $P,h,g,rho$. Compute $dot{V}$.
Addendum:
$rho$ is the density of water, we usually take $rho = 1000 kg/m^3$.
$g$ is the acceleration due to gravity, we usually take $g = 9.81 m/s^2$.
In the above, $h=3 m$, $P= 50 times 10^6 W$.
The dot as in $dot{m}$ usually means the rate of change with respect to time.
So, $dot{V}$ is the change of volume per unit time, or flow rate.
answered Jan 9 at 22:08
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
Thank you, I don't understand what half of these Letters mean, but I will google them.
$endgroup$
– Tim
Jan 9 at 22:11
1
$begingroup$
It is more than a little surprising to me that you are doing physics and the letters do not mean anything to you.
$endgroup$
– copper.hat
Jan 9 at 22:13
$begingroup$
Well, they do say me something. but I'm not sure about conversions and all the stuff. I'm just a normal student in 8th grade.
$endgroup$
– Tim
Jan 9 at 22:20
$begingroup$
Do you know the formula for potential energy of a mass $m$ at height $h$ with acceleration due to gravity $g$?
$endgroup$
– copper.hat
Jan 9 at 22:26
$begingroup$
I added an explanation of terms, I assumed that you were at the university level, sorry.
$endgroup$
– copper.hat
Jan 9 at 22:38
|
show 1 more comment
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1 Answer
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oldest
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1 Answer
1
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oldest
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active
oldest
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active
oldest
votes
$begingroup$
This is physics, not math.
If a mass $m$ of water falls through a height $h$ then the potential energy change is
$mgh$.
If $dot{m}$ of water passes through a height $h$ then the potential energy change per unit time is $dot{m}gh$.
If water has an average density of $rho$, and the volume flow rate is $dot{V}$ then the
mass flow rate is $dot{m} = rho dot{V}$.
If all the potential energy is converted to usable work $P$ in the turbine then we
have $P = rho dot{V} gh$.
You have $P,h,g,rho$. Compute $dot{V}$.
Addendum:
$rho$ is the density of water, we usually take $rho = 1000 kg/m^3$.
$g$ is the acceleration due to gravity, we usually take $g = 9.81 m/s^2$.
In the above, $h=3 m$, $P= 50 times 10^6 W$.
The dot as in $dot{m}$ usually means the rate of change with respect to time.
So, $dot{V}$ is the change of volume per unit time, or flow rate.
answered Jan 9 at 22:08
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
Thank you, I don't understand what half of these Letters mean, but I will google them.
$endgroup$
– Tim
Jan 9 at 22:11
1
$begingroup$
It is more than a little surprising to me that you are doing physics and the letters do not mean anything to you.
$endgroup$
– copper.hat
Jan 9 at 22:13
$begingroup$
Well, they do say me something. but I'm not sure about conversions and all the stuff. I'm just a normal student in 8th grade.
$endgroup$
– Tim
Jan 9 at 22:20
$begingroup$
Do you know the formula for potential energy of a mass $m$ at height $h$ with acceleration due to gravity $g$?
$endgroup$
– copper.hat
Jan 9 at 22:26
$begingroup$
I added an explanation of terms, I assumed that you were at the university level, sorry.
$endgroup$
– copper.hat
Jan 9 at 22:38
|
show 1 more comment
$begingroup$
This is physics, not math.
If a mass $m$ of water falls through a height $h$ then the potential energy change is
$mgh$.
If $dot{m}$ of water passes through a height $h$ then the potential energy change per unit time is $dot{m}gh$.
If water has an average density of $rho$, and the volume flow rate is $dot{V}$ then the
mass flow rate is $dot{m} = rho dot{V}$.
If all the potential energy is converted to usable work $P$ in the turbine then we
have $P = rho dot{V} gh$.
You have $P,h,g,rho$. Compute $dot{V}$.
Addendum:
$rho$ is the density of water, we usually take $rho = 1000 kg/m^3$.
$g$ is the acceleration due to gravity, we usually take $g = 9.81 m/s^2$.
In the above, $h=3 m$, $P= 50 times 10^6 W$.
The dot as in $dot{m}$ usually means the rate of change with respect to time.
So, $dot{V}$ is the change of volume per unit time, or flow rate.
answered Jan 9 at 22:08
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
Thank you, I don't understand what half of these Letters mean, but I will google them.
$endgroup$
– Tim
Jan 9 at 22:11
1
$begingroup$
It is more than a little surprising to me that you are doing physics and the letters do not mean anything to you.
$endgroup$
– copper.hat
Jan 9 at 22:13
$begingroup$
Well, they do say me something. but I'm not sure about conversions and all the stuff. I'm just a normal student in 8th grade.
$endgroup$
– Tim
Jan 9 at 22:20
$begingroup$
Do you know the formula for potential energy of a mass $m$ at height $h$ with acceleration due to gravity $g$?
$endgroup$
– copper.hat
Jan 9 at 22:26
$begingroup$
I added an explanation of terms, I assumed that you were at the university level, sorry.
$endgroup$
– copper.hat
Jan 9 at 22:38
|
show 1 more comment
$begingroup$
This is physics, not math.
If a mass $m$ of water falls through a height $h$ then the potential energy change is
$mgh$.
If $dot{m}$ of water passes through a height $h$ then the potential energy change per unit time is $dot{m}gh$.
If water has an average density of $rho$, and the volume flow rate is $dot{V}$ then the
mass flow rate is $dot{m} = rho dot{V}$.
If all the potential energy is converted to usable work $P$ in the turbine then we
have $P = rho dot{V} gh$.
You have $P,h,g,rho$. Compute $dot{V}$.
Addendum:
$rho$ is the density of water, we usually take $rho = 1000 kg/m^3$.
$g$ is the acceleration due to gravity, we usually take $g = 9.81 m/s^2$.
In the above, $h=3 m$, $P= 50 times 10^6 W$.
The dot as in $dot{m}$ usually means the rate of change with respect to time.
So, $dot{V}$ is the change of volume per unit time, or flow rate.
answered Jan 9 at 22:08
copper.hatcopper.hat
127k559160
$endgroup$
This is physics, not math.
If a mass $m$ of water falls through a height $h$ then the potential energy change is
$mgh$.
If $dot{m}$ of water passes through a height $h$ then the potential energy change per unit time is $dot{m}gh$.
If water has an average density of $rho$, and the volume flow rate is $dot{V}$ then the
mass flow rate is $dot{m} = rho dot{V}$.
If all the potential energy is converted to usable work $P$ in the turbine then we
have $P = rho dot{V} gh$.
You have $P,h,g,rho$. Compute $dot{V}$.
Addendum:
$rho$ is the density of water, we usually take $rho = 1000 kg/m^3$.
$g$ is the acceleration due to gravity, we usually take $g = 9.81 m/s^2$.
In the above, $h=3 m$, $P= 50 times 10^6 W$.
The dot as in $dot{m}$ usually means the rate of change with respect to time.
So, $dot{V}$ is the change of volume per unit time, or flow rate.
answered Jan 9 at 22:08
copper.hatcopper.hat
127k559160
edited Jan 9 at 22:34
answered Jan 9 at 22:08
copper.hatcopper.hat
127k559160
answered Jan 9 at 22:08
copper.hatcopper.hat
127k559160
answered Jan 9 at 22:08
copper.hatcopper.hat
127k559160
127k559160
$begingroup$
Thank you, I don't understand what half of these Letters mean, but I will google them.
$endgroup$
– Tim
Jan 9 at 22:11
1
$begingroup$
It is more than a little surprising to me that you are doing physics and the letters do not mean anything to you.
$endgroup$
– copper.hat
Jan 9 at 22:13
$begingroup$
Well, they do say me something. but I'm not sure about conversions and all the stuff. I'm just a normal student in 8th grade.
$endgroup$
– Tim
Jan 9 at 22:20
$begingroup$
Do you know the formula for potential energy of a mass $m$ at height $h$ with acceleration due to gravity $g$?
$endgroup$
– copper.hat
Jan 9 at 22:26
$begingroup$
I added an explanation of terms, I assumed that you were at the university level, sorry.
$endgroup$
– copper.hat
Jan 9 at 22:38
|
show 1 more comment
$begingroup$
Thank you, I don't understand what half of these Letters mean, but I will google them.
$endgroup$
– Tim
Jan 9 at 22:11
1
$begingroup$
It is more than a little surprising to me that you are doing physics and the letters do not mean anything to you.
$endgroup$
– copper.hat
Jan 9 at 22:13
$begingroup$
Well, they do say me something. but I'm not sure about conversions and all the stuff. I'm just a normal student in 8th grade.
$endgroup$
– Tim
Jan 9 at 22:20
$begingroup$
Do you know the formula for potential energy of a mass $m$ at height $h$ with acceleration due to gravity $g$?
$endgroup$
– copper.hat
Jan 9 at 22:26
$begingroup$
I added an explanation of terms, I assumed that you were at the university level, sorry.
$endgroup$
– copper.hat
Jan 9 at 22:38
$begingroup$
Thank you, I don't understand what half of these Letters mean, but I will google them.
$endgroup$
– Tim
Jan 9 at 22:11
$begingroup$
Thank you, I don't understand what half of these Letters mean, but I will google them.
$endgroup$
– Tim
Jan 9 at 22:11
1
1
$begingroup$
It is more than a little surprising to me that you are doing physics and the letters do not mean anything to you.
$endgroup$
– copper.hat
Jan 9 at 22:13
$begingroup$
It is more than a little surprising to me that you are doing physics and the letters do not mean anything to you.
$endgroup$
– copper.hat
Jan 9 at 22:13
$begingroup$
Well, they do say me something. but I'm not sure about conversions and all the stuff. I'm just a normal student in 8th grade.
$endgroup$
– Tim
Jan 9 at 22:20
$begingroup$
Well, they do say me something. but I'm not sure about conversions and all the stuff. I'm just a normal student in 8th grade.
$endgroup$
– Tim
Jan 9 at 22:20
$begingroup$
Do you know the formula for potential energy of a mass $m$ at height $h$ with acceleration due to gravity $g$?
$endgroup$
– copper.hat
Jan 9 at 22:26
$begingroup$
Do you know the formula for potential energy of a mass $m$ at height $h$ with acceleration due to gravity $g$?
$endgroup$
– copper.hat
Jan 9 at 22:26
$begingroup$
I added an explanation of terms, I assumed that you were at the university level, sorry.
$endgroup$
– copper.hat
Jan 9 at 22:38
$begingroup$
I added an explanation of terms, I assumed that you were at the university level, sorry.
$endgroup$
– copper.hat
Jan 9 at 22:38
|
show 1 more comment
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$begingroup$
How is this about mathematics?
$endgroup$
– copper.hat
Jan 9 at 21:41
$begingroup$
@copper.hat I can't quite follow what your question is. Do you mean why I posted it on Mathematics? As I didn't see a Physics SE. And physic belongs to math. The math is calculating the waterflow.
$endgroup$
– Tim
Jan 9 at 21:44
$begingroup$
And what is the difficulty? It is not about math. It is a physics homework question.
$endgroup$
– copper.hat
Jan 9 at 21:45
$begingroup$
I don't know how I can get the water flow out of generated Power and fall height.
$endgroup$
– Tim
Jan 9 at 21:46
1
$begingroup$
Mhm. 1 m^3 = 1000 Liters = 1000kg so W=1000*3=3000 So if I understand right, 50'000'000 / 3000=16666.6.. Does this mean a waterflow of 16'666L/s or 16qm of water per seconde?
$endgroup$
– Tim
Jan 9 at 21:51