Getting the rate of drain from a tank [closed]
A tank with a top radius of 1m, a bottom radius of 0.5m and a height of 2m is initially filled with water. Water drains through a square hole of side 3cm in the bottom.
How do I get the rate of drain,
begin{equation}
frac{partial V}{partial t}
end{equation}
of the tank?
calculus
asked Nov 25 at 6:24
Gabriel Henry J. Lopez
31
closed as off-topic by Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz Nov 25 at 9:06
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A tank with a top radius of 1m, a bottom radius of 0.5m and a height of 2m is initially filled with water. Water drains through a square hole of side 3cm in the bottom.
How do I get the rate of drain,
begin{equation}
frac{partial V}{partial t}
end{equation}
of the tank?
calculus
asked Nov 25 at 6:24
Gabriel Henry J. Lopez
31
closed as off-topic by Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz Nov 25 at 9:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
A tank with a top radius of 1m, a bottom radius of 0.5m and a height of 2m is initially filled with water. Water drains through a square hole of side 3cm in the bottom.
How do I get the rate of drain,
begin{equation}
frac{partial V}{partial t}
end{equation}
of the tank?
calculus
asked Nov 25 at 6:24
Gabriel Henry J. Lopez
31
A tank with a top radius of 1m, a bottom radius of 0.5m and a height of 2m is initially filled with water. Water drains through a square hole of side 3cm in the bottom.
How do I get the rate of drain,
begin{equation}
frac{partial V}{partial t}
end{equation}
of the tank?
calculus
calculus
asked Nov 25 at 6:24
Gabriel Henry J. Lopez
31
asked Nov 25 at 6:24
Gabriel Henry J. Lopez
31
asked Nov 25 at 6:24
Gabriel Henry J. Lopez
31
asked Nov 25 at 6:24
Gabriel Henry J. Lopez
31
asked Nov 25 at 6:24
Gabriel Henry J. Lopez
31
31
closed as off-topic by Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz Nov 25 at 9:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz Nov 25 at 9:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.
Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.
Bernouli s principle says
$frac{1}{2}v^2=gh$
$v=sqrt{2gh}$
Area=side length of square hole squared.
Keep in mind h varies with time.
Volume in tank:
$V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$
So :
$Asqrt{2gh}=frac{dV}{dt}$
Will give you a differential equation in h as function of t.
Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.
answered Nov 25 at 6:53
TurlocTheRed
828311
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.
Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.
Bernouli s principle says
$frac{1}{2}v^2=gh$
$v=sqrt{2gh}$
Area=side length of square hole squared.
Keep in mind h varies with time.
Volume in tank:
$V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$
So :
$Asqrt{2gh}=frac{dV}{dt}$
Will give you a differential equation in h as function of t.
Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.
answered Nov 25 at 6:53
TurlocTheRed
828311
add a comment |
Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.
Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.
Bernouli s principle says
$frac{1}{2}v^2=gh$
$v=sqrt{2gh}$
Area=side length of square hole squared.
Keep in mind h varies with time.
Volume in tank:
$V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$
So :
$Asqrt{2gh}=frac{dV}{dt}$
Will give you a differential equation in h as function of t.
Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.
answered Nov 25 at 6:53
TurlocTheRed
828311
add a comment |
Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.
Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.
Bernouli s principle says
$frac{1}{2}v^2=gh$
$v=sqrt{2gh}$
Area=side length of square hole squared.
Keep in mind h varies with time.
Volume in tank:
$V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$
So :
$Asqrt{2gh}=frac{dV}{dt}$
Will give you a differential equation in h as function of t.
Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.
answered Nov 25 at 6:53
TurlocTheRed
828311
Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.
Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.
Bernouli s principle says
$frac{1}{2}v^2=gh$
$v=sqrt{2gh}$
Area=side length of square hole squared.
Keep in mind h varies with time.
Volume in tank:
$V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$
So :
$Asqrt{2gh}=frac{dV}{dt}$
Will give you a differential equation in h as function of t.
Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.
answered Nov 25 at 6:53
TurlocTheRed
828311
edited Nov 25 at 7:31
answered Nov 25 at 6:53
TurlocTheRed
828311
answered Nov 25 at 6:53
TurlocTheRed
828311
answered Nov 25 at 6:53
TurlocTheRed
828311
828311
add a comment |
add a comment |
