Mixing
Speed of water jet driven by 300 bar
In a setting where portals can be opened between one location and another,
Suppose you open a portal of area one square meter, with one end at the surface and the other end at the bottom of an ocean, three kilometers down, where the pressure is three hundred bar. Obviously a jet of water will come through the portal.
What will be the speed of the jet? Equivalently, how many cubic meters per second will come through?
science-based transportation
asked Nov 20 '18 at 0:16
rwallace
719415
add a comment |
In a setting where portals can be opened between one location and another,
Suppose you open a portal of area one square meter, with one end at the surface and the other end at the bottom of an ocean, three kilometers down, where the pressure is three hundred bar. Obviously a jet of water will come through the portal.
What will be the speed of the jet? Equivalently, how many cubic meters per second will come through?
science-based transportation
asked Nov 20 '18 at 0:16
rwallace
719415
6
Obligatory XKCD
– Cort Ammon
Nov 20 '18 at 7:31
4
Bit of a bigger hole, but obligatory What If XKCD
– CalvT
Nov 20 '18 at 10:10
There would be no high pressure water jet. The portal acts on water particles with force enough to balance the pressure. The second principle of teleportation is that energy required for a body to enter a portal is equal of resulting gain of its potential energy - thus the energy conservation principle is not being violated.
– abukaj
Nov 22 '18 at 9:46
add a comment |
In a setting where portals can be opened between one location and another,
Suppose you open a portal of area one square meter, with one end at the surface and the other end at the bottom of an ocean, three kilometers down, where the pressure is three hundred bar. Obviously a jet of water will come through the portal.
What will be the speed of the jet? Equivalently, how many cubic meters per second will come through?
science-based transportation
asked Nov 20 '18 at 0:16
rwallace
719415
In a setting where portals can be opened between one location and another,
Suppose you open a portal of area one square meter, with one end at the surface and the other end at the bottom of an ocean, three kilometers down, where the pressure is three hundred bar. Obviously a jet of water will come through the portal.
What will be the speed of the jet? Equivalently, how many cubic meters per second will come through?
science-based transportation
science-based transportation
asked Nov 20 '18 at 0:16
rwallace
719415
asked Nov 20 '18 at 0:16
rwallace
719415
asked Nov 20 '18 at 0:16
rwallace
719415
asked Nov 20 '18 at 0:16
rwallace
719415
asked Nov 20 '18 at 0:16
rwallace
719415
719415
6
Obligatory XKCD
– Cort Ammon
Nov 20 '18 at 7:31
4
Bit of a bigger hole, but obligatory What If XKCD
– CalvT
Nov 20 '18 at 10:10
There would be no high pressure water jet. The portal acts on water particles with force enough to balance the pressure. The second principle of teleportation is that energy required for a body to enter a portal is equal of resulting gain of its potential energy - thus the energy conservation principle is not being violated.
– abukaj
Nov 22 '18 at 9:46
add a comment |
6
Obligatory XKCD
– Cort Ammon
Nov 20 '18 at 7:31
4
Bit of a bigger hole, but obligatory What If XKCD
– CalvT
Nov 20 '18 at 10:10
There would be no high pressure water jet. The portal acts on water particles with force enough to balance the pressure. The second principle of teleportation is that energy required for a body to enter a portal is equal of resulting gain of its potential energy - thus the energy conservation principle is not being violated.
– abukaj
Nov 22 '18 at 9:46
6
6
Obligatory XKCD
– Cort Ammon
Nov 20 '18 at 7:31
Obligatory XKCD
– Cort Ammon
Nov 20 '18 at 7:31
4
4
Bit of a bigger hole, but obligatory What If XKCD
– CalvT
Nov 20 '18 at 10:10
Bit of a bigger hole, but obligatory What If XKCD
– CalvT
Nov 20 '18 at 10:10
There would be no high pressure water jet. The portal acts on water particles with force enough to balance the pressure. The second principle of teleportation is that energy required for a body to enter a portal is equal of resulting gain of its potential energy - thus the energy conservation principle is not being violated.
– abukaj
Nov 22 '18 at 9:46
There would be no high pressure water jet. The portal acts on water particles with force enough to balance the pressure. The second principle of teleportation is that energy required for a body to enter a portal is equal of resulting gain of its potential energy - thus the energy conservation principle is not being violated.
– abukaj
Nov 22 '18 at 9:46
add a comment |
2 Answers
2
active
oldest
votes
Pressure from depth
Hydrostatic pressure is the pressure felt due to the weight of things above it. The pressure at any depth in the ocean can be calculated by the equation for hydrostatic pressure:
$$p - p_0 = rho gh.$$
Here, $p_0$ is atmospheric pressure, and $p$ is pressure at the desired depth. The density of the fluid in the ocean is $rho$, $g$ is the acceleration due to gravity, and $h$ is the height of the column of fluid.
The density of water changes slightly with temperature, and even more slightly with pressure. We will assume a standard value of 1030 kg/m$^3$, which is accurate enough to three significant digits.
$$begin{align}p &= 1030text{ kg/m}^3cdot9.81text{ m/s}^2cdot3000text{ m} + 101000text{ Pa}\
&=30.4 text{ MPa}end{align}$$
Flow velocity from pressure
Bernoulli's equation governs incompressible flow of fluids at a low Mach number. Since our flow will be well below the speed of sound (as we'll see) and since we've already made an incompressibility assumption by using constant density in the last equation, we can use Bernoulli's equation here. Bernoulli's equation is
$$c = frac{v^2}{2} + gz + frac{p}{rho}$$
where $c$ is a constant, $v$ is the velocity of fluid flow, $z$ is elevation above reference, and $g$, $p$, and $rho$ are as before.
The constant can be factored out of the equation by rewriting the equation in terms of total head, which has units of meters. Before we re-write, we will cancel out $z$ as zero. You are asking for flow through a portal with no depth, there is no net elevation change between the pressure and low pressure zones. The energy head equation is
$$H = frac{p}{rho g}+frac{v^2}{2g}.$$
For the zero-velocity situation, i.e. on the ocean side of the portal, we set
$$H = frac{p_O}{rho g}.$$
On the atmosphere side of the portal, we set
$$H = frac{p_a}{rho g}+frac{v^2}{2g}.$$
Now we can set these two quantities equal to each other to solve for the velocity of the fluid flowing through the portal.
$$begin{align}frac{p_O}{rho g} &= frac{p_a}{rho g} + frac{v^2}{2g}\
frac{30400000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2} &= frac{101000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2}+frac{v^2}{2cdot9.81text{ m/s}^2}\
v^2 &= 2cdot9.81cdotleft(frac{304000000}{10100}-frac{101000}{10100}right)text{ m}^2text{/s}^2\
v&= 243 text{ m/s}
end{align}$$
Volumetric flow from flow velocity
Volumetric flow is expressd as
$$dot{V} = vA,$$ where $A$ is the area of the portal. Sinc $A = 1text{ m}^2$, we have
$$dot{V} = 243 text{ m}^3text{/s}.$$
This is equivalent to the flow of the Tiber river at Rome.
edited Nov 20 '18 at 3:50
Davislor
2,906714
answered Nov 20 '18 at 1:59
kingledion
72.8k26244431
4
To put this into perspective: That is ~870km/h it is already 70% of Mach1/Speed of sound in air. If you jump off a plane the terminal velocity is about 60 m/s, this will hit you a lot harder. The power of the jet will hit like a full-blown tsunami and devastate most things in its path.
– Falco
Nov 20 '18 at 10:27
4
@Falco It is 70% of speed of sound in air, but it is about 16% of speed of sound in water, so kingledion's assertion that the speed is much less than the speed of sound is correct. "The power of the jet will hit like a full-blown tsunami" - err, I think you vastly underestimated how dangerous the water will be. It is moving at roughly the speed from a black-powder musket, and it is a lot heavier than a musket ball.
– Martin Bonner
Nov 20 '18 at 13:17
@MartinBonner Yeah, I meant to put that in the post but looks like I forgot. Mach number of fluid flow is relative to the speed of sound in the fluid.
– kingledion
Nov 20 '18 at 13:37
@MartinBonner I don't know about "heavier" since the water is not as dense as a musket ball or stone flung from a catapult. Water Jet Cutters operate at an order of magnitude higher MPa so this water Jet will probably not cut through stone or steel, but will probably destroy a castle wall? I would be interested if one could calculate the destructive power comperatively? (Maybe N/cm² ?)
– Falco
Nov 20 '18 at 13:38
2
Obligatory: this breaks conservation of energy in a BIG way. Since it hasn't been said yet.
– MindS1
Nov 20 '18 at 13:47
|
show 4 more comments
This is easy. Just use Bernoulli equation. For such situation, it is like this water is falling from $3 , mathrm{km} .$
$$
V_{text{exit}}
= sqrt{2 H g}
= sqrt{2 times 3000 times 9.81}
= 242.61 , frac{mathrm{m}}{mathrm{s}}
$$
edited Nov 20 '18 at 6:30
Nat
3971412
answered Nov 20 '18 at 1:29
Artemijs Danilovs
1,674112
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Pressure from depth
Hydrostatic pressure is the pressure felt due to the weight of things above it. The pressure at any depth in the ocean can be calculated by the equation for hydrostatic pressure:
$$p - p_0 = rho gh.$$
Here, $p_0$ is atmospheric pressure, and $p$ is pressure at the desired depth. The density of the fluid in the ocean is $rho$, $g$ is the acceleration due to gravity, and $h$ is the height of the column of fluid.
The density of water changes slightly with temperature, and even more slightly with pressure. We will assume a standard value of 1030 kg/m$^3$, which is accurate enough to three significant digits.
$$begin{align}p &= 1030text{ kg/m}^3cdot9.81text{ m/s}^2cdot3000text{ m} + 101000text{ Pa}\
&=30.4 text{ MPa}end{align}$$
Flow velocity from pressure
Bernoulli's equation governs incompressible flow of fluids at a low Mach number. Since our flow will be well below the speed of sound (as we'll see) and since we've already made an incompressibility assumption by using constant density in the last equation, we can use Bernoulli's equation here. Bernoulli's equation is
$$c = frac{v^2}{2} + gz + frac{p}{rho}$$
where $c$ is a constant, $v$ is the velocity of fluid flow, $z$ is elevation above reference, and $g$, $p$, and $rho$ are as before.
The constant can be factored out of the equation by rewriting the equation in terms of total head, which has units of meters. Before we re-write, we will cancel out $z$ as zero. You are asking for flow through a portal with no depth, there is no net elevation change between the pressure and low pressure zones. The energy head equation is
$$H = frac{p}{rho g}+frac{v^2}{2g}.$$
For the zero-velocity situation, i.e. on the ocean side of the portal, we set
$$H = frac{p_O}{rho g}.$$
On the atmosphere side of the portal, we set
$$H = frac{p_a}{rho g}+frac{v^2}{2g}.$$
Now we can set these two quantities equal to each other to solve for the velocity of the fluid flowing through the portal.
$$begin{align}frac{p_O}{rho g} &= frac{p_a}{rho g} + frac{v^2}{2g}\
frac{30400000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2} &= frac{101000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2}+frac{v^2}{2cdot9.81text{ m/s}^2}\
v^2 &= 2cdot9.81cdotleft(frac{304000000}{10100}-frac{101000}{10100}right)text{ m}^2text{/s}^2\
v&= 243 text{ m/s}
end{align}$$
Volumetric flow from flow velocity
Volumetric flow is expressd as
$$dot{V} = vA,$$ where $A$ is the area of the portal. Sinc $A = 1text{ m}^2$, we have
$$dot{V} = 243 text{ m}^3text{/s}.$$
This is equivalent to the flow of the Tiber river at Rome.
edited Nov 20 '18 at 3:50
Davislor
2,906714
answered Nov 20 '18 at 1:59
kingledion
72.8k26244431
4
To put this into perspective: That is ~870km/h it is already 70% of Mach1/Speed of sound in air. If you jump off a plane the terminal velocity is about 60 m/s, this will hit you a lot harder. The power of the jet will hit like a full-blown tsunami and devastate most things in its path.
– Falco
Nov 20 '18 at 10:27
4
@Falco It is 70% of speed of sound in air, but it is about 16% of speed of sound in water, so kingledion's assertion that the speed is much less than the speed of sound is correct. "The power of the jet will hit like a full-blown tsunami" - err, I think you vastly underestimated how dangerous the water will be. It is moving at roughly the speed from a black-powder musket, and it is a lot heavier than a musket ball.
– Martin Bonner
Nov 20 '18 at 13:17
@MartinBonner Yeah, I meant to put that in the post but looks like I forgot. Mach number of fluid flow is relative to the speed of sound in the fluid.
– kingledion
Nov 20 '18 at 13:37
@MartinBonner I don't know about "heavier" since the water is not as dense as a musket ball or stone flung from a catapult. Water Jet Cutters operate at an order of magnitude higher MPa so this water Jet will probably not cut through stone or steel, but will probably destroy a castle wall? I would be interested if one could calculate the destructive power comperatively? (Maybe N/cm² ?)
– Falco
Nov 20 '18 at 13:38
2
Obligatory: this breaks conservation of energy in a BIG way. Since it hasn't been said yet.
– MindS1
Nov 20 '18 at 13:47
|
show 4 more comments
Pressure from depth
Hydrostatic pressure is the pressure felt due to the weight of things above it. The pressure at any depth in the ocean can be calculated by the equation for hydrostatic pressure:
$$p - p_0 = rho gh.$$
Here, $p_0$ is atmospheric pressure, and $p$ is pressure at the desired depth. The density of the fluid in the ocean is $rho$, $g$ is the acceleration due to gravity, and $h$ is the height of the column of fluid.
The density of water changes slightly with temperature, and even more slightly with pressure. We will assume a standard value of 1030 kg/m$^3$, which is accurate enough to three significant digits.
$$begin{align}p &= 1030text{ kg/m}^3cdot9.81text{ m/s}^2cdot3000text{ m} + 101000text{ Pa}\
&=30.4 text{ MPa}end{align}$$
Flow velocity from pressure
Bernoulli's equation governs incompressible flow of fluids at a low Mach number. Since our flow will be well below the speed of sound (as we'll see) and since we've already made an incompressibility assumption by using constant density in the last equation, we can use Bernoulli's equation here. Bernoulli's equation is
$$c = frac{v^2}{2} + gz + frac{p}{rho}$$
where $c$ is a constant, $v$ is the velocity of fluid flow, $z$ is elevation above reference, and $g$, $p$, and $rho$ are as before.
The constant can be factored out of the equation by rewriting the equation in terms of total head, which has units of meters. Before we re-write, we will cancel out $z$ as zero. You are asking for flow through a portal with no depth, there is no net elevation change between the pressure and low pressure zones. The energy head equation is
$$H = frac{p}{rho g}+frac{v^2}{2g}.$$
For the zero-velocity situation, i.e. on the ocean side of the portal, we set
$$H = frac{p_O}{rho g}.$$
On the atmosphere side of the portal, we set
$$H = frac{p_a}{rho g}+frac{v^2}{2g}.$$
Now we can set these two quantities equal to each other to solve for the velocity of the fluid flowing through the portal.
$$begin{align}frac{p_O}{rho g} &= frac{p_a}{rho g} + frac{v^2}{2g}\
frac{30400000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2} &= frac{101000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2}+frac{v^2}{2cdot9.81text{ m/s}^2}\
v^2 &= 2cdot9.81cdotleft(frac{304000000}{10100}-frac{101000}{10100}right)text{ m}^2text{/s}^2\
v&= 243 text{ m/s}
end{align}$$
Volumetric flow from flow velocity
Volumetric flow is expressd as
$$dot{V} = vA,$$ where $A$ is the area of the portal. Sinc $A = 1text{ m}^2$, we have
$$dot{V} = 243 text{ m}^3text{/s}.$$
This is equivalent to the flow of the Tiber river at Rome.
edited Nov 20 '18 at 3:50
Davislor
2,906714
answered Nov 20 '18 at 1:59
kingledion
72.8k26244431
4
To put this into perspective: That is ~870km/h it is already 70% of Mach1/Speed of sound in air. If you jump off a plane the terminal velocity is about 60 m/s, this will hit you a lot harder. The power of the jet will hit like a full-blown tsunami and devastate most things in its path.
– Falco
Nov 20 '18 at 10:27
4
@Falco It is 70% of speed of sound in air, but it is about 16% of speed of sound in water, so kingledion's assertion that the speed is much less than the speed of sound is correct. "The power of the jet will hit like a full-blown tsunami" - err, I think you vastly underestimated how dangerous the water will be. It is moving at roughly the speed from a black-powder musket, and it is a lot heavier than a musket ball.
– Martin Bonner
Nov 20 '18 at 13:17
@MartinBonner Yeah, I meant to put that in the post but looks like I forgot. Mach number of fluid flow is relative to the speed of sound in the fluid.
– kingledion
Nov 20 '18 at 13:37
@MartinBonner I don't know about "heavier" since the water is not as dense as a musket ball or stone flung from a catapult. Water Jet Cutters operate at an order of magnitude higher MPa so this water Jet will probably not cut through stone or steel, but will probably destroy a castle wall? I would be interested if one could calculate the destructive power comperatively? (Maybe N/cm² ?)
– Falco
Nov 20 '18 at 13:38
2
Obligatory: this breaks conservation of energy in a BIG way. Since it hasn't been said yet.
– MindS1
Nov 20 '18 at 13:47
|
show 4 more comments
Pressure from depth
Hydrostatic pressure is the pressure felt due to the weight of things above it. The pressure at any depth in the ocean can be calculated by the equation for hydrostatic pressure:
$$p - p_0 = rho gh.$$
Here, $p_0$ is atmospheric pressure, and $p$ is pressure at the desired depth. The density of the fluid in the ocean is $rho$, $g$ is the acceleration due to gravity, and $h$ is the height of the column of fluid.
The density of water changes slightly with temperature, and even more slightly with pressure. We will assume a standard value of 1030 kg/m$^3$, which is accurate enough to three significant digits.
$$begin{align}p &= 1030text{ kg/m}^3cdot9.81text{ m/s}^2cdot3000text{ m} + 101000text{ Pa}\
&=30.4 text{ MPa}end{align}$$
Flow velocity from pressure
Bernoulli's equation governs incompressible flow of fluids at a low Mach number. Since our flow will be well below the speed of sound (as we'll see) and since we've already made an incompressibility assumption by using constant density in the last equation, we can use Bernoulli's equation here. Bernoulli's equation is
$$c = frac{v^2}{2} + gz + frac{p}{rho}$$
where $c$ is a constant, $v$ is the velocity of fluid flow, $z$ is elevation above reference, and $g$, $p$, and $rho$ are as before.
The constant can be factored out of the equation by rewriting the equation in terms of total head, which has units of meters. Before we re-write, we will cancel out $z$ as zero. You are asking for flow through a portal with no depth, there is no net elevation change between the pressure and low pressure zones. The energy head equation is
$$H = frac{p}{rho g}+frac{v^2}{2g}.$$
For the zero-velocity situation, i.e. on the ocean side of the portal, we set
$$H = frac{p_O}{rho g}.$$
On the atmosphere side of the portal, we set
$$H = frac{p_a}{rho g}+frac{v^2}{2g}.$$
Now we can set these two quantities equal to each other to solve for the velocity of the fluid flowing through the portal.
$$begin{align}frac{p_O}{rho g} &= frac{p_a}{rho g} + frac{v^2}{2g}\
frac{30400000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2} &= frac{101000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2}+frac{v^2}{2cdot9.81text{ m/s}^2}\
v^2 &= 2cdot9.81cdotleft(frac{304000000}{10100}-frac{101000}{10100}right)text{ m}^2text{/s}^2\
v&= 243 text{ m/s}
end{align}$$
Volumetric flow from flow velocity
Volumetric flow is expressd as
$$dot{V} = vA,$$ where $A$ is the area of the portal. Sinc $A = 1text{ m}^2$, we have
$$dot{V} = 243 text{ m}^3text{/s}.$$
This is equivalent to the flow of the Tiber river at Rome.
edited Nov 20 '18 at 3:50
Davislor
2,906714
answered Nov 20 '18 at 1:59
kingledion
72.8k26244431
Pressure from depth
Hydrostatic pressure is the pressure felt due to the weight of things above it. The pressure at any depth in the ocean can be calculated by the equation for hydrostatic pressure:
$$p - p_0 = rho gh.$$
Here, $p_0$ is atmospheric pressure, and $p$ is pressure at the desired depth. The density of the fluid in the ocean is $rho$, $g$ is the acceleration due to gravity, and $h$ is the height of the column of fluid.
The density of water changes slightly with temperature, and even more slightly with pressure. We will assume a standard value of 1030 kg/m$^3$, which is accurate enough to three significant digits.
$$begin{align}p &= 1030text{ kg/m}^3cdot9.81text{ m/s}^2cdot3000text{ m} + 101000text{ Pa}\
&=30.4 text{ MPa}end{align}$$
Flow velocity from pressure
Bernoulli's equation governs incompressible flow of fluids at a low Mach number. Since our flow will be well below the speed of sound (as we'll see) and since we've already made an incompressibility assumption by using constant density in the last equation, we can use Bernoulli's equation here. Bernoulli's equation is
$$c = frac{v^2}{2} + gz + frac{p}{rho}$$
where $c$ is a constant, $v$ is the velocity of fluid flow, $z$ is elevation above reference, and $g$, $p$, and $rho$ are as before.
The constant can be factored out of the equation by rewriting the equation in terms of total head, which has units of meters. Before we re-write, we will cancel out $z$ as zero. You are asking for flow through a portal with no depth, there is no net elevation change between the pressure and low pressure zones. The energy head equation is
$$H = frac{p}{rho g}+frac{v^2}{2g}.$$
For the zero-velocity situation, i.e. on the ocean side of the portal, we set
$$H = frac{p_O}{rho g}.$$
On the atmosphere side of the portal, we set
$$H = frac{p_a}{rho g}+frac{v^2}{2g}.$$
Now we can set these two quantities equal to each other to solve for the velocity of the fluid flowing through the portal.
$$begin{align}frac{p_O}{rho g} &= frac{p_a}{rho g} + frac{v^2}{2g}\
frac{30400000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2} &= frac{101000text{ Pa}}{1030text{ kg/m}^3cdot9.81text{ m/s}^2}+frac{v^2}{2cdot9.81text{ m/s}^2}\
v^2 &= 2cdot9.81cdotleft(frac{304000000}{10100}-frac{101000}{10100}right)text{ m}^2text{/s}^2\
v&= 243 text{ m/s}
end{align}$$
Volumetric flow from flow velocity
Volumetric flow is expressd as
$$dot{V} = vA,$$ where $A$ is the area of the portal. Sinc $A = 1text{ m}^2$, we have
$$dot{V} = 243 text{ m}^3text{/s}.$$
This is equivalent to the flow of the Tiber river at Rome.
edited Nov 20 '18 at 3:50
Davislor
2,906714
answered Nov 20 '18 at 1:59
kingledion
72.8k26244431
edited Nov 20 '18 at 3:50
Davislor
2,906714
edited Nov 20 '18 at 3:50
Davislor
2,906714
edited Nov 20 '18 at 3:50
Davislor
2,906714
2,906714
answered Nov 20 '18 at 1:59
kingledion
72.8k26244431
answered Nov 20 '18 at 1:59
kingledion
72.8k26244431
answered Nov 20 '18 at 1:59
kingledion
72.8k26244431
72.8k26244431
4
To put this into perspective: That is ~870km/h it is already 70% of Mach1/Speed of sound in air. If you jump off a plane the terminal velocity is about 60 m/s, this will hit you a lot harder. The power of the jet will hit like a full-blown tsunami and devastate most things in its path.
– Falco
Nov 20 '18 at 10:27
4
@Falco It is 70% of speed of sound in air, but it is about 16% of speed of sound in water, so kingledion's assertion that the speed is much less than the speed of sound is correct. "The power of the jet will hit like a full-blown tsunami" - err, I think you vastly underestimated how dangerous the water will be. It is moving at roughly the speed from a black-powder musket, and it is a lot heavier than a musket ball.
– Martin Bonner
Nov 20 '18 at 13:17
@MartinBonner Yeah, I meant to put that in the post but looks like I forgot. Mach number of fluid flow is relative to the speed of sound in the fluid.
– kingledion
Nov 20 '18 at 13:37
@MartinBonner I don't know about "heavier" since the water is not as dense as a musket ball or stone flung from a catapult. Water Jet Cutters operate at an order of magnitude higher MPa so this water Jet will probably not cut through stone or steel, but will probably destroy a castle wall? I would be interested if one could calculate the destructive power comperatively? (Maybe N/cm² ?)
– Falco
Nov 20 '18 at 13:38
2
Obligatory: this breaks conservation of energy in a BIG way. Since it hasn't been said yet.
– MindS1
Nov 20 '18 at 13:47
|
show 4 more comments
4
To put this into perspective: That is ~870km/h it is already 70% of Mach1/Speed of sound in air. If you jump off a plane the terminal velocity is about 60 m/s, this will hit you a lot harder. The power of the jet will hit like a full-blown tsunami and devastate most things in its path.
– Falco
Nov 20 '18 at 10:27
4
@Falco It is 70% of speed of sound in air, but it is about 16% of speed of sound in water, so kingledion's assertion that the speed is much less than the speed of sound is correct. "The power of the jet will hit like a full-blown tsunami" - err, I think you vastly underestimated how dangerous the water will be. It is moving at roughly the speed from a black-powder musket, and it is a lot heavier than a musket ball.
– Martin Bonner
Nov 20 '18 at 13:17
@MartinBonner Yeah, I meant to put that in the post but looks like I forgot. Mach number of fluid flow is relative to the speed of sound in the fluid.
– kingledion
Nov 20 '18 at 13:37
@MartinBonner I don't know about "heavier" since the water is not as dense as a musket ball or stone flung from a catapult. Water Jet Cutters operate at an order of magnitude higher MPa so this water Jet will probably not cut through stone or steel, but will probably destroy a castle wall? I would be interested if one could calculate the destructive power comperatively? (Maybe N/cm² ?)
– Falco
Nov 20 '18 at 13:38
2
Obligatory: this breaks conservation of energy in a BIG way. Since it hasn't been said yet.
– MindS1
Nov 20 '18 at 13:47
4
4
To put this into perspective: That is ~870km/h it is already 70% of Mach1/Speed of sound in air. If you jump off a plane the terminal velocity is about 60 m/s, this will hit you a lot harder. The power of the jet will hit like a full-blown tsunami and devastate most things in its path.
– Falco
Nov 20 '18 at 10:27
To put this into perspective: That is ~870km/h it is already 70% of Mach1/Speed of sound in air. If you jump off a plane the terminal velocity is about 60 m/s, this will hit you a lot harder. The power of the jet will hit like a full-blown tsunami and devastate most things in its path.
– Falco
Nov 20 '18 at 10:27
4
4
@Falco It is 70% of speed of sound in air, but it is about 16% of speed of sound in water, so kingledion's assertion that the speed is much less than the speed of sound is correct. "The power of the jet will hit like a full-blown tsunami" - err, I think you vastly underestimated how dangerous the water will be. It is moving at roughly the speed from a black-powder musket, and it is a lot heavier than a musket ball.
– Martin Bonner
Nov 20 '18 at 13:17
@Falco It is 70% of speed of sound in air, but it is about 16% of speed of sound in water, so kingledion's assertion that the speed is much less than the speed of sound is correct. "The power of the jet will hit like a full-blown tsunami" - err, I think you vastly underestimated how dangerous the water will be. It is moving at roughly the speed from a black-powder musket, and it is a lot heavier than a musket ball.
– Martin Bonner
Nov 20 '18 at 13:17
@MartinBonner Yeah, I meant to put that in the post but looks like I forgot. Mach number of fluid flow is relative to the speed of sound in the fluid.
– kingledion
Nov 20 '18 at 13:37
@MartinBonner Yeah, I meant to put that in the post but looks like I forgot. Mach number of fluid flow is relative to the speed of sound in the fluid.
– kingledion
Nov 20 '18 at 13:37
@MartinBonner I don't know about "heavier" since the water is not as dense as a musket ball or stone flung from a catapult. Water Jet Cutters operate at an order of magnitude higher MPa so this water Jet will probably not cut through stone or steel, but will probably destroy a castle wall? I would be interested if one could calculate the destructive power comperatively? (Maybe N/cm² ?)
– Falco
Nov 20 '18 at 13:38
@MartinBonner I don't know about "heavier" since the water is not as dense as a musket ball or stone flung from a catapult. Water Jet Cutters operate at an order of magnitude higher MPa so this water Jet will probably not cut through stone or steel, but will probably destroy a castle wall? I would be interested if one could calculate the destructive power comperatively? (Maybe N/cm² ?)
– Falco
Nov 20 '18 at 13:38
2
2
Obligatory: this breaks conservation of energy in a BIG way. Since it hasn't been said yet.
– MindS1
Nov 20 '18 at 13:47
Obligatory: this breaks conservation of energy in a BIG way. Since it hasn't been said yet.
– MindS1
Nov 20 '18 at 13:47
|
show 4 more comments
This is easy. Just use Bernoulli equation. For such situation, it is like this water is falling from $3 , mathrm{km} .$
$$
V_{text{exit}}
= sqrt{2 H g}
= sqrt{2 times 3000 times 9.81}
= 242.61 , frac{mathrm{m}}{mathrm{s}}
$$
edited Nov 20 '18 at 6:30
Nat
3971412
answered Nov 20 '18 at 1:29
Artemijs Danilovs
1,674112
add a comment |
This is easy. Just use Bernoulli equation. For such situation, it is like this water is falling from $3 , mathrm{km} .$
$$
V_{text{exit}}
= sqrt{2 H g}
= sqrt{2 times 3000 times 9.81}
= 242.61 , frac{mathrm{m}}{mathrm{s}}
$$
edited Nov 20 '18 at 6:30
Nat
3971412
answered Nov 20 '18 at 1:29
Artemijs Danilovs
1,674112
add a comment |
This is easy. Just use Bernoulli equation. For such situation, it is like this water is falling from $3 , mathrm{km} .$
$$
V_{text{exit}}
= sqrt{2 H g}
= sqrt{2 times 3000 times 9.81}
= 242.61 , frac{mathrm{m}}{mathrm{s}}
$$
edited Nov 20 '18 at 6:30
Nat
3971412
answered Nov 20 '18 at 1:29
Artemijs Danilovs
1,674112
This is easy. Just use Bernoulli equation. For such situation, it is like this water is falling from $3 , mathrm{km} .$
$$
V_{text{exit}}
= sqrt{2 H g}
= sqrt{2 times 3000 times 9.81}
= 242.61 , frac{mathrm{m}}{mathrm{s}}
$$
edited Nov 20 '18 at 6:30
Nat
3971412
answered Nov 20 '18 at 1:29
Artemijs Danilovs
1,674112
edited Nov 20 '18 at 6:30
Nat
3971412
edited Nov 20 '18 at 6:30
Nat
3971412
edited Nov 20 '18 at 6:30
Nat
3971412
3971412
answered Nov 20 '18 at 1:29
Artemijs Danilovs
1,674112
answered Nov 20 '18 at 1:29
Artemijs Danilovs
1,674112
answered Nov 20 '18 at 1:29
Artemijs Danilovs
1,674112
1,674112
add a comment |
add a comment |
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6
Obligatory XKCD
– Cort Ammon
Nov 20 '18 at 7:31
4
Bit of a bigger hole, but obligatory What If XKCD
– CalvT
Nov 20 '18 at 10:10
There would be no high pressure water jet. The portal acts on water particles with force enough to balance the pressure. The second principle of teleportation is that energy required for a body to enter a portal is equal of resulting gain of its potential energy - thus the energy conservation principle is not being violated.
– abukaj
Nov 22 '18 at 9:46