Current Density of water through a pipe

[Colts]

Homework Statement

(a) Water flows through a pipe of radius 4 cm at a rate of 3.5 m/s. Suppose each water molecule is singly ionized (highly unrealistic). What is the electric current density associated with the flow of these ions?
b) What is the electric current in the pipe?

Homework Equations

Not sure

The Attempt at a Solution

I don't know the charge of water and don't understand what a singly ionized molecule is? I tried to look it up and there is an electron on the outside of the molecule. Does this mean I can use the density of water to get the charge of the water? and use that to find the current?

 

[Dick]

Homework Statement

(a) Water flows through a pipe of radius 4 cm at a rate of 3.5 m/s. Suppose each water molecule is singly ionized (highly unrealistic). What is the electric current density associated with the flow of these ions?
b) What is the electric current in the pipe?

Homework Equations

Not sure

The Attempt at a Solution

I don't know the charge of water and don't understand what a singly ionized molecule is? I tried to look it up and there is an electron on the outside of the molecule. Does this mean I can use the density of water to get the charge of the water? and use that to find the current?

Yes, it sounds like that's what you are expected to do. A singly ionized water molecule will have lost one of its electrons. So the charge on the molecule is +1.

 

[Colts]

I got density of water to be 1000 kg/m^3 and 1 mole for every 18 grams of water.
So I did:
$\frac{1000 kg}{m^{3}}$ 2∏ $(.04m)^{2}$ $\frac{1 gram}{1000 kg}$ $\frac{1 mole}{18 grams}$ $\frac{6.022x10^{23}}{1 mole}$$1.6x10^{-19}$C
So I got 53.81 C/M
Then I multiplied this by 3.5 M/s to get
188.35 C/s

 

[SteamKing]

Why do you have 2pi in the area of the pipe?

 

[Dick]

I got density of water to be 1000 kg/m^3 and 1 mole for every 18 grams of water.
So I did:
$\frac{1000 kg}{m^{3}}$ 2∏ $(.04m)^{2}$ $\frac{1 gram}{1000 kg}$ $\frac{1 mole}{18 grams}$ $\frac{6.022x10^{23}}{1 mole}$$1.6x10^{-19}$C
So I got 53.81 C/M
Then I multiplied this by 3.5 M/s to get
188.35 C/s

You've got the right idea. But you've got an extra 2 in that formula. The area of a circle is pi*r^2, not 2*pi*r^2. And even if I include that the number doesn't come out right. Check that you've got the decimal point in the right place.

 

[rude man]

I got density of water to be 1000 kg/m^3 and 1 mole for every 18 grams of water.
So I did:
$\frac{1000 kg}{m^{3}}$ 2∏ $(.04m)^{2}$ $\frac{1 gram}{1000 kg}$ $\frac{1 mole}{18 grams}$ $\frac{6.022x10^{23}}{1 mole}$$1.6x10^{-19}$C
So I got 53.81 C/M
Then I multiplied this by 3.5 M/s to get
188.35 C/s

What happened to flow rate of 3.5 m/s? And area = pi r^2, not 2 pi r^2.

 

[Colts]

Ok, I messed up the area so without the 2 in there I get current density to be 26.9 C/M

Rude Man:
I used the flow rate to get the current.

 

[Dick]

Ok, I messed up the area so without the 2 in there I get current density to be 26.9 C/M

Rude Man:
I used the flow rate to get the current.

That seems ok.

 

[rude man]

Gee, fellas, I get an astronomically higher number than anything I've seen on this thread so far.

Question to the OP: what is the meaning of his "1gm/1000kg" coefficient in his post #6?
How about 1000g/kg instead? Only a factor of 10^6 different ...

 

[rude man]

Ok, I messed up the area so without the 2 in there I get current density to be 26.9 C/M

Rude Man:
I used the flow rate to get the current.

Not sure I undertsand that. The only difference between current and current density is the area of the pipe: current = current density x pipe area.

You seem to have computed charge per unit length, not current density.

 

[Dick]

Gee, fellas, I get an astronomically higher number than anything I've seen on this thread so far.

Question to the OP: what is the meaning of his "1gm/1000kg" coefficient in his post #6?
How about 1000g/kg instead? Only a factor of 10^6 different ...

Yeah, you are right. I goofed. There is a slip up in the units.