how much current to run anode?

[JBI]

I am getting a platinum coated anode that will have a surface area of 26
cm2 (100 mm long x 8 mm diameter). For my application, the maximum
recommended current density is 200 mA/cm. Based on that, how much
current should be flowing through the anode? I calculate square root of
26 (5.1) times 0.2 which is 1.02 A, is this correct? Thanks.

 

[amdx]

On 6/26/2018 9:37 AM, JBI wrote:

I am getting a platinum coated anode that will have a surface area of 26
cm2 (100 mm long x 8 mm diameter).  For my application, the maximum
recommended current density is 200 mA/cm.  Based on that, how much
current should be flowing through the anode?  I calculate square root of
26 (5.1) times 0.2 which is 1.02 A, is this correct?  Thanks.

I'm probably confused but, if it's 8mm in diameter, isn't it 25mm* in
circumference? And, if it's 25mm, then 25mm x 100mm = 2500mm^2,
convert to cm^2 and get 25cm^2. then multiply 200ma x 25cm = 5 amps.
I'll let an adult correct me. I have never seen a platinum coated
electrode.



* have not included areas on end.

 

[JBI]

On 06/26/2018 11:16 AM, amdx wrote:

On 6/26/2018 9:37 AM, JBI wrote:
I am getting a platinum coated anode that will have a surface area of
26 cm2 (100 mm long x 8 mm diameter).  For my application, the maximum
recommended current density is 200 mA/cm.  Based on that, how much
current should be flowing through the anode?  I calculate square root
of 26 (5.1) times 0.2 which is 1.02 A, is this correct?  Thanks.

  I'm probably confused but, if it's 8mm in diameter, isn't it 25mm* in
circumference?  And, if it's 25mm, then 25mm x 100mm = 2500mm^2, convert
to cm^2 and get 25cm^2. then multiply 200ma x 25cm = 5 amps.
  I'll let an adult correct me. I have never seen a platinum coated
electrode.



* have not included areas on end.

You're at the same place I was originally when I did the calculation,
which made me question whether or not I should take the square root of
the area and then multiply that by current density. When I took the
square, I got the 1.02A above, but much closer to your value when I
didn't.

Platinum coated electrodes have many applications:

https://tinyurl.com/yaufchs3

 

[pfjw@aol.com]

On Tuesday, June 26, 2018 at 11:16:09 AM UTC-4, amdx wrote:

On 6/26/2018 9:37 AM, JBI wrote:
I am getting a platinum coated anode that will have a surface area of 26
cm2 (100 mm long x 8 mm diameter).  For my application, the maximum
recommended current density is 200 mA/cm.  Based on that, how much
current should be flowing through the anode?  I calculate square root of
26 (5.1) times 0.2 which is 1.02 A, is this correct?  Thanks.

I'm probably confused but, if it's 8mm in diameter, isn't it 25mm* in
circumference? And, if it's 25mm, then 25mm x 100mm = 2500mm^2,
convert to cm^2 and get 25cm^2. then multiply 200ma x 25cm = 5 amps.
I'll let an adult correct me. I have never seen a platinum coated
electrode.



* have not included areas on end.

Lemme see- surface area of a cylinder = pi x D x H.

3.14159 x 8 = 25.133

25.133 x 100 = 2513.3

0.0200 x 2513.3 = 502.7

Divide by 100 (mm - cm) = 5.02 A.

add the end-cap, if relevant = pi x r^2 = 3.14159 x 16 = 50.27 = 0.01A

Or, close enough.

Peter Wieck
Melrose Park, PA

 

[amdx]

On 6/26/2018 10:54 AM, JBI wrote:

On 06/26/2018 11:16 AM, amdx wrote:
On 6/26/2018 9:37 AM, JBI wrote:
I am getting a platinum coated anode that will have a surface area of
26 cm2 (100 mm long x 8 mm diameter).  For my application, the
maximum recommended current density is 200 mA/cm.  Based on that, how
much current should be flowing through the anode?  I calculate square
root of 26 (5.1) times 0.2 which is 1.02 A, is this correct?  Thanks.

   I'm probably confused but, if it's 8mm in diameter, isn't it 25mm*
in circumference?  And, if it's 25mm, then 25mm x 100mm = 2500mm^2,
convert to cm^2 and get 25cm^2. then multiply 200ma x 25cm = 5 amps.
   I'll let an adult correct me. I have never seen a platinum coated
electrode.



* have not included areas on end.

You're at the same place I was originally when I did the calculation,
which made me question whether or not I should take the square root of
the area

I think the mistake your making is the 26 is already in cm^2, no need to
take the square root. Just 26 x 0.2 = 5.2 amps.

OR, another way, your anode is 100mm x 8mm dia or 100mm x 25mm,
convert to cm for 10cm x 2.5cm this equals 25cm^2. 25cm^2 x 0.2 amps = 5
amps.


and then multiply that by current density.  When I took the

square, I got the 1.02A above, but much closer to your value when I didn't.

Platinum coated electrodes have many applications:

https://tinyurl.com/yaufchs3

 

[amdx]

On 6/26/2018 11:06 AM, pfjw@aol.com wrote:

On Tuesday, June 26, 2018 at 11:16:09 AM UTC-4, amdx wrote:
On 6/26/2018 9:37 AM, JBI wrote:
I am getting a platinum coated anode that will have a surface area of 26
cm2 (100 mm long x 8 mm diameter).  For my application, the maximum
recommended current density is 200 mA/cm.  Based on that, how much
current should be flowing through the anode?  I calculate square root of
26 (5.1) times 0.2 which is 1.02 A, is this correct?  Thanks.

I'm probably confused but, if it's 8mm in diameter, isn't it 25mm* in
circumference? And, if it's 25mm, then 25mm x 100mm = 2500mm^2,
convert to cm^2 and get 25cm^2. then multiply 200ma x 25cm = 5 amps.
I'll let an adult correct me. I have never seen a platinum coated
electrode.



* have not included areas on end.

Lemme see- surface area of a cylinder = pi x D x H.

3.14159 x 8 = 25.133

25.133 x 100 = 2513.3

0.0200 x 2513.3 = 502.7

Divide by 100 (mm - cm) = 5.02 A.

add the end-cap, if relevant = pi x r^2 = 3.14159 x 16 = 50.27 = 0.01A

Or, close enough.

Peter Wieck
Melrose Park, PA

Sometimes it would be nice for an OP to say, Hey, thanks for the help,
or you're F'ing nuts and don't have clue. Just something, so we know
they didn't get hit by a car!

 

[JBI]

On 06/28/2018 10:22 AM, amdx wrote:

On 6/26/2018 11:06 AM, pfjw@aol.com wrote:
On Tuesday, June 26, 2018 at 11:16:09 AM UTC-4, amdx wrote:
On 6/26/2018 9:37 AM, JBI wrote:
I am getting a platinum coated anode that will have a surface area
of 26
cm2 (100 mm long x 8 mm diameter).  For my application, the maximum
recommended current density is 200 mA/cm.  Based on that, how much
current should be flowing through the anode?  I calculate square
root of
26 (5.1) times 0.2 which is 1.02 A, is this correct?  Thanks.

    I'm probably confused but, if it's 8mm in diameter, isn't it
25mm* in
circumference?  And, if it's 25mm, then 25mm x 100mm = 2500mm^2,
convert to cm^2 and get 25cm^2. then multiply 200ma x 25cm = 5 amps.
    I'll let an adult correct me. I have never seen a platinum coated
electrode.



* have not included areas on end.

Lemme see- surface area of a cylinder = pi x D x H.

3.14159 x 8 = 25.133

25.133 x 100 = 2513.3

0.0200 x 2513.3 = 502.7

Divide by 100 (mm - cm) = 5.02 A.

add the end-cap, if relevant = pi x r^2 = 3.14159 x 16 = 50.27 = 0.01A

Or, close enough.

Peter Wieck
Melrose Park, PA

 Sometimes it would be nice for an OP to say, Hey, thanks for the help,
or you're F'ing nuts and don't have clue. Just something, so we know
they didn't get hit by a car!

Quite right, Thank you ALL!