• Re: Amperage for 5HP table saw?

    From Ripsaw@21:1/5 to All on Mon Nov 8 19:01:32 2021
    I have the exact same Unisaw. I connect my saw to my 30 amp dryer connector via a 30 amp twist lock converter and I have never experienced any issue.
    I also have a 24" 5hp delta x5 planner that I also connect through my dryer outlet and also have never had an issue.

    --
    For full context, visit https://www.homeownershub.com/woodworking/amperage-for-5hp-table-saw-218022-.htm

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From krw@notreal.com@21:1/5 to 9dd8514b1cf2e9993897409011a5bd99@ex on Mon Nov 8 15:59:20 2021
    On Mon, 8 Nov 2021 19:01:32 +0000, Ripsaw <9dd8514b1cf2e9993897409011a5bd99@example.com> wrote:

    I have the exact same Unisaw. I connect my saw to my 30 amp dryer connector via a 30 amp twist lock converter and I have never experienced any issue.
    I also have a 24" 5hp delta x5 planner that I also connect through my dryer outlet and also have never had an issue.

    After 17 years he's probably figured it out or burned down his house.
    Do you feel better after trolling?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Bob Davis@21:1/5 to Joe McDonald on Mon Nov 8 16:07:24 2021
    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>, <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
    at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow" breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From krw@notreal.com@21:1/5 to All on Mon Nov 8 20:59:41 2021
    On Mon, 08 Nov 2021 20:30:47 -0500, Joe Gwinn <joegwinn@comcast.net>
    wrote:

    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrobertdavis@gmail.com> wrote:

    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
    at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be
    borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the >>> > startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow"
    breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.

    Sure it does, for RMS volts and amps, and pure resistances (like metal
    wires and components).

    Motors and inductors and capacitors are a different kettle of fish. An >ordinary ohmmeter measures DC resistance, which for an inductor is the >resistance of the copper wire winding, but AC current and voltage are >connected by AC reactance., which will be far larger than the DC
    resistance at the usual operating frequency of that inductor.

    A motor is a little different than an inductor. A motor is doing
    "work". If the motor were 100% efficient, it would look like a pure
    resistor. They aren't, of course, but the more efficient they are,
    the closer to a resistor they appear.

    80% efficiency for fully loaded small motors isn't too far off but
    probably a little high. They get worse when lightly loaded but for
    figuring circuit size, the fully loaded numbers are used. A safety
    factor of another 20% is a good idea (if not code). Motors can be
    overloaded. ;-)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Joe Gwinn@21:1/5 to wrobertdavis@gmail.com on Mon Nov 8 20:30:47 2021
    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrobertdavis@gmail.com> wrote:

    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
    at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be
    borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the
    startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow"
    breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.

    Sure it does, for RMS volts and amps, and pure resistances (like metal
    wires and components).

    Motors and inductors and capacitors are a different kettle of fish. An
    ordinary ohmmeter measures DC resistance, which for an inductor is the resistance of the copper wire winding, but AC current and voltage are
    connected by AC reactance., which will be far larger than the DC
    resistance at the usual operating frequency of that inductor.

    Joe Gwinn

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Clare Snyder@21:1/5 to wrobertdavis@gmail.com on Mon Nov 8 21:28:50 2021
    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrobertdavis@gmail.com> wrote:

    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
    at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be
    borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the
    startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow"
    breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.
    Yes it does - the R just means reactance instead of resistance -
    wheather inductive or capacitive. With resistive loads it's the same
    as with DC - read the resistance on your ohm-meter. With "inductive
    loads" the reactance will always be higher than the resistance (draws
    less current)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Clare Snyder@21:1/5 to All on Mon Nov 8 21:25:44 2021
    On Mon, 08 Nov 2021 20:30:47 -0500, Joe Gwinn <joegwinn@comcast.net>
    wrote:

    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrobertdavis@gmail.com> wrote:

    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
    at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be
    borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the >>> > startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow"
    breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.

    Sure it does, for RMS volts and amps, and pure resistances (like metal
    wires and components).

    Motors and inductors and capacitors are a different kettle of fish. An >ordinary ohmmeter measures DC resistance, which for an inductor is the >resistance of the copper wire winding, but AC current and voltage are >connected by AC reactance., which will be far larger than the DC
    resistance at the usual operating frequency of that inductor.

    Joe Gwinn
    You will want a "high magnetic" breaker - which, fortunately for you
    most "twinned" breakers are (at leat all Square D QO breakers are ). I
    ended up using a twinned breaker for my central vac (120 volt 12 amp)
    because it quite occaisionally tripped the single standard 15 amp
    breaker on start-up and a single high magnetic breaker was not
    readilly available locally and cost more to order in than my local
    supplier charged for the twinned breaker. Hasn't tripped since (about
    3 years) - and it tripped on 3 different single breakers - - -

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Bob Davis@21:1/5 to Clare Snyder on Wed Nov 10 05:22:37 2021
    On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:
    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrober...@gmail.com> wrote:
    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
    at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be
    borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the >> > startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow"
    breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.
    Yes it does - the R just means reactance instead of resistance -
    wheather inductive or capacitive. With resistive loads it's the same
    as with DC - read the resistance on your ohm-meter. With "inductive
    loads" the reactance will always be higher than the resistance (draws
    less current)

    What is a real world example of purely resistive load supplied by AC?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Bob Davis@21:1/5 to Joe Gwinn on Wed Nov 10 05:19:37 2021
    On Monday, November 8, 2021 at 7:30:56 PM UTC-6, Joe Gwinn wrote:
    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrober...@gmail.com> wrote:

    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
    at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be
    borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the >> > startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow"
    breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.
    Sure it does, for RMS volts and amps, and pure resistances (like metal
    wires and components).

    Motors and inductors and capacitors are a different kettle of fish. An ordinary ohmmeter measures DC resistance, which for an inductor is the resistance of the copper wire winding, but AC current and voltage are connected by AC reactance., which will be far larger than the DC
    resistance at the usual operating frequency of that inductor.

    Joe Gwinn
    You are right in the theory. In practice, doing calculations about motor current with simple ohm's law does not work because a motor is not a pure resistive load.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Scott Lurndal@21:1/5 to Bob Davis on Wed Nov 10 14:42:33 2021
    Bob Davis <wrobertdavis@gmail.com> writes:


    What is a real world example of purely resistive load supplied by AC?

    An incandescent light bulb.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From krw@notreal.com@21:1/5 to wrobertdavis@gmail.com on Wed Nov 10 12:14:36 2021
    On Wed, 10 Nov 2021 05:22:37 -0800 (PST), Bob Davis
    <wrobertdavis@gmail.com> wrote:

    On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:
    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrober...@gmail.com> wrote:
    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >> >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be
    borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the >> >> > startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow" >> >> breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.
    Yes it does - the R just means reactance instead of resistance -
    wheather inductive or capacitive. With resistive loads it's the same
    as with DC - read the resistance on your ohm-meter. With "inductive
    loads" the reactance will always be higher than the resistance (draws
    less current)

    What is a real world example of purely resistive load supplied by AC?

    Your oven. Water heaters, space heaters, light bulbs (sorta, though temperature dependent).

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From J. Clarke@21:1/5 to wrobertdavis@gmail.com on Wed Nov 10 16:42:30 2021
    On Wed, 10 Nov 2021 05:22:37 -0800 (PST), Bob Davis
    <wrobertdavis@gmail.com> wrote:

    On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:
    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrober...@gmail.com> wrote:
    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >> >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be
    borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the >> >> > startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow" >> >> breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.
    Yes it does - the R just means reactance instead of resistance -
    wheather inductive or capacitive. With resistive loads it's the same
    as with DC - read the resistance on your ohm-meter. With "inductive
    loads" the reactance will always be higher than the resistance (draws
    less current)

    What is a real world example of purely resistive load supplied by AC?

    I believe an oven or (non-induction) stove would come pretty close.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Markem618@21:1/5 to jclarke.873638@gmail.com on Wed Nov 10 16:33:54 2021
    On Wed, 10 Nov 2021 16:42:30 -0500, J. Clarke
    <jclarke.873638@gmail.com> wrote:

    On Wed, 10 Nov 2021 05:22:37 -0800 (PST), Bob Davis
    <wrobertdavis@gmail.com> wrote:

    On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:
    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrober...@gmail.com> wrote:
    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >>> >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be >>> >> > borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the >>> >> > startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous >>> >> load above 16 amps. It seems to me, that with the proper type of "slow" >>> >> breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.
    Yes it does - the R just means reactance instead of resistance -
    wheather inductive or capacitive. With resistive loads it's the same
    as with DC - read the resistance on your ohm-meter. With "inductive
    loads" the reactance will always be higher than the resistance (draws
    less current)

    What is a real world example of purely resistive load supplied by AC?

    I believe an oven or (non-induction) stove would come pretty close.

    Electric baseboards, electric resistor elements in a heat pump system
    for when the temp is to cold.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Bob Davis@21:1/5 to Bob Davis on Wed Nov 10 15:29:04 2021
    On Wednesday, November 10, 2021 at 7:22:39 AM UTC-6, Bob Davis wrote:
    On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:
    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrober...@gmail.com> wrote:
    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be >> > borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the >> > startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous
    load above 16 amps. It seems to me, that with the proper type of "slow" >> breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.
    Yes it does - the R just means reactance instead of resistance -
    wheather inductive or capacitive. With resistive loads it's the same
    as with DC - read the resistance on your ohm-meter. With "inductive
    loads" the reactance will always be higher than the resistance (draws
    less current)
    What is a real world example of purely resistive load supplied by AC?

    Thanks to everyone who responded.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Clare Snyder@21:1/5 to jclarke.873638@gmail.com on Thu Nov 11 00:30:25 2021
    On Wed, 10 Nov 2021 16:42:30 -0500, J. Clarke
    <jclarke.873638@gmail.com> wrote:

    On Wed, 10 Nov 2021 05:22:37 -0800 (PST), Bob Davis
    <wrobertdavis@gmail.com> wrote:

    On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:
    On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
    <wrober...@gmail.com> wrote:
    On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:
    In article <1078204646.761648@smirk>,
    <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:
    What amperage breaker does one need for a 5HP table saw on a 240V
    single phase circuit?

    My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >>> >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
    (look it up in the NEC sometime), this means a 25 A breaker would be >>> >> > borderline sufficient, and that a 30A breaker should be generous.
    With a slow breaker (not a fuse), there should be no problems with the >>> >> > startup surge of the motor tripping the breaker.
    Using Ohm's law this is a relatively simple question to answer...
    1 horsepower = 745.7 watts
    Power (watts) = I (amperage) x E (voltage)
    When you have two values for the variables in the equation you can
    figure out the third.
    P 3728.5 (watts 1 HP x 5)
    Therefore I = --- or I = ------
    E 230 volts

    230 volts = 16.21 amps
    240 volts = 15.53 amps

    Using the "80% rule", a 20 amp breaker should not exceed a continuous >>> >> load above 16 amps. It seems to me, that with the proper type of "slow" >>> >> breaker, you should be OK with a 20 amp breaker.

    Joe

    PS
    I am not an electrician or an electrical engineer. In fact, I have
    never even played one on TV.

    Ohm's law is for DC. It does not apply to AC.
    Yes it does - the R just means reactance instead of resistance -
    wheather inductive or capacitive. With resistive loads it's the same
    as with DC - read the resistance on your ohm-meter. With "inductive
    loads" the reactance will always be higher than the resistance (draws
    less current)

    What is a real world example of purely resistive load supplied by AC?

    I believe an oven or (non-induction) stove would come pretty close.
    99% close enough? A baseboard heater or any calrod heater, an
    incandescent lamp. So that covers an electric water heater - and
    heat-chaser cables - and electric in-floor heating cables - even an
    electric blanket. (Yes there IS a small inductive element if it has a
    coiled heat element - but virtually none if carbon fiber)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)