Discharge Resistor Power Rating

Volts that the line is charged up to = 60kV
Resistor Ohmic value = 250kOhm
Capacitance of line to earth = 5.4microF
therefore total energy stored = 0.5*C*V^2=9.72kJ
time constant = C*R = 1.35 seconds
now what i need to know is what power rating does the resistor need to be if it's temperature isn't to go over let's say 50 degrees C based on a duty cycle of 1 discharge every 2 minutes.
I'd appreciate ideas to solve it....I've got some ideas but if anyones done exactly this type of thing I'd appreciate ur input.
many thx
Danny
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Danny, it kinda sounds like an exam question for a test.
i think we need to know if the resistor is in free air or heat sunk and if so how big a heat sink and/or if it is forced air cooled (and if so how many CFM) and probably a bunch of other thermodynamic properties. ( and and assumption of ambiant temp)
i can give you some real world examples of "bleeder" resistors in HV equipment if you think it will be useful. my stuff is in the 9 kV and less range at present.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
yes it does sound rather like homework....but it isn't. we just had an incident where our hipotronics bleeder resistor exploded and I need to a) determine the rating according to hipotronics and b) determine what it should be by calculation.
I think i can do the calculations but just wondered if the 'wheel' might already be invented...
the resistor is wire wound around a hollow ceramic core (plugged at both ends) with ceramic/glass outer.
I've done these calcs before for 'on' 'off' duty cycles but never with exponential shaped voltage and current pulses. it complicates it a bit. assuming adiabatic conditions for the 'on' part will simplify it a lot methinks. Once I have established radiation and convection levels required to keep the temp under the desired temp...say 60 degrees C I can then extrapolate the steady state power that would reach the same temperature (which'll be much lower of ourse) and that'll be the rating required (plus a safety margin of some reasonable value).
I'll have a go tomorrow.

many
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Power = v*v/r = 60,000 *60,000 / 250,000 = 3,600,000/250 = 14,400 watts @ a duty cycle of 1.125% = 162 watts continuous
But there is no thermal definition of your R, need the theta J/C and the rest of the thermal constants to do anything about temperature. Also If you put a 162 watt resistor in, it may still detonate as the duty cycle to average power is assumed under certain conditions(like the R can stand the initial voltage, and/or initial current).

if
less
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

determine
by
required
(plus
Hipotronics makes a variety of HV power supplies and HV test equipment. http://www.hipotronics.com / you should refer this issue directly to their engineering dept.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Peak power = V(max)^2/R = 60,000^2/250,000 .4 kW
average power = 9720 / 240 = 40.5 W
It's a case of how much over the 40.5 W you need to go to avoid the pulse raising the temperature of the resistor to high, for this you need to know the resistors thermal capacity ( Joules / degree ). It also depends on how fast it dissipates heat, as this determines the temperature it starts at if it's being cycled repeatedly, this will be a function of temperature dependant on the resistors physical characteristics and environment.
Under pulse loads temperature will not be even over the resistor so your 50 C limit could leave cool parts while other bits go over their limit.
Why use such a low temperature... a nice red glow from a suitable resistance wire each time the system discharges should not be a problem.
You need a lot more information than you have given the group to answer your problem.
--
Jonathan

Barnes's theorem; for every foolproof device
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
forgive me but I believe i gave all the infroamtion neccessary. If indeed the duty cycle is 1.125% then for an equivalent steady state power of 162 watts one can evaluate the thermal time constant , the surface area required to dissipate the energy, which in turn will give you the continuous power rating required.
Danny

characteristics
50
resistance
your
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
I think the pulse with a 14.7 kW peak could fry a resistor before it has a chance to dissipate the heat unless it's fairly large.
Your power pulse would raise the temperature of a 500 g chunk of aluminium by about 20 degrees C.
A 50 W rated resistor ( at 32 g ) would go up by 300 C.
BTW, what sort of resistor are you thinking of using at 60 kV ?

pulse
know
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

This is something I was wondering about too. Although the *average* power dissipated might be less than 50 watts, one should also look at the behavior during the 'pulse'. Extremely high power for very short periods can *average* out to a mild power level. But a pulse of 9 720 Joules of energy can raise the temperature appreciably before convection can carry it away from the resistor. Even if the exterior is kept cool by a heat sink, the slower heat transfer through the resistor body itself (before it reaches the heat sink) could allow a high central core temperature.
For example a 1 oz resistor (28.375 g) made of ceramic (specific heat capacity of 0.670 J/g-K), the temperature rise for this much energy deposited uniformly in the ceramic in a very short pulse would be...
9 720 J / 28.375g / (0.670 J/g-K) = 511 K (511 C)
Of course, the outer surface would not heat up this much since it's in contact with a heat sink, but considering the ceramic's ability to transfer heat from the core to the outer surface, the inner regions could approach this kind of temperature rise. Just the thermal expansion alone might crack/shatter the material. Of course, the time constant of 1.35 seconds means the 'pulse' isn't all that narrow. But I haven't seen anyone else bring this point up.
Such high-pulse power is studied in nuclear fuel pellets to ascertain the survivability of certain accidents. The pulse is very short duration but very high power. As a result, although the 10-second average power level is quite low, the central parts of the fuel can actually reach melting points (~1500C). Even while the outer surface of the pellet is kept cool. Of course this isn't nuclear fuel, but some of the dynamics are interestingly similar.
daestrom
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Danny,
You may want to consider replacing the defective component with either a Kanthal Globar or HVR Advanced Power ceramic tubular or solid ceramic resistor that is truly rated for pulsed HV duty. This style resistor dissipates heat more uniformly through the entire body of the resistor, providing greater pulse handling capability. Wirewound ceramic resistors often fail (very spectacularly!) in a high energy pulse environment since the thermal mass of the actual resistive element is comparatively small.
Both of the suggested manufacturers conveniently specify pulsed power ratings in joules and maximum peak voltage.
http://www.hvrapc.com/index.html (BTW, while you're there, check out HVR's thermal modeling tool to confirm your selection):
http://www.globar.com/ec/tubular.php.html
Good luck,
-- Bert --
--
---------------------------------------------------------------------
We specialize in UNIQUE items! Coins shrunk by huge magnetic fields,
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

required
No, This has to be measured by MFGR of the resistor. You cannot calculate it with any accuracy..
You can just try and find out what happens and tell us later.
Resistors have Voltage ratings, and Current ratings, Theta J/C ratings, as well as power ratings.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Peak power = V(max)^2/R = 60,000^2/250,000 .4 kW
average power = 9720 / 240 = 40.5 W
It's a case of how much over the 40.5 W you need to go to avoid the pulse raising the temperature of the resistor to high, for this you need to know the resistors thermal capacity ( Joules / degree ). It also depends on how fast it dissipates heat, as this determines the temperature it starts at if it's being cycled repeatedly, this will be a function of temperature dependant on the resistors physical characteristics and environment.
Under pulse loads temperature will not be even over the resistor so your 50 C limit could leave cool parts while other bits go over their limit.
Why use such a low temperature... a nice red glow from a suitable resistance wire each time the system discharges should not be a problem.
You need a lot more information than you have given the group to answer your problem.
--
Jonathan

Barnes's theorem; for every foolproof device
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.