From v03.n1078 Fri Oct 3 13:22:28 1997 From: Anthony Jackson Date: Thu, 15 May 1997 10:17:53 -0700 (PDT) Subject: V2E: first draft of heat rules. This was supposed to go out last night, but I typed gurpnet-l. Anyway, here's an early draft of my heat rules, feel free to make comments. This is a general set of rules for heat in GURPS Vehicles; some attempt has been made at realism, but significant simplifications have been made. Heat Eliminators: There are basically four ways to eliminate heat: a) radiant cooling b) conductive cooling c) enhanced conductive cooling (i.e. fans) d) evaporative cooling 1) What produces heat: Technically, any system which produces or uses power produces heat. In the case of routine power requirements, this can be ignored. If you want to deal with this, you may assume that the heat output of systems is 1% of power requirement at TL 13+, 2% at TL 12, (12-TL)*5% at lower TLs; in addition, you may assume that _in atmosphere_ a power plant (12-TL)*1% of its power production in heat (thus, a power plant can run at 20% of power and never produce heat). As a special case, fans, rotors, and the like are assumed to produce zero heat (this is not actually true...but they have enough air flow to make the issue ignoreable). In addition, crew produce 0.15 kilowatts per crewmember. As a rule, this is more complexity than is really wanted; as a simplification, ignore the ability of the power plant to remove heat, and assume no systems except weapons produce heat (few other systems use enough power to make this an important factor). Rather than trying to compute actual waste heat (which is generally huge) assume that energy weapons have a heat output of (15-TL)*5%, -5% per TL after introduction to a max of -15%. Thus, a TL9 laser has heat output of 25%, not 50%... This is actually being fairly generous; assume that additional heat is being flushed as loss of beam cohesion. 2) Where Heat Goes: Heat normally has the effect of increasing the temperature of the location the system is in (optionally, you may have it strictly increase the temperature of whatever created the heat). For a conventional location, assume that it takes (weight/4) kilojoules of waste heat to increase temperature by 1%. If temperature rises above 100 degrees, systems must start making HT rolls (at +5, -1/10 degrees); roll once/minute and once/10 degrees of heat gained. A failure means that a system stopped operating, a critical failure disables it. At TL 6+, Systems may be hardened against heat; increase weight by 50% and cost by 100%; heat-hardened systems may ignore temperatures up to TL*20 degrees, with normal penalties after that. Drivetrains, engines, and energy weapons are normally hardened against heat, but are also assumed to run (TL-5)*20 degrees warmer than the system. 3) Eliminating Heat: Heat may be eliminated in the following ways: a) it may be pumped into special heat-absorbing materials. It must eventually be flushed from here. b) it may be used to evaporate coolant (which is lost) c) it may be radiated or conducted away, via surface and heat vanes. d) the system may have a fan, which pumps in cold air and pumps out hot air. 4) Heat-absorbing materials At TL 4-6, the standard heat-absorbing material is water. Water can absorb 1 kilojoule of heat per pound-degree; once it reaches 212 degrees (or lower, for typical mixes) it will evaporate. It is assumed that most systems which produce substantial heat already have cooling channels. Water tanks only absorb heat if they are at least 5 degrees cooler than the system being cooled. Water coolant usually includes antifreeze, and typically costs about $0.1 per gallon; water coolant may be set to have a boiling point as low as 160 degrees. At TL 7+, specialized heat-absorbing materials are available; these materials undergo a reversible change of state at somewhere near the operating temperature of the system, and thus can absorb quite large amounts of heat. Assume that a heat-absorbtion system has a specific heat of (TL-6)*2 degrees, over a temperature range of about (TL-3)*20 degrees. 5) Evaporative materials At TL 7-, the usual evaporative cooling material is water. Water will absorb 1000 kilojoules of heat per pound by evaporating; any water heat absorber can be designed to boil at a temperature of 160 degrees or higher; any further heat added to the tank will boil water instead of increasing the temperature of the tank. At TL 8+, specialized heat-absorbing materials are available, which will undergo a non-reversible change of state at somewhere near the maximum operating temperature of the system. Assume that these systems will absorb (TL-6)*2000 kilojoules per pound. Like water coolant, heat absorbing materials can be set to destructively melt at some point, doing this. 6) Radiant and Conductive cooling: Radiant cooling uses the blackbody emission of an object; blackbody radiation is equal to 5.67E-8 W/(m^2K^4). Radiant cooling is free if you are radiating heat from the component which contains the heat; if you are pumping heat into a thermal vane, this requires power (equal to (vane temp/body temp - 1)*input power. Conductive cooling is more or less linear in temperature differential, as well as being heavily influenced by air speed. Conductive cooling does not work in a vacuum, and is _extremely_ efficient in water. Simplified R/C cooling: take (inside temperature - outside temperature), multiply by 0.004. In a vacuum, treat outside temperature as 0 (F, not K) and multiply by 0.0005. This is close enough to accurate for the range of temperatures a thermal vane is normally at. A thermal pump has a power requirement of (temperature difference)*(pump rate)*0.0002 kW. A pump must be rated for temperature difference, (minimum 25; maximum (TL-4)*50) and kilowatts of output; it has a weight equal to (max power required)*10* (10-TL, minimum 1). Systems with IR or emissions cloaking may not eliminate heat in this way. In addition, find the 'range/speed' modifier; for every - -1 for speed, increase cooling by 20% (thus, at 25 mph cooling is doubled). Thermal vanes: a thermal vane is simply a subassembly with no components; is is normally bought with a very light frame. The main advantage of a thermal vane is that it can operate at very high temperatures without the risk of damaging delicate components. 7) Fans: From within an atmosphere, by far the easiest way to cool a vehicle. A fan has a weight of (10-TL, minimum 1) per kilowatt of heat it eliminates, requires 0.01 kilowatts per pound, and costs $20/lb. Fans increase their cooling by 10% * range/speed modifier. 8) Example: A TL 8 LAV has a 10 megajoule blaster cannon on it (ROF 1), which weighs 156 lb; the only other systems in the location containing it are its mounting (stabilized, universal; 94 lb) and whatever heat storage we apply. - -base power required is 20,000 (per shot). - -Heat is (15-8-0)*.05*20,000 = 7000 kWs. - -Heat capacitance = (156+94)/4 = 62. One shot = 110 degrees. So, assuming a typical base operating temperature of 80 degrees, firing the weapon once will hit 190 degrees, requiring 3 HT rolls at -3 to avoid shutting off. We decide that we don't want a temperature jump of more than 10 degrees per shot, which means a heat capacitance of 700 kj/degree. Adding 640 in TL8 thermal plastic weighs 160 lb (total in location is now 410 lb). - -Surface area is now 25 sf. Conductive cooling is thus 0.1 kW/degree. If we are willing to accept operating temperatures of 40 over background, heat from one shot can be dumped in about half an hour. We decide that we want to be able to clear heat from a shot in one minute, which will require 116 kilowatts of cooling. We add 120 kilowatts of fans for 240 lb (total is now 650 lb); area has now increased to 35, giving surface cooling of 0.14 kW/degree, which is still insignificant.