what size of boiler do i need if i have a calorifire of size 6000 litres to heat water?
470685
You must calculate difference of enthalpy between first and second condition.  Enthalpy depends on  temperature and pressure.And it can be find from thermodynamic tables.\n\nI accept 1 bar for water's pressure \n\n1 bar and 20 degree water's enthalpy = 84 kj/kg \n1 bar and 55 degree water's enthalpy = 104,725 kj/kg \n\ndifferent between inlet enthalpy and outlet enthalpy = 104,725-84 = 20 kj / kg (this heat necessary to increase of 1 kg water's degree from 20 C to 55 C ) \n\namount of water = 6000 lt = 6 m3\nspecific density of water which is at 20 C and 1 bar = 0,0010017 m3/kg\nmass of water = 6 / 0,0010017 = 5990 kg \n\ntotal heat requirement = 5990 kg x 20 kj/kg = 119800 kj \n\nXXX if this proses time is 1 hour then \n\npower or capacity of boiler for perfect situation ( perfect situation = no losses and boiler effiency 100 %) = 119800 / 3600 second = 33,2 kw \nif boiler effiency = 85 % then \nreal capacity of boiler for proces time is 1 hour = 33,2 x (100/ 85) = 39,05 kw \n\nXXX if this proses time is a half hour then \n\npower or capacity of boiler for perfect situation ( perfect situation = no losses and boiler effiency 100 %) = 119800 / 1800 second = 66,55 kw \nif boiler effiency = 85 % then \nreal capacity of boiler for proces time is 1 hour = 66,55 x (100/ 85) = 78,3 kw \n\nif you want to calculate capacity of boiler ,  first you must know proses time and pressure.
1009332
Convert everything to inches, then multiply.  LxHxW.  They need to be the same units, though.
765230
EER = Q in (BTU/hr) divided by W in (watts)\n\n600/500= 1.2
210952
Are you speaking of a Heat Pipe and its conversion?\n\n\nCheck out the following links:
1055920
There are several techniques for measuring heat on small lab vessels. The only practical solution for larger systems (above 1 litre) however is to measure the heat gained or lost by the heating/cooling medium The idea is that you measure the mass flow and temperature change of cooling/heating fluid (plus corrections for heat loses and temperature changes). The problem with this method however is that it does not work well with tradition cooling systems. To measure heat on large systems, you need to use a cooling jacket which operates at constant flux.
1116381
ASHRAE is the organization that publishes the standards by which heating equipment is sized and vented.  They charge for their publications however.\n\nTalk with a reputable boiler manufacturer like Paterson Kelly or Hydrotherm and they will probably give you some guidelines for no charge.
108765
hi there,\nits like, you need to know the temp and the mass flow rate of the bottom product of the distillation column. also you need to know the boiling point of the bottom product. the mass flow rate depends on your design calculation. temp of the bottom product you should already know. and boiling point of the bottom product is readily availble.\n\nnow use m.cp.dt to find heat required to raise the temp of the product to its boiling point. where cp is the specific heat capacity of the bottom product and is available in handbooks or sometimes expressed as functions of temperature itself. and dt stands for differnce of boiling point of bottom product and the current temp of the bottom product.call this value as h1 (say).\n\nnow you know the latent heat of vapourisation of the bottom product again refer to data handbooks for the values. multiply this by the mass flow rate. and this is the heat required to vapourise the bottoms. lets call this h2 (say)\n\nnow h1+h2, gives you the heat load for the reboiler. use this heat load to calculate the steam requirement etc...\nyou can mail me sanjaykumar_16@yahoo.com for any clarifications.
682836
You need to do a heat load calculation on your house to determine your heating and cooling needs. Then you size your heat pump equipment. where ever you get the equipment from should be able to tell you diameter and length of tube. You need to decide how much room you have for the installation. If your room is unlimited go with larger tube and 2 pipe system. What type of soil do you have? You can save some time and effort by using different installation methods.
1023774
450 watts min or 600 watts to max it out
914327
A fired heater will probably be the cheapest way.  You could try electric heaters (you can buy the elements from water heaters at appliance repair shops - they use 240V), but the electricity cost will probably be more than for the fuel (if you were thinking gas/petrol, maybe propane or natural gas will be cheaper).  I also don't know if I'd be comfortable putting electric elements in 3500 gallons of biodiesel - I've seen a few kitchen fires, and burning oil is never pleasant.\n\nYou could also consider solar heating - it would be cheaper in the long run, but any efficient installation would have to involve heating panels, piping, pumps, and a lot of water.  You could try painting your tank black, but that probably won't work very well unless you live in the desert or high plains.
318461
Depends on what you need.  I work for a manufacturer of heat transfers.  Tell me what you need and maybe I can help you out.
925329
You must approach this question with an understanding of energy conservation. Change in Energy = mass*heat capacity*change in temperature, where heat capacity = 1.00 for pure water. By evaluating your conditions with the above definition of thermal energy, it will show that you must supply 2 calories of heat to the system.\n\nIf you were to add another substance to the original system. You would once again approach the problem using energy conservation. The energy of a hot liquid (mass1*heat capacity*change in temperature) + the energy of a cold liquid (mass2*heat capacity*change in temperature) = mass(1+2)*heat capacity*change in temperature.\n\n-Christopher
784078
The optimum thickness is infinity. You need to factor in the cost for various thicknesses and weigh that against how much heat you are willing to lose.
13598
Your heat load is equivalent to 100,000 100 Watt light bulbs or 10,000 1,000 Watt electric heaters.  When all equipment is operating, your maximum heat load will be 10,000 kilowatts for the heat sources you mention (are there any other loads?).\n\nAir conditioners are often rated in Btu/hr, therefore you must convert Watts to Btu/Hr.\n\n     Watts x 3.415 = Btu/Hr (See source)\n     100,000 sq. ft. x 100 Watts/Sq. ft. = \n     10,000,000 Watts\n\n     10,000,000 x 3.415 = 34,150,000 Btu/Hr\n\n Large air conditioners may also be rated in tons (cooling effect required to produce a ton of ice in 24 hours)  See second source.\n\nA ton (of air conditioning) is equal to 12,000 Btu/Hr.\n\n   34,150,000Btu/Hr / 12,000Btu/Hr/ton = \n   2,845.8 ton capacity.\n\nCalculations are for sensible heat unless water vapor is to be condensed which depends on the desired humidty.\n\nYou should have a licensed professional engineer (PE) make final calculations and look for other important considerations before funding an air conditioning plant.\n\nAddition: How much heat results from power consumption in a closed room will depend on what the power is used for.  For example if an electric heater is used to melt a block of ice no heat will be available to heat the air.  In like manner, if heat is used to warm something that is then removed from the room, some of the heat (supplied to the object) will become unavailable.  But heat of friction or heat used to warm anything that remains in the room will be available to heat the air in the room.  Open doors or ventilation can exhaust heat.  Large motors with power factors may be a special case because of phase angles.  However, most equipment such as small motors, heaters, TV sets, computers and light bulbs should convert almost all electrical energy into heat.  Therefore the Watts of electricity listed on the equipment label should all be converted to the equivalent amount of heat.
494718
I think you're referring to a closed-cycle (i.e., water condensing) steam engine. The first picture on the ref. 1 page is a diagram of the Rankine cycle engine. The exhaust steam passes into a condenser where it gives up heat and returns to the liquid state (water). Then it is pumped back into the boiler. A small, high-pressure, low-displacement pump is used, so the power needed is far less that that generated by the engine. Since water is scarce you may have to use air cooling for the condenser rather than water cooling. And (unlike the representation in the diagram) the condenser must be vented to avoid excessive pressure, so some steam may be lost. The ref. site will give you some ideas for engine construction, including links to engine parts and kits. Unfortunately these tend to be expensive. If you can find a junked engine or piston pump you might be able to adapt it to steam use, but adapting a gasoline engine would be difficult and my knowledge ends at this point. Good luck.
476810
http://ushotstuff.com/Heat.Scale.htm
978594
Acetone       2.13kJ/kg K or  0.51  btu/lb F\nWater, fresh 4.19kJ/kg K or  1.0   btu/lb F\n\nsee also http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html