fizzy | 06/02/2017 17:21:36 |
![]() 1860 forum posts 121 photos | Im currently trying to accurately determine the effect of a small superheater addition to one of my small boilers. It dawned on me that it might actually be cooling the steam instead of heating it! Any ideas as to how I can accurately and cheaply determine the exiting steam jet temperature? Thanks |
JA | 06/02/2017 17:47:36 |
![]() 1605 forum posts 83 photos | Fizzy What is the steam pressure and how much superheat do you hope to have? I should add I do not think this is an easy measurement. At work we measured the temperatures of gas and the usual question was "what have you actually measured?" If a thermocouple had been used you could say with complete confidence "the temperature of the thermocouple bead". JA |
J Hancock | 06/02/2017 17:54:56 |
869 forum posts | If a comparison is all you need , then perhaps you could try wrapping a tiny piece of low melting point metal round the pipe and see which steam melts it ? |
SillyOldDuffer | 06/02/2017 18:01:05 |
10668 forum posts 2415 photos | You could try one of these . The hardest part (I think) is getting the thermocouple leads safely inside the boiler. The penetrator would have to be electrically insulated and capable of standing the heat and pressure. Some sort of screw fitting stuffed with heat hardening modelling clay perhaps? Dave |
DrDave | 06/02/2017 18:10:42 |
264 forum posts 52 photos | Or could you make use of an infrared thermometer if you have a clear view of the pipes both before and after the superheater (and they are not lagged)? |
duncan webster | 06/02/2017 19:18:33 |
5307 forum posts 83 photos | Weld (or silver solder) the hot junction of a thermocouple to the steam pipe after the super heater and then wrap it in insulation. Then weld the cold junction to the boiler shell, and similarly wrap with insulation. The output might have to be amplified using an instrument amp (something like an AD623), but will give you the increase in temperature from the super heater without having to mess with cold junction compensation. I'm told you can weld thermocouple wires to the surface by discharging a capacitor across the joint, if you get it to work tell us how, I've never tried it but it sounds useful. |
fizzy | 06/02/2017 19:24:16 |
![]() 1860 forum posts 121 photos | thanks all - i had thought of just blasting steam at a digital thermometer, would this work? |
fizzy | 06/02/2017 19:31:26 |
![]() 1860 forum posts 121 photos | pressure is set to blow off at 60psi which from memory would be around 120 deg c. I just neet to see if the steam leaving the superheater is hotter than that going in. It is open i.e. not connected to any pistons at the moment.
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JA | 06/02/2017 19:33:22 |
![]() 1605 forum posts 83 photos | One problem with a thermocouple is that the wires conduct heat as well as electricity. Therefore the thermocouple is likely to read colder than the steam. To a degree this difference can be reduced by careful design. JA |
Neil Wyatt | 06/02/2017 20:25:04 |
![]() 19226 forum posts 749 photos 86 articles | DrDave's idea of an IR thermometer is a good one, just paint a black patch on the pipes to take a reading from, they don't work on shiny surfaces. Neil |
J Hancock | 06/02/2017 21:36:02 |
869 forum posts | Blasting 'steam' at a thermometer won't tell you a lot about the temperature of the steam down the pipe, only how cool it is as it expands out of the pipe ! |
Hopper | 07/02/2017 01:41:43 |
![]() 7881 forum posts 397 photos | Posted by J Hancock on 06/02/2017 21:36:02:
Blasting 'steam' at a thermometer won't tell you a lot about the temperature of the steam down the pipe, only how cool it is as it expands out of the pipe ! How does steam cool as it expands? (other than from hitting the cooler atmosphere). I know refrigerant does, as it expands and changes state from liquid to gas and needs to absorb latent heat from the surrounds as a result. But steam starts as steam and stays as steam. Where does any expansion-related temperature reduction come from? |
Russell Eberhardt | 07/02/2017 07:59:52 |
![]() 2785 forum posts 87 photos | Posted by Hopper on 07/02/2017 01:41:43:But steam starts as steam and stays as steam. Where does any expansion-related temperature reduction come from?
Charles' Law from basic physics. Russell |
Hopper | 07/02/2017 08:21:35 |
![]() 7881 forum posts 397 photos | Posted by Russell Eberhardt on 07/02/2017 07:59:52:
Posted by Hopper on 07/02/2017 01:41:43:But steam starts as steam and stays as steam. Where does any expansion-related temperature reduction come from?
Charles' Law from basic physics. Russell Isn't Charles's law that gas volume is directly proportional to temperature? As one increases, so does the other. So when the steam expands to a greater volume, the temperature should also be greater in direct proportion? |
Brian Oldford | 07/02/2017 08:27:03 |
![]() 686 forum posts 18 photos | On this occasion Wobblipedia seems to be right on the money. https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect
|
Howi | 07/02/2017 09:01:20 |
![]() 442 forum posts 19 photos | Posted by Hopper on 07/02/2017 08:21:35:
Posted by Russell Eberhardt on 07/02/2017 07:59:52:
Posted by Hopper on 07/02/2017 01:41:43:But steam starts as steam and stays as steam. Where does any expansion-related temperature reduction come from?
Charles' Law from basic physics. Russell Isn't Charles's law that gas volume is directly proportional to temperature? As one increases, so does the other. So when the steam expands to a greater volume, the temperature should also be greater in direct proportion? I think you may have that the wrong way round, if you compress a gas it's temperature will increase, if a gas is allowed to expand by definition it's temperature will decrease. Boyle's law/Charle's law |
Hopper | 07/02/2017 09:41:50 |
![]() 7881 forum posts 397 photos | And yet steam that goes through a pressure reducing valve is superheated when it goes out the other side, as in a building heating system. IE, is at the same (or near same) temperature as before but is now at a lower pressure, therefore the temperature is above the boiling temp of water at that pressure. And does not Boyle's law apply only to a gas at constant temperature? And Charles's to a gas at constant pressure? Both of which are not constant as steam comes out the end of a super heater pipe and supposedly cools as it expands? And in the example of compressing a gas and its temperature increases, is not the temperature increase due to the work (energy) added to the system by compressing it, forcing the molecules closer together and so on?
Edited By Hopper on 07/02/2017 09:43:15 |
Ian P | 07/02/2017 09:47:47 |
![]() 2747 forum posts 123 photos | Posted by Hopper on 07/02/2017 01:41:43:cool as it expands? (other than from hitting the cooler atmosphere).
. But steam starts as steam and stays as steam. I think that means you have discovered perpetual motion!
Ian P |
Ian S C | 07/02/2017 10:12:14 |
![]() 7468 forum posts 230 photos | There will be a chart some where, then all you need is a pressure gauge, as the pressure goes up, so does the temperature. Ian S C Edited By Ian S C on 07/02/2017 10:12:52 |
Russell Eberhardt | 07/02/2017 11:08:35 |
![]() 2785 forum posts 87 photos | Posted by Hopper on 07/02/2017 08:21:35:Isn't Charles's law that gas volume is directly proportional to temperature? As one increases, so does the other. So when the steam expands to a greater volume, the temperature should also be greater in direct proportion?
Sorry, Ishould have said it's a combination of Charles' Law and Boyle's Law. If you "blast steam at a thermometer" it will expand as a result of the pressure drop and thus the temperature will fall. When it falls sufficiently condensation will occur and the gas laws no longer apply as it is no longer only steam but a mixture of steam and water droplets. Russell. |
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