degree of super heayed steam?
408165
The degree of superheat is a matter of economics. Many boilers and steam systems are limited to a maximum temperature of 750 degs F because they are made from mild steel. Prolonged exposure to temperatures in this range can cause the carbon in to steel to change. I don't know the metallurgy just that it is a code limitation.\n\nYou can go higher in temperature but you must then go to different steels. Large steam power plants do this but I don't know the details. You need to talk with a boiler manufacturer.
981893
It goes into energy storage modes of the water molecule.  One is vibration, another is translation or kinetic energy.  The water molecule travels faster.  Steam at 100 C burns much worse than water at 100 C because although the temperature is the same the energy is much higher.
741934
This question is not very difficult, but you have to be careful with the unit.\n\nFirst, to solve this question, you have to make the assumption that there is no energy lost to the surrounding. So, this reduces to a two system question: one system releases energy while the other one absorbs energy. In this particular case, the energy released by one system equals to the energy absorbs by another system.\n\nTo picture this, consider the following, one system goes through the following steps:\n\n[100 degC H2O (g)] --- (step 1) ---> [100 degC H2O (l)] --- (step 2) ----> [79 deg C H2O (l)]\n\nThe other system goes through:\n\n[260g, 19 degC H2O (l)] --- (step 3) ----> [260g, 79 degC H2O (l)]\n\nIt is obvious that both the energy involved in step 1 and 2 is negative (release energy) while the energy involved in step 3 is positive (absorb energy). We will now solve this question backwards. Starts from finding out the energy required by step 3, then find out the mass of steam that is needed to do the job.\n\nEnergy absorbed during step 3:\nE3 = m * c * (delta T) = 260 * 4.18 * (79-19) = 65208 J\n\nEnergy released in step 1:\n(you have to first convert heat of vaporization to unit of J/g: 40.7*1000/18 = 2261.111 J/g)\nE1 = m' * l = m' * 2261.111 = 2261.111m'\n(since we don't know how much steam is needed, let's just leave it as a variable)\n\nEnergy released in step 2:\n(the temperature of the same mass as above will be lowered to 79 degC)\nE2 = m' * 4.18 * (100-79) = 87.78m'\n\nTotal energy released = (2261.111 + 87.78)m' = 2348.89m'\n\nEnergy released = energy absorbed\n2348.89m' = 65208\nm' = 27.76 g
363447
this re minds me of a kind of heat called latent heat.this is the heat a subtance requires to change it state without changing it temperature,so water changes from liquid to steam at the same temperatutre if it assumes it latent heat of vapourization.thus water could be liquid at 100*c and also steam at the same temperature.
578840
buckets of steam? You can put steam in buckets? All this time , my teachers lied...oh the shock of it all!!!
264462
Because steam is hotter than boiling water.
98850
Yes. It will superheat and have a violent reaction when something touches it.
947952
Highlight the data and put your cursor at the bottom right of the lowset box, the cursor should turn into a small +. hold the cursor down and drag your mouse doun a few rows.  Once you let up on your mouse you should see the info you want.\n\nYes, if you increase pressure you can increase the temperature of the water, but it needs to be in a closed system, it's called super heating.
915075
I don't have the constant for the change of steam to water.  But 373.15 is the boiling point of water.  So simply multiply 42.3 g/h20 by the constant for the conversion of steam to liquid and you will get it.
1112226
Looking at the phase diagram of water,\nat 3000 psi (~20684 kPa, ~2.1 E6 Pa), water can exist in the vapor phase at rougly any temperature at or above 500 Kelvin (rough estimate based on graph).\n\nThere is no one specific, constant, temperature of steam at a given pressure.
1032149
Q=MC (delta) T....(5000 G)(.377 J/G degree C)(200 Degree C) which should be.....i get 377 J, which isn't your answer....i just had an exam on this last week, so i thought for sure i am right....the only thing i am not sure about for certain is the specific heat of coal, i got it from a website.....need more info if i am able to help, surely they provided the specific heat of coal....also if latent heat of fusion of vaporization has a part, the answer above is going to be different....sorry dude
831873
Is it referring to the heat load calulation for air con
62689
specific heat of steam is 2.02 J/g*K\n\nU steam= 215000J - 75.0g*2.02J/g*K(25K)= 211213J\n\nU liquified = 211213 - 40,700J/mol*mol/18g*75g = 41629J\n\nU water @ 0 C = 41629J - (4.18J/g*K)*75g*100K = 10279J\n\nU freezing = 10279J - 6000J/mol * mol/18g * 75g = -14721\n\nso you have some water and some ice, or liquid and solid phases.
679372
Its so easy to do a search\n\nhttp://www.taftan.com/thermodynamics/STURBINE.HTM\n\nhttp://en.wikipedia.org/wiki/Steam_turbine
361244
first we must know turbine's inlet pressure and temperature (depend on design ) , mass flow of steam to turbine (for 2,5 MW)  and effiency of turbine (for calculate of exhaust conditions )\n\nexample : 100 MW double exhaust steam turbine need to 160 ton steam at 500 C and 50 bar for high pressure   and 40 ton steam at 200 C and 5 bar for low pressure. \n\nthere are many kind of boiler . if you dont know some require value then you can not calculate boiler pressure , temperarture and  stages (eco , eva , superheater) .\n\ni can say only about capacity of boiler  ; \n\ni f we accept turbine effiency = 70 % \nthen amount of approximately steam power = 2,5 . 100 / 70 = 3,571 MW \nand \nİf we accept boiler effiency = 85 % then real capacity of   \nboiler = 4,2 MW \nthis value near capacity of boiler. \n\nboiler inlet condition = 1 bar 50 C (i accept) \nenthalpy of inlet water = (from compressed water tables) = 200 kj/kg \nmass of flow = 5 ton/hour (i accept) = 5000 kg/hour = 5000/3600 kg/sn= 1,39 kg/sc.\n\n3,571 MW = 3571 kw = 1,39 . (hç-200) \n\nhç= outlet enthalpy of boiler \n\nhç = 2769 kj/kg (this value must stay at superheated region) \nwe must look to superheated region at thermodynamic tables. \nand we can find best outlet conditions for 2769 Kj/kg. and temperature can not be high than 500 C for 2769 kj/kg. \nwe must take care to pipe line length because some losses cause to change superheated steam to saturated steam or near conditions. \nif superheated steam early change conditions then blades of turbine can be damaged. at the end of turbine must be saturated steam or little bit big superheated steam for protection of blades. \nfinally \n4 bar , 160 C conditions enough for 2769 kj/kg.\n\n\nsummary : \nboiler inlet 1 bar 50 C \nboile r outlet 4 bar , 160 C \nmass of flow = 5 ton / hour \n\nalso next step , calculate of boiler's economizer , evaporatör , superheater , main steam line pipe dia. , pumps (circulating and feed water) , drums , recirculation lines , condenser , auxilary cooling water system (for cooling of condenser) , gland steam line , regulating system , control syst. , vacuum sys. (if necessary for condenser) , water treatment sys. , chemical dosing points ,  etc. etc. etc.\n\nturbine inlet conditions = 4 bar , 160 C 2769 kj/kg\n\nnet power = 2,5 mw = 2500 kw \n\n2500 = 1,39.(2769-hs) \nhs = 970 kj/kg (enthalpy of at the end of turbine ) \n\nthis entalphy shown turbine outlet conditions \n\n0,5 bar 85,4 C or 0,2 bar 85 C (vacuum)\n\nwater must be cooled up to 50 C (inlet of boiler)
933894
Hi. If you took a large can that could be sealed with a cap and put a small amount of water it, then heated the water until it boiled and was steaming from the top, then sealed the cap, the steam would condense. This would allow the outside pressure on the can due to the atmosphere to crush the can right before your eyes.
978582
These answers start answering themselves as you read.  I think you missed this key data while studing.\nhumidity is the amount of steam(heated water) in the air\na drop in heat makes condensation(the water is no longer steam but water)\nthat is the point of all these questions keep that in mind while answering them.\n\nhope this helps\n-David-