pranj5
- 385
- 5
As per this http://www.criticalprocesses.com/Use%20of%20enthalpies%20to%20calculate%20energy%20needed.htm, it will take around 135.1 kJ/kg to compress steam from 2.536 kPa to 1 bar (20C saturation level to 100C saturation). And if the machine is 70% efficient, then the gross power consumption would be 193 kW for producing 1 kg/sec steam flow. And, that means the gross enthalpy stored in steam at 100C is around 2674.95 kW. Now, if we consider that the electricity comes from a 33% efficient plant, that means 193 kW of electricity equals to 579 kW of heat. If we subtract that from 2674.95, that equals to 2095.95 kW of heat. Not a bad amount at all.Mech_Engineer said:I think its time you calculate the alleged savings before making any more claims. Use a steam chart of your choosing, maybe the h-s chart we were looking at before.
All the data i have from this site.
Simple. I just want to be more confident and want to see how much correct I am. After all, we all are human and anybody can make mistakes.Chestermiller said:You seem so very confident about all this. And, since you are so confident, what was your reason for starting this thread in the first place?
Last edited: