6 replies to this topic

[gajus21]

Welcome
I have a question: how to calculate the amount of carbon needed to recive 800 Mg / h steam.
Parameters: Water: 180 ° C, 14 MPa,
Steam: 540oC, 9,6MPa

 

Coal calorific value: 23 000 kJ / kg

I tried to count from the enthalpy difference of steam and water, as well as evaporation + heated to the desired temperature, but I recive very differents values.

 

Total heat needed to warm to that temperature. 2 176 160 000 kJ ->95 Mg carbon
Enthalpy difference: 1 556 800 000 kJ ->68Mg carbon

 

I'll be very grateful for Your help

 

[Cooks]

Honestly, your question is unclear.

 

Be clearer so one can point you aright

[gajus21]

Honestly there is more than enough informations. I found informacion how to calculate this maybe someone will be needed it

BQW = D(ip – iw) + SQS

B - mass carbon

Qw - calorific value

D - mass of steam

ip and iw - enthalpy

SQS - lost of heat other

There was only one thing to check: enthalpy steam and water

[gegio1960]

with your data and a steam table:

h water @ 180°C & 14 MPa = 182.3 kcal/kg

h steam @ 540°C & 9.6 MPa = 851.7 kcal/kg

coal NHV = 23000 / 4.186 = 5494.5 kcal/kg

DH of the whole process = 800000 * (851.7 - 182.3) = 5.4*10E8 (ideal... you should consider an efficiency)

weight of coal = 5.4*10E8 / 5494.5 / 1000 = 97.5 t .... similar to 95 t of your 1st calculation

[gajus21]

gegio1960 thank you but I have one more question from where You have taken information aboute steam becasue I can not find parameteter for steam in 540 oC

[breizh]

http://www2.spiraxsa...eated-steam.asp

 

Consider this resource and key in your parameters .

Breizh

[gajus21]

Thank you