1 / 36

Energy Balance Analysis of A Steam Generator

Energy Balance Analysis of A Steam Generator. BY P M V Subbarao Associate Professor Mechanical Engineering Department I I T Delhi. A Criteria for performance Rating ……. First Law for SG:Steady State Steady Flow. Q. W fans. Q steam. m fuel. m air. Q loss.

spencer
Download Presentation

Energy Balance Analysis of A Steam Generator

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Energy Balance Analysis of A Steam Generator BY P M V Subbarao Associate Professor Mechanical Engineering Department I I T Delhi A Criteria for performance Rating ……..

  2. First Law for SG:Steady State Steady Flow Q

  3. Wfans Qsteam m fuel m air Qloss First Law Analysis of Furnace:SSSF

  4. Some Valid Assumptions • The role of a furnace is to promote combustion and generate high enthalpy (Temperature) gas products. • Not for accelerating or decelerating the fluid. ⇒ Vi = Ve • Not for lifting or dropping the fluid. ⇒ Zi = Ze • There is some amount of heat transferred to boiling water. • The loss to the ambient should be minimum.

  5. Wfans Qsteam m fuel m air Qloss First Law Analysis of Furnace:SSSF

  6. Direct Method of SG Performance Analysis • Energy balance: • Fuel Energy = Steam Enthalpy + Losses. • Measurements: • Steam Flow Rate • Steam properties • Fuel flow rate. • Difficulties: • Measurement of steam flow rate. • Measurement of fuel flow rate. • Errors in measurements.

  7. Performance Testing of SG Air Flow Rate Dry Flue gas Analysis Ex. Gas Flow Rate

  8. Indirect Method of SG Performance Analysis • For every 100 kg of Coal. But A gas analyzer measures dry volume percentages of individual gases.

  9. Output of A Gas Analyzer • Total Dry Exhaust gases: P +R + T + U + V kmols. • Volume of gases is directly proportional to number of moles. • Volume fraction = mole fraction. • Volume fraction of CO2 : x1

  10. Output of A Gas Analyzer • Volume fraction of CO2 : x1 = P * 100 /(P +R + T + U + V) • Volume fraction of CO : x2= V * 100 /(P +R + T + U + V) • Volume fraction of SO2 : x3= R * 100 /(P +R + T + U + V) • Volume fraction of O2 : x4= U * 100 /(P +R + T + U + V) • Volume fraction of N2 : x5= T * 100 /(P +R + T + U + V) • These are dry gas volume fractions. • Emission measurement devices indicate only Dry gas volume fractions.

  11. Measurements: • Volume flow rate of air. • Volume flow rate of exhaust. • Dry exhaust gas analysis. • x1 +x2 +x3+ x4 + x5 = 100 or 1 • Ultimate analysis of coal. • Combustible solid refuse. nCXHYSZOK +en 4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash → x1 CO2 +x6 H2O +x3 SO2 + x5 N2 + x4 O2 + x2 CO + x7 C + Ash

  12. Stoichiometry for 100 kmols of Exhaust Gas • nCXHYSZOK +en 4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash & Moisture in fuel → x1CO2 +x6 H2O +x3 SO2 + x5 N2 + x4 O2 + x2 CO + x7C + Ash • x1, x2,x3, x4 &x5 : These are dry volume fractions or percentages. • Conservation species: • Conservation of Carbon: nX = x1+x2+x7 • Conservation of Hydrogen: nY = 2 x6 • Conservation of Oxygen : nK + 2 ne (X+Y/4+Z-K/2) = 2x1 +x2 +2x3 +2x4+x6 • Conservation of Nitrogen: e n 3.76 (X+Y/4+Z-K/2) = x5 • Conservation of Sulfur: nZ = x3

  13. nCXHYSZOK +en 4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash & Moisture in fuel → x1CO2 +x6 H2O +x3 SO2 + x5 N2 + x4 O2 + x2 CO + x7C + Ash • Re arranging the terms (Divide throughout by n): CXHYSZOK +e 4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash & Moisture in fuel → (x1/n)CO2 +(x6/n) H2O +(x3/n) SO2 + (x5/n) N2 + (x4/n) O2 + (x2/n) CO + (x7/n) C + Ash CXHYSZOK +e 4.76 (X+Y/4+Z-K/2) AIR + Moisture in Air + Ash Moisture in fuel → P CO2 +Q H2O +R SO2 + T N2 + U O2 + V CO + W C + Ash

  14. Specific Flue Gas Analysis • For each kilogram of fuel: • Air : e 4.76 (X+Y/2+Z-K/2) * 29.9 /100kg. • CO2 : P * 44/100 kg. • CO : V * 28/100 kg. • Oxygen in exhaust : 32 * U/100 kg. • Unburned carbon: 12*12/100 kg.

  15. Various Energy Losses in A SG • Heat loss from furnace surface. • Unburned carbon losses. • Incomplete combustion losses. • Loss due to hot ash. • Loss due to moisture in air. • Loss due to moisture in fuel. • Loss due to combustion generated moisture. • Dry Exhaust Gas Losses.

  16. Loss due to moisture in air. • Loss due to moisture in fuel. • Loss due to combustion generated moisture. • Dry Exhaust Gas Losses • ~ 4.5% Heat gained by boiling water 40% Fuel Energy 100% Hot gas Flue gas • Heat loss from furnace surface. • Unburned carbon losses. • Incomplete combustion losses. • Loss due to hot ash. Heat gained by superheater & reheater 40% Heat gained by economizer & air preheater 12%

  17. Energy Credits • Chemical Energy in the fuel. • Energy credit supplied by sensible heat in entering air (recycling of energy). • Energy credit supplied by sensible heat in the fuel(Recycling of energy). • Energy credit supplied by auxiliary drives.

  18. Wfans Qsteam n fuel n fluegas n air Qfans Furnace Energy Balance • First Law for Furnace in SSSF Mode (in molar form):

  19. Dry Exhaust Gas Losses • As gasses are leaving at temperature higher than ambient temperature. • For 100 kg of fuel. • QDEGL =S n fluegasDhfluegas • QDEGL = n CO2DhCO2 + n CODhCO +n O2DhO2 +n N2DhN2 + n SO2DhSO2 kJ. • QDEGL =PDhCO2 + RDhSO2+ TDhN2 + U DhO2+ V DhCOkJ. • Alternate method: • Total number of moles of dry exhaust gas nex.gas = P+R+T+U+V • QDEGL = nex. Gas Cp,exgas (Tex.gas - Tatm) • Cp.exgas = 30.6 kJ/kmol. K • Typical value of DEGL ~ 4.5%

  20. Accurate Calculation of Gas Enthalpy • For any gas

  21. Properties of Gases

  22. Unburned carbon losses. • For 100 kg of fuel • QUCL = W * MC * Calorific Value of Carbon : kJ • QUCL = W * 12 * 33820 kJ.

  23. Incomplete combustion losses • For 100 kg of fuel: • QICL = V * MCO * CV of CO. kJ. • QICL = V * 28 * 23717 kJ.

  24. Loss due to moisture in Combustion air • For 100 kg of fuel: • QMCAL = e 4.76 (X+Y/2+Z-K/2) * 29.9 * w * Csteam * (Tg – 25) kJ • Where w is absolute or specific humidity : kg of moisture per kg of dry air. • Csteam is the specific heat of steam at constant pressure. • Tg is the temperature of exhaust gas.

  25. Losses due to moisture in fuel & combustion generated moisture. • For 100 kg of fuel: • QML = ( M +9* Y) {2442 + Csteam * (Tg – 25) } kJ. • M is the moisture content in the fuel, %. • Y is the combustible hydrogen atoms in the fuel.

  26. Loss due to hot ash or Slag • For 100 kg of fuel • QASL = A * Cp,ash * Tash • Where Cp.ash, is the specific heat of ash, 0.5 – 0.6 kJ/kg K. • Tash is the temperature of the ash or slag. • Tash = Varies from 300 to 800 oC

  27. Heat loss from furnace surface • Loss due to Surface Radiation and Convection. • QRCL = As ( hs) (Tsurface - Tamb) kW • As = Total surface area, m2 • hs = Surface heat transfer coefficient. • For 100 kg of fuel: • Rate of heat loss/fuel flow rate * 100

  28. Off Design Performance

More Related