5aebc28a01febafb333dd2edfdb5dfc8.ppt
- Количество слайдов: 44
Draught Systems BY Dr. P M V Subbarao Mechanical Engineering Department I I T Delhi Proper combustion requires sufficient Breathing.
Resistance to Air & Gas Flow Through Steam Generator System
Draft Required to Establish Air Flow Air in Flue as out
Natural Draft Patm Hchimney p. A = patm + atm *g *Hchimney p. B = patm + gas *g *Hchimney Tgas Tatm A B
Natural Draft • Natural Draft across the furnace, • pnat = p. A – p. B • pnat = patm + atm *g *Hchimney - (patm + gas *g *Hchimney) • pnat = ( atm - gas )*g *Hchimney • Natural draft establishes the furnace breathing by – Continuous exhalation of flue gas – Continuous inhalation of fresh air. • • The amount of flow is limited by the strength of the draft. Flow introduces a resistance and weakens the draft. At steady state flow resistance = Natural draft. pnat = kres * V 2
Pressure Variation in a Variable Density Fluid Column
Natural Draft Zref p. A = pref + p Hchimney Tgas Tatm A B
Zref, , pref p. A = pref + p Hchimney Tgas Tatm A B
Pressure variations in Troposphere: Linear increase towards earth surface Tref & pref are known at Zref. : Adiabatic Lapse rate : 6. 5 K/km
Reference condition: At Zref : T=Tref & p = pref
Pressure at A: Pressure variation inside chimney differs from atmospheric pressure. The variation of chimney pressure depends on temperature variation along Chimney. Temperature variation along chimney depends on rate of cooling of hot gas Due to natural convection. Using principles of Heat transfer, one can calculate, Tgas(Z). If this is also linear: T = Tref, gas + g s(Zref-Z). Lapse rate of gas, gas is obtained from heat transfer analysis.
Natural Draft • Natural Draft across the furnace, • pnat = p. A – p. B The difference in pressure will drive the exhaust. • Natural draft establishes the furnace breathing by –Continuous exhalation of flue gas –Continuous inhalation of fresh air. • The amount of flow is limited by the strength of the draft.
Pressure Difference Generated in natural draft gas path The pressure of natural draft is pnd = ( a - g) g H where pnd = head of natural draft, Pa a = ambient air density, kg/m 3 g = gas density in the flue, kg/m 3 H = height difference between the beginning and the end of the section, m The flue gas density, is g is calculated as Tg = gas temperature 0 C g = gas density in the flue under the standard atmospheric condition (00 C, 1 atm) Nm 3/kg
where 0= excess air ratio in the flue gas V 0 = theoretical air requirement for unit weight of fuel, Nm 3/kg Vg = flue gas produced per unit weight of fuel, Nm 3/kg V 0 A = ash content of the fuel in %
Local Gas Temperature in the Stack
Mechanical (Artifical)Draft : Induced Draft Essential when Natural Draft cannot generate required amount of breathing Hchimney p. A = patm + atm *g *Hchimney Tatm A p. B = pfan, s B B Tgas
Mechanical (Artifical)Draft : Forced Draft Hchimney p. A = pfan Tatm A p. B = patm + gas *g *Hchimney Tgas B
Mechanical (Artifical)Draft : Balanced Draft Hchimney p. A = pfan. b Tatm A p. B = pfan, s B B Tgas
Resistance to Air & Gas Flow Through Steam Generator System
+ve -ve
Pressure drop in Air and Gas Duct Systems Bernoulli equation – pressure drop across a flow passage Frictional resistance along flow path: where f = coefficient of friction L = length of the duct, m ddl = equivalent diameter of the duct, m ρ = density of air or gas calculated at the mean gaso temperature, kg/m 3 u = cross section average velocity of air or gas in the duct, m/sec A = cross section area of duct, m 2
Equivalent diameter for rectangular duct is given as where a and b are sides of the duct, mm. The coefficient of friction for flow through tubes can be approximated as shown below, for 5000 < Re<108, 10 -6< (e/ddl)<0. 01
Minor Losses Calculation of Local pressure drops: where Δp = local pressure drop K = local resistance factor, = density of air or gas at the position of the pressure drop calculated, kg/m 3 u = velocity of air through the fittings m/s.
pressure drop through air ducts Average Flow area, Aav , Volume flow rate of hot air after the air heater can be calculated as
Pressure drop across a burner pa K = 1. 5 for tangential burner 3. 0 for swirl burner
Pressure drop across heating surfaces Pressure drop across tube bundles: Inline arrangement: K = n K 0 Where n = number of tube rows along the flow direction K 0 = loss coefficient for one row of tubes K 0 depends on σ1 = s 1/d, σ2 = s 2/d , Φ = (s 1 - d ) Where s 1 is lateral pitch & s 2 is longitudinal pitch If σ1 <= σ2 : – 0. 5 Φ – 0. 2 Re – 0. 2 K 0 = 1. 52 (σ1 – 1) If σ1 > σ2 : K 0 = 0. 32 (σ1 – 1) – 0. 5 (Φ – 0. 9) – 0. 2 Re – 0. 2/Φ S 1 S 2
Staggered Arrangement The loss coefficient is obtained as K = K 0 (n+1) Where K 0 is the coefficient of frictional resistance of one row of tubes K 0 depends on σ1 = s 1/d, Φ = (s 1 - d ) / (s 2 l - d ) Where s 2 l is the diagonal tube pitch given by s 2 l = √ ( 0. 25 s 12 + s 22) and K 0 can be written as, K 0 = Cs Re-0. 27 Cs is design parameter of the staggered banks S 1 S 2
For 0. 17 <= Φ <= 1. 7 and σ1 >= 2. 0, Cs = 3. 2 If σ1 < 2. 0, then Cs given as Cs = 3. 2 + (4. 6 – 2. 7 Φ)(2 - σ1) For Φ = 1. 7 – 5. 2, Cs = 0. 44(Φ+1)2
Cross-Flow over Finned Tubes Inline arrangement for round fins where σ1 l = (pitch of fin, Pf / diamter of tube, d) σ2 ll = (height of fin, hf / diamter of tube, d) Re = ( u pf / γ) For square fine with = 0. 33
Staggered Arrangement 1) for round fins s 1=s 2=d+2 hf for round fins s 1=s 2=2 d
Gas side pressure drop in finned-tube economizers
Pressure drop in tubular air heaters where Δpmc is the pressure drop in the tube Kin and Kout are local resistance factors at inlet and outlet Pressure drop through rotary air heater Corrugated plate-corrugated setting plate Re >= 2. 8 x 103 f = 0. 78 Re-0. 25 Re < 2. 8 x 103 f = 5. 7 Re-0. 5
Corrugated plate- plane setting plate Re >= 1. 4 x 103 f = 0. 6 Re-0. 25 Re < 1. 4 x 103 f = 33 Re-0. 8 Plane plate- plane setting plate Re >= 1. 4 x 103 f = 0. 33 Re-0. 25 Re < 1. 4 x 103 f = 90/ Re
Pressure drop in ducts joining air heater and dust collector The volume flow rate of gases at the induced draft fan is determined by , where Vgf = volume flow rate of gases at the exit of the duct, m 3/s Tg = temperature of flue gas leaving the duct, 0 C = leakage air ratio behind the air heater B = fuel firing rate, kg/s where e is the excess air ratio in the flue gas at the duct exit T 0 is the cold air temperature, 0 C
Pressure drop through convective section Mass conservation for unchanged density, u A = u 1 A 1 = u 2 A 2 local pressure loss, p 1 = total loss, p = p 1 + p 2 + …………………. + pn = (k 11 + k 22 …………. . + knn) where k 11 = k 1 (A/A 1)2, k 12 = k 2 (A/A 1)2
Ash Collectors • Following Table is used to estimate the pressure drop in Ash collectors. • Cyclone: 15 – 20 m/s 70 – 90% 500 – 1000 Pa • ESP: 1– 2 m/s 99% 100 – 200 Pa
Pressure Drop through Stack where pst = stack pressure drop, Pa f = friction factor Lst = height of the chimney, m D= diamter of the chimney , m Kc = resistance factor at the stack outlet = gas density in the stack, kg/m 3 uc= gas velocity at the chimney outlet, m/s
Total gas side pressure drop Pa where p 1 = total pressure drop from the furnace outlet to the dust collector, Pa p 2 = pressure drop after the dust collector, Pa = ash content in the glue gas, kg/kg pa v = average pressure of the gas, Pa pg o = flue gas density at standard conditions, kg/Nm 3
The ash fraction of the flue gas calculated as, where f h = ratio of fly ash in flue gas to total ash in the fuel A = ash content of working mass, % Vg = average volume of gas from furnace to dust collector calculated from the average excess air ratio, Nm 3/kg of fuel
The pressure drop from the balance point of the furnace to the chimney base is prest = pexit + pgas – pnd where pexit = pressure drop up to the boiler outlet
Air Pressure Losses p Total losses Burner Losses APH Losses Ducts & dampers losses Percent Boiler Rating
Draught Losses p Total losses Furnace, SH & RH Losses Economizer Losses Ducts & dampers losses Percent Boiler Rating
5aebc28a01febafb333dd2edfdb5dfc8.ppt