# Less then 1Equals to 1Not Present PTVfVgVfgUfUgUfghfhg hfg SfSgSfg 0.01 1 10 100 So with the help of steam table we select the value of needed item at.

## Presentation on theme: "Less then 1Equals to 1Not Present PTVfVgVfgUfUgUfghfhg hfg SfSgSfg 0.01 1 10 100 So with the help of steam table we select the value of needed item at."— Presentation transcript:

Less then 1Equals to 1Not Present

PTVfVgVfgUfUgUfghfhg hfg SfSgSfg 0.01 1 10 100 So with the help of steam table we select the value of needed item at given pressure.

Q1. Calculate the value of enthalpy if the steam is at 10 bar and 1. Wet at x=0.90, 2. dry and saturated 3. superheated at 300 c ? PTVfVgVfgUfUgUfghfhghfgSfSgSfg 10 putting the values in the formula h= Similary done with dry. h=hg(at 10bar) = superheated steam :- From the steam table tsat =, Cp=, h=hg + cp(tsup-tsat) ANS  h(wet)=, h(dry) =, h(sup)=

1.La-Mont Boiler 2.Benson Boiler 3.Loeffler Boiler 4.Velox Boiler

Q. A boiler produces 10kg of steam per kg of fuel from feed water temperature at 30c, at 9bar absolute pressure, What is the Equialent evaporation and at 100c per kg of fuel and factor of evaporation, if the steam produces is 0.9dry.and boiler efficiency if Calorific value is 34000 KJ/kg Given :- me= 10kg, tf=30c, P=9bar, x=0.9 To find :- E, Fe,Efficiency of boiler. Solution:-,E= me(h-hf1)/2257, So we need h and hf1 value, h f1= 4.187 multiply Feed water temperature(tf) = 4.187 multiply 30 = 125.66 X is less than 1 so, the steam is WET so h=hf +xhfg From the steam table value of hf and hfg at 9bar taken h=hf + xhfg = 742.64 + 0.9 in to 2029.5 = 2569.19 Factor of Evaporation(Fe ) = (h-hf1)/2257 = 1.083 E= me in to Fe = 10.83

BOILERECONOMIZERSUPERHEATER Heat used in Boiler = me( h-hf2) hf2= 4.187 in to tf2 Heat used in Economizer= Me.cpw(t2-t1) Heat used in superheater = h(sup) – h(wet) T1 T2 T3 T4

1. Heat lost in raising steam x1= me (h-hf1) 2.Heat lost to chimney or flue gases x2 = mgcpg(tg-tb) 3.Heat lost by combustion x3 = 9h2+mm (2676 + cp(tg-100)-hb) 4. Heat lost in moisture x4 = mm (2676 + cp(tg-100)-hb) 5. Heat lost due to unburnt carbon x5 = m1(C.V. of carbon) 6. Heat lost due to incomplete combustion x6 = m2(C.V. of carbonmono oxide) 7. Heat lost due to Radiation x7 = X – ( X1+X2+X3+X4+X5+X6 ) Mg= mass of dry flue gases Cpg= Specific heat of dry flue gases Tg= Temp. of flue gases Tb= temperature of boiler room Mm= mass of moisture Cp= specific heat of superheated steam Hb= enthalpy of water at boiler room M1= mass of carbon in ash pit M2= mass of carbon monoxide hb = 4.187 tb X = (1- moisture %) C.V.

HEAT Supply kJHeat consumed KJ% 12345671234567 X1 X2 X3 X4 X5 X6 X7 TOTALx100 Q. In a Boiler trial, the following observations were made-pressure of the steam 12 bar, Rate of steam evaporation is 620kg/hr, Fuel used is 65kg/hr, moisture in fuel is 2% by mass, mass of dry glue gases is 10kg/kg, of fuel, Lower C.V. is 33000kj/kg, Temp. of flue gasesis 320c, Temp. of boiler house is 26c, Feed water temp. is 60c, cp is 1.005kj/kg k, Dryness fracyion is 0.96, Draw heat balance sheet.

Given:- P=12bar, ms=620kg/hr, mf=65kg/hr, mm= 2%or 0.02kg/kg, mg=10kg/g, C.V. = 33000 kJ/kg, tg=320c, t1=60c, cpg=1.005, x=0.96. To find:- Heat Balance Sheet Solution :- me= ms/mf  620/65 =9.54 hf1= 251.09,(at 60c), at (12bar)- hf=798.43, hfg= 1984.3 X = (1- moisture %) C.V.  (1-0.02)33000 = 32340Kj/kg 1.Heat lost in raising steam x1= me (h-hf1) – me(hf+xhfg – hf1) = 23394.64KJ 2.Heat lost to chimney or flue gases x2 = mgcpg(tg-tb) – 2954.7kJ 3. Heat lost in moisture x3 = mm (2676 + cp(tg-100)-hb) - 60.741KJ 4. Heat lost due to Radiation x4 = X – (X1+X2+X3+X4) - 5929.93 KJ

HEAT Supply kJHeat consumed KJ% Heat supplied 32340x1 x2 x3 X4 23394.64 2954.7 60.741 5929.93 72.34 9.14 0.188 18.34 TOTAL32340 100

Download ppt "Less then 1Equals to 1Not Present PTVfVgVfgUfUgUfghfhg hfg SfSgSfg 0.01 1 10 100 So with the help of steam table we select the value of needed item at."

Similar presentations