Answer to Question #108245 in Thermal Power Engineering for Jayapal Reddy Chittep

Question #108245
2 boilers r connected with parallel to chimney.each boiler is 250 kg/s steam flow with enthalpy 3450 KJ/KG n efficiency is 88.05%,fuel used is 34.3MJ/kg n 8% sulphur.Calculate the chimney height.
1
Expert's answer
2020-04-14T01:44:47-0400

Solution.

H=14(Q)0.3;

H - height chimney , Q=(Fuel consumption(Kg/Hr)×Sulphur content(%)×2)/100;

Sulphur content=8%;

=(Q1/Q2)×100%;

Q1=m× H;

m=250 kg/s;

H= 3450 KJ/KG=3.45×103J/kg;

Q1=250 kg/s×3.45×103J/kg=862.5×103J/s;

Q2=(Q1/ )×100%;

=88.05%;

Q2=((862.5×103J/s)/88.05%)×100%=979.6×103J/s;

Q2=q×m; m=Q2/q; q=34.3MJ/kg=34.3×106J/kg;

m=(979.6×103J/s)/34.3×106J/kg=28.56×10-3kg/s=102.8 kg/Hr;

Q=(102.8 kg/Hr×8%×2)/100%=16.45Kg/Hr;

H1=14×(16.45)0.3=14×2.32=32.48 m;

Since the two boilers are connected in parallel, then H=2×H1=64.96 meters.

Answer: H = 64.96 meters.


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