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

Question

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.

Answer 1

Solution.

H=14(Q)^0.3;

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

Sulphur content=8%;

$\eta$ =(Q₁/Q₂)×100%;

Q₁=m×$\Delta$ H;

m=250 kg/s;

$\Delta$H= 3450 KJ/KG=3.45×10³J/kg;

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

Q₂=(Q₁/$\eta$ )×100%;

$\eta$ =88.05%;

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

Q₂=q×m; $\implies$ 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.