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.

Expert's answer

**Solution.**

H=14(Q)^{0.3};

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

Sulphur content=8%;

=(Q_{1}/Q_{2})×100%;

Q_{1}=m× H;

m=250 kg/s;

H= 3450 KJ/KG=3.45×10^{3}J/kg;

Q1=250 kg/s×3.45×10^{3}J/kg=862.5×10^{3}J/s;

Q_{2}=(Q_{1}/ )×100%;

=88.05%;

Q_{2}=((862.5×10^{3}J/s)/88.05%)×100%=979.6×10^{3}J/s;

Q_{2}=q×m; m=Q_{2}/q; q=34.3MJ/kg=34.3×10^{6}J/kg;

m=(979.6×10^{3}J/s)/34.3×10^{6}J/kg=28.56×10^{-3}kg/s=102.8 kg/Hr;

Q=(102.8 kg/Hr×8%×2)/100%=16.45Kg/Hr;

H_{1}=14×(16.45)^{0.3}=14×2.32=32.48 m;

Since the two boilers are connected in parallel, then H=2×H_{1}=64.96 meters.

**Answer: **H = 64.96 meters.

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