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shreddinglicks

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**Summary::**I'm going through my homework. I am confident I have everything correct except for my boiler efficiency. It seems too high at 99%. I would like to think my logic is sound but it can't be right.

A boiler is designed to generate steam at 5 MPa and 400oC. A fuel is selected with a heating value of 35,000 kJ/kg burned at a rate of 600 kg/hr. Feed water into an economizer inserted in the boiler is raised from 45 oC to 150oC, while the flue gases are cooled at the same time from 400oC to 225oC. The flue gas then enters an air preheater in which the temperature of combustion air is raised by 100oC. A forced draught fan delivers the air to the air preheater at a pressure of 1 bar and a temperature of 20oC with a pressure rise across the fan of 20 cm of water. The power input to the fan is 15 kW and it has mechanical efficiency of 80%. Neglecting heat losses and assuming the flue gases as ideal gas with constant specific heat of

*Cp*= 1.01 kJ/kg-K, calculate: (a) the mass flow rate of the air, (b) the temperature of flue gases leaving the plant, (c) the mass flow rate of the steam, and (d) the efficiency of the boiler.

For air mass flow rate:

w ̇_fan= (m ̇_air*ν*ΔP)/η_fan

where,

w ̇_fan = 15 kj/s

ν = .816 kj/kg*k

ΔP = 2 kpa

η_fan = .8

m ̇_air = 7.3 kg/s

For flue gas Temperature:

c_(p,air)*m ̇_air*ΔT=c_(p,gas)*m ̇_gas*ΔT

where,

m ̇_air = m ̇_gas

c_(p,air) = c_(p,gas) approximately

T_gas = 125 Celsius

For steam mass flow rate:

Q_in=m ̇_steam*Δh

Q_in = 5833 kj/s

Δh = 3196 - 632 = 2564 kj/kg From steam table

m ̇_steam = 2.27 kg/s

For boiler efficiency:

η= Q_out/Q_in

Q_in = 5833 kj/s

This is where I am confused. Is Q_out = m ̇_steam * Δh?