Steam boiler efficiency calculations

[shreddinglicks]

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?

 

[Twigg]

m ̇_air = m ̇_gas

Does this really account for the whole mass balance in the boiler? What else besides air is used up in combustion?

Q_in = 5833 kj/s

To be exactly accurate, don't forget the 15kW from the fan. That being said, it's a small contributor.

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

Yep.

I haven't gone over every step you did with a fine toothed comb but I think that will help.

 

[shreddinglicks]

1. That energy balance is taken at the air preheater. There is only flue gas and air.
2. OK
3. If that is true than, my Q_in = Q_out. Isn't that unrealistic?

 

[Twigg]

1. The energy balance for the preheater is fine. I'm saying the mass balance you came up with for the preheater isn't right. At the preheater, the two gases' mass flow rates are not equal. If you do a mass balance at the boiler, that will tell you the mass flow rate of the steam.
3. If you calculate the steam mass flow rate following my suggestion in 1, it won't be. The reason you are getting Q_in = Q_out is a case of circular reasoning. I think you'll find everything will make a lot more sense after you do a mass balance for the boiler.

 

[shreddinglicks]

balance at boiler

m_fuel * HV = m_air*c_air* ΔT + m_steam* Δh

η= Q_out/Q_in = .65

 

[Twigg]

What I mean by mass balance is a conservation of mass equation that only includes $\dot{m}_{fuel}$, $\dot{m}_{air}$, and $\dot{m}_{steam}$. Remember, air and fuel go into the boiler, and only the flue gases come out. What does that tell you about those three quantities?

 

[shreddinglicks]

What I mean by mass balance is a conservation of mass equation that only includes $\dot{m}_{fuel}$, $\dot{m}_{air}$, and $\dot{m}_{steam}$. Remember, air and fuel go into the boiler, and only the flue gases come out. What does that tell you about those three quantities?

I don’t follow. The air and fuel mix but they don’t mix with the water.

 

[Twigg]

Sorry, I mixed up the steam and the flue gases in my previous post. I meant to say that the mass flow rate of the flue gases (which are what exchange heat in the preheater, not the steam) can be calculated from $\dot{m}_{air}$ and $\dot{m}_{fuel}$. Then you'll have the air flow correct and you can calculate part (b) correctly.

You'll need to calculate the mass flow of the steam by thinking about energy balance in the boiler. How much energy does it take to produce one kilogram of steam at 5 MPa and 400C starting from post-economizer feed water at 150C? You already know how much heat per second is provided by combustion, so that will tell you the mass flow rate of the steam.

 

[shreddinglicks]

Sure, and the mass flow of steam would be from

m_fuel * HV = m_air*c_air* T_in + m_steam* Δh + m_gas*c_gas*T_out

correct?

Also
m_fuel = .16 kg/s
m_air = 7.3 kg/s
m_gas = 7.5 kg/s

If I use the first equation I get

m_steam = 1.45 kg/s

efficiency = .64

 

[Twigg]

The enthalpy of the air goes on the other side of the equation. The air is consumed in combustion so it's energy is made available. For more concrete proof, by the equation you have above with the air on the right side, the hotter the air going into the boiler, the lower the efficiency, which isn't right.

 

[shreddinglicks]

The enthalpy of the air goes on the other side of the equation. The air is consumed in combustion so it's energy is made available. For more concrete proof, by the equation you have above with the air on the right side, the hotter the air going into the boiler, the lower the efficiency, which isn't right.

Yes, I miswrote that.

m_fuel * HV = -m_air*c_air* T_in + m_steam* Δh + m_gas*c_gas*T_out