Calculating the Mass Flow Rate of a boiler

[sci0x]

Homework Statement

Natural gas is burned in a shell and tube boiler to raise 5 tonnes per hr of dry saturated steam at 4.0 bar g. Condensate is returned to boiler at 4.0 bar g., 100 degrees C.

i) calculate mass flow rate of natural gas required (kgs-1)
ii) the steady state heat supplied to the heating application (MW)

Net enthalpy of combustion of natural gas = 45.6Mj kg-1
Boiler Effeciency = 76%
Enthalpy dry saturated steam at 4 bar g =2748 kJ kg-1
Enthalpy of dry saturated steam at 0 bar g. = 2676 kJ kg-1
Enthalpy of dry saturated water (152 degrees) = 637 kJ kg-1
Enthalpy of dry satirated waterat 0 bar g. (100 degrees C) = 419 kJ kg-1

Relevant Equations

Q = UAdT

Heat lost by steam = Heat taken up by water

Im trying to do some past exam papers. If i can get help with the method on how to solve and have correct formulas, it will help me.

I) There is one formula in the notes to get mass flow rate:
Heat lost by steam = Heat taken up by water
GsHfg = (GwCpw)(change in temp)
Gs x 2748 x 10x103 = (5000/3600) x 419 x (152-100)
Gs = 0.011 kg/s
However the past paper says answer is 0.09 kg/s

If I can get help from an expert I would appreciate it.

 

[Chestermiller]

Homework Statement:: Natural gas is burned in a shell and tube boiler to raise 5 tonnes per hr of dry saturated steam at 4.0 bar g. Condensate is returned to boiler at 4.0 bar g., 100 degrees C.

i) calculate mass flow rate of natural gas required (kgs-1)
ii) the steady state heat supplied to the heating application (MW)

Net enthalpy of combustion of natural gas = 45.6Mj kg-1
Boiler Effeciency = 76%
Enthalpy dry saturated steam at 4 bar g =2748 kJ kg-1
Enthalpy of dry saturated steam at 0 bar g. = 2676 kJ kg-1
Enthalpy of dry saturated water (152 degrees) = 637 kJ kg-1
Enthalpy of dry satirated waterat 0 bar g. (100 degrees C) = 419 kJ kg-1
Homework Equations:: Q = UAdT

This is not the correct equation to use. The correct equation to use is the open system (control volume) version of the 1st law of thermodynamics. Please write down the form of this equation that applies to the water/steam stream flowing between the entrance and exit of the tubular heat exchanger.

 

[sci0x]

I think I am on the right track, although I am confused by all the enthalpy options given and this isn't the correct final answer for mass flow rate but anyway:

In 1 second 5000kg/3600s = 1.38kg

Final Enthalpy:
H2 = (m)(h2)
= (1.38)(2748)
= 3792.24

Initial Enthalpy:
H1 = (1.38)(419)
= 578.22

Final - Initial = 3214.02kJ = Q
Boiler has 76% effeciency so need to burn (3214)(100/76) = 4228.97 = Q

Water heat in = - steam heat out
(M)(637)(152-100) = - - 4228.97
M = 0.127kg/s

The answer should be 0.09 kgs-1

 

[Chestermiller]

Maybe you should use units. Your gas energy requirement is 4229 kJ/s = 4.22 MJ/s. The heating value of the natural gas is 45.6 MJ/kg. So the gas rate is 4.22/45.6 = 0.093 kg/s

 

[sci0x]

Thank you, can you help with the last part:

Ii) the steady state heat supplied to the boiler:

(Kg of steam)(steam enthalpy)(temp) / time
(5000)(2748)(52) / 3600
= 198466.6 W = 1.98MW

The answer should be 3.23 MW

Thanks

 

[Chestermiller]

Thank you, can you help with the last part:

Ii) the steady state heat supplied to the boiler:

(Kg of steam)(steam enthalpy)(temp) / time
(5000)(2748)(52) / 3600
= 198466.6 W = 1.98MW

The answer should be 3.23 MW

Thanks

$$\frac{(5000)(2748-419)}{(3600)(1000)}$$