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A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and sanitizing the brewing tanks. 90.0% of the heat produced by the boiler is used to create saturated steam at 6.0 bars (absolute) from liquid water at 18° C. If 115 m3/h of steam is needed, what rate of heat does the boiler output?

Respuesta :

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = [tex]\frac{0.03194}{0.315}[/tex]

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m  

here h is heat required and m is mass flow rate

H = 2680.36  × 0.10139

H =  271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = [tex]\frac{H}{0.90}[/tex]

net boiler heat = [tex]\frac{271.75}{0.90}[/tex]

net boiler heat = 301.94 kW

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