Question

Propane burns with 15% excess air. Before the propane-air mixture enters the burner, air preheats from 32 F to 575 F. Determine the requirements of the preheater (BTU / h) if the amount of propane fed to the burner is 1.35x105 SCFH (standard cubic feet per hour)

Note: In this case, take standard conditions in the natural gas industry, such as 60 F and 1 atm.

Report: a) Flow diagram of the complete process and its table of material balance with each one of the streams, b) Q preheater.

Answer 1

Propane (C₃H₈)

The reactions occuring are

Propane fed = 1.35×10⁵ SCFH

= 1.35×10⁵ ft³/h = 3822.774 m³/h

SCFH conditions

T = 60°F = 15.55° C = 288.55 K

P = 1 atm = 1.013×10⁵ Pa

n = PV/RT

n = (1.013×10⁵× 3822.77 ) /(8.314×288.55)

n = 161419.873 mol/h

n = 161.4198 Kmol/h

According to stiochiometry amount of O₂ required = 161.4198(5) = 807.099 Kmol/h

Supplies O₂ is 15% excess

So O₂ supplies = 807.099(1.15) =928.163 Kmol/h

Mol% of O2 in air = 21%

Air supplied = 928.163/(0.21) =

4419.8278 kmol/h

A)

The flow diagram is given below

The feed analysis in burner

Component	kmol/h	mol%
Propane	161.4198	3.52
O2	928.163	20.26
N2	3491.660	76.20
Total	4581.2428	100

Product analysis

Component	kmol/h	mol%
CO2	161.4198(3) = 484.2594	10.21
H2O	161.4198(4)= 645.6792	13.61
O2	928.163-807.099= 121.064	2.55
N2	3491.660	73.62
Total	4742.6626	100

B)

Air to be heated from

32°F to 575°F

T₁ = 32°F = 0°C = 273 K

T₂ = 575°F = 301.66°C = 574.66 K

Average temperature for C_p = (273+574.66) /2 = 423.833 K

From handbook

Cp of air (423.833 K) = 29.48 KJ/kmol K

Heat required for preheating

Q = nCp(∆T)

Q = (4419.8278) (29.48) (574.66-273) = 39305249.29 KJ/h

1 BTU= 1.055 KJ

Q = 37256160.47 BTU/h

Q = 3.72561×10⁷ BTU/h

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