Question

Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kPa. Determine:
a. the air-fuel ratio
b. the temperature at which the water vapor in the products will start condensing.

Answer 1

Answer:

44.59°c

Explanation:

Given data :

Total pressure = 105 kpa

complete combustion

A) Determine air-fuel ratio

A-F = $\frac{N_{air} }{N_{fuel} } = \frac{(Nm)_{air} }{(Nm)_{c} (Nm)_{n} }$

N = number of mole

m = molar mass

A-F = $\frac{(6.75*4.76)kmole * ( 29kg/mol)}{(3kmole)* 12kg/mol + (6kmol)*(1kg/mol)}$  =  22.2 kg air/fuel

hence the ratio of Fuel-air = 1 : 22.2

B) Determine the temperature at which water vapor in the products start condensing

First we determine the partial  pressure of water vapor before using the steam table to determine the corresponding saturation temp

partial pressure of water vapor

Pv = $\frac{(N_{water vapor}) }{N_{pro} } * ( P_{ro} )$

N watervapor ( number of mole of water vapor ) = 3

N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol

Pro = 105

hence Pv = ( 3/33.53 ) * 105 =  9.39kPa

from the steam pressure table the corresponding saturation temperature to 9.39kPa =  44.59° c

Temperature at which condensing will start = 44.59°c

An equation showing the products of propylene with their mole numbers is attached below