A mixture of n-hexane vapor and air leaves a solvent recovery unit and flows thr

Question

A mixture of n-hexane vapor and air leaves a solvent recovery unit and flows through a 70-cm diameter duct at a velocity of 3.00 m/s. An a sampling point in the duct the temperature is 40 degree C, the pressure is 850 mm Hg, and the dew point of the sampled gas is 25 degree C. The gas is fed to a condenser in which it is cooled at constant pressure, condensing 60% of the hexane in the feed. Perform a degree of-freedom analysis to show that enough information is available to calculate the required condenser outlet temperature (degree C) and cooling rate (kW). Perform the calculations. If the duct diameter were 35 cm for the same molar flow rate of the feed gas, what would the gas velocity be?

Explanation / Answer

From the given data, molar flow rate of mixture can be calculated. The dew point gives the partial pressure of hexane vapro. From the equation

Mole of hexane/total moles = partial pressure of hexane/ total pressure

Moles of hexane vapor can be calculated.

Since 60% of hexane is condensed, moles of hexane at out let can be calculated. Hence the partial pressure of hexane ( which in turn will be equal to vapor pressure , sicne air is saturated at the outlet of condenser),    temperature corresponding to this partial presure = vapor presure can be calculated.

Volumetric flow rate of the gas = cross sectional area* velocity = (PI/4)* (70/100)2* 3=1.155 m3/s

Molar flow rate of gas, n= PV/RT

P= 850/760 atm =1.118 atm, T= 40+273.15K= 313.15K, R =0.0821 L.atm/mole.K

Total moles = 1.118*1.155*1000L/(0.0821*313.15)=50.23 moles/sec

Dew point at the inlet= 25 deg.c

Dew point is the temperature at which partial pressure of vapor= vapor pressure of hexane liquid

So vapor pressure of hexane liquid = 151.2 mm Hg at 25 deg,c= partial pressure of hexane

At inlet

Mole of hexane/total moles = partial pressure of hexane/ total pressure

Moles of hexane = 50.23* 151.2/(850-151.2)=10.87 moles /s of hexane at inlet

Mole of air = 50.23-10.87=39.36 moles/s. This does not change during condensation

60% of the hexane is condensed in the condenser, mole of hexane condensed ( unknown)= 10.87*0.6= 6.522 moles/sec

Moles of hexane uncondensed = 10.87-6.522 =4.348 moles/s

Total moles at outlet = 4.348+39.36=43.71 moles/se

At outlet of condenser , partial pressure of hexane = (moles of hexane/ total moles)* total pressure

= (4.348/43.71)* 850 =84.55 mm Hg

At the out let of condenser, the air is assumed to be saturated with hexane. So 84.55 mm Hg partial pressure of hexane correspond to vapor pressure at that temperature.

So temperature corresponding to 84.55 mm Hg partial pressure = 12.1 deg.c

Load on the condenser = (moles of   hexane condensed/s * latent heat of vaporizatino f hexane = 6.522*43.85 Kj/mol=286 Kilo joules/sec = 286 Kw

From mass flow rate = density *V* cross sectional area

Since density remained constant

V1* cross sectionala rea= V2* cross sectional area

3*(70/100)2= V2*( 35/100)2

V2= velcoity = 3*4=12 m/sec

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