The feed (F) is an equimolar liquid mixture of n-pentane and n-hexane at 20 o C

Question

The feed (F) is an equimolar liquid mixture of n-pentane and n-hexane at 20o C that is fed to a vessel in which the mixture is heated in the vessel to 50 oC. The total flowrate of the feed (F) is 100 kmol/h. The vapor product (V) is all vapor and the total flow rate of V = 16.14 kmol/h. The mole fraction of n-pentane in Stream V is 0.71. The mole fraction of n-hexane in Stream V is 0.29. The liquid product (L) is all liquid and contains n-pentane and n-hexane. The entire system operates at 1 atm absolute pressure.

Is this an open or closed system? Why?

What are the references you will use for this problem? Tell me each components temperature, phase, and pressure.

Determine the amount of heat (Q) needed in kJ/h. Does heat need to be added or removed? Why?

Explanation / Answer

1) It is open system because feed is entering into the vessel while vapor and liquid stream are leaving the system/ vessel.

2)

Following references can be used:

Boiling point of pentane = 36.1oC

Boiling point of hexane = 68oC

Heat capacity of liquid hexane = 197.66 J/ mol. K

Heat capacity for vapor hexane = 142.6 J/mole.K

Heat of vaporization of hexane = 167.2 KJ/ mol

Heat capacity of liquid pentane = 167.19 J/mol.K

Heat capacity of vapor pentane = 120 J/mol. K

Heat of vaporization of pentane = 146.8 KJ/mol

Pentane liquid phase temperature is 35oC (below boiling point).

Pentane vapour phase temperature is 50oC (since vessel is heated to 50o C).

Hexane liquid phase temperature is 50oC (since vessel is heated to 50o C).

Hexane vapor phase temperature is 68oC (otherwise hexane vapors won't be formed).

For calculating vapour phase pressure, Dalton's law will be used.

Pi = yiPT

yi = vapour phase mole fraction for that component

Pt = total pressure = 1 atm

Vapour phase pressure for pentane = 0.71 *1 = 0.71 atm

Vapor phase pressure for hexane = 0.29*1 = 0.29 atm

Since pressure is due to vapor phase only, liquid phase pressure of both components can be neglected.

3)

Feed moles of pentane = 50 kmol

Feed moles of hexane = 50 kmol

Vapor moles of pentane = 16.14 * 0.71 = 11.5 kmol

Vapor moles of hexane = 16.14 * 0.29 = 4.7 kmol

Total heat required = heat to raise liquid pentane temperature (20o to 36o) + heat due to latent of vaporization (pentane) + heat to raise tempeerature 36 to 50 (gas pentane) + heat to raise liquid hexane temperature (20 to 68) + latent heat of vaporization (hexane) = 50*167.19*16 + 146.8*11500 + 11.5*120*14 + 50*197.66*48 + 167.2*4680 (heat needed to raise temp = n*Cp*delT, heat of vaporization = C*moles)

= 133752 + 1688200 + 19320 + 474384 + 782496 = 3098152 kJ = 3.09 * 106 kJ/hr

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