n-pentane is partially vaporized in a boiler using steam. The n-pentane enters t

Question

n-pentane is partially vaporized in a boiler using steam. The n-pentane enters the boiler as a liquid at 20 degree C and 5 atm at a rate of 150 kmol/hr. 50% of the pentane entering is to be vaporized at 5 atm pressure. The steam fed to the boiler is saturated vapor at 2 bar and is condensed to saturated liquid at 2 bar. Calculate the rate at which steam is fed to the condenser in kg/hr. Construct an enthalpy table to do the energy balance. Make sure you state the reference states for your enthalpy calculations.

Explanation / Answer

flow rate of liquid pentane= 150 kmol/hr. amount to be vaporized = 150*0.5= 75 kmol/hr

molar mass of n-pentane = 72, mass of n-pentane to be vaporized = 75*72 kg/hr=5400 kg/hr

boiling poiny of n-pentane at 1 atm =36.11deg.c =36.11+273= 309.11K

at 5 atm

latent heat of vaporization of n-pentane = 357 Kj/Kg= 357*72 j/mole=25704 J/mol

frrom classius clayperon equation ln (P2/P1)= (deltaH/R)* (1/T1-1/T2)=

P2= 5 atm P1= 1 atm, T1= 309.11 and T2=?

ln5= (25704/8.314)*(1/309.11-1/T2)

T2=368 K =368=273= 95 deg.c

specific heat of liquio= 167.19 j/mol. K

hence from 20 deg.c to 95 deg.c, it is sensible heat = 150*1000 mol/hr *167.19*(95-20) j/hr=1.88*109 j/hr

at 95 deg.c ii is latent heat for 50% of n-pentane = 150*0.5*1000*25740j j/hr =1.93*109 j/hr

total heat to be added = (1.88+1.93)*109 joules/hr = 3.81*109 joules/hr

frpm steam tables.

enthalpy of saturated steam at 2 bar = 2529 j/gm and enthapy of saturated liquid = 503.5 j/gm

heat to be removed from steam = 2529-503.5 =2025.5 j/gm

hecne stteam required = 3.81*109/ 2025.5=1.88*106 g/hr = 1.88*103 kg/hr

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.