Question

A heat exchanger is used to heat acetone. 10 kg/min of steam enters at 550 C and 20.0 bar and leaves as saturated steam at 1 bar. The acetone enters at 20 C and leaves at 70 C. The heat exchanger loses 3000 kJ/min to the surroundings. The acetone is under a constant pressure of 1 atm.

Determine the volumetric flow rate (L/h) of acetone leaving the heat exchanger.

Answer 1

Lets make enthalpy balance ,

Heat in =Heat out

Here heat is supplied by steam and that heat is used to heat acetone and in the process 300 Kj/Min of heat is lost to the surrounding

dH_steam=dH_acetone+ H_lost

Enthalpy at 550 oC and P=20 bar = 3577.6 Kg/Kj ( From steam table)

Enthalp of saurated steam at 1 bar = 2675.4 KJ/Kg ( From steam table)

dHsteam = 3577.6-2675.4 =902.2 Kj/Kg

Cp of acetone = 77.28 KJ/KMole = 77.28/58 K =1.3324 Kj/Kg K ( taken from standard references)

Let the flow rate of aceone be 'm' Kg/min

dHacetone = m*Cp*dT = m*1.3324*(70-20)=m*66.62 Kj/min

Hence subsituting all the known data in ethalp equation

10*902.2= m*66.62 + 3000

value of 'm' from above equation ,

m=90.393 Kg/min =90.393/58=1.5585 Kmole/min =1.5585*60 = 93.510 Kmole/hr=93.510*10³ Mole/hr

m=93.510*10³ Mole/hr

Now we will use ideal gas law to convert mole per hr to Litre /hr

PV=NRT

Where,

P = 1 atm

N= moles/hr

V= Volume per hr

R= 0.0821 Litre atm Mole^-1 K^-1

Toutlet = 70 oC =343.15 k

1*V= 93.510*10³ *0.0821*343.15

V= 2.634*10⁶ Litre/hr

Note : The accuracy of solution depends on the accuracy of Cp and enthalpy values of steam. More accurate the values of enthalpy and heat capacity , more acccurate the result.