Question

26 mol/hr of stream 1 (gas, at 1.1 atm and 0°C) containing 30% (mol basis) propane and rest butane is mixed with a second stream (gas, at 1.1 atm and 45°C) containing 60% (mol basis) propane and rest butane. The output stream is 70 mol/hr (gas, at 1.1 atm). The relative enthalpies as a function of temperature (T is in °C) are given as: H (J/mol) for propane = 91*T and H (J/mol) for butane= 121*T a) What is the composition of the output stream? b) What is the temperature of the output stream, assuming no heat exchange with the surrounding?

Answer 1

Given data:

Enthalpy of propane H1(J/mol) = 91*T(C)

Enthalpy of butane H2(J/mol) = 121*T(C)

Assumptions : staedy state process and well mixed.

Let us assume inlet flow rate stream 1:

Molar flow rate n1 = 26 mol/h

Mole fraction of propane x1 = 0.3

Temperature T1 = 0 C

Stream 2:

Molar flow rate = n2

Mole fraction of propane x2 = 0.6

Temperature T2 = 45 C

calculations :

a) material balance :

Overall material balance :

n1 + n2 = n3

26 + n2 = 70

n2 = 44 mol/h.

Propane composition material balance :

n1*x1 + n2*x2 = n3*x3

26*0.3 + 44*0.6 = 70*x3

x3 = 0.488

Composition of product stream :

Mole fraction of propane x3 = 0.488

Mole fraction of butane = 1- 0.488 = 0.512

propane = 48.8%

butane = 51.2%

b) enthalpy balance :

Inlet stream 1 enthalpy H1 = 26*[ 91*0.3*0 + 91*0.7*0] = 0

Inlet stream 2 enthalpy H2 = 44mol/h*[ 0.6*91*45 + 0.4*121*45]

H2 = 203940 J/h

Output product stream H3 = 70*[ 0.488*91*T + 0.512*121*T] = 7445.2T J/h

Assuming no heat transfer then applying enthalpy balance,

H1 + H2 = H3.

0 + 203940 = 7445.2*T

T = 27.39 C

Outlet product temperature T = 27.39 C

Ans: 27.39 C