Question

a)

With the following data,
how the energy of this process is calculated?

- 1

b) Make the process heat transfer diagram (Kj / s)

Answer 1

a)

In the process, a feed stream F is entering a preheater as stream 1 passig through a valve. The outlet stream from pre heater = 2.

let, molar flow rate of stream F = mf moles/hr, average specific heat = Cpf, temperature of stream = Tf.

Then energy of stream F = mf * Cpf * (Tf - To)

To = bottom temperature = 273K = 0 C.

Now, as in the valve no energy transfer is occuring, temperature of stream 1 = Tf. let molar flow rate of stream 1 = m1.

No compositional change is occuring, so average specific heat = Cpf.

energy of stream 1 = m1* Cpf * (Tf - To)

In the preheater, temperature of stream rises to T2. No mass transfer occured, so molar flow rate = m1, specific heat = Cpf.

energy of stream 2 = m1 * Cpf * (T2 - To)

Energy inlet in the preheater = m1 * Cpf * (T2 - To) - m1*Cpf * (Tf - To) = m1 * Cpf * (T2 - Tf)

Then stream 2 enters distillation column, where Q is the constant heat supply.

overhead product is V, molar flow rate = mv, average specific heat = Cpv, temperature = Tv,

bottom product is L, molar flow rate = ml, average specific heat = Cpl, temperature = Tl,

Energy of overhead product = mv * Cpv * (Tv - To)

Energy of bottom product = ml * Cpl * (Tl - To)

By mass balance -

m2 = mv + ml

By overall energy balance -

mv* Cpv * (Tv - To) + ml * Cpl * (Tl - To) = Q + m2 * Cpf * (T2 - To)

b) Given, feed molar flow rate = mf = m1 = m2 = 100 kmol/sec. with 20 mol% Ethanol (l), 80 mol% water (l).

specific heat of water = 75.2 J/mol.K specific heat of ethanol (l) = 112.4 J/mol.K

average specific heat of stream F= 0.8 * 112.4 + 0.2 * 75.2 J/mol.K = 104.96 J/mol.K = Cpf

Also given ,feed temperature = Tf = T1 = T2 = 25 C = 298 K.

so, energy of stream F = energy of stream 1 = energy of stream 2

= 100 * 10^3 mol/sec * 104.96 J/mol.K * (298 - 273)K  

= 262.4 *10^6 J/sec = 262.4 *10^3 kJ/sec

so no energy input is done in the preheater, as no temperature change.

We are considering the overhead product is vapor and bottom product is in liquid phase.

given, overhead product molar flow rate, mv = 42.276 kmol/sec. composition -35 mol% ethanol (vp), 65 mol% H2O (vp). Temperature of overhead product = Tv = 90 C = 363 K

specific heat of H2O(vp) = 36.5 J/mol.K, specific of ethanol(vp) = 78.28 J/mol.K

so, average specific heat of overhead product = 36.5 * 0.65 + 78.28 * 0.35 J/mol.K = 51.123 J/mol.K

Energy of overhead product = 42.276 kmol/sec*1000 mol/kmol * 51.123 J/mol.K * (363 - 273)K = 194514.8353 kJ/s

bottom product molar flow rate = ml = 51.724 kmol/s, composition - 6% ethanol(l), 94% H2O(l)

temperature of bottom product = Tl = 90 C = 363 K.

average specific heat of bottom product = 112.4 * 0.06 + 75.2 * 0.94 J/mol.K = 77.432 J/mol.K

energy of bottom product = 51.724 kmol/sec * 1000mol/kmol *77.432 J/mol.K * 90K = 360458.3491 kJ/sec

so, Q + 262.4*10^3 kJ/sec = 360458.3491 kJ/sec + 194514.8353 kJ/sec,

or Q = 554973.1844 - 262.4*10^3 = 292573.184 kJ/sec.

Process heat transfer diagram -

F= 262.4*10^3 kl/sec 1 = 262.4.10^3 kl/sec 2 = 262.4.10^3 kl/sec V = 194514.835 kl/sec Flash distillation column Valve preheater Q= 292573.184 k]/sec L = 360458.35 kl/sec