Please show me how to do every single step of this thermodynamics problem: 3.5.

Question

Please show me how to do every single step of this thermodynamics problem: 3.5. Allyl chloride (A) symthesis provides a simple prototype of many petrochemical processes. 869 kg per hour of propylene (C3) are fed with a l% excess of chlorine (C12) at 25OC. The entire process operates at roughly 10 bar. C12 is recycled to achieve a 50% excess of C12 at the reactor inlet. The reactor conversion is 100% of the propylene to AC and hydrochloric acid (HCl) at 511 C. The reactor effluent is cooled to 35 Cand sent to a distillation column where 98% of the entering AC exits the bottom with l% ofthe entering Cl2 and no HCl. This AC product stream exits at 57OC. The tops of the first column are sent to a second column where 99% of the entering C12 exits the bottom at 36°C, along with 1% of the entering HCl, all of the AC, and is recycled. The tops of the second column exit at -31°C and are sent for waste treatment. Using the method of Heat of Formation method for the energy balance and ideal gas reference states with Eqn 45 to estimate the heat of vaporization, complete the following. 12, HCI C12, HCI, AC C3, C12 coli C12, HCI, AC AC, C12 a. Write a balanced stoichiometric equation for this reaction. (Hint: Check the NIST WebBook for chemical names and formulas.) b. Perform a material balance to determine compositions and flow rates for all Streams c. Using only streams (1), (6), (7), calculate the energy balance in MJhof the entire process. Does the process involve a net energy need or surplus? d. Determine the heat load on the reactor in MJ/h. e. Calculate the enthalpies in MJ/h ofthe feed stream 1 Calculate the enthalpy in MU h of the stream 4 entering the first distillation colum. g. Calculate the enthalpy in MJ/h of the AC product stream 6 h. Calculate the enthalpy in MI/h of the C12 recycle stream 8 i. Calculate the enthalpy in MJ/h of the HCl waste stream7.

Explanation / Answer

(a)

Balnced equation:

C3H6 +Cl2----->C3H5Cl+HCl

(b)

Feed:

Propylene =869 kg/hr

M.W .of propylene = 42.079

propylene =869 /42.079 =20.65 kmol/hr

Cl2 is 1% in excess

As per stoichiometry 1 mol propylene=1 mol Cl2

But Cl2 is 1% in excess.So 1+1/100=1.01 mol Cl2

So Cl2 mole =1.01*20.65 =20.858 kmol

Cl2 MW=70.9

So Cl2 =20.858*70.9=1478.84 kg/hr

Now define all stream as per problem statement:

Stream 1: propylene =869 kg/hr and Cl2 1478.84 kg/hr

(2) stream:

we guess x kmol of propylene is fed to reactor

So Cl2 is 50% in excess =1.5x kmole

100% conversion is there.So as per stoichiometry

moles of Allyl chloride =x kmol

HCl = x kmol

(3) and (4) stream Allyl chloride = x mole ,HCl =x mole,Cl2 =1.5x-x=0.5 x mole

(6) stream: Cl2 =1% of enetring=00.01*0.5=0.0005 x

AC=98% of entering =0.98x

(5) stream: HCl=x mol,AC=0.02x,Cl2 = 0.99*0.5x=0.495x

(8)stream:Cl2 =99% of entering =0.99*0.495 x=0.49005x,AC=0.02x,HCl=0.01x

(7) stream:Cl2=0.00495 x,HCl=0.99x

Now we do balnce:

As per flowsheet

Reactor feed

(1)+(8)=(2)

So take Cl2 balance

20.858+0.49005 x=1.5 x

So, x=20.858/(1.5-0.49005)=20.65 kmol

stream 2:

C3=20.65 kmol /hr= 20.65*42.079=869 kg/hr

Cl2=1.5*20.65=30.975 kmol/hr=30.975*70.9=2196.1275 kg/hr

stream 3 &4:

AC=20.65 kmol/hr=20.65*76.52=1580.138 kg/hr

Cl2=0.5*20.65=10.325 kmol/hr=10.325*70.9=732.0425 kg/hr

HCl=20.65 kmol/hr=20.65*36.5=753.725 kg/hr

stream 5:

HCl=20.65 kmol/hr=753.725 kg/hr

AC=0.02x=0.02*20.65=0.413 kmol/hr=0.413*76.52=31.60 kg/hr

Cl2 = 0.495*20.65=10.22 kmol/hr =724.72 kg/hr

stream 6:

Cl2 =0.0005*20.65=0.010325 kmol/hr=0.732 kg/hr

AC=0.98*20.65=20.237 kmol/hr=1548.53 kg/hr

stream 7:

Cl2=0.00495*20.65=0.1022 kmol/hr=7.247 kg/hr

HCl=0.99*20.65=20.4435 kmol/hr=746.18 kg/hr

stream 8:

Cl2 =0.49005*20.65=10.12 kmol/hr=717.47 kg/hr

AC=0.02*20.65=0.413 kmol/hr=31.60 kg/hr

HCl=0.01*20.65=0.2065 kmol/hr=7.53725 kg/hr

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