MASS & ENERGY BALANCEFUNDAMENTAL MATERIAL BALANCE

CALCULATIONS (WEEK 3)

ADILAH IMANANIS SYAHEERAHASYIQIN AFIFA

BY:

Question 1

A liquid mixture of benzene and toluene contains 55.0% benzene by mass. The mixture is to be partially evaporated to yield a vapor containing 85.0% benzene and a residual liquid containing 10.6% benzene by mass.

b) Next, suppose the process is to be carried out in a closed container that initially contains 100.0 kg of the liquid. Let mv(kg) and ml(kg) be the masses of the final vapor and liquid phases respectively. Draw and label a process flowchart, then write and solve integral balances on total mass and on benzene to determine mv and ml. For each balance, state which terms of the general balance equation (accumulation = input + generation - output - consumption) you discarded and why you discarded them.

evaporator

55kg/hB45kg/hT100kg/h

VapormBv = 0.85 x mvmTv = 0.15 x mv

liquidmBl = 0.106 x mlmTl = 0.895 x ml

Input + generation - output - consumption = accumulation0 0 0

*Assume generation, consumption and accumulation is zero because it is a non-reactive process and steady state.

Degree of freedom: *unknown on the chart is 6 and the material balance is 4.*We can determine mv and ml by: mv = mbv + mtv ml = mbl + mtl

Hence,

ndf = nunknown - nindependent equation =6-6 =0

Problem can be solved!

So, Input = output

B: 55 kg/h = 0.85 x mv + 0.106 x ml 1

100 kg/h = mv + ml 2

55 kg/h = 0.85 (100 kg/h - ml ) + 0.106 ml55 kg/h = 85 kg/h - 0.85 ml + 0.106 ml0.744 ml = 30 kg/hml = 40.32 kg/h

mv = 100 kg/h - 40.32 kg/h =59.68 kg/h

Sub 2 to 1 :

Sub ml to 2 :

C) Returning to the continuous process, suppose the evaporator is built and started up and the product stream flow rates and composition are measured.The measured percentage of benzene in the vapor stream is 85% and the product stream flow rates have the values calculated in part (a) but the liquid product stream is found to contain 7% benzene instead of 10.6%. One possible explanation is that a mistake was made in the measurement.

Give at least 5 others:

● Benzene is being consumed as a reactant within the process unit.

● Benzene is accumulating in the process unit.- Possibly adsorbing on the walls or other surfaces of the evaporator.

● Benzene is leaking from the unit.● Process happen at unsteady state● Feed composition is not correct

Three hundred gallons of a mixture containing 75.0 wt% ethanol (ethyl alcohol) and 25% water (mixture specific gravity = 0.877) and a quantity of a 40.0 wt% ethanol-60% water mixture (SG = 0.952) are blended to produce a mixture containing 60.0 wt% ethanol. The object of this problem is to determine, the required volume of the 40% mixture.

a) Draw and label a flowchart of the mixing process and do the degree-of-freedom analysis.

b) Calculate V40

Question 2

Three hundred gallons of a mixture containing 75.0 wt% ethanol (ethyl

alcohol) and 25% water (mixture specific gravity = 0.877)

and a quantity of a 40.0 wt% ethanol-60% water mixture (SG = 0.952)

are blended to produce a mixture containing 60.0 wt% ethanol. The object

of this problem is to determine, the required volume of the 40% mixture.

Mixture A:

Mixture B:

Mixture C:

Total VA

MIXER

40.0 wt% ethanol60.0 wt% water

60.0 wt% ethanol40.0 wt% water

75.0 wt% ethanol25.0 wt% water

Mixture A

Mixture B

Mixture C

VA = 300 galMA = ?SGA = 0.877

VB = ?MB = ?SGB = 0.952

VC = VA + VBMC = ?

Step 1: Draw & label the flowchart.

MIXER

40.0 wt% ethanol60.0 wt% water

60.0 wt% ethanol40.0 wt% water

75.0 wt% ethanol25.0 wt% water

Mixture A

Mixture B

Mixture C

VA = 300 galMA = ?SGA = 0.877

VB = ?MB = ?SGB = 0.952

VC = VA + VBMC = ?

Step 2: Do the degree-of-freedom analysis.

Ndf = N unknown - N independent equation

N unknown variables (4) MA, MB, MB, V40

N independent equations (4) SGA, SGB, wt% ethanol of mixture C, wt% water of mixture C

Ndf = 4 - 4 = 0 ✔ The problem is solvable.

pA = SGA x pnet = 0.877 x 1000 kg/m³ = 877 kg/m³

MA = 300 gal 877 kg 1 lbm 1 m³ 1 m³ 0.453593 kg 264.17 gal

= 2195.680 lbm

MIXER

40.0 wt% ethanol60.0 wt% water

60.0 wt% ethanol40.0 wt% water

75.0 wt% ethanol25.0 wt% water

Mixture A

Mixture B

Mixture C

VA = 300 gal MA = ?SGA = 0.877

VB = ?MB = ?SGB = 0.952

VC = VA + VBMC = ?

Step 3: Find the unknown variables, & do conversion of unit.

Input + generation - output - consumption = accumulation

Input + generation - output - consumption = accumulation

Input = output

Assumption: 1) No reaction occurs:

consumption and generation = 0

2) Steady state: accumulation = 0

Step 4: Write a balanced equation.

MIXER

40.0 wt% ethanol60.0 wt% water

60.0 wt% ethanol40.0 wt% water

75.0 wt% ethanol25.0 wt% water

Mixture A

Mixture B

Mixture C

VA = 300 gal MA = 2195.680 lbmSGA = 0.877

VB = ?MB = ?SGB = 0.952

VC = VA + VBMC = ?

Step 5: Solve for the desired variable.

Input = Output (VA) (SGA) (Ethanol,A) + (VB) (SGB) (Ethanol,B) = (VC) (Ethanol,C) (VA) (SGA) (Ethanol,A) + (VB) (SGB) (Ethanol,B) = (VA + VB) (Ethanol,C) (300 gal) (0.877) (0.75) + VB (0.952) (0.40) = [(300 gal) (0.877) + VB (0.952)] (0.6) 197.325 gal + 0.381 VB = 157.86 gal + 0.5712 VB 39.465 gal = 0.1902 VB VB = 207.49 gal = V40

✔ Balanced equation

✔ Correct

MIXER

40.0 wt% ethanol60.0 wt% water

60.0 wt% ethanol40.0 wt% water

75.0 wt% ethanol25.0 wt% water

Mixture A

Mixture B

Mixture C

VA = 300 gal MA = 2195.680 lbmSGA = 0.877

VB = ?MB = ?SGB = 0.952

VC = VA + VBMC = ?

pB = SGB x pnet= 0.925 x 1000 kg/m³= 925 kg/m³

MB = 207.215 gal 952 kg 1lbm 1 m³ 1 m³ 0.453593 kg 264.17 gal

= 1646.3 lbm

MC = MA + MB = 2195.680 lbm + 1646.3 lbm = 3841.98 lbm

Step 6: Check for the solutions.

THANK YOU FOR YOUR ATTENTION, HAVE A GREAT DAY AHEAD !