INTRODUCTION TO ENGINEERING CALCULATION

UNIT & DIMENSION

1) DENSITY

• where: = density• M = mass• V = volume• Unit SI : kg/m3

• American Unit : Ibm/ft3 H2O = 1000 kg/m3

= 1 g/cm3

= 62.4 Ibm/ft3

V

M

2) CONCENTRATION

• Concentration of A in a mixture of A and B:

• SI unit = kg/m3

• Widely used = mg/L• ppm (parts per million) = mg/L (for water)• H2O : 1 mL = 1 g by weight (when water = 1 g/cm3)

• Percentage: mass ratio or volume ratio

VolumeBV

VolumeAV

MassAM

ionAConcentratC

Where

LmgVV

MC

B

A

A

A

BA

AA

:/

ppmg

g

g

g

cm

g

mL

g

L

mg1

1000000

1

1000

001.0

1000

001.0

1000

001.01

3

massBM

massAM

Apercentage

Where

xMM

M

B

A

A

BA

AA

100

3) Flow Rate:• Gravimetric or Mass flow rate, QM (kg/s, Ibm/s)• Volumetric (volume) flow rate, Qv (m3/s, ft3/s)• QM = Qv• Where = density• Relationship between mass flow of A, concentration of A & total volume flow (A+B)

• QMA

= CA x QV(A + B)

• QMA = Mass flow rate component A• CA = Density component A• Qv(A+B) = Total volume flow A + B

• Example: Wastewater treatment plant discharges a flow of 1.5 m3/s (water & solids) at a solids concentration of 20 mg/L. How much solids is the plant discharging each day?

• Solution:• QMA = CA x QV(A+B) •

%101.0100

1

101

10110000

6

4

x

xppm

Example: A wastewater sludge has a solids concentration of 10,000 ppm. Express this in percent solids (mass basis), assuming that the density of the solids is 1 g/cm3.

Solution:

daykg

day

sx

m

Lx

s

mx

mg

kgxx

L

mgQMA

/2592

8640010

5.1101

203

336

Retention / Detention /Residence Time• Average particle of fluid spend in a container through

which the fluid flow (exposed to treatment or a reaction)

• Or the time it takes to fill a container.

• t = V / Q

where t = residence time

V= volume of container (L3)

Q= flow rate into the container (L3/t)

Example:

A lagoon has a volume of 1500 m3, and the flow into the lagoon is 3 m3/hour. What is the retention time in this lagoon.

t = 1500 m3/ (3 m3/hour) = 500 hours

CONCEPT OF MATERIAL BALANCE

OBJECTIVE

• To obtain quantitative relationship between inflow and outflow of a process:• System WITH chemical reaction

• System WITHOUT chemical reaction

PROCESSInfluent effluent

System Boundary

Process: An operation which causes physical or chemical transformation

System: Partly or Whole of a specified process for material balance

Law of Mass conservation: New materials will not be produced in a system, and the

present materials in the system will not be destructed.

A black box with one inflow (influent) and one outflow

(effluent)

Material Balance Equation

Accumulation Rate = Input Rate + Generation Rate – Output Rate – Consumption Rate

Example: Every year 50, 000 persons enter city A, while 75, 000 persons leave the city, 22, 000 babies are born and 19, 000 die. Give the material balance equation to show the population at city A.

A (person/yr) = 50, 000 p/yr + 22, 000 p/yr – 75, 000 p/yr – 19, 000 p/yr

A = -22, 000 persons/year

Steady-State Process

• System WITHOUT chemical reaction• Inflow Rate = Outflow Rate

• System WITH chemical reaction• Inflow Rate + Generation Rate – Output Rate –

Consumption Rate = 0

Steady-state : Flow not changing with time

Accumulation Rate = 0

System WITHOUT chemical reaction

• Example:

• A gravity thickener is used in the thickening process of sewage sludge as in the figure. What is the Cu value.

GRAVITY THICKERNER

Qi = 40 m3/hrCi = 5000 mg/L

Qo = 30 m3/hrCo = 25 mg/L

Qu = 10 m3/hrCu = ???

Answer• Volume accumulated = Volume in – Volume out +

Volume produced – Volume consumed

• 0 = 40 – (30 + Qu) + ) – 0

• Qu = 10 m3/hr

• For solid mass balance:

0 = (CiQi – [(CuQu) + (CoQo) + 0 – 0

0 = (5000mg/L)(40 m3/h) – [Cu(10m3/hr) + (25mg/L)(30 m3/h)

Cu = 19,900 mg/L

Example:

Bandar Baru Bangi generates domestic waste of approximately 120 tons/day. All wastes are sent to a transfer station before being transported to a landfill. 20 tons/day of the wastes are recycled and the rest are sent to the landfill. How much waste will have to be sent to the landfill?

ANSWER

[Mass per unit of refuse IN] = [Mass per unit of refuse OUT]

120 = 20 + M

M = 100 tons/day mass of refuse to the landfill

TRANSFERSTATION

120 ton/dayRecycle

20 ton/day

Landfill???

Questions1) An air pollution control device is used to remove particles with

concentration of 125, 000 ug/m3 at a flow rate of 180 m3/s. The device has successfully removed 0.48 metric ton/day. What is the exhaustion rate of the particles to the air? Given 1 metric ton = 106g.

2) Sludge contains 80 % water weight. It is required to dry so that the sludge weight is now 80%. Estimate how much water needed to be removed.

3) A river with flow rate of 10 m3/s has heavy metal (zinc) concentration approximately 20 mg/L. An electronic factory released zinc concentration at 40 mg/L to the river and the flow rate of 5 m3/s. With the assumption of complete mixing between the two flows, estimate zinc concentration at the river downstream.

Question 1

QCz1= 0.48 metric ton/hari = 5.6 g/s

Air Pollution ControlDeviceQ = 180 m3/s

Czi = 125 000 g/m3

= 0.125g/m3

Removal

Effluent to air

QCz2 = ???

Influent Rate = Effluent Rate(180 m3/s)(0.125 g/m3) = 5.6 g/s + QCz2

QCz2 = 16.9 g/s (exhaustive rate of particles to air)

Influent

Question 2

Drying Equipment

Wet Sludge (100 kg)

80% water weight 20% sludge weight

Water removed (Y kg) ????

Dry Sludge (X kg)

20% Water Weight80% Sludge Weight

Sludge Balance:Influent (sludge) = Effluent (sludge)

0.2 (100) = 0.8 (X)X = 25 kg

Water Balance:Influent (water) = Effluent (water)

0.8 (100) = 0.2 (25) + Y

Y = 75 kg

Question 3

QS = 10 m3/s = 10 000 L/sCS = 20 mg/L

QE = 5 m3/s = 5000 L/sCE = 40 mg/L

QM = 15 m3/s = 15 000 L/sCM = ???

Influent Rate = Effluent Rate(10 000)(20) + (5000)(40) = (15 000)(CM)

CM = 26.67 mg/L

EXAMPLE OF MULTIPLE SYSTEMFigure 1 shows a flow of sludge thickening process by using a centrifuge . The sludge solid concentration is, C0 = 4% and is required to be thickened to a concentration, CE = 10% by using the centrifuge. However, the centrifuge is capable to produce sludge with solid concentration of 20% from sludge with solid concentration of 4%. Thus, the facility operator decided to make a bypass at the inflow to the centrifuge, and later mix it with the outflow from the centrifuge with solid concentration of 20% to produce sludge with solid concentration of 10%. Assuming the solid density is 1 g/cm3, which is similar to water density. Determine the flow rate in each flow.

Bypass

QB = ??CB = C0 = 4%Q0 = 1 gal/min

C0 = 4 %

QA = ??CA = 4%

QC = ??Cc = 0.1%

CENTRIFUGE

QK = ??CK = 20%

QE = ??CE = 10%

FIGURE 1

SOLUTION

• Assuming Steady State condition

• 1) Overall Balance

Bypass

QB = ??CB = C0 = 4%Q0 = 1 gal/min

C0 = 4 %

QA = ??CA = 4%

QC = ??Cc = 0.1%

CENTRIFUGE

QK = ??CK = 20%

QE = ??CE = 10%

Q0 = Qc + QE Equation 1

Q0C0 = QCCC + QECE Equation 2

Answer: QC = 0.606 gal/min QE = 0.394 gal/min

• 2) Balance at the mixing point

QB = ??CB = 4% solid

QK = ??CK = 20% solid

QE = 0.394 gal/minCE = 10 % solid

QB + QK = QE Equation 1

QBCB + QKCK = QECE Equation 2

Answer: QK = 0.1478 gal/min QB = 0.2462 gal/min

• 3) Balance at the centrifuge

QA = ??CA = 4% solid

QK = 0.1478 gal/minCK = 20% solid

QE = 0.394 gal/minCE = 0.1% solid

QA = QK + QE

= 0.7538 gal/min

System WITH chemical reaction • Kinetic Reaction:

• Mathematical expression describing a rate at which the mass or volume of some material A is changing with time, t is:dA/dt = r where r is reaction rate.

• Zero-order Reaction• r = k where k = constant reaction rate (mass/time)

• First-order Reaction• The change of component A is proportional to the quantity of the component itself.• r = kA where unit for k = time-1

• dA/dt = kA• Second-order Reaction

• The change is proportional to the square of the component A.• r = kA2 where unit for k = (time x mass)-1

• dA/dt = kA2

• If material A is being USED or DESTROYED, hence:• r = -k, r = -kA, r = -kA2

• First Order Reaction:• dA/dt = r = -kA

• The equation is integrated between A0 (t=0) and A (t=t)

ktAA

eA

A

atau

ktA

A

dtkA

dA

kt

tA

A

0

0

0

0

lnln

ln

0

• Reactor:• Tank or container used to undergo a reaction (chemical or

biological).

• Reactor is classified according to the flow characteristics and mixing conditions:

• Mixed-Batch Reactor

• Plug Flow Reactor (PFR)

• Completely Mixed Flow (CMF)/Continuous Stirred Tank Reactor (CSTR)

QCA0

QCA

QCA0

QCA

V

V

V

CA0 = concentration of A at time t=0 = mass/volumeA0 = Mass of material A at time t=0CA = Concentration of A at any time t = mass/volumeQ = Flow rateV = Reactor volume

Example 1 (CSTR)

A volume of CSTR is required to change component A to 98%. Kinetic reaction is rA = kCA where k = 0.10s-1. The inflow rate is 75 liter/s with the initial concentration, CA0 is 0.05 mol/L. There is no volume difference in the reaction.

Inflow Rate + Generation Rate – Outflow Rate – Consumption Rate = Accumulation Rate

Q = 75 L/sCA0 = 0.05 mol/L

Q = 75 L/sCA = 0.001mol/L(??)

V

LkC

CCQV

VkCCCQ

VkCQCQC

A

AA

AAA

AAA

750,36)001.0(1.0

)001.005.0(75)(

)(

0

0

0

0

Vdt

dCVrQCVrQC A

AA 210

Example 2 Mixed Batch ReactorAn industrial wastewater treatment process is using activated carbon to remove colour from water. The reduction in colour is according to first-order reaction in batch-adsorption system. If k value is 0.35/day, how much time is required to remove 90% of the colour in the water?

• CA0 = intial colour concentration

• CA = colour concentration at time t

• To remove 90% colour, it is required to achieve 0.1CA0

dayst

t

tC

C

ktC

C

A

A

A

A

58.635.0

302.2

35.01.0ln

35.01.0

ln

ln

0

0

0

Example 3

Disinfection process is required to destroy coliform organisms in drinking water. This is a first-order reaction, with k value, k = 1.0/day. The influent concentration, C0 is 100 coli/mL. The volume of reactor, V is 400 L, with the flow rate, Q is 1600 L/day. Determine the coliform concentration in the effluent.

• Inflow Rate + Generation Rate – Outflow Rate – Consumption Rate = Accumulation Rate

QC0 + 0 – QC – rV = 0

Where r = kC

(1600)(100) - (1600)(C) – (1)(C)(400) = 0

C = 80 coli/mL