The Balance Equation and TheThe Balance Equation and The Mass BalanceMass Balance Chapter 3

General balance equation, Average velocity, Velocity distribution, The mass balance), Steady-state, , ), y ,

Unsteady-state balances, Mass balances for mixtures.

The General Balance EquationqA balance is a method of accounting for something (here mass or material).

General Balance:IN - OUT + GENERATION - CONSUMPTION = ACCUMULATION

EXAMPLEProcess unit

OutputInput

EXAMPLE:26 students signed up [IN]

3 added [GENERATION]

1 dropped [CONSUMPTION]

no incompletes [ACCUMULATION]

What is the number that finishes the course [OUT]?[ ]

Mass and Volume Flow RatesMass flow rate: The amount of mass flowing through a cross section per unit time

.mDenoted by my

A fluid flows into or out of a control volume, usually through pipes or ducts

Consider a differential surface dAc; a differentialConsider a differential surface dAc; a differentialmass flow rate of fluid flowing across a small areaelement dAc, in a cross section of the pipe isproportional to dA itself the density ρ and theproportional to dAc itself, the density ρ, and thecomponent of the flow velocity normal to dAc,which is denoted as Vn. This is expressed as:

.nm V dAδ ρ=

The mass flow rate through the entire cross-sectional area of a pipe or duct:sectional area of a pipe or duct:

. .m mδ= ∫ n cV dAρ= ∫A A

∫

Velocity Distribution and Average velocityIn general compressible flow, both ρ and Vn vary across the pipe; however, ρin many application is constant over the pipe cross section. Thu, we can take ρ outside the integral. ρ g

Velocity, however, is never uniform over a cross section of a pipe; i.e. no slip condition

The velocity varies from zero at the wall to some maximum value at or near the centerline of the pipecenterline of the pipe.

Average velocity is defined as:

1avg n c

A

V V dAA

= ∫ where Ac is the area of the cross section normal to the flow direction

A

if speed were Vavg all through the cross section, the mass flow rate would be identical to that obtained by integration of the velocity profile

Thus, for incompressible flow or even forcompressible flow where ρ is uniform across A :

.m V Aρ= (kg/s)compressible flow where ρ is uniform across Ac: avg cm V Aρ

Volume flow rate: The volume of the fluid fowing through a cross section per unit time, denoted as Q, and is given by:

n c avg c cQ V dA V A VA= = =∫ (m3/s)A∫

The mass and volume flow rates are related by:.

m Qρ= Qρ

Examplep

A typical self-service gasoline pump puts 15 gal of fuel into our tank in 2 min The inside diameter of the nozzle is 1 0 in What are the volumetricmin. The inside diameter of the nozzle is 1.0 in. What are the volumetric flow rate, mass flow rate, and average velocity?

The Mass BalanceThe conservation of mass principle for a control volume can be expressed as:

The net mass transfer to or from acontrol volume during a time intervalΔt is equal to the net change (increaseor decrease) of the total mass withinthe control volume during Δt. That is,Total mass entering Total mass leaving Net change of mass⎛ ⎞ ⎛ ⎞ ⎛ ⎞Total mass entering Total mass leaving Net change of massthe CV during t the CV during t within the CV during t⎛ ⎞ ⎛ ⎞ ⎛ ⎞

− =⎜ ⎟ ⎜ ⎟ ⎜ ⎟Δ Δ Δ⎝ ⎠ ⎝ ⎠ ⎝ ⎠m m m− = ΔO in out CVm m m= Δ

This can also expressed in rate form as:.. CV

in outdmm m

dt− =

Or, MassBalance

dt

For more than one inlet or outles:.. CV

outdmm m

dt− =∑ ∑

in out dt

The Steady-State BalanceDuring a steady-flow process, the total amount of mass contained within a control volume does not change with time (m = constant)change with time (mCV = constant)

⇒ Total mass entering = total mass leaving

Example: Steady flow of water through garden hose nozzle

0CVdmdt

∴ =

The conservation of mass principle for a general steady-flow system withmultiple inlets and outlets is expressed in rate as:

dt

p p

..out

in out

m m=∑ ∑in out

The Steady-State Flow, One Dimensional Mass BalanceMany engineering devices such as nozzles, diffusers, turbines, compressors,

and pump involve single stream (only one inlet and one outlet.

For such cases, the inlet state denoted by the subscript 1 and the outlet state by the subscript 2, and the summation is dropped in this case

I h d h i l d fl b l iIn other words, the single stream steady-flow mass balance is:. .

1 2m m= 1 1 1 2 2 2V A V Aρ ρ=

Special case: Incompressible Flow

1 1 1 2 2 2

In this case ρ1 = ρ2 1 1 2 2V A V A=

ExampleIn a natural-gas pipe line at station 1 the pipe diameter is 2 ft and the flow conditions are 800 psia, 60oF, and 50 ft/s. At station 2 the pipe di t i 3 ft d th fl diti 500 i 60oF h t idiameter is 3 ft and the flow conditions are 500 psia, 60oF, what is the velocity at station 2? What is the mass flow rate?

i fl i i i i h i id di i

ExampleWater is flowing in a pipe. At point 1 the inside diameter is 0.25 m and the velocity is 2 m/s. What are the mass flow rate and the volumetric flow rate? What is the velocity at point 2 where thevolumetric flow rate? What is the velocity at point 2 where the inside diameter is 0.125?

Unsteady-State Mass BalanceCharacteristics of steady and unsteady-state processes are compared below:

Examples• Power plant operationp p• Start up and shutdown

Unstead state mass balances do not introd ce ant ne ideas be ond thoseUnsteady-state mass balances do not introduce ant new ideas beyond those seen so far; but generally lead to more complicated mathematics

ExampleA 4-ft high, 3-ft-diameter cylindrical water tank whose top is open to the

atmosphere is initially filled with water. Now the discharge plug near thebottom of the tank is pulled out, and a water jet whose diameter is 0.5 inp jstream. The average velocity of the jet is approximated as V = (2gh)1/2,where h is the height of water in the tank measured from the center of thehole (a variable) and g is the gravitational acceleration. Determine how longhole (a variable) and g is the gravitational acceleration. Determine how longit takes for the water to drop to 2 ft from the bottom

ExampleThe microchip diffusion furnace shown in the drawing contains air, which may

be considered an ideal gas. The vacuum pump is pumping air out prior tobeginning the thermal diffusion set up. During the pump out process theg g p g p p pheating coils in the tank hold the temperature in the tank constant at 68oF.The volumetric flow rate at the inlet of the pump, independent of thepressure, is 1.0 ft3/min. How long does it take the pressure to fall from 1 atmpressure, is 1.0 ft /min. How long does it take the pressure to fall from 1 atmto 0.0001 atm

ExampleIn the previous example, we considered a vacuum system with zero leak. No

real vacuum systems are totally leak-free; engineers work very hard to keepthe leakage rate as low as possible. If the tank in the previous example has ag p p pleak of 0.0001 lbm/min of air, what will be the pressure-time look like, andwhat will be the final pressure?

Mass Balances for Mixtures

In this case, more than one streams are involved in the flowing process, where the summation in the mass balance equation is considered.

ExampleA paint shop uses a special paint that contains benzene as a solvent. In the coursepa s op uses a spec a pa a co a s be e e as a so ve . e cou se

of an 8-h day the paint evaporates 5 kg (22 lb) of benzene (q = 5 kg/hr). Theshop dimensions are 10 m × 4 m × 4m. To protect the health of workers, theconcentration of benzene in the shop air is limited to less than or equal to theconcentration of benzene in the shop air is limited to less than or equal to theindustrial hygiene standard for benzene, which is 1.3 mg/m3. If it is required tokeep the concentration c of benzene in the shop at or below this permittedconcentration how large a flow of ventilating air must be supplied?concentration, how large a flow of ventilating air must be supplied?

JUSTDepartment of Chemical Engineering Fluid Mechanics – ChE 343 1-14

Problem Solvingg