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PROBLEMS 53

in Table 3.5. The term y is used in engineering practice to represent mole (or volume) fraction of gases; the term x is often used for liquids and solids.

3.19 PartialPressure I

The exhaust to the atmosphere from an incinerator has a S02 concentration of0.12 mm Hg partial pressure. Calculate the parts per million of S02 in the exhaust.Solution: First calculate the mole fraction y. By Dalton's law

y= Pso2p (3.27)

Since the exhaust is discharged to the atmosphere, the atmospheric pressure,7 60 mm Hg, is the total pressure P.

Thus,

y= (0.12mmHg)/(760mmHg) = 1.58 x 10-4

ppm = (y)(l06) = (1.58 X 10-4) (106)

= 158 ppm

3.20 PartialPressure II

A storage tank contains a gaseous mixture consisting of 30% C02, 5% CO, 5% H20, 50% N2, and 10% 02, by volume. What is the partial pressure of each component if the total pressure is 2 atm? What are their pure-component volumes if the total pressure is 2 atm? What are the pure-component volumes if the total volume is 10 ft3? What are the concentrations in ppm (parts per million)?

Solution: Dalton's law states that the partial pressure Pa of an ideal gas is given by

Pa = YaP (3.27)

where Ya= mole fraction of component aP = total pressure

Thus

Pco2 = 0.30(2) = 0.60 atm

Peo = 0.05(2) = 0.10 atm

PH20 = 0.05(2) = 0.10 atm

PN2 = 0.50(2) = 1.00 atm

Po2 = 0.10(2) = 0.20 atm

P = 2.00atm

5 FUNDAMENTA GAS 5PROBLE

Amagat' s law states that the pure component volume Va of an ideal gas is given by

Va= Ya V (3.29)

where V is the total volume.

Thus

Vco2 = 0.30(10) = 3.00 ft3

Veo = 0.05(10) = 0.50 ft3

VH20 = 0.05(10) = 0.50 ft3

VN2 = 0.50(10) = 5.00 ft3

Vo2 = 0.10(10) = 1.00 ft3

V = 10.00 ft3

By definition, the parts per million (ppm) is given by

ppma = Yal06

Thus

ppmc02 = 0.30(106) = 3.00 x 105 ppm

ppmc0 = 0.05(106) = 0.50 x 105 ppm

ppmH2o = 0.05(106) = 0.50 x 105 ppm

ppmN2 = 0.50(106) = 5.00 x 105 ppm

ppm02 = 0.10(106) = 1.00 x 105 ppm

Unless otherwise stated, ppm for gases is by mole or volume and is usually designated as ppmv. When applying the term to liquids or solids, the basis is almost always by mass; the notation may appear as ppmw or ppmm.

3.21 Vapor PressureCalculation

Two popular equations that are used to estimate the vapor pressure of compounds are the Clapeyron and Antaine equations. The Clapeyron equation is given by

lnp' =A-- B T (3.40)


where p' and Tare the vapor pressure and temperature, respectively, and A and B are the experimentally determined Clapeyron coefficients. The Antaine equation is given by

Inp' =A B T+C

(3.41)

where A, B, and C are the experimentally determined Antaine coefficients.Use the Clapeyron and Antaine equations to estimate the vapor pressure of

acetone at OºC. The Clapeyron coefficients have been experimentally determined to be

A= 15.03

B = 2817

for p' and T in mm Hg and K, respectively. The Antaine coefficients are

A= 16.65

B = 2940

e= -35.93

with p' and T in the same units.

Solution: First, calculate the vapor pressure p' of acetone at OºC using theClapeyron equation:

In p' = 15.03 - [2817/(O+ 273)]

= 4.7113

p' = 111.2mm Hg

Calculate the vapor pressure of acetone at OºC using the Antaine equation:

lnp' = 16.65 - [2940/(273 - 35.93)]

= 4.2486

p' = 70.01 mm Hg

The Clapeyron equation generally overpredicts the vapor pressure at or near ambient conditions. The Antaine equation is widely used in industry and usually provides excellent results. Also note that, contrary to statements appear- ing in the Federal Register and sorne EPA publications, vapor pressure is nota function of pressure.

3.22 Reynolds Number

The Reynolds nurnber (Re)

(a) Describes fluid flow and is equal to µ.,CP/ pDQ(b) Equals 6.02 x 1023

e


(e) Describes how a fluid behaves while flowing and is defined as the inertial forces divided by the viscous forces (Dvp/ µ,)

(d) Is generally used only for liquids

Solution: Answers (a), (b), and (d) are obviously incorrect. The definition of Re is given in Equation (3.13). The correct answer is therefore (e).

3.23 Reynolds Number Calculation

Calculate the Reynolds number for a gas flowing through a 5-inch-diameter pipe at 10 fps (feet per second) with a density of 0.050 lb/Ir' and a viscosity of 0.065 cP? Is the flow turbulent or laminar?

Solution: By definition

Re=-Dvp

µ,(3.13)

Substitution yields

Re= e~:) (1~ft) (º·º!~lb)/ (o.0:5cP) (6.720 x ~~~41b/ft s)= (5/12 ft) (lOft/s) (0.0501b/ft3)/(0.065 x 6.72 x 10-41b/ft · s)

=4770

The Reynolds number is >2100; therefore, this gas flow is turbulent. Generally, moving gases are in the turbulent flow regime.

3.24 Process CalculationAn externa} gas stream is fed into an air pollution control device at a rate of10,000 lb/hr in the presence of 20,000 lb/hr of air. Because of the energyrequirements of the unit, 1250 lb /hr of a vapor conditioning agent is added toassist the treatment of the stream. Determine the rate of product gases exiting the unit in pounds per hour (lb/hr). Assume steady-state conditions.

Solution: Apply the conservation law for mass to the control device on a rate basis:

Rate of mass in - rate of mass out + rate of mass generated

= rate of mass accumulated (3.34)

Note that mass is not generated and steady conditions (no accumulation) apply. Rewrite this equation subject to the conditions in the problem statement.

Rate of mass in = rate of mass out

or


Now refer to the problem statement for the three inlet flows:

m¡n = 10,000 + 20,000 + 1250 = 31,250 lb/hr

Determine mout, the product gas flow rate. Since m¡n = mout, it follows that

lnout = 31,250 lb /hr

As noted earlier, the conservation law for mass may be written for any compound whose quantity is not changed by chemical reaction and for any chemical element whether or not it has participated in a chemical reaction. It may be written for one piece of equipment around several pieces of equipment, or around an entire process. It may be used to calculate an unknown quantity directly, to check the validity of experimental data, or to express one or more of the independent relationships among the unknown quantities in a particular problem situation.

3.25 Collection Efficiency

Given the following inlet loading and outlet loading of an air pollution control unit, determine the collection efficiency of the unit:

Inlet loading = 0.02 gr/ft3

Outlet loading = 0.001 gr /ft3

Solution: Collection efficiency is a measure of the degree of performance of a control device; it specifically refers to the degree of removal of a pollutant and may be calculated through the application of the conversation law for

mass. Loading, generally refers to the concentration of pollutant. The equation describing collection efficiency ( fractional) E in terms of inlet and outletloading is

E = _(i_nl_e_t _lo_a_d_in_g_) _-_(_o_ut_le_t_l_oa_d_in_g_)

inlet loading(3.42)

Calculate the collection efficiency of the control unit in percent for the rates provided:

2-0.1E =-2-100 = 95%

The term TJ is also used as a symbol for efficiency E. The reader should also note that the collected amount of pollutant by the control unit is the product of E and


the inlet loading. The amount discharged to the atmosphere is given by the inlet loading minus the amount collected.

3.26 Penetration Definition

Define penetration.

Solution: By definition, the penetration Pis given byP = 100 - E; percent basis

P = 1.0 - E; fractional basis(3.43)

Note that there is a 10-fold increase in P as E goes from 99.9 to 99%. For a multiple series of n collectors, the overall penetration is simply given by

(3.44)

For particulate control in air pollution units, penetrations and/ or efficiencies canbe related to individual size ranges. The overall efficiency ( or penetration) is then given by the contribution from each size range, i.e., the summation of the product of mass fraction and efficiency for each size range. This is treated in more detail in the chapters on particulates (Chapters 7-12).

3.27 Spray Tower Application

A proposed incineration facility design requires that a packed column and a spray tower be used in series for the removal of HCI from the flue gas. The spray tower is to operate atan efficiency of 65% and the packed column at an efficiency of98%. Calculate the mass flow rate of HCl leaving the spray tower, the mass flow rate of HCl entering the packed tower, and the overall efficiency of the removal system if 76.0 lb of HCl enters the system every hour.

Solution: As defined in Problem 3.25

Then,

E= lnin ~ lnout

m¡n(3.45)

mout = (1 - E) (m¡n)

For the spray tower

lnout = (1 - 0.65) (76.0) = 26.6 lb /hr HCl

The mass flow rate of HCl leaving the spray tower is equal to the mass flow rate of HCl entering the packed column. For the packed column

lnout = (1 - 0.98) (26.6) = 0.532 lb/hr HCl


The overall efficiency can now be calculated:

lrl¡n - lrlout 76.0 - 0.532E= =-----m¡n 76.0

= 0.993

= 99.3%

3.28 Compliance Determination

A proposed incinerator is designed to destroy a hazardous waste at 2100ºF and 1 atm. Current regulations dictate that a minimum destruction and removal efficiency (DRE) of 99.99% must be achieved. The waste flow rate into the unit is 960 lb/hr while that flowing out of the unit is measured as 0.08 lb/hr. Is the unit in compliance?

Solution: Select as a basis the 1 hourly rate of operation. The mass equation employed for efficiency may also be used to calculate the minimum destruction and removal efficiency.

DRE = m¡n ~ mout (IOO) = 960 - 0.08 (IOO)m¡n 960

= 99.992% (3.45)

Thus the unit is operating in compliance with present regulations. The answer is yes.

3.29 Velocity Determination

Given 20,000 fr'/min of air at ambient conditions exiting a system through a pipe whose cross-sectional area is 4 ft2, determine the mass flow rate in lb/rnin and the exit velocity in ft/s.Solution: The continuity equation is given by

where p = liquid densityS = cross-sectional areav = velocity

m=pSv (3.46)

Since 20,000 ft3 / min of air enters the system, then

q1 = S1v1

= 20,000 ft3/min

where subscript 1 refers to inlet conditions. Now, assuming that p = 0.075 lb/ft", then

m = p1S1v1 = (0.075) (20,000) = 15001b/min


3.30 Outlet Temperature

Heat at 18.7 x 106 Btu/hr is transferred from the flue gas of an incinerator. Calculate the outlet temperature of the gas stream using the following information:

Average heat capacity cP of gas= 0.26 Btu/(lb · ºF) Gas mass flow rate m = 72,000 lb /hrGas inlet temperature T1 = 1200ºF

Solution: The first law of thermodynamics states that energy is conserved. For a

flow system, neglecting kinetic and potential effects, the energy transferred Q to or from the flowing medium is given by the enthalpy change t:..H of the medium. The enthalpy of an ideal gas is solely a function of temperature; enthalpies of liquids and most real gases are almost always assumed to depend on temperature alone. Changes in enthalpy resulting from a temperature change for a single- phase material may be calculated from the equation

(3.47)

or

where t:..H = enthalpy changem = mass of flowing mediumcp= average heat capacity per unit mass of flowing medium across the

temperature range of t:..Tt:..i!. = enthalpy change per unit timem = mass flow rate of flowing medium

(Note: The symbol t:.. means "change in.")

Solve the conservation law for energy for the gas outlet temperature T2:

where Q is the rate of energy transfer:

The gas outlet temperature is therefore

T2 = (-18.7 X 106 / { (72,000) (0.26)}] + 1200

= 200ºF

This equation is based on adiabatic conditions, i.e., the entire heat load is transferred from the flowing gas. The unit is assumed to be perfectly insulated so that no heat is transferred to the surroundings. However, this is not the case in a real-world application. As with mass balances, an enthalpy balance may be

=


performed within any properly defined boundary, whether real or imaginary. For example, an enthalpy balance can be applied across the entire unit or process. The enthalpy of the feed stream(s) is equated with the enthalpy of the product stream(s) plus the heat loss from the process. All the enthalpy terms must be based on the same reference temperature. Finally, the enthalpy has two key properties that should be kept in mind:

l. Enthalpy is a point function, i.e., the enthalpy change from one state (say,200ºF, 1 atm) to another state (say 400ºF, 1 atm) is a function only of the two states and not the path of the process associated with the change.

2. Absolute values of enthalpy are not important. The enthalpy of water at 60ºF,

1 atm, as recorded in sorne steam tables is O Btu/lbmol. This choice of zero is arbitrary. However, another table may indicate a different value. Both are correct! Note that changing the temperature of water from 60 to IOOºF results in the same change in enthalpy using either table.

Enthalpy changes may be expressed with units (English) of Btu, Btu/lb, Btu/lbmol, Btu/scf, or Btu/time depending on the available data and calculation required.

3.31 Process Cooling Water RequirementDetermine the total flow rate of cooling water required for the services listed below assuming that a cooling tower system supplies the water at 90ºF with a retum temperature of 115ºF. How much freshwater makeup is required if 5% of the retum water is sent to "blowdown?" Note that the cooling water heat capacity is 1.00 Btu/(lb · ºF), the heat of vaporization at cooling tower operating conditions is 1030 Btu/lb, and the density of water at cooling tower operating conditions is 62.0 lb/ft3.

TABLE 3.6 Cooling Water-Heat Duty Data

Process Unit Heat Duty, Btu/hr Required Temperature, ºF

12,000,000 2502 6,000,000 200-2763 23,500,000 130-1764 17,000,000 3005 31,500,000 150-225

Solution: The required cooling water flow rate, rhcw is given by the following equation:

. QHLme»

(ilT)(cp)(p) (3.48)

where Q8L = heat load, Btu/minD..T = change in temperature = 115ºF - 90ºF = 25ºF

7


cP = heat capacity = 1.00 Btu/(lb · ºF)p = density of water = (62.0 lb/ft3) (0.1337 ft3/ gal) = 8.289 lb/ gal

rhcw = cooling water flow rate, mass/time

The heat load is

QHL = (12 + 6 + 23.5 + 17 + 31.5) (106Btu/hr)/ (60 min/hr)

= 1,500,000Btu/min

Thus,

_ 1,500,000 Btu/min _ mqcw (25ºF) (l.OOBtu/lb · ºF)(8.2891b/gal) -

50gp

where qcw = water volumetric flow rate, gal/min. The blowdown flow q80 is given by

qBD = (BDR)(qcw)

where BDR is the blowdown rate = 5% = 0.05.

Thus

qBD = (0.05) (7250 gpm) = 362.5 gpm

The amount of water vaporized by the cooling tower qv is given by

where Hv is the heat of vaporization = 1030 Btu/lb. Substitution yields

(1,500,000 Btu/min)q = = 175 7 gpm v (8.2891b/gal) (1030Btu/lb) ·

(3.49)

3.32 Steam Requirement Options

Determine how many pounds per hour of steam are required for the following two cases: (1) if steam is provided at 500 psig and (2) if steam is provided at both 500 and 75 psig pressures. The plants heating requirements are listed in Table 3.7.

TABLE 3.7 Process Heat Duty Data

Process Unit Unit Heat Duty (UHD), Btu/hr Required Temperature, ºF

10,000,000 2502 8,000,000 4503 12,000,000 4004 20,000,000 300


The properties of saturated steam are listed in Table 3.8.

TABLE 3.8 Steam Data

Pressure Provided, psig

Saturation Ternperature,ºF

Enthalpy of Vaporization, filiv, Btu/lb

75 320 894500 470 751

Solution: The total required flow rate of 500 psig steam mBT is given by

For this equation:

mBI (mass flow rate of 500 psig steam through unit 1)= UHDi/ dHv = 13,320 lb/hr

mB2 (mass flow rate of 500 psig steam through unit 2)= UHD2/ dHv = 10,655 lb/hr

mB3 (mass flow rate of 500 psig steam through unit 3)= UHD3/dHv = 15,980lb/hr

m84 (mass flow rate of 500 psig steam through unit 4)= UHD4/ dHv = 26,635 lb/hr

Thus,

lnBT = 66,590 lb/hr

The required combined total flow rate of 500 and 7 5 psig steam mcT

is given by

For this situation

m75.1 (mass flow rate of 75 psig steam through unit 1)= UHD/ ~Hv = 11,185 lb/hr

m75.4 (mass flow rate of 75 psig steam through unit 4)= UHD/ sn; = 22,371.4 lb/hr

6PROBLE6 FUNDAMENTA GAS

straight with negative slopes. The value of Twa corresponds to the value of the abscissa at the point of intersection of this line with the saturation curve.

Based on the problem statement, calculate the flow rate of water in the air. Note that both the given flow rate and humidity are on a dry basis.

Water flow rate = (0.0805) (10,000) = 805 lb/hr

Calculate the total flow rate by adding the dry gas and water flow rates:

Total flow rate = 10,000 + 805 = 10,805 lb/hr

The molar rate of water and dry gas are thus

Moles gas= 10,000/30 = 333.3 lbmol/hr

Moles water= 805/18 = 44.7lbmol/hr

Calculate the mole fraction of water vapor using the above two results.

= 44.7 = o 12Ywater (44.7 + 333.3) .

The average molecular weight of the mixture becomes

MW = (1.0 - 0.12) (30) + (0.12) (18)

= 28.6 lb/lbmol

The molar flow rate of the wet gas may now be determined:

n = 10805/28.6 = 378 lbmol/hr

The ideal gas law may be applied to calculate the volumetric flow rate of the wet gas:

q = nRT/P

= (378) (0.73) (460 + 180)/1.0

= 1.77 X 105 ft3 /hr

(3.22)

The following are sorne helpful points on the use of psychrometric charts:

1. In problems involving the use of the humidity chart, it is convenient to choose a mass of dry air as a basis since the chart uses this basis.

2. Heating or cooling at temperatures above the dew point (temperature at which the vapor begins to condense) corresponds to a horizontal movement on the


chart. As long as no condensation occurs, the absolute humidity stays constant.

3. If the air is cooled, the system follows the appropriate horizontal line to the left until it reaches the saturation curve and follows this curve thereafter.

3.34 Saturated Water Discharge

A flue gas is discharged at 120ºF from an HCl absorber in a hazardous waste incinerator (HWI) facility in which carbon tetrachloride (CC14) is being inciner- ated. If 9000 lb/hr (MW = 30) of gas enters the absorber essentially dry (negli- gible water) at 560ºF, calculate the moisture content, the mass flow rate, and the volumetric flow rate of the discharged gas. The discharge gas from the absorber may safely be assumed to be saturated with water vapor. The discharge humidity of the flue gas is approximately.

'Hout = 0.0814lb H20/lb bone-dry air

Solution: The 'Hout represents the moisture content of the gas at outlet conditions in lb H20 /lb dry air. If the gas is assumed to have the properties of air, the discharge water vapor rate is

liZHzo = (0.0814) (9000)

= 733 lb/hr

The total flow rate leaving the absorber is

11'ltotal = 733 + 9000

= 9733 lb/hr

The volumetric (or molar) flow rate can be calculated only if the molecular weight of the gas is known. The average molecular weight of the discharge flue gas must first be calculated from the mole fraction of the flue gas (fg) and water vapor (wv).

= 9000/30 = O 88Yfg (9000/30) + (733/18) .

= 733/18 = O 12Ywv (733/18) + (9000/30) .

MW = (0.88) (30) + (0.12) (18) = 28.6 lb


The ideal gas law is employed to calculate the actual volumetric flow rate, qª.

mqª = -=-(RT/P)

MW9733 (0.73) (460 + 140)

qa = 28.6 1.0

= 1.49 X 105 ft3 /hr (3.22)

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title.

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