Chapter 4
Mass and Energy Balances
In this chapter we will apply the conservation of mass and conservation of energy laws to open systems or control volumes of interest. The balances will be applied to steady and unsteady system such as tanks, turbines, pumps, and compressors.
The general balance equation can be written as
Accumulation = Input + Generation - Output - Consumption
The terms (Generation - Consumption) are usually combined to call Generation with positive value for net generation and negative value for net consumption.
Let mcv = total mass (kg) of A within the system at any time.
m" in = rate (kg/s) at which A enters the system by crossing the boundaries.
m" out = rate (kg/s) at which A leaves the system by crossing the boundaries.
r"gen = rate (kg/s) of generation of A within the system by chemical reactions.
r"cons = rate (kg/s) of consumption of A within the system by chemical reactions.
Then the mass balance on species A can be written as
dm cv = m" + r" − m" − r" (4.1-1) dt in gen out cons
If there is no chemical reaction, the mass balance equation is simplified to
dm cv = m" − m" (4.1-2) dt in out
Example 4.1-1 ------Balance on an interest earning checking account during month of February
Withdrawn = $525.00 Deposit = $1000.00 Interest = $3.00 Fees = $5.00 Solution ------
Accumulation = $1000 + $3 - $525 - $5 = $477.00 /month (February) 4-1 Example 4.1-2 ------Snake-Eyes Maggoo is a man of habit. For instance, his Friday evenings are all alike-into the joint with his week’s salary of $180, steady gambling at “2-up” for two hours, then home to his family leaving $45 behind. Snake-Eyes’ betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictable - at a rate proportional to his cash at hand. This week Snakes-Eyes received a raise, so he played for three hours, but as usual went home with $135. How much was his raise? Solution ------
Step #1: Define the system.
System: Snake-Eyes’ money.
Step #2: Find equation that contains M, Snake-Eyes’ money.
We can make a balance on his money.
Step #3: Making a balance on Snake-Eyes’ money.
dm cv = − output dt
where output = loss at any time = kmcv, k = proportional constant, hence
dmcv = − kmcv (E-1) dt
Step #4: Specify the boundary conditions for the differential equation.
The constant k can be obtained by integrating the differential equation from t = 0 when Snake-Eyes usually starts with $180 to t = 2 hours when Snake-Eyes has $135 left.
≈135 ’ ln∆ ÷ = − 2k or k = 0.14384 /hour «180 ◊
Step #5: Solve the resulting equation and verify the solution.
Equation (E-1) is integrated again from t = 0, mcv = mo to t = 3 hours, M = $135
mo = 135 exp( 3k) = 135 exp( 3 × 0.14384) = $207.85
So, Snake-Eyes’ raise is $207.85 - $180.00 = $27.85
4-2 Example 4.1-3. ------A tank contains 2 m3 of pure water initially as shown in Figure E4.1-3. A stream of brine containing 25 kg/m3 of salt is fed into the tank at a rate of 0.02 m3/s. Liquid flows from the tank at a rate of 0.01 m3/s. If the tank is well mixed, what is the salt concentration (kg/m3) in the tank when the tank contains 4 m3 of brine.
Fi , ρAi V(t = 0) = 2 cubic meter 3 Fi = 0.02 m /s 3 ρAi = 25 kg/m ρA F , ρA Figure E4.1-3 A tank system with input and output.
Solution ------
Step #1: Define the system.
System: salt and water in the tank at any time.
Step #2: Find equation that contains ρA, the salt concentration in the tank at any time. The salt balance will contain ρA.
Step #3: Apply the salt balance around the system.
d(Vρ A ) = FiρAi - FρA = 0.5 – 0.01ρA (E-1) dt
where V is the brine solution in the tank at any time. Need another equation to solve for V and ρA.
dV 3 = Fi – F = 0.02 – 0.01 = 0.01 m /s (E-2) dt
Step #4: Specify the boundary conditions for the differential equation.
At t = 0, V = 2 m3, ρA = 0, at the final time t, V = 4 m3
Step #5: Solve the resulting equations and verify the solution. Integrate Eq. (E-2)
V t dV = 0.01 dt , to obtain V = 2 + 0.01t —2 —0
When V = 4 m3, t = 200 sec
4-3
The LHS of Eq. (E-1) can be expanded to
dρ A dV V + ρA = 0.5 – 0.01ρA dt dt
Hence
dρ A (0.01t + 2) + 0.01ρA = 0.5 - 0.01ρA dt
The above equation can be solved by separation of variables
dρ dt A = 0.5 − 0.02ρ A 0.01t + 2
1 1 - ln (0.5 – 0.02ρA) = ln(0.01t + 2) + C1 0.02 0.01
1 - ln (0.5 – 0.02ρA) = ln(0.01t + 2) + C (E-3) 2
at t = 0, ρA = 0, hence
1 - ln(0.5) = ln(2) + C (E-4) 2
Eq. (E-3) - Eq. (E-4)
1 ≈ 0.5 − .02ρ ’ ≈ 0.01t + 2 ’ - ln∆ A ÷ = ln∆ ÷ 2 « .5 ◊ « 2 ◊
1 - 0.04ρA = (1 + 0.005 t)-2 Finally
25 ρA = 25 - (E-5) (1+ 0.005t)2
at t = 200 sec 4-4
25 ρA = 25 - = 18.75 kg/m3 (1 + 0.005 × 200)2
Verify the solution
3 At t = 0, from (E-5); ρA = 0, as t → ∞, ρA = 25 kg/m
------The following example requires numerical integration.
Example 4.1-4. ------A gas storage tank with a floating roof receives a steady input of 540 m3/h of a natural gas. The rate of withdrawal of gas from the tank, Fw (m3/min), varies more or less randomly during the day and is recorded at 10-min intervals. At 8:00 A.M. one morning the volume of stored gas is 3.00x103 m3. The withdrawal rate data for the next 4 h for Fw are as follows: ------11.4 (at 8:00), 11.9, 12.1, 11.8, 11.5, 11.3, 11.4, 11.1, 10.6, 10.8, 10.4, 10.2 10.2, 9.8, 9.4, 9.5, 9.3, 9.4, 9.5, 9.3, 9.6, 9.6, 9.4, 9.9, 9.8 (at 12:00) ------The temperature and pressure of the inlet, stored, and outlet gases are equal and nearly constant throughout the given time period. Calculate the stored gas volume at noon, using Simpson’s rule to evaluate the integral. Solution ------Step #1: Define the system.
System: gas in the storage tank.
Fi , ρi
F = 540 m3/min i 60 ρ F , ρ W
Step #2: Find equation that contains V, the stored gas volume at any time.
The mass balance will contain V.
Step #3: Apply the gas mass balance around the system.
d(ρV ) 540 = ρiFi − ρFw = ρ − ρFw (E-1) dt 60
Since the temperature and pressure of the inlet, stored, and outlet gases are equal and nearly constant throughout the given time period, ρ = ρi, Eq. (E-1) becomes
4-5
dV = 9 − Fw (E-2) dt
Step #4: Specify the boundary condition for the differential equation.
At t = 0 (8:00 AM), V = 3.00x103 m3.
Step #5: Solve the resulting equation and verify the solution.
Integrate Eq. (E-2) to obtain
t 3 V = 3.00×10 + 9.00t - Fwdt (E-3) —0
t 240 at t = 240 min, Fwdt = Fwdt —0 —0
Simpson’s Rule - Odd number of equally spaced data points, h = interval between successive x values.
x2 h » ÿ f (x) dx ≈ … y1 + 4 ƒ yi + 2 ƒ yi + yn Ÿ —x 1 3 i=2,4,... i=3,5,... ⁄
240 Fwdt = Width×Average Height —0 1 = 240× [ 11.4+ 4 (11.9 + 11.8 + 11.3 + 11.1 + 10.8 + 10.2 6 ×12 + 9.8 + 9.5 + 9.4 + 9.3 + 9.6 + 9.9)
+ 2 (12.1 + 11.5 + 11.4 + 10.6 + 10.4 + 10.2
+ 9.4 + 9.3 + 9.5 + 9.6 + 9.4) + 9.8] 240 3 Fwdt = 2488 m —0
V = 3.00×103 + 9.00×240 - 2488 = 2672 m3
Verify the solution: at t = 0, from Eq. (E-3) V = 3.00×103 m3
4-6 Example 4.1-5. ------Water is flowing through a large circular conduit with inside radius R and a velocity profile given by the equation
» ≈ r ’2 ÿ v(fps) = 8 …1 − ∆ ÷ Ÿ … « R ◊ ⁄Ÿ Determine the mass flow rate through the pipe and the average water velocity in the 2.0 ft pipe.
r 6 ft 2 ft
Solution ------
Since v is a function of r, we first need to determine the mass flow rate through the differential area dA = 2πrdr
d m" = (ρv)(dA) = (ρv)( 2πrdr)
The mass flow rate through the area πR2 is then obtained by integrating over the area
2 R » ≈ r ’ ÿ m" = 2πρ — 8…1 − ∆ ÷ Ÿ rdr (E-1) 0 … « R ◊ ⁄Ÿ
r Let z = Ω dr = Rdz, equation (E-1) becomes R
1 ≈ 2 4 ’ 1 z z 2 2 2 ∆ − ÷ m" = 16πρR — (1− z ) zdz Ω m" = 16πρR 2 4 0 « ◊ 0
m" = 4πρR2 = 4π×62.4×32 = 7057 lb/s
The average velocity in the 2-ft pipe is
m" 7057 vave = = = 36 ft/s ρπR 2 62.4 ×π ×12
4-7 Water Treatment
Nature has been purifying (reclaiming) and recycling sewage for years using lakes, streams, and rivers. Dependable, clean water is crucial to maintaining the growth, economy and lifestyle of people living in urban areas. In arid Southern California where most of the water must be imported from Northern California, water reclamation is one of the methods used by the Sanitation Districts of Los Angeles County to meet the demand for water. In fact, the Districts’ water reclamation and recycling system is one of the largest in the world.
The San Jose Water Reclamation Plant at Whittier, California, receives wastewater from all of Los Angeles county and treats 100 million gallons of water per day. The main part of the plant is the primary, secondary, and tertiary refinement systems as shown in Figure 2.4-1.
Primary Secondary Tertiary
Chemical Air Chemical
addition compressor addition Chlorine SO2
Primary Aeration Final settling tank settling tank tank Chlorine contact Gravity tank Primary filter Influent solids Filter pumps Waste backwash activated recovery solid tank to joint water pollution control plant
Figure 4.1-1 Schematic of the San Jose Creek Water Reclamation Plant
The raw sewage enters the primary tanks with the addition of a polymer to help coagulate the solids. The major solids settle out and drain to an outlet to a joint water pollution control plant. The mixture then proceeds to the secondary refinement system where microorganisms are added to the slurry and air pumped to the bottom of the tank to aerate the system. The microbes break down the solids in the “activated slurry” with a residence time of about 4 to 6 hours. Most of the microorganisms are reused, and the slurry proceeds to the final settling tanks. The partially clarified water then passes through chlorinators and coagulant injectors and finally proceeds through a series of sand, gravel and coal filter, by gravity, to chlorine contact tanks. After the chlorine disinfects the water, sulfur dioxide is introduced to de- chlorinate the water before reuse. Water leaving the reclamation plants is carefully monitored to meet State and Federal drinking water standards.
4-8 Example 4.1-6.1 ------Figure E4.1-6 shows the schematic of a process for treating residential sewage. In this simplified process, sewage at a rate of 5000 gal/min is pumped into a well-mixed aeration tank where the concentration of bacteria CB,aration is maintained at 0.2 lb/gal. The treated sewage is then pumped to a settling tank where the bacterial is separated and recycled back to the aeration tank. The treated sewage leaving the settling tank has no bacteria in it while the recycle sewage contains a bacterial concentration of 1.0 lb/gal. Both the aeration and the settling tanks have the same volume of 5×106 gallons. You can assume the liquid (sewage) density remains constant throughout the process and neglect the mass loss due to the generation of CO2 leaving the aeration tank. 1) If 5000 gal/min of sewage enters and leaves the treatment facility, determine the two volumetric flow rates Qtreated and Qrecycle. 2) If the recycle pump fails so that the flow rate through the process pump is reduced to 5000 gal/min, determine the time it takes before the sewage treated by the process is unsafe to release. A minimum level of 0.1 lb/gal of bacteria in the aeration tank is necessary to assure safe levels of sewage in the discharge.
Air Sewage CO2 Qin Process pump Settling Qtreated Aeration tank tank Qout
Air Recycle pump
Qrecycle
Figure E4.1-6 A process for treating residential sewage.
Solution ------
1) Determine the two volumetric flow rates Qtreated and Qrecycle.
Let choose the system to be the aeration tank since we have the most information about this unit. Applying the conservation of mass to the aeration tank at steady-state yields
Qinρin + Qrecycleρrecycle = Qtreatedρtreated
Since the densities are equal, we have
5000 + Qrecycle = Qtreated (E-1)
A steady-state bacteria balance around the aeration tank yields
QinCB,in + QrecycleCB,recycle = QtreatedCB,treated
Qrecycle(1.0) = Qtreated(0.2)
1 Duncan and Reimer, Chemical Engineering Design and Analysis, 4-9
Substituting Qrecycle = 0.2Qtreated into equation (E-1) gives
5000 + 0.2Qtreated = Qtreated Ω Qtreated = 6250 gal/min
Qrecycle = 0.2Qtreated = 1250 gal/min
2) The time it takes before the sewage treated by the process is unsafe to release
Applying the conservation of bacteria to the aeration tank yields
d (VCB) = QinCB,in − QtreatedCB dt
In this equation, V is the volume of the aeration tank, and CB is the bacteria concentration inside the tank.
Since Qin = Qtreated, V is a constant, therefore,
dCB Qtreated = − CB dt V
V The ratio gives the average time a molecule resides in the tank and is called the Qtreated residence time of water in the tank
V Fluid volume = = residence time = τ (E-2) Qtreated Volumetric flow rate
Equation (E-2) can be written as
dC C B = − B dt τ
Separating the variables and integrating over the limits
0.1 dC 1 t B = − dt —0.2 —0 CB τ
Upon integrating and substituting the limits, we obtain
≈ 0.1’ t 5 ×106 ln ∆ ÷ = − Ω t = τ ln(2) = ln(2) = 693 min. « 0.2 ◊ τ 5000
4-10