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Chapter 4 Mass and Energy Balances

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Chapter 4 Mass and Energy Balances 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. 4.1 Conservation of Mass 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Balance on an interest earning checking account during month of February Withdrawn = $525.00 Deposit = $1000.00 Interest = $3.00 Fees = $5.00 Solution ------------------------------------------------------------------------------------------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 v(fps) = 8 1 − ∆ ÷ « R ◊ 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.
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