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## Acid Neutralization Reactor

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**Acid Neutralization Reactor**Module 4: Acid neutralization reactor Lecture 2: Analysis of the feed tank and the reactor for the case of no reaction Mark J. McCready Chemical Engineering**Outline for today**• Quick review of mass balance equations • Analysis of the reactor for 2 feeds but no reaction • Expectation of a Steady State ... • Analysis of the feed tank that is draining by gravity • How does the depth of liquid affect the flow rate? • Bernoulli equation to relate effects of gravity, pressure and velocity within a fluid**Reactor, what will the exit concentration be?**Acid in, 1 Control volume Base in, 2 3 Flow out, 3**Feed Tank, how fast does it drain?**We will use this control volume for the tank h**Start of new material**• First we will analyze the reactor using the mass balance equations. • Today, there will be no reaction. • But, we will allow for inlet streams of different concentration. • We will see that if the inlet concentrations and flows are constant, a steady - state is expected where there is no change in the concentration with time in the tank.**Today, we will not do reaction,just use the tank as a mixer**Control volume Salt solution in, 1 Another salt solution in, 2 3 Flow out, 3**Recall Mass Balance**• General mass balance equation for a fixed control volume • Rate of Accumulation = • Rate In - Rate Out + Production by reaction- • Consumption by reaction • Overall • Component mass (mole) balance • r j- density of stream j, (mass/length3) • qj -- volumetric flow rate of stream j, (length3 /time) • V -- active volume of reactor,(length3) • cji-- molar concentration of species i in stream j, (moles/length3) • ri -- molar reaction rate per volume (moles/(length3 -time))**Today, we will not do reaction,just use the tank as a mixer**Salt solution in, 1 Control volume Another salt solution in, 2 Flow out, 3**Mass Balance**1 A sketch of our problem looks like: 2 3 • We see two inputs and one output • Overall mass balance • Component mass balance, for salt • qj -- volumetric flow rate of stream j (m3/s) • V -- active volume of reactor (m3) • cjsalt-- molar concentration of salt in stream j (moles/m3) 3 The reaction term is 0!!**Simplifications**• Flowrates in are not changing in time • Reactor is filled at the beginning • Thus, overall mass balance tells us nothing we don’t find obvious. • What about the salt balance? We expect that it will tell us what comes out, if we know what goes in.**Balance equation**• The salt balance equation# • Can be solved to give… • You can solve this equation by numerical integration #A green background slide means that we don’t expect you to get the answer, because we used mathematics you may not yet understand. But, the answer will be insightful.**Plot of the concentration**Initial concentration =0 • We see that there is an initial transient (exponential) that depends on reactor volume and then a steady state is reached after which there is no further time variation. (If the inlets remain constant!) • Steady state answer: Note different volumes and abscissa scales Initial concentration =0**Steady state concentration**• For this example we have • q1 = 10 m3/s, c1 = 2 moles/ m3 • q2 = 5 m3/s, c2 = 3 moles/m3 • Thus: • q3 = (10 + 5)= 15 m3/s Note different volumes and abscissa scales**Steady state behavior?**• Is there always a steady state if we have steady inputs to a reactor? • Maybe this is obvious ?? • Should we have even bothered to integrate? • Think of some examples….**Feed Tank, how fast does it drain?**Now let’s examine a feed tank We need a new control volume This tank has an exit stream, but no inlet streams. h**Draining tank**A sketch of our problem looks like: Control valve • We have just one output • Overall mass balance: since we can use the chain rule to get But…, how do we get u3? h u3,A3 area of exit pipe, velocity of fluid leaving in stream 3. we know the flowrate and velocity are related by thus**Draining tank**Factors that affect exit liquid flowrate • ATank --not really • h-- yup! • A3-- yes, consider a 4”pipe versus a hypodermic needle • How open the valve is (as denoted by K) • g, gravity -- well of course • can’t drain a tank on the space station with gravity! Control valve, K h h=0 u3,A3 area and velocity**Pressure-depth relation**• Common occurrence in the summer • Basic equation of hydrostatics: Wow,my ears hurt r =density g=gravitation constant P=pressure h=depth of liquid**Effect of depth**• So we expect that if the depth is greater, the flow rate will be faster • Can we quantify this? • Recall from Physics, • Consider conversion ofpotential to kinetic energy for a fluid blob. • First we take the case of no “friction” or drag**KE--PE relation,we get velocity**• Consider a blob of fluid in our tank. It will follow the path shown with no friction Control valve, K h a m g h = PE h=0 u3,A3 area and velocity h DKE+ DPE = 0 1/2 m (ub2 -ua2 )+mg (hb-ha) = 0 ub2 = 2 g Dh h=0 b KE = 1/2 m u2**Draining tank with control valve**• We see that the velocity will not depend on the area of exit pipe. • Now for the real system we have a control valve that can open and close, the easiest way to deal with this is to consider that it causes a “loss” of energy. • DKE + DPE + “losses” = 0 • 1/2 m (ub2 -ua2 )+mg(hb-ha) + K/2 ub2 = 0 • (1+K) ub2 = 2 g Dh As K increases, velocity decreases. As the valve is closed, K increases!**WHITE BOARD STUFF**• Notre Dame law of wind direction • How momentum of fluid is converted to an increase in pressure as it impinges on a wall? • Student--University paradox • How the pressure must increase if the fluid is to be slowed down. • Work--Energy Principle from Physics • Bernoulli Equation**Bernoulli equation**• The relation between the pressure, the velocity, the change in height and frictional losses: • For our draining tank, there is no pressure change, and the relation between u and h is • Now we can go back to the mass balance and finish solving the problem**Draining tank**Recall the mass balance • You can solve this numerically to find how h changes in time. Control valve, K We use our relation, note that the “b” subscript is now “3” h h=0 u3,A3 area and velocity To get a final equation that can be solved ...**Draining tank**This one has a rather ugly analytical solution… h(t)= Here is a plot of some results K varies from 0 to 12 K=12 K=0**Filling/draining tank (for homework)**2 1 • What do the equations for this tank look like? • This last equation can be easily solved numerically to get height versus time. Substitute for the unknown flow rate and the liquid depth 3 3 Now use the Bernoulli equation**Recap (mixing tank)**1 2 3 • Component mass balance for “mixing tank” • The behavior is: Steady-state answer**Recap (draining tank)**Control valve, K • Overall mass balance for draining tank u3,A3 area and velocity K varies from 0 to 12**RecapBernoulli equation**• Bernoulli equation • Useful engineering equation to describe large-scale fluid flows. It relates changes in pressure, height and velocity and accounts for frictional losses.