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[Music].[Music].hello everyone and welcome to our next.lesson.in this video we will now use the.continuity and general energy balance.equations.to solve unsteady state or transient.problems.for you to practice you are highly.encouraged to solve the sample problems.together as i discuss.these solutions let's begin as discussed.previously.the energy balance for an open system is.shown here.we are using the version without.expansion work that's why we are using.enthalpy.in the second term of the equation and.the work term.in the right hand side refers to shaft.work.this energy balance can be used in.conjunction with the mass balance.shown here to complete our solution for.unsteady state processes.the first term of both equations stays.and must be integrated to yield the.internal energy or mass.as a function of time let's consider.some common cases.first we have transient single flow.systems.in these types of systems a storage.vessel is either being filled.or being emptied therefore we have one.flow stream.either positive that's going in or.negative.going out for a tank being filled let us.simplify the continuity equation.we assume that our tank is the control.volume.from here on it is important that you.know how to recognize and draw.the initial and final states of the.system.our state 1 here is the initial states.it is at time.t1 and initial mass m1 our final state.is state 2.with time t2 and final mass m2.the delta mass flow rate in the second.term of the equation.is output minus input such that if we.transpose this to the other side of the.equation.this becomes input minus output.since we only have an input our right.hand side of the equation becomes.the mass flow rate of the input stream.separating the variables we now have an.integrable equation.we are integrating with respect to time.because this is dt.so the limits of integration are values.of time this is t1 to t2.i have here illustrated states 1 and.state 2..remember that state 1 is at a time equal.to t1 and state 2 is at a time.equal to t2 integrating.dm becomes m or mass.and applying the limits of integration.we have m2 minus.m1 or the mass at state 2 minus the mass.at state 1..on the right hand side assuming that the.input.mass flow rate is constant we can move.this out of the integral and we now have.the integral of dt.integrating this becomes t or time and.applying the limits of integration this.becomes t2 minus t1 or.delta t delta t is the time elapsed from.state 1 to state 2..finally the input mass flow rate.multiplied by delta t.yields the mass of the substance that.was added inside the tank.we now arrive at our final mass balance.equation.that is the final mass of the contents.of the tank that is.m2 is equal to the mass added.to the tank plus the initial mass of the.contents of the tank.this should make sense because the final.mass of the system is simply the sum.of the initial mass and the added mass.similarly for a tank being emptied or.drained.the continuity equation simplifies to dm.over the control volume over dt is equal.to the negative.mass flow rate of the output stream.following the same steps we did earlier.namely separation of variables and then.integrating.we arrive at the mass balance equation.m2 is equal to m1.minus m out this also makes sense.because the final mass of the substance.inside the tank.is equal to the initial mass minus the.mass.taken out of the system now let's go.back to the case of a tank.being filled but now we consider the.energy balance.this is our energy balance equation now.please keep in mind that the small.letter u on the first term of the.equation refers to the specific.internal energy while the small u.here at the middle that is not.italicized refers to the velocity of the.stream.okay that disambiguation is very.important because we are using small.letters to denote specific properties.and we are using capital letters to.denote absolute properties.and it is important that we disambiguate.between internal energy and velocity.assuming an insulated tank that is our.rate of heat transfer is equal to zero.assuming no shaft work that is shaft.perk is equal to zero.and assuming negligible kinetic and.potential energies.we can cancel this term here as well as.this term here.representing the kinetic and potential.energies respectively.and of course there is no heat transfer.and there is no shaft work.our energy balance becomes the rate of.change of the internal energy.is equal to the inlet mass flow rate.multiplied to the specific enthalpy.of the inlet stream or the input stream.okay now separating the variables again.and then integrating.we will arrive with our simplified.energy balance equation.that is m2 times u2 minus m1 times u1.equals mn times h in this is the final.mass.multiplied to the final specific.internal energy.minus the initial mass times the initial.internal energy.is equal to the mass of the added.substance.or the mass of the inputs multiplied by.the specific enthalpy.of the input and since when we multiply.a specific property.to the mass we get the absolute property.so simplifying further we have.the final internal energy is equal to.the initial internal energy.plus the enthalpy of the input this is.our simplified energy balance equation.now we can use this equation together.with a simplified mass balance equation.to then solve for the unknown variables.in the problem.let us demonstrate that with an example.steam.at 1.4 megapascals and 300 degree.celsius.is flowing in a pipe connected to a.valve and into an evacuated tank.the valve is opened and steam is allowed.to flow through the tank.until the pressure inside the tank.reaches 1.4 megapascals.then the valve is closed the process is.adiabatic.and neglect the kinetic end potential.energies.that are mean the final temperature of.the steam inside.the tank.here are the given of the problem so i.have here illustrated state one and.state two.in state one our initial pressure is.zero because.the problem states that it was initially.an evacuated tank.therefore it has no contents inside okay.by that logic the initial mass of the.system is equal to.zero for state two it was stated that.the valve will be closed once the.pressure inside reaches 1.4 megapascal.and we are to solve for the final.temperature of the system.for our input stream labeled here as.mass flow rate.input its pressure is 1.4 megapascals.and its temperature is 300 degrees.celsius.for our assumptions it was stated that.the system is adiabatic therefore q is.equal to zero.there is no shaft work therefore ws is.equal to zero.and we are assuming that kinetic and.potential energies are negligible.now let us go through the continuity.equation and then the general energy.balance equation.so that we can simplify the problem.let's start with the continuity.equation for the continuity equation.that is.dm over the control volume over dt.is equal to negative delta.mass flow rates of the flow streams now.let us represent our tank.as our control volume and since we only.have one flow stream and that is our.input stream.and input streams are negative on the.right hand side of the equation.our continuity equation becomes d m.control volume over dt is equal to.mass flow rates in now as we have done.earlier we can integrate this expression.after the separation of variables.with our limits of integration from.state 1 to state 2.we will be left with final mass minus.initial mass is equal to the mass.of the input rearranging we have.the final mass is equal to initial mass.plus the mass of the input this is now.our simplified.mass balance so for now we will set this.aside and then we will simplify our.general energy balance equation.for our first law of thermodynamics.expression or the energy balance.expression.we retain the first term that is d m.u cv over dt.plus delta of mass flow rate multiplied.by.the specific enthalpy for the flow.streams is equal to.zero now this is already the simplified.version because there's no kinetic.energy.there's no potential energy and there.are no heat and work.terms transposing the delta flow streams.to the other side we now have.d mu cv.over dt is equal to negative delta.mass flow rates specific enthalpy of the.flow streams.again we only have one flow stream that.is our input stream.so our equation becomes d mu.cv over dt is equal to.the mass flow rate of the inlet stream.multiplied by the enthalpy.of the inlet stream that is specific.enthalpy.separating the variables and then.integrating with limits of integrations.from state 1 to state 2.our energy balance now becomes m2.u2 minus m1 u1.is equal to mass inputs multiplied by.the.specific enthalpy input this is our.simplified.first law of thermodynamics expression.going back to the problem.we are looking for the final temperature.of the system or.t2 now if you take a look at state 2 we.have.one defined property and that is the.final pressure.for us to be able to completely define a.states of a system we need at least.two independent properties pressure is.one of them.and from our equations here we can also.determine the internal energy.of the final state or state two of the.system.therefore from our energy balance we.simply have to solve for the expression.of u2 and then substitute the continuity.equation.and then solve for the other parameters.before we do so.we can now cancel the terms for m1 in.both the continuity and the energy.balance equations because initially our.tank was evacuated.so therefore m1 is zero m1 is.zero solving for u2 we now have.u2 is equal to m in.h in over m2.from our continuity equation we now have.learned that m2.is equal to m in therefore our u2.therefore our u2 simplifies to just.becoming.the enthalpy input for our simplified.equation.we are saying that the final internal.energy of the system.is equal to the enthalpy of the inlet.stream or more specifically saying the.final specific.internal energy of the system is equal.to the specific enthalpy.of the inlet stream if you take a look.at our inlet stream.we have two independent properties that.is pressure and temperature.we can determine the specific enthalpy.from those values.all we have to do is to take a look at.the steam tables to determine.h in for that let us take a look at some.steam tables.here we have the steam tables from the.book fundamentals of thermodynamics by.sontag.and we are currently looking at the.properties for saturated water.so our inlet stream is at 1.4.megapascals or 1400 kilopascals.and the temperature of 300 degrees.celsius if we take a look at the.pressure of 1400 kilopascals.the saturation temperature is only.195.07 degrees celsius.since our given temperature 300 degree.celsius.is greater than the saturation.temperature we therefore conclude that.our given.is not saturated and it is a superheated.steam so let us look for the entries for.superheated steam.here we have superheated vapor water let.us look for pressure.1400 kilopascals here we have the.pressure 1400 kilopascals.and then at a temperature of 300 degrees.celsius we can immediately determine.the specific enthalpy and that is.3040.35.kilojoules per kilogram by the.determination of that specific enthalpy.this is also equal to the final specific.internal energy of our system let's go.back to our board.again since we have already identified.the enthalpy of the inlet that is.equivalent to.the final specific internal energy of.the system.now for state 2 we now have two.independent properties that is.that the final pressure is 1.4.megapascals.and that the final internal energy is.3040.35 kilojoules per kilogram.now the only thing left for us to do is.to go back to the steam tables.and determine the temperature at state 2.okay remember the values for the final.pressure and the final.internal energy let's take a look at.this superheated table since we are.already here.at the final pressure of 1400.kilopascals.our internal energy is 3040.35.3040.35 lies between 2952.5 and 3121.10.therefore we have to interpolate between.these two values of the internal energy.and these two values of the temperature.interpolating that is 3040.35.minus 2900.divided by the difference between the.two table values that's three one.two one point one minus.two nine five two 0.5.we then multiply this to the difference.between 500.and 400 and then add this to 400 degrees.celsius our final temperature.is 452 degrees celsius.and that is how we perform a first law.of thermodynamics analysis.to solve a transient problem just.remember some of the concepts that we.have discussed in this problem.such as we need two independent.properties from each stream or each.state in order to fully describe it.and we need to combine the simplified.continuity equation and first law of.thermodynamics energy balance.in order for us to have our final.working equation.just keep those in mind and we won't.have any problems in our future examples.let's go back let us now consider the.case of transients multiple input.and output streams unlike the previous.case.we are now looking at more than one flow.stream it could either be multiple.inputs.multiple outputs or multiple inputs and.outputs.the most common type of this is you have.one inputs and one.output stream our continuity equation.simply involves.summation terms and summarizes all flow.streams into either inputs or.outputs the final form is the final mass.minus the initial mass is equal to the.summation of all the input masses.minus the summation of all the output.masses.so basically it's the same as what we.have previously discussed but we are now.including summation terms.we apply the same simplification to the.energy balance equation.so for the simple case of no heat and.work transfer.our energy balance becomes the final.internal energy.minus the initial internal energy is.equal to.the summation of the total energies of.the input streams.minus the summation of the total.energies of the output streams.the approach of solving such kinds of.problems are the same.establish first the material balance.from the continuity equation.next establish the energy balance from.the first slow analysis.and then combine the two equations then.look for fluid properties to solve for.the unknown.let's try to solve this example a water.heater tank.receives 0.2 kilograms per second of.cold water at 15 degrees celsius it.delivers.warm water at 50 degrees celsius at 0.2.kilograms per second there is.30 liters of water pulled inside the.heater.since the inlet and outlet flow rates.are the same.the water inside the tank stays at 30.liters.if the electricity suddenly goes out and.the inlet and outlet streams are still.constantly flowing.determine the time it takes before the.outlet temperature drops to 25 degrees.celsius assume a well-insulated tank.and neglect kinetic and potential.energies.here are the given to our problem so the.scenario is we have a water heater.that is being fed with cold water at 15.degrees celsius.and you are obtaining hot water at 50.degrees.celsius that is when there is.electricity because the electricity is.what is heating the water.but when the electricity goes out we now.enter our state.2. that is our input is still continuous.at a temperature of 15 degrees celsius.but our output now has decreased to 25.degrees.celsius what we are now looking for is.how long.will this process take how long will it.take before my outlets.temperature will become 25 degrees.celsius.okay so for a simple analysis of the.problem.we have our tank that has an equal.amount of flow rate input and flow rate.output.therefore the volume of water inside the.tank would remain the same and that is.30 liters let us assume that the density.of water.remains constant at 1 000 kilograms per.cubic meter or 1 kilogram per liter.therefore we can say that our m1 is.equal to.m2 and that is 30 kilograms.let us now define our control volume and.that.is the entire tank itself.this is our cv next we have the.continuity equation.we write dm control volume.over dt is equal to negative.delta mass flow rate of the flow streams.our negative delta mass flow rate of the.flow streams is simply.outputs minus input we now have d.m cv over dt is equal to.mass flow rate's output minus mass flow.rate.input since we have established that the.mass flow rate input and output are the.same.this are both cancelled and we are left.with dm.cv over dt is equal to zero.therefore we are saying that the mass.inside.remains constant this simplifies to m1.is equal to m2 which we have already.stated.earlier that's because the flow rate of.the input is equal to the flow rate of.the output.okay now we can go to the energy balance.for our first law analysis we have the.mu.under the control volume over dt is.equal to.negative delta mass flow rates.multiplied by the enthalpies.of the flow streams kinetic energy.potential energy.heat and work are all zero therefore.this is written as.d mu cv.over dt is equal to mass flow rate.in enthalpy in minus.mass flow rate out enthalpy.out we only have one inlet stream and.one outlet stream.now since the mass flow rates are.constant we can factor out our mass flow.rate.this becomes d m u cv.over dt is equal to mass flow rate which.is 0.2 kilograms per second.multiplied by h inlet minus h.outlet this is our simplified first law.of thermodynamics equation.by this point before we integrate we.have to define our internal energy.and our enthalpy let us start with the.internal energy.our internal energy is defined as cv.times delta t.wherein the delta t will be the.temperature of the tank.or the temperature of the water inside.the system minus a reference.temperature we can set the value of the.reference temperature later.now since our internal energy term is.inside of the differential of time.we need to get du over dt for this.equation cv is constant.therefore we have cv multiplied by dt.over dt the numerator dt is the.differential in temperature.inside of the tank and the denominator.is the differential.in time the second term dt ref over dt.is equal to zero.because our reference temperature is a.constant its differential.is zero we can then substitute this cv.dt over dt to our equation later.next the enthalpy by the same token is.defined as cp.times delta t this becomes t in.minus t ref this is our.enthalpy inputs our enthalpy output can.be defined.the same way that this enthalpy out is.equal to cp.times temperature out minus a reference.temperature under this circumstances our.enthalpy inlet is constant that is.because.the mass flow rate of the inlet is.constant and the temperature of the.inlet.is constant at 15 degrees celsius the.only temperatures that change.are the temperature inside of the tank.that is your t.and the temperature of the outlet stream.that is t.out therefore for the simplicity of this.problem.let us let the value of t reference.as 15 degrees celsius what this allows.us to do.is to cancel the value of the enthalpy.inlets because t in is 15 degrees.celsius.t reference is also 15 degrees celsius.that gives us an inlet.enthalpy of zero and that simplifies our.equation.substituting the definitions of internal.energy and.enthalpy we now have m cv.times the temperature over d time.is equal to negative mass flow rate.times cp times p out minus.e reference we now have a differential.equation that is a function of.temperature.and time which is what we want because.we are solving for the time it takes for.the temperature to drop.from 50 to 25 degrees celsius we have.another problem here and that is.the temperature in our differential is.the temperature.inside the tank while our temperature.term outside of the differential.is the temperature of the outlet stream.that is t.out one way on how we can reconcile this.is.if we assume that there is perfect.mixing.inside of the tank such that the.temperature.of the water inside of the tank is equal.to the temperature of the water on the.outlet stream.that is a very reasonable assumption.because the outlet stream.also comes from the water from the.inside of the tank.okay therefore we can now write this as.m.cv dt out over dt.is equal to negative mass flow rate.times cp.times t out minus t reference another.simplifying case.we can apply is that for liquids we can.say that.the heat capacity at constant pressure.is almost equal to the heat capacity at.constant volume.because liquids are not as affected as.pressure compared to.gases so this is a reasonable assumption.what that allows us.is we can now cancel the heat capacity.terms.we are now left with mass times the.temperature out over the time.is equal to negative mass flow rate.times temperature out minus temperature.reference separating the variables we.have.the temperature out over temperature.out minus temperature reference is equal.to.negative mass flow rate over mass.d d finally we can integrate this.expression.our limits of integration are from state.1 to state.2. for the outlet temperature that would.be the outlet temperature.at state 1 to the outlet temperature at.state 2. so that is from 50 degrees.celsius to 25 degrees celsius.now for time we are going to assume that.at state 1.time is equal to 0 and at state 2.time is equal to t and t is the time.needed for the temperature to go down.from 50 to 25 degrees.celsius integrating we now have the.natural logarithm.of t out minus t reference.is equal to negative mass flow rate over.mass.this are both constants times t.for the right hand side of the equation.the limits are t.out one to t.out two for the right hand side of the.equation the limits are from 0.to t substituting and solving for time.we now have our final expression.time is equal to negative.mass over mass flow rates multiplied by.the natural logarithm of t.out to minus t reference.over t out one.minus t reference this is now our final.expression.for determining the amount of time.needed to bring down the temperature of.the outlets from 50 to 25 degrees.celsius all that's left now is to.substitute all the given into our.working equation.that is negative mass is 30 kilograms.mass flow rate is 0.2 kilograms per.second.multiplied by the natural logarithm of t.out to that is the outlet temperature at.the second state.that's 25 degrees celsius minus the.reference temperature of 15 degrees.celsius.divided by outlet temperature of the.first state or the initial state.50 degrees celsius minus the reference.temperature 15 degrees.celsius our time is 188.seconds or in minutes that is.3.13 minutes.our interpretation is if you are taking.a warm shower.and the power suddenly stops you have.3.13 minutes.for the water to go from 50 degrees.celsius which is a little bit warm to 25.degrees celsius which could be a little.cold.so you might want to end your shower.early based on our computations.now you see the value of thermodynamics.it lets you know when your water will be.too cold for you to shower.that is it for this lesson i hope you.have learned something.after this lesson we all should know how.to use the first law of thermodynamics.equation.for unsteady state systems please do.watch out for separate videos detailing.the solutions to more.problems thank you for listening and as.always keep safe.[Music].[Music].[Music].you.

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