3) Dynamic processes This is the general case in which both the holdups within equipment as well as the flow rates and compositions of the input and output streams can vary with time. This type of mass balance is discussed in more detail in Section XI Examples a. Discrete process Let us now consider the application of the principle of the conservation of mass to a discrete process. For each of the conserved species, it can be stated as follows I-1) Change in holdup= additions to the control volume withdrawals from the control volume Consider the following example We have a tank that initially holds 100 Kg of a solution containing 40% by weight of salt in water (1) We add 20 Kg of salt to the tank and allow it to dissolve What do we now have in the tank? First we have to identify the conserved species. Since there are no chemical reactions involved, both salt and water are conserved species. Next we have to define the control volume It seems natural to choose the salt solution in the tank. our basis is the amount of solution originally contained in the tank Now we can define a material balance for each conserved species as follows: Water Initial holdup of water=(1-0.4 (100)=60 Kg Change in holdup of water additions of water to the control volume withdrawals of water from the control volume Since no water is either added or withdrawn, the change in this holdup is zero. Therefore the holdup of water after the salt addition is still 60 K: Initial holdup of salt=(0.4)(100)=40 Kg
-8- 3) Dynamic processes. This is the general case in which both the holdups within equipment as well as the flow rates and compositions of the input and output streams can vary with time. This type of mass balance is discussed in more detail in Section XI. Examples a. Discrete process Let us now consider the application of the principle of the conservation of mass to a discrete process. For each of the conserved species, it can be stated as follows: (I-1) Change in holdup = additions to the control volume - withdrawals from the control volume Consider the following example: We have a tank that initially holds 100 Kg of a solution containing 40% by weight of salt in water. (1) We add 20 Kg of salt to the tank and allow it to dissolve. What do we now have in the tank? First we have to identify the conserved species. Since there are no chemical reactions involved, both salt and water are conserved species. Next we have to define the control volume. It seems natural to choose the salt solution in the tank. Our basis is the amount of solution originally contained in the tank. Now we can define a material balance for each conserved species as follows: Water Initial holdup of water = (1 - 0.4)(100) = 60 Kg Change in holdup of water = additions of water to the control volume - withdrawals of water from the control volume Since no water is either added or withdrawn, the change in this holdup is zero. Therefore the holdup of water after the salt addition is still 60 Kg. Salt Initial holdup of salt = (0.4)(100) = 40 Kg
Change in holdup of salt= additions of salt to the control volume withdrawals of salt from the control volume 'e add 20 Kg of salt and withdraw no salt. Therefore the change in the holdup of salt =+20 Kg and the holdup of salt after the addition is 40+ 20=60 Kg. A simple calculation shows that the concentration of salt in the tank is now 50 weight % (2)We withdraw 40 Kg of the solution now in the tank Since the solution in the tank is 50 weight salt and 50 weight water, in withdrawing 40 Kg of solution, we will withdraw 20 Kg of salt and 20 Kg of water. This will leave 60-20 40 Kg of each component in the tank. The composition has not changed from Step 1 b. Continuous steady-state process Let us consider a continuous mixer which has two input streams and, of course, one output stream(See Figure 6 for a diagram). Suppose the first input stream has a flow rate of 10000 lb/hr of a 40 wt. solution of salt in water while the second input stream has a flow rate of 20000 lb/hr of a 70 wt %solution of salt in water. What is the flow rate and composition of the output stream? Since the system is now characterized in terms of rates of flow into the control volume (additions) and rates of flow out of the system(withdrawals), we need to restate the principle of conservation of mass as follows: (1-2) Rate of change of holdup rate of additions to the control volume rate of withdrawals from the control volume For a continuous system operating in the steady state, the holdup does not change with time Therefore, the rate of change of holdup is zero and Eqn. 1-2 becomes -3) Rate of withdrawals from the control volume Rate of additions to the control volume Let us apply this to the mixer problem. The control vol the contents of the mixer(er though these do change) and the basis is the total rate of flow to the mixer. As in the previous example, the conserved species are salt and water
-9- Change in holdup of salt = additions of salt to the control volume - withdrawals of salt from the control volume We add 20 Kg of salt and withdraw no salt. Therefore the change in the holdup of salt = +20 Kg and the holdup of salt after the addition is 40 + 20 = 60 Kg. A simple calculation shows that the concentration of salt in the tank is now 50 weight %. (2) We withdraw 40 Kg of the solution now in the tank. Since the solution in the tank is 50 weight % salt and 50 weight % water, in withdrawing 40 Kg of solution, we will withdraw 20 Kg of salt and 20 Kg of water. This will leave 60 - 20 = 40 Kg of each component in the tank. The composition has not changed from Step 1. b. Continuous steady-state process Let us consider a continuous mixer which has two input streams and, of course, one output stream (See Figure 6 for a diagram). Suppose the first input stream has a flow rate of 10000 lb/hr of a 40 wt. % solution of salt in water while the second input stream has a flow rate of 20000 lb/hr of a 70 wt. % solution of salt in water. What is the flow rate and composition of the output stream? Since the system is now characterized in terms of rates of flow into the control volume (additions) and rates of flow out of the system (withdrawals), we need to restate the principle of conservation of mass as follows: (I-2) Rate of change of holdup = rate of additions to the control volume - rate of withdrawals from the control volume For a continuous system operating in the steady state, the holdup does not change with time. Therefore, the rate of change of holdup is zero and Eqn. I-2 becomes (I-3) Rate of withdrawals from the control volume = Rate of additions to the control volume Let us apply this to the mixer problem. The control volume is the contents of the mixer (even though these do change) and the basis is the total rate of flow to the mixer. As in the previous example, the conserved species are salt and water. Salt
Rate of withdrawal of salt rate of additions of salt to the mixer Rate of additions=(10000)0.4)+(20000)(0.7) 1 8000 lb/hr of salt Water Rate of withdrawal of water rate of additions of water to the mixer Rate of additions=(10000(0.6)+(20000(03) 12000lb/hr Thus the stream leaving the mixer has a flow rate of 18000 lb/hr of salt and 12000 lb/hr of water, for a total of 30000 lb/hr. This is exactly the total flow rate of the mixer output we get by adding up the total flow rates to the mixer. Also, the composition of salt of the stream leaving the mixer (100(1800030000)=60wt% Dynamic Process Consider the surge tank shown in Figure I-1 ater flows into the tank with a flow rate Fin lb/hr. It flows out at a rate fout lb/hr. The flow rates in and out can be adjusted by means of the valves in the inlet and outlet piping. The tank has the form of an upright cylinder that has a cross section area of s ft The liquid level in the tank is z ft We know from experience that if the flow rates in and out are not exactly equal, the level in the tank will change with time If the inlet flow rate exceeds the outlet flow rate. then the level will rise and vice versa. Now, the purpose of a surge tank is to absorb changes in the inlet flow rate while maintaining a relatively constant outlet flow rate CThe reservoirs that supply water to a town or city are surge tanks where the inlet flow is the run -off from rainstorms and the outlet flow Figure 1. Surge Tank is the daily consumption by the town or city. )Thus, a question that designers of surge tanks must ask is, given an estimate of the variations of inlet and outlet flow rates as functions of time, how
-10- F in F out z Rate of withdrawal of salt = rate of additions of salt to the mixer Rate of additions = (10000)(0.4) + (20000)(0.7) = 18000 lb/hr of salt Water Rate of withdrawal of water = rate of additions of water to the mixer Rate of additions = (10000)(0.6) + (20000)(0.3) = 12000 lb/hr Thus the stream leaving the mixer has a flow rate of 18000 lb/hr of salt and 12000 lb/hr of water, for a total of 30000 lb/hr. This is exactly the total flow rate of the mixer output we get by adding up the total flow rates to the mixer. Also, the composition of salt of the stream leaving the mixer = (100)(18000)/(30000) = 60 wt %. c. Dynamic Process Consider the surge tank shown in Figure I-1. Water flows into the tank with a flow rate Fin lb/hr. It flows out at a rate Fout lb/hr. The flow rates in and out can be adjusted by means of the valves in the inlet and outlet piping. The tank has the form of an upright cylinder that has a cross section area of S ft2 . The liquid level in the tank is z ft. We know from experience that if the flow rates in and out are not exactly equal, the level in the tank will change with time. If the inlet flow rate exceeds the outlet flow rate, then the level will rise and vice versa. Now, the purpose of a surge tank is to absorb changes in the inlet flow rate while maintaining a relatively constant outlet flow rate. (The reservoirs that supply water to a town or city are surge tanks where the inlet flow is the run-off from rainstorms and the outlet flow Figure I- 1. Surge Tank is the daily consumption by the town or city.) Thus, a question that designers of surge tanks must ask is, given an estimate of the variations of inlet and outlet flow rates as functions of time, how
big must the surge tank be so that it never runs dry(town loses its water supply)or never overflows(area surrounding the reservoir is flooded ) In general this is a complex design problem but let us look at a simple example to at least illustrate the concept Suppose that under normal conditions the level in the tank is to be half the height of the tank. If the flow rate into the tank becomes zero for a period of time(no rain), how long will it take for the tank to run dry if the outlet flow rate is maintained at its usual value? Specifically, suppose that the cross section area of the tank is 10 ft and its height is 10 ft and the normal outlet flow rate is 12 480 lb/hr First, let us take the volume of liquid in the tank as the control volume. The holdup of water in the control volume will be Holdup= Szp, where p is the density of water( 62.4 Ib/ft) Consider an interval of time At. Suppose that over that time interval the inlet and outlet flow rates are constant but not necessarily equal. Then, by the conservation of mass the change in the holdup will be given by (I-1) Holdup-t- Holdup==Fn△t-Fou△t Now, if we divide both sides of the mass balance equation(Eqn. l-1) by At and take the limit as At->0, we get the differential form of the mass balance, to wit. (-2) d szp/dt= Fin-Fout If we assume that is constant (a reasonable assumption if the temperature is also reasonably constant), then our mass balance equation becomes (-3) dz/dt=(Fin- Fout /S For our problem Fin=0 and Fout=12, 480 lb/hr, both constant. We can calculate dz/dt, that is dz/dt=(0-12480(10)(624)=-20f/h Since the nominal level is 5.0 ft(half the tank height of 10ft), it will take 0. 25 hour or 15 minutes for the tank to run dry Further examples of the application of the principle of conservation of mass, particularly for reacting systems, will be found in the subsequent sections of these notes l1-
-11- big must the surge tank be so that it never runs dry (town loses its water supply) or never overflows (area surrounding the reservoir is flooded). In general this is a complex design problem but let us look at a simple example to at least illustrate the concept. Suppose that under normal conditions the level in the tank is to be half the height of the tank. If the flow rate into the tank becomes zero for a period of time (no rain), how long will it take for the tank to run dry if the outlet flow rate is maintained at its usual value? Specifically, suppose that the cross section area of the tank is 10 ft2 and its height is 10 ft and the normal outlet flow rate is 12,480 lb/hr. First, let us take the volume of liquid in the tank as the control volume. The holdup of water in the control volume will be Holdup = Szr, where r is the density of water (62.4 lb/ft3 ). Consider an interval of time Dt. Suppose that over that time interval the inlet and outlet flow rates are constant but not necessarily equal. Then, by the conservation of mass the change in the holdup will be given by (I-1) Holdup|t= t - Holdup|t=0 = Fin Dt - Fout Dt Now, if we divide both sides of the mass balance equation (Eqn. I-1) by Dt and take the limit as Dt -> 0, we get the differential form of the mass balance, to wit, (I-2) d[Szr]/dt = Fin - Fout If we assume that is constant (a reasonable assumption if the temperature is also reasonably constant), then our mass balance equation becomes (I-3) dz/dt = (Fin - Fout)/Sr For our problem Fin = 0 and Fout = 12,480 lb/hr, both constant. We can calculate dz/dt, that is dz/dt = (0 - 12480)/(10)(62.4) = -20 ft/hr. Since the nominal level is 5.0 ft (half the tank height of 10ft), it will take 0.25 hour or 15 minutes for the tank to run dry. Further examples of the application of the principle of conservation of mass, particularly for reacting systems, will be found in the subsequent sections of these notes
I. PROCESSES Revised October 2. 1999 The Concept of a process
-12- II. PROCESSES Revised October 2, 1999 A. The Concept of a Process