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additions of water to the control volume withdrawals of water from the control volume Since no water is either added or withdrawn, the change in this holdup is zero.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