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