I need help answering only the 2.10 questions with the equations shown/ worked out. I have provided the worked out example. Please don't copy old answers that have been posted, as they are wrong/ do not help. Thanks. 2.10 Example 2.1 plots responses for changes in input flows for the stirred tank blending system. Repeat part (b) and plot it. Next, relax the assumption that V is constant, and plot the response of x(t) and V(t) for the change in w_1 for t=0 to 15 minutes. Assume that w_2 and w remain constant.EXAMPLE 2.1 A stirred-tank blending process with a constant liquid holdup of 2 m^3 is used to blend two streams whose densities are both approximately 900 kg / m^3. The density does not change during mixing. (a) Assume that the process has been operating for a long period of time with flow rates of w_1=500 kg / min and w_2=200 kg / min, and feed compositions (mass fractions) of x_1=0.4 and x_2=0.75. What is the steadystate value of x ? (b) Suppose that w_1 changes suddenly from 500 to 400 kg / min and remains at the new value. Determine an expression for x(t) and plot it. (c) Repeat part (b) for the case where w_2 (instead of w_1 ) changes suddenly from 200 to 100 kg / min and remains there. (d) Repeat part (c) for the case where x_1 suddenly changes from 0.4 to 0.6 (in addition to the change in w_2 ). (e) For parts (b) through (d), plot the normalized response x_N(t), \ x_N(t)=x(t)-x(0)/x(\\\\infty )-x(0)\ where x(0) is the initial steady-state value of x(t) and x(\\\\infty ) represents the final steady-state value, which is different for each part. SOLUTION (a) Denote the initial steady-state conditions by x, w, and so on. For the initial steady state, Eqs. 2-4 and 2-5 are applicable. Solve Eq. 2-5 for x : \ x=w_1x_1+w_2x_2/w=(500)(0.4)+(200)(0.75)/700=0.5\ (b) The component balance in Eq. 2-3 can be rearranged (for constant V and \\\ ho ) as \ [ \\\\tau d x/d t+x=w_1 x_1+w_2 x_2/w; x(0)=x=0.5 ]\ where \\\\tau V / w. In each of the three parts, (b)-(d), \\\\tau =3 min and the right side of Eq. 2-19 is constant for this example. Thus, Eq. 2-19 can be written as \ [ 3 d x/d t+x=C^*; x(0)=0.5 ]\ where \ C^*w_1x_1+w_2x_2/w\ The solution to Eq. 2-20 can be obtained by applying standard solution methods (Kreyszig, 2011): \ x(t)=0.5 e^-t / 3+C^*(1-e^-t / 3)\ For case (b), \ C^*=(400 kg / min)(0.4)+(200 kg / min)(0.75)/600 kg / min=0.517\ Substituting C^* into Eq. 2-22 gives the desired solution for the step change in w_1 : \ x(t)=0.5 e^-t / 3+0.517(1-e^-t / 3)\ (c) For the step change in w_2, \ C^*=(500 kg / min)(0.4)+(100 kg / min)(0.75)/600 kg / min=0.458\ and the solution is \ x(t)=0.5 e^-t / 3+0.458(1-e^-t / 3)\ (d) Similarly, for the simultaneous changes in x_1 and w_2, Eq. 2-21 gives C^*=0.625. Thus, the solution is \ x(t)=0.5 e^-t / 3+0.625(1-e^-t / 3)\ (e) The individual responses in Eqs. 2-22-2-24 have the same normalized response: \ x(t)-x(0)/x(\\\\infty )-x(0)=1-e^-t / 3\ The responses of (b)-(e) are shown in Fig. 2.2. The individual responses and normalized response have the same time dependence for cases (b) -(?) because \\\\tau =V\\\ ho / w=3 min for each part. Note that \\\\tau is the meanresidence time of the liquid in the blending tank. If w changes, then \\\\tau and the time dependence of the solution also change. This situation would occur, for example, if w_1 changed from 500 kg / min to 600 kg / min. These more general situations will be addressed in Chapter 4. Figure 2.2 Exit composition responses of a stirred-tank blending process to step changes in (b) Flow rate w_1 (c) Flow rate w_2 (d) Flow rate w_2 and inlet composition x_1 (e) Normalized response for parts (b)-(d).