Solution to Question 14 in the morning session of the 2014 Chemistry Exam – Discussion
Thread Content
This post was last edited by yyuucc on 2016-1-19 at 23:18. A rotary vacuum filter is used to filter a certain suspension; the processing capacity is 50 m3/h. It is known that for every 1 m3 of filtrate produced, 0.05 m3 of filter cake is obtained. The immersion depth of the drum is required to be 0.35. The thickness of the filtration surface shall be no less than 7 mm. The filter cake is incompressible. The resistance of the filtering medium can be ignored. The measured filtration constant K = 8*10^-4 m2/s. What is the filtration area of the filter? Inspired by the 【security door】, I redid the calculation: V is the volume of the filtrate in m3, and n is the rotation speed ； Q = nV = 50/1.05/3600 = 5/378 m3/s; (1) Q2 = K*A2*0.35*n. Thus, (5/378)2 = 0.0008*A*n*0.35*A. For mass balance, with a time of 1 second: 50/1.05/3600*(0.05/1) = n*LA = n*0.007*A. (3) Solving this gives nA = 0.0945; substituting this value into equation (2) yields A = 6.61 ； What is the correct answer?
I calculated it to be 7.29, as follows: Considering a filter cake thickness of 7 mm: n x 0.007m x A = 50m3/3600s x (0.05/1.05) > A x n = 0.0945 m2/s. Q2 = K x A2 x 0.35 x n; by substituting the values, we get: A = 7.29m2
This post was last edited by yyuucc on 2016-1-19 22:44. There is a question regarding the first formula； It should be n x 0.007m x A = 0.05*V. n is the rotational speed. Within one cycle, V is unknown, rather than 50 m3/3600 s /1.05