MASS AND VOLUMETRIC FLOWRATES Fluid flows between these flat plates. What is the volume rate of flow? v 1 In time t, the fluid that crosses surface 1 will go distance L which is V x  t. L H x y If the width (in the z-direction) is W, then the volume of fluid that has passed across surface 1 is H x W x L. The volume rate of flow will be this volume divided by the time period or Q = HW L/ t = HWV. In other words, this is the (normal) component of velocity times the area of surface 1. Now suppose the velocity is not constant, but varies with y. How do we find the volumetric flowrate? (See the next slide.)

Fluid flows between these flat plates. What is the volume rate of flow Fluid flows between these flat plates. What is the volume rate of flow? Note that the velocity varies with y. v1 H1 H = H1 + H2 + H3 v2 y v3 x H2 In time +t, the fluid that crosses surface 1 will go distance L1 =V1 x +t at the top, distance L2 =V2 x +t in the middle, and distance L3 =V3 x +t at the bottom. H3 L1 1 If the width (in the z-direction) is W, then the volume of fluid that has passed across surface 1 is H1 x W x L1 + H2 x W x L2 + H3 x W x L3. The volume rate of flow will be this volume divided by the time period or Q = [ H1 x W x L1 + H2 x W x L2 + H3 x W x L3]/ +t or Q = H1WV1 + H2WV2 + H3WV3. In other words, this is sum of the (normal) velocity elements times the corresponding partial areas of surface 1. Suppose there are N different values of velocity with y. Then Q = S HiWVi or Q = W S HiVi , where Vi depends on y. As N®¥, Hi®dy and the velocity becomes a H/2 continuous function V(y). In this case, Q = W ò-H/2 V(y) dy. Find Q if V(y) = c(1-y2), c=const.

The volume rate of flow is independent of the surface 1 or 1` that we Fluid flows between these flat plates. What is the volume rate of flow across surface 1`? y h H x q In time +t, the fluid that crosses surface 1 will go distance L which is V x +t. v n L 1` 1 The volume rate of flow is independent of the surface 1 or 1` that we use to do the calculation. As before, the volume rate of flow will be Q = HWV. This can be rewritten as Q = (h cos q)WV or hWVcos q. Note that Vcos q is the same as ½V ½½n½cos q since ½n½= 1, but this is just V×n. For an arbitrary surface, the volumetric flowrate across that surface will be the area times the normal component of velocity. In this case, it is Q = hW(V×n) for any planar surface 1`. For a nonplanar surface, we have Q = òò (V×n) dA where the contributions are “added” for all elements of area dA.

In the first slide, the volume rate of flow was shown to be Q = HWV. The units are length x length x length/unit time or volume/unit time. If we know the volume per unit time flowing across surface 1, what is the mass flow rate? The mass flowrate will simply be the density times the volumetric flowrate or M = rQ = rHWV. What are the units for the right hand side? Does it make sense? For the most general case, the mass flowrate will be: M = òòr(V×n) dA If we are examining some unit operation, then we will need to find M for each inlet and outlet. Rationalize the units on the right hand side of this integral equation. Why are there two integral signs? In what fundamental law or principle does the integral expression above appear?