Unformatted text preview: 54. 55. 56. (e) We see above that k = (a/rrb)./2g = 14/3. Substitution of a 3W2, b = L g = (338600):2 ft/hr2 yields r = (1/60)\/7/12 ftz0.15 in for the radius of the
bottom—hole. With g = 32 ft/secz and a = int/12f, Equation (24) sim lifies to
p If 2: denotes the distance from the center of the cylinder down to the ﬂuid surface, then y :2 3—z and Aty) = 10(9—22)"2. Hence the equation above becomes
dz 7:
10 9__Z21r2_ : __ 3—Z H2,
( ) dt 18( ) 180(3+z)mdz : arch, and integration yields
120(3+z)”2 = m+ C. Now 2 = 0 when t = 0, so C = 120(3)”. The tank is empty when 2 = 3 (that is,
when y = 0) and thus after r = (120110633 — 3“) = 362.90 see.
It therefore takes about 6 min 3 sec for the fluid to drain completely. A(y) = tr(8y—y2) as in Example 7 in the text, but now a
so the initial value problem is Jr/144 in Equation (24), 18(8y+ yzbt’ = 4J— , 32(0) = 8.
We seek the value of I when y = 0. The answer is I: 869 see = 14 min 29 sec. The crosssectional area function for the tank is A = 3(1— yz) and the area of the
bottomhole is a 1047:". so Eq. (24) in the text gives the initial value problem myﬁ)? = —10““2r,l2><9i.8y, y(0) = 1. Simplification gives (you _ym)% : _ 1,4X 104m so integration yields Section 1.4 21 ...
View
Full Document
 Spring '11
 Page

Click to edit the document details