312 Mr Harrison, The pressure in a viscous liquid etc. 



We have 



i| = |-V^ [cot «./'+/"], 

 1 dp _2v ., 



Hence eliminating f, 



^ + ,-4 [/'" +/" cot d -/' cosec2 e + 6/'] = 0. 



Therefore ff = 0, and 



/'" +/" cot 9-f' cosec2 d+6f' = 0. 

 Hence/' (6) = 0, and the boundary conditions cannot be satisfied^ 

 since u becomes independent of 6. 



For slow motion, or any motion in which the inertia terms can 

 be neglected, we have 



/'" +/" cot e -f cosec2 d + 6/' = (1). 



A first integral is 



/"+/'cot0+6/+C = O (2). 



The solution of (2) suitable for the present purpose is 



/(6') = Z)(2-3sin2^)-iC. 

 Let f{e)-^u„ 6=0, 



f{e) = 0, d^a. 

 We have D = uJ3 sin^ a, 



(7 = 2 (2 — 3 sin^ a) ^o/sin^ a. 

 Hence u = Uq (sin^ a — sin^ 6)/r^ sin^ a. 



Integrating the equations of motion, we have 



f=-|^(/'cot^+r) + i^x(^) 



^"d P^=p+F,ir). 



Hence ,^ ^ '^V(^) + ^^^ + 5, 



and Vm^_19^_b. 



p 3 r* 



The lateral pressure will continually increase as the channel 



widens if C be negative, that is, if sin a > (f)^, or a > 54° 45'. If 

 a < 54° 45', for sufficiently small values of Uq the pressure will 

 continually diminish. 



