Based on attached material, please solve following problem.A journal bearing of length 60 mm is to support a steady load of 20 kN. The journal, of diameter 50 mm, runs at 10 revolutions per second. Assuming c/R = 0.001 and ε = 0.6, determine a suitable viscosity of the oil, using the theory of (a) the very long bearing with the full Sommerfeld condition, (b) the very short bearing with the half Sommerfeld condition. For (a) determine the power required to overcome friction, and the best position for the oil supply hole relative to the load line. Would this position also be suitable according to theory (b)? We consider first a bearing of infinite length, so that the lubricant flow may be Bearing of infinite lengthtaken as two-dimensional (in the plane of the diagram). We assume also thatconditions are steady, that the viscosity and density are constant throughout,and that the axes of journal and bush are exactly straight and parallel. Asthe film thickness h is so small compared with the radii, the fact that thefilm is curved may be neglected, and, for a small length &x in the directionof flow, the flow is governed essentially by eqn 6.35. Hencedp12μ (ΩRb6µ2R(h – ho)h39 =(6.48)dxwhere ho = 2q/NR, which, for steady conditions, is constant and may beregarded as representing the value of h at which dp/dx = 0. By puttingx = RO and h = c(1 + ɛ cos 0) we transform eqn 6.48 to1 dpR de6µ2R1ho(6.49)(1+8 cos 0)2c(1+ɛ cos 0)3The pressure p may now be determined by integrating eqn 6.49 with respectto 0. This may be achieved by using a new variable a such that1- 821- e cos a1+ɛ cos 0 =(6.50)and thus de = (1 – e2)'/²da/(1 - e cos a). Then pc² /6µ2R² = Cp, say, sothata - e sin aCp(1 - 82)3/2 c(1 – 82)5/2ho%3D-2ɛ sin a +82·sin 2a + Co+(6.51)24where Co is an integration constant. 232 Laminar flow between solid boundariesNow if the lubricant occupies the entire space between journal and bush,eqn 6.51 must give the same value forp at the maximum clearance whetherwe set 0 = 0 or 27, that is, a = 0 or 27. Hence2лho0+ Co =(1 - 82)3/2 c(1 – 82)5/22л | 1 ++ Coand soc(1 – e2)ho1+82/2(6.52)Substituting this relation in eqn 6.51 we obtaine(2 – 82 – e cos a) sin a CoCp(1+82)3/2(2 + 82)E(2 + ɛ cos 0) sin 0+ Co(2 + 82)(1+ ɛ cos 0)2(6.53)This expression is plotted in Fig. 6.23, which shows that when e isbetween a and 2n, the pressure drops considerably below the value ate = 1. The result was first obtained by the German physicist ArnoldSommerfeld (1868-1951), and so the existence of these symmetrical high-and low-pressure sections of the graph is termed the Sommerfeld boundarycondition.

Given- l=60 mm =0.06mm W=20 KN d=50 mm=0.05m N=10 rpm oil viscosity Z=0.011 Kg/m-sBearing Length- l=1.5d =1.5×0.05l=75 m Bearing pressure- P=WA =Wl.d 2=20×10375×50 P=5.534 N/mm2Coefficient friction- μ=331042Npdc+k =331040.011×6005.33500.15+0.002μ=0.002 Rubbing velocity;- v=πdN60=π×0.05×60060 v=15.7 m/sHeat loss in friction:- Qg=μ.W.v=0.02×20×103×15.7=6280 Wpower wasted in frictionQg=6.28 kw