A steel belt has one side immersed in a viscous liquid of density = 1840 kg m-3 and viscosity = 25.54 mPas. The semi-infinite body of liquid is bounded by the belt horizontal surface (xz-plane) as illustrated in Figure 1.

Analyse the motion equations for this fluid system and derive the differential equation (A-1).

And Calculate the force exerted on the belt (at y = 0) incurred by the belt moving in the viscous fluid at time t = 0.01 s

(a) A steel belt has one side immersed in a viscous liquid of density =1840kgm3and viscosity =25.54mPas. The semi-infinite body of liquid is bounded by the belt horizontal surface ( xz-plane) as illustrated in Figure 1. The immersion area of belt in the fluid is 2.0m1.0m. The belt is set in motion in the positive x-direction only at a constant velocity of v0=0.5ms1(vy=0 and vz=0). The pressure gradients along x, y, and z, and the gravity forces can be ignored. This Newtonian fluid is incompressible. (i) The velocity is a function of time ( t ) and distance (y) to the belt surface, which can be expressed by the partial differential equation (A1) below, tvx=y22vx n equations: (tvx+vxxvx+vyyvx+vzzvx)=xp[xxx+yyx+zzx]+gx(tvy+vxxvy+vyyvy+vzzvy)=yp[xxy+yyy+zzy]+gy(tvz+vxxvz+vyyvz+vzzvz)=zp[xxz+yyz+zzz]+gz Analyse the motion equations for this fluid system and derive the differential equation (A1)