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. The immersion area of belt in the fluid is 2.0 m x 1.0 m. The belt is set in motion in the positive x-direction only at a constant velocity of v0 = 0.5 m s-1 (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.

(a) A steel belt has one side immersed in a viscous liquid of density =1840kgm3 and 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 Motion 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). (ii) The velocity varying with time t and distance y is expressed as equation (A-2) below, v0vx(y,t)=1erf4ty=erfc4ty (A2) Calculate the force exerted on the belt (at y=0 ) incurred by the belt moving in the viscous fluid at time t=0.01s. [10 MARKS] (b) A student studied heat transfer in a semi-infinite solid as shown in Fig.(a). The solid surface is exposed to a heating source with a constant heat flux input qB. The heat conduction flux qA at the given time t and the distance x in the solid can be expressed by the equation (B1). qA=qB[1erf(2tx)] where Ti is the initial temperature of the solid. is the thermal diffusivity of the solid. According to equation (B-1), the student derived equation (B-2) for the temperature distribution T(x,t) T(x,t)=Ti+kqB{2te4tx2x[1erf(2tx)]}(B2) Verify the equation (B-2) (you must show the problem-solving procedure). [10 MARKS] Additional information: erf(x)=20xet2dt;erf(0)=0;erf()=1;erf(2)=0.99;dxderf(u)=2exp(u2)dxdu;y(1erf(y))dy=yerf(y)+ey2y