For (a), the following relationships may be helpful: exer=ex(cosex+siney)=coseex=(sinex+tosey)ex=siner=rrer+re+rzez (b) Obtain an expression for FD. (c) Find the drag coefficient. The general equation for the drag coefficient CD from Problem 1 is still valid, but here the projected area will be different. Problem 2 (15 points): Drag on a cylinder at low Reynolds number Consider creeping flow at velocity v=Ux past a long cylinder of radius R and length L that is perpendicular to the approaching fluid. Figure 1 is repeated below - except now imagine instead of a sphere it is a cross-section of a cylinder. Unlike the case of flow pas a sphere, there are no simple expressions for the velocity and pressure through the fluid. However, adequate approximations for the region near the cylinder when Re0 are shown below (Batchelor, 1967, pp. 244-246). You will use these results to obtain an expression for the drag force on a the cylinder. vr(r,)=2C[1(rR)22ln(Rr)]cosv(r,)=2C[1(rR)2+2ln(Rr)]sinP(r,)=r2CcosC=ln(Re7.4)U,Re=v2RU (a) Starting from Eqn 6.6-26 (Deen) and Eqn A 539 (for dSr ) show that the drag on the cylinder would be: FD=0102[(exerP(R,))+re0ex]Rddz