Assignment # 1

Subject Code: MTH642

Subject Name: Fluid Mechanics

Student ID: BCXXXXXXXXX

Student Name: zaniklawnak

Solution

Question 1

Given:$$\vec{V}=(u,v)= (U_0+bx)\hat{i}-by\hat{j}$$V=(u,v)=(U0​+bx)i^−byj^​

For steady flow,$$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}$$ax​=u∂x∂u​+v∂y∂u​ $$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$$ay​=u∂x∂v​+v∂y∂v​

Now,$$\frac{\partial u}{\partial x}=b,\quad \frac{\partial u}{\partial y}=0$$∂x∂u​=b,∂y∂u​=0 $$\frac{\partial v}{\partial x}=0,\quad \frac{\partial v}{\partial y}=-b$$∂x∂v​=0,∂y∂v​=−b

Therefore,$$a_x=(U_0+bx)(b)+(-by)(0)$$ax​=(U0​+bx)(b)+(−by)(0) $$\boxed{a_x=bU_0+b^2x}$$ax​=bU0​+b2x​

Similarly,$$a_y=(U_0+bx)(0)+(-by)(-b)$$ay​=(U0​+bx)(0)+(−by)(−b) $$\boxed{a_y=b^2y}$$ay​=b2y​

Hence acceleration vector is$$\boxed{ \vec a=(bU_0+b^2x)\hat i+b^2y\hat j }$$a=(bU0​+b2x)i^+b2yj^​​

Question 2

Given pressure field$$P=P_0-\frac{\rho}{2} \left[ 2U_0bx+b^2(x^2+y^2) \right]$$P=P0​−2ρ​[2U0​bx+b2(x2+y2)]

Material derivative of pressure:$$\frac{DP}{Dt} = u\frac{\partial P}{\partial x} + v\frac{\partial P}{\partial y}$$DtDP​=u∂x∂P​+v∂y∂P​

First calculate derivatives:$$\frac{\partial P}{\partial x} = -\rho(U_0b+b^2x)$$∂x∂P​=−ρ(U0​b+b2x) $$\frac{\partial P}{\partial y} = -\rho b^2y$$∂y∂P​=−ρb2y

Substituting values:$$\frac{DP}{Dt} = (U_0+bx)\Big[-\rho(U_0b+b^2x)\Big] + (-by)(-\rho b^2y)$$DtDP​=(U0​+bx)[−ρ(U0​b+b2x)]+(−by)(−ρb2y) $$\boxed{ \frac{DP}{Dt} = -\rho b(U_0+bx)^2 + \rho b^3y^2 }$$DtDP​=−ρb(U0​+bx)2+ρb3y2​

This is the rate of change of pressure following a fluid particle.

Question 3

Given:$$u(0)=u_{entrance}$$u(0)=uentrance​ $$u(L)=u_{exit}$$u(L)=uexit​

Since velocity decreases parabolically, assume$$u(x)=Ax^2+Bx+C$$u(x)=Ax2+Bx+C

At $x=0$x=0$$C=u_{entrance}$$C=uentrance​

Assuming zero slope at entrance,$$B=0$$B=0

Using $u(L)=u_{exit}$u(L)=uexit​$$AL^2+u_{entrance}=u_{exit}$$AL2+uentrance​=uexit​ $$A= \frac{u_{exit}-u_{entrance}}{L^2}$$A=L2uexit​−uentrance​​

Therefore,$$\boxed{ u(x)=u_{entrance} + \left( u_{exit}-u_{entrance} \right) \frac{x^2}{L^2} }$$u(x)=uentrance​+(uexit​−uentrance​)L2x2​​

This is the required centerline velocity equation.

Question 4

Given velocity field$$u=U_0+bx$$u=U0​+bx $$v=-by$$v=−by

Equation of streamline:$$\frac{dy}{dx} = \frac{v}{u} = \frac{-by}{U_0+bx}$$dxdy​=uv​=U0​+bx−by​

Separating variables:$$\frac{dy}{y} = -\frac{b\,dx}{U_0+bx}$$ydy​=−U0​+bxbdx​

Integrating both sides:$$\ln y = -\ln(U_0+bx)+C$$lny=−ln(U0​+bx)+C $$\ln[y(U_0+bx)] = C$$ln[y(U0​+bx)]=C

Taking exponential:$$\boxed{ y(U_0+bx)=C_1 }$$y(U0​+bx)=C1​​

or$$\boxed{ y=\frac{C_1}{U_0+bx} }$$y=U0​+bxC1​​​

Hence the analytical expression for streamlines is obtained.