Question

For the speed equation along centerline of a diffuser, calculate the fluid acceleration along the diffuser centerline as a function of x and the given parameters. For L = 1.56 m, uentrance = 24.5 m/s, and uexit = 17.5 m/s, calculate the acceleration at x = 0 and x = 1.0 m.

Answer 1

Answer:

$a = v\cdot \frac{dv}{dx}$, $v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}$, $\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}$

Explanation:

Let suppose that fluid is incompressible and diffuser works at steady state. A diffuser reduces velocity at the expense of pressure, which can be modelled by using the Principle of Mass Conservation:

$\dot m_{in} - \dot m_{out} = 0$

$\dot m_{in} = \dot m_{out}$

$\dot V_{in} = \dot V_{out}$

$v_{in} \cdot A_{in} = v_{out}\cdot A_{out}$

The following relation are found:

$\frac{v_{out}}{v_{in}} = \frac{A_{in}}{A_{out}}$

The new relationship is determined by means of linear interpolation:

$A (x) = A_{in} +\frac{A_{out}-A_{in}}{L}\cdot x$

$\frac{A(x)}{A_{in}} = 1 + \left(\frac{1}{L}\right)\cdot \left( \frac{A_{out}}{A_{in}}-1\right)\cdot x$

After some algebraic manipulation, the following for the velocity as a function of position is obtained hereafter:

$\frac{v_{in}}{v(x)} = 1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1\right) \cdot x$

$v(x) = \frac{v_{in}}{1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x}$

$v (x) = v_{in}\cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x \right]^{-1}$

The acceleration can be calculated by using the following derivative:

$a = v\cdot \frac{dv}{dx}$

The derivative of the velocity in terms of position is:

$\frac{dv}{dx} = -v_{in}\cdot \left(\frac{1}{L}\right) \cdot \left(\frac{v_{in}}{v_{out}}-1 \right) \cdot \left[1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}} -1 \right) \cdot x \right]^{-2}$

The expression for acceleration is derived by replacing each variable and simplifying the resultant formula.

Answer 2

Answer:

At x = 0, acceleration = 0

At x = 1.0, Acceleration = - 124.08m/s²

Explanation:

Given Data;

L = 1.56m

Entrance (u)= 24.5m/s

exit (u) = 17.5m/s

x = 1.0m

The speed along the centreline of a diffuser is given as;

u =u entry + ((u exit - u entry)x²)/L²-------------------------1

For acceleration in the x-direction, we have

ax = udu/dx + vdu/dy + wdu/dz + du/dt ------------------2

Since it's one dimensional flow, equation 2 reduces to

ax = udu/dx -----------------------------------3

substituting equation 1 into equation 3, we have

ax =  2Uentry (Uexit - Uentry)x/L² + 2(Uexit - Uentry)²*x³/L⁴  ---4

At x = 0, substituting into equation 4, we have

a(0) = 2uentry(uexit-uentry) (0)/L² + 2 (uexit - u entry)²(0)³/L⁴

a(0) = 0

At x = 1.0m, equation 4 becomes

a(1) = 2 *24.5(17.5 -24.5)(1)/1.56² + 2(17.5-24.5)²(1)³/1.56⁴

     =( 49 * -2.87) + 16.547

     = -140.63

    = - 124.08m/s²