Answer:
[tex]a = v\cdot \frac{dv}{dx}[/tex], [tex]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}[/tex], [tex]\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}[/tex]
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:
[tex]\dot m_{in} - \dot m_{out} = 0[/tex]
[tex]\dot m_{in} = \dot m_{out}[/tex]
[tex]\dot V_{in} = \dot V_{out}[/tex]
[tex]v_{in} \cdot A_{in} = v_{out}\cdot A_{out}[/tex]
The following relation are found:
[tex]\frac{v_{out}}{v_{in}} = \frac{A_{in}}{A_{out}}[/tex]
The new relationship is determined by means of linear interpolation:
[tex]A (x) = A_{in} +\frac{A_{out}-A_{in}}{L}\cdot x[/tex]
[tex]\frac{A(x)}{A_{in}} = 1 + \left(\frac{1}{L}\right)\cdot \left( \frac{A_{out}}{A_{in}}-1\right)\cdot x[/tex]
After some algebraic manipulation, the following for the velocity as a function of position is obtained hereafter:
[tex]\frac{v_{in}}{v(x)} = 1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1\right) \cdot x[/tex]
[tex]v(x) = \frac{v_{in}}{1 + \left(\frac{1}{L}\right)\cdot \left(\frac{v_{in}}{v_{out}}-1 \right)\cdot x}[/tex]
[tex]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}[/tex]
The acceleration can be calculated by using the following derivative:
[tex]a = v\cdot \frac{dv}{dx}[/tex]
The derivative of the velocity in terms of position is:
[tex]\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}[/tex]
The expression for acceleration is derived by replacing each variable and simplifying the resultant formula.
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
