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

At x = 0, acceleration = 0

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

Step-by-step 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

Answer 2

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

Step-by-step 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:

`(v_(out))/(v_(in)) = (A_(in))/(A_(out))`

The new relationship is determined by means of linear interpolation:

`A (x) = A_(in) +(A_(out)-A_(in))/(L)\cdot x`

`(A(x))/(A_(in)) = 1 + \left((1)/(L)\right)\cdot \left( (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:

`(v_(in))/(v(x)) = 1 + \left((1)/(L)\right)\cdot \left((v_(in))/(v_(out))-1\right) \cdot x`

`v(x) = (v_(in))/(1 + \left((1)/(L)\right)\cdot \left((v_(in))/(v_(out))-1 \right)\cdot x)`

`v (x) = v_(in)\cdot \left[1 + \left((1)/(L)\right)\cdot \left((v_(in))/(v_(out))-1 \right)\cdot x \right]^(-1)`

The acceleration can be calculated by using the following derivative:

`a = v\cdot (dv)/(dx)`

The derivative of the velocity in terms of position is:

`(dv)/(dx) = -v_(in)\cdot \left((1)/(L)\right) \cdot \left((v_(in))/(v_(out))-1 \right) \cdot \left[1 + \left((1)/(L)\right)\cdot \left((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.