Reconsider Prob. 11-68. Using EES (or other) software, investigate the effect of acceleration on

Reconsider Prob. 11-68. Using EES (or other) software,...

Key Concepts

Fluid Statics

This is the branch of fluid mechanics that deals with fluids at rest and the forces in equilibrium acting on them. It covers the study of pressure distribution within a fluid due to gravitational forces and other influences, providing the basis for understanding how fluids behave when disturbed by external accelerations.

Non-Inertial Reference Frames

When analyzing systems in accelerated frames, additional pseudo-forces must be introduced to account for the observed phenomena. These inertial forces, which are not present in inertial frames, are crucial for understanding the apparent changes in fluid behavior, such as the modification of the free surface profile under acceleration.

Effective Gravity

Effective gravity is the resultant acceleration acting on a fluid when both gravitational acceleration and any additional acceleration (due to the movement of the system) are combined. This concept is key to determining how the free surface of a fluid orients itself in a dynamically accelerating environment.

Free Surface Behavior

The free surface of a fluid reorients itself to be perpendicular to the direction of effective gravity. In an accelerating system, this means that the surface can tilt, and its slope becomes a function of the ratio of the applied acceleration to the gravitational acceleration.

Numerical Simulation

Numerical simulation involves the use of computational tools to model and solve physical systems where analytical solutions may be complex or unavailable. Software like EES enables the systematic investigation of parameters, such as acceleration, and their effects on phenomena like the slope of a fluid's free surface, through tabulated data and visual plots.

Transcript

00:01 Hi everyone, so what we have is nitrogen gas that is going through a diffuser.

00:06 It has an inlet pressure of 60 kilopascals and a temperature of 7 degrees c.

00:10 And at the exit of this diffuser, it has a pressure of 85 kilopascals and a temperature of 27 degrees c.

00:16 What we're asked to find is the exit velocity and the inlet over exit area ratio given while we're varying the inlet velocity from 210 to 350 meters per second.

00:29 So the first thing we're going to do is we're just going to state our assumptions.

00:34 So first thing we're going to assume is that this is a steady -state problem, so nothing is going to be changing with respect to time.

00:41 Next, we're going to assume that nitrogen is an ideal gas.

00:45 Therefore, that's going to allow us to use constant -specific heats.

00:50 And we're also going to assume that any changes in potential energy are negligible and that the device is adiabatic.

01:00 Therefore any heat transfer is also negligible and we're also going to assume that there are no work interactions that's also going to be zero so what we can do now is go ahead and look at the properties of our gas so we know that we're looking at nitrogen nitrogen has a molar mass of m which is equal to 28 kilograms kilmole we can also look at the enthalpyes relevant to each temperature so we know the initial temperature is 7 degrees c which is 280 degrees kelvin therefore that gives us an h bar enthalpy equal to 8141 kilojoules per kilomol similarly for exit temperature t2 of 27 degrees c which is 300 kelvin that gives us an h2 bar equal to 8 ,723 kilojoules per kilomol.

01:57 So since there is only one inlet and one outlet, therefore the mass flow rate going through the entire system is going to be the same.

02:06 We're going to take the diffuser as our system, and that's also going to be the control volume because mass is crossing that boundary.

02:12 So when we write our energy balance equation, we'll see that the energy coming in minus the energy leaving is equal to the change in system energy but because of our steady state assumption the system energy is not changing the perspective time so therefore we can say that e .d .n is equal to e.

02:34 Dot out.

02:36 Now we can go ahead and expand upon the constituents within this term.

02:41 What we can say is that m.

02:45 H1 plus b1 squared over 2 is equal to m.

02:50 Dot h2 plus v2 squared over 2.

02:55 As you can see, the m dots will cancel out.

02:57 And what we're left with is 0 is equal to h2 minus h1 plus v1 squared, or v2 squared minus v1 squared, all over 2...

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