MACHINE PROBLEM 1 Elajah Mae A. Zaragoza BS Chemical Engineering III Principles of Transport Processes

FLOW THROUGH A CIRCULAR PIPE

Consider a steady-state, laminar flow of an incompressible fluid of constant density () and viscosity () in a vertical cylindrical tube of height L and radius R. The fluid here is moving downward because of both the gravity force and pressure (Pz) acting to the z-direction. The shell in which we will consider to use for the momentum balance has a shell thickness of r.

Assumptions/Postulates:

1. The fluid is flowing downwards and it is assumed as a fluid flowing in a steady, laminar flow, vz is the only velocity that has a value. , and both v and vr has no value. ,2. The velocity acting in the z-direction (vz ), is a function of r only. That means that vz is dependent to the value of r but is independent to the values of z and .3. There is a pressure change only in the z-direction. Hence,

Vanishing And Non-Vanishing Components Of

From the definition of combined momentum balance (

For convective transport part

From the postulate, vz has a value , and

is a tensor that has 9 components and out of 9, only 1 component left with value.

Therefore, all components except for has no value.

For molecular transport part From the postulate, vz ,, , and the assumption that the fluid is an incompressible fluid = 0 (Since vr = 0, and , then

= 0 (Since v = 0, and , then

= 0 (Here, vz 0, but vz is not a function of z, the term = 0 , then

= 0 (Since v = 0 and vr = 0, then and )

= 0 (Since v = 0 and vz is not a function of , then and

(Since vz has a value and is a function of r, then and has a value.

is a 9 component tensor and out of 9, only 2 of them only has a value which is rz and zr.

zz is the z-directed moving flux exerted by the moving fluid downwards so it have pressure Pz and rz is the z-directed flux exerted by the fluid flowing from the radial direction. Since vr = v = 0, the vanishing components can be removed from the equation.

For a steady state, the total momentum balance is the rate of the combined momentum in minus the combined momentum out plus the force of gravity acting on the system is equal to 0. It is better to solve the rate of z-directed momentum in and out separately. Also the surface area is multiplied to which the momentum flux acts upon.

Across annular surface: (in)

(out)

Across cylindrical surface: (in)

(out)

Gravity force acting in z direction:

Shell momentum balance at steady state:

Dividing the equation by 2 and since vz is not a function of z, is zero.

Simplifying and taking the limit as r approaches zero

Since = 0 and , we can add that to the equation so that it can be simplified to become .

Integrating,

To get the value of, we need to have a no-slip boundary condition that we can apply to the system. Using the boundary condition at r = 0,, the constant will be equal to zero. If is infinite, the momentum flux would be infinite at the axis of the tube.

Divide both sides by The momentum flux distribution is:

Then substitute to to get the velocity distribution of the system and integrate the equation.

The boundary condition to get the value of the constant is the liquid-solid interface. At B.C 2, the velocity of the solid is equal to the velocity of the fluid at the surface of the solid. Since the wall of the tube is not moving at (r = R), then the velocity of the fluid is zero (

The velocity distribution is parabolic.

At r = 0 the velocity will be at its maximum, while at that point the z momentum will be zero because the velocity will be moving from higher to lower velocities. From the figure the velocity is distributed from r = 0 moving towards the walls of the tube.

FLOW THROUGH AN ANNULUS

An annulus is comparative to cylinder but it has another smaller cylindrical pipe from which the fluid is flowing upwards from to R. It is assumed that there will be a maximum velocity at some plane which is at .

Assumptions/Postulates:

1. The fluid is flowing downwards and it is assumed as a fluid flowing in a steady, laminar flow, vz is the only velocity that has a value. , and both v and vr has no value. ,2. The velocity acting in the z-direction (vz ), is a function of r only. That means that vz is dependent to the value of r but is independent to the values of z and .3. There is a pressure change only in the z-direction. Hence,

Vanishing And Non-Vanishing Components Of

From the definition of combined momentum balance (

For convective transport part

From the postulate, vz has a value , and

is a tensor that has 9 components and out of 9, only 1 component left with value.

Therefore, all components except for has no value.

For molecular transport part From the postulate, vz ,, , and the assumption that the fluid is an incompressible fluid

= 0 (Since v = 0, and , then

= 0 (Here, vz 0, but vz is not a function of z, the term = 0 , then

= 0 (Since v = 0 and vr = 0, then and )

= 0 (Since v = 0 and vz is not a function of , then and

(Since vz has a value and is a function of r, then and has a value.

is a 9 component tensor and out of 9, only 2 of them only has a value which is rz and zr.

We will arrive at the same shell momentum balance. The only difference is that the flow of fluid is opposing the gravitational force. That makes the negative.

vz is not a function of z, so is zero.

Since = 0 and , we will get the same results as the cylindrical pipe.

From the figure, somewhere within surface area and there will be a maximum velocity at where momentum flux is zero ). That would be the boundary condition of the equation.boundary coposing the gravitational forces that there is anothe at that point the mo

We change to and put the value of into the equation.

Since,

Substitute

Using the solid no slip interface conditions, at solid surface at and, the velocity of the fluid is zero. Getting the value of .

Substituting the value of , we have the value of with respect to the fixed plane .

Substituting the value of we now have the velocity distribution of the system.

Getting the equation for the shear stress,

From the equation, the at to is negative. Since stress has two directions, the z momentum exerted by the moving fluid moving towards r direction (toward the z-axis) will make negative. From the plane where the maximum velocity occurs, the momentum transfer will go towards the walls of the outer and smaller tube.