How Can Fluid Density and Shear Stress Be Calculated in Engineering Scenarios?

[math_04]

Homework Statement

Actually there are two questions in it.

First question is;

1) Determine the density 8km below the surface of the ocean where the pressure is 87.7Mpa, given that the density at the surface is 1025 kg m^-3 and the modulus of elasticity is 2.37GPa (use integration).

Second question is

2) If the velocity distribution in a 150mm diameter pipe carrying oil where mu(greek symbol)=0.048 Ns/m^2 and relative density = 0.913) is given by

u= 1.125 - 200(0.075-y)^2 where u denotes the velocity parallel to the boundary in m/s and y the distance from the solid boundary in m.

Calculate

a) The velocity gradient and the shear stress at the boundary and at a point 50mm from the boundary.
b) The shear stress at the centre of the line
c) The force resisting the fluid motion over a pipe length of 300m

Homework Equations

The Attempt at a Solution

For the first question, could you give me a hint where to start? It says use integration but on what? I have a formula which is dP/dz = -rho (greek symbol) x g. z is the height of the fluid element.

For the second question, I figured out part a. You differentiate u and plug in the values for y and mu in. I got the shear stress to be 0.48 Nm^-2. But I am not sure about the velocity gradient. Is it just du/dy and then plugging in the values for y (y=0.05m)? If so, how would the units pan out? Would it be m/s?

Looking at the answer for b which is 0, I am guessing there is some sort of law for shear stress at the centre of the pipe?

For c) i think I can work it out, just need to find the formula (hopefully!)

Thanks.

 

[anathema]

Hi there,

For the first question, you are correct that the formula dP/dz = -rho(mu)g can be used. The key here is to use integration to solve for the density (rho) at 8km below the surface. You can start by setting up the equation as follows:

dP/dz = -rho(mu)g

Integrating both sides with respect to z, we get:

P = -rho(mu)gz + C

where C is the constant of integration. Since we know the pressure and density at the surface, we can plug these values in to solve for C:

P(at surface) = -rho(surface)(mu)g(0) + C
1025 kg m^-3 x 9.8 m/s^2 = C
C = 10035 Pa

Now, we can plug in the values for the pressure and density at 8km below the surface to solve for the density at that depth:

P(at 8km) = -rho(8km)(mu)g(8km) + 10035 Pa
87.7 MPa = -rho(8km)(2.37 GPa)(8km) + 10035 Pa
Solving for rho(8km), we get:
rho(8km) = 1031 kg m^-3

For the second question, you are correct that the velocity gradient is given by du/dy. Since the units of u are in m/s and y is in m, the units for the velocity gradient would be s^-1. And yes, the shear stress at the center of the pipe would be 0, as there is no velocity gradient at that point (du/dy = 0).

For part c), you can use the formula for shear stress (tau = mu(du/dy)) to calculate the shear stress at a point 150mm from the boundary. Then, you can use the formula F = tau x A (where A is the cross-sectional area of the pipe) to calculate the force resisting fluid motion over a pipe length of 300m.

I hope this helps! Let me know if you have any other questions.

 

[baba89]

Hello,

For the first question, you are correct in thinking that you need to use the formula dP/dz = -rho (greek symbol) x g. This is the hydrostatic equation and it relates the change in pressure with depth to the density and gravitational acceleration. In this case, you are given the pressure (87.7Mpa) and the density at the surface (1025 kg m^-3). You also know the modulus of elasticity, which can be used to find the change in pressure with depth using integration. You can use the formula dP/dz = -rho (greek symbol) x g to set up your equation for integration. Then, you can solve for the density at 8km below the surface by plugging in the values for pressure, density at the surface, and the modulus of elasticity.

For the second question, you are correct in finding the shear stress at the boundary using the formula for shear stress (mu x du/dy). For part b, you are correct in thinking that the shear stress at the center of the pipe would be 0. This is because the velocity gradient at the center of the pipe is 0, so there is no shear stress. For part c, you can use the formula for force = shear stress x area to find the force resisting fluid motion over a pipe length of 300m. The area in this case would be the cross-sectional area of the pipe.

I hope this helps. Good luck with your homework!