Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ )
/ 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
Answer:
robotic technology
Explanation:
Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.
Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.
One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.
Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.
Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.
Answer:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
Explanation:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
Answer:
![250\ \text{lbm/min}](https://tex.z-dn.net/?f=250%5C%20%5Ctext%7Blbm%2Fmin%7D)
![625\ \text{ft/min}](https://tex.z-dn.net/?f=625%5C%20%5Ctext%7Bft%2Fmin%7D)
Explanation:
= Area of section 1 = ![10\ \text{ft}^2](https://tex.z-dn.net/?f=10%5C%20%5Ctext%7Bft%7D%5E2)
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 = ![4\ \text{ft}^3/\text{lbm}](https://tex.z-dn.net/?f=4%5C%20%5Ctext%7Bft%7D%5E3%2F%5Ctext%7Blbm%7D)
= Density of fluid = ![0.2\ \text{lb/ft}^3](https://tex.z-dn.net/?f=0.2%5C%20%5Ctext%7Blb%2Fft%7D%5E3)
= Area of section 2 = ![2\ \text{ft}^2](https://tex.z-dn.net/?f=2%5C%20%5Ctext%7Bft%7D%5E2)
Mass flow rate is given by
![m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}](https://tex.z-dn.net/?f=m%3D%5Crho%20A_1V_1%3D%5Cdfrac%7BA_1V_1%7D%7Bv_1%7D%5C%5C%5CRightarrow%20m%3D%5Cdfrac%7B10%5Ctimes%20100%7D%7B4%7D%5C%5C%5CRightarrow%20m%3D250%5C%20%5Ctext%7Blbm%2Fmin%7D)
The mass flow rate through the pipe is ![250\ \text{lbm/min}](https://tex.z-dn.net/?f=250%5C%20%5Ctext%7Blbm%2Fmin%7D)
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1
![m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}](https://tex.z-dn.net/?f=m%3D%5Crho%20A_2V_2%5C%5C%5CRightarrow%20V_2%3D%5Cdfrac%7Bm%7D%7B%5Crho%20A_2%7D%5C%5C%5CRightarrow%20V_2%3D%5Cdfrac%7B250%7D%7B0.2%5Ctimes%202%7D%5C%5C%5CRightarrow%20V_2%3D625%5C%20%5Ctext%7Bft%2Fmin%7D)
The speed at section 2 is
.
Answer:
![X_B = 1.8 \times 10^{-3} m = 1.8 mm](https://tex.z-dn.net/?f=X_B%20%3D%201.8%20%5Ctimes%2010%5E%7B-3%7D%20m%20%3D%201.8%20mm)
Explanation:
Given data:
Diffusion constant for nitrogen is ![= 1.85\times 10^{-10} m^2/s](https://tex.z-dn.net/?f=%3D%201.85%5Ctimes%2010%5E%7B-10%7D%20m%5E2%2Fs)
Diffusion flux ![= 1.0\times 10^{-7} kg/m^2-s](https://tex.z-dn.net/?f=%3D%201.0%5Ctimes%2010%5E%7B-7%7D%20kg%2Fm%5E2-s)
concentration of nitrogen at high presuure = 2 kg/m^3
location on which nitrogen concentration is 0.5 kg/m^3 ......?
from fick's first law
![J = D \frac{C_A C_B}{X_A X_B}](https://tex.z-dn.net/?f=J%20%3D%20D%20%5Cfrac%7BC_A%20C_B%7D%7BX_A%20X_B%7D)
Take C_A as point on which nitrogen concentration is 2 kg/m^3
![x_B = X_A + D\frac{C_A -C_B}{J}](https://tex.z-dn.net/?f=x_B%20%3D%20X_A%20%2B%20D%5Cfrac%7BC_A%20-C_B%7D%7BJ%7D)
Assume X_A is zero at the surface
![X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}](https://tex.z-dn.net/?f=X_B%20%3D%200%20%2B%20%28%2012%5Ctimes%2010%5E%7B-11%7D%20%29%20%5Cfrac%7B2-0.5%7D%7B1%5Ctimes%2010%5E%7B-7%7D%7D)
![X_B = 1.8 \times 10^{-3} m = 1.8 mm](https://tex.z-dn.net/?f=X_B%20%3D%201.8%20%5Ctimes%2010%5E%7B-3%7D%20m%20%3D%201.8%20mm)