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podryga [215]
3 years ago
7

What can be used as a tracing powder when checking for small oil leaks

Engineering
1 answer:
dangina [55]3 years ago
6 0

Answer:

Talcum powder!

Explanation:

How to find the source of an oil leak:

This method simply involves using Talcum Powder as a visual aid to help pinpoint the location of an oil leak,

so you use Talcum powder for small oil leak :]

Have a wonderful day!

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When the psychologist simply records the relationship between two variables...
Wewaii [24]
When a psychologist simply records the relationship between two variables without manipulating them, it is called a correlational study.

The observed relationship does not by itself reveal which variable causes the other. This is the directionally problem. Also, the relationship may be due to a third variable controlling both of the observed variables.
8 0
3 years ago
A fluid has a dynamic viscosity of 0.048 Pa.s and a specific gravity of 0.913. For the flow of such a fluid over a flat solid su
sattari [20]

Answer:

Explanation:

First we should recall how Newton's laws relates shear stress to a fluid's velocity profile:

\tau = \mu \cfrac{\partial v}{\partial y}

where tau is the shear stress, mu is viscosity, v is the fluid's velocity and y is the direction perpendicular to flow.

Now, in this case we have a parabolic velocity profile, and also we know that the fluid's velocity is zero at the boundary (no-slip condition) and that the vertex (maximum) is at y=75 \, mm and the velocity at that point is 1.125 \, m/s

We can put that in mathematical terms as:

v(y)= A+ By +Cy^2 \\v(0) = 0\\v(75 \, mm) = 1.125 \, m/s\\v'(75 \, mm) = 0\\

From the no-slip condition, we can deduce that A=0 and so we are left with just two terms:

v(y) = By + C y ^2 \\

We know that the vertex is at y= 75 \, mm and so we can rewrite the last equation as:

v(y) = k(y-75 \, mm) ^2+h

where k and h are constants to be determined. First we check that v( 75 \, mm) = 1.125 \,  m/s :

v( 75 \, mm) = k(75 \, mm -75 \, mm) ^2+h = h = 1.125 \, m/s\\\\h= v_{max} = 1.125 \,  m/s

So we found that h was the maximum velocity for the fluid, now we have to determine k, for that we need to make use of the no-slip condition.

v( 0) = k( -75 \, mm) ^2+  1.125 \,  m/s= 0 \quad (no \, \textendash slip)  \\\\k= - \cfrac{ 1.125 \, m/s }{(75 \, mm ) ^2} = - \cfrac{ 1125 \, mm/s }{(75 \, mm ) ^2}\\\\k= -  \cfrac{0.2}{mm \times s}

And thus we find that the final expression for the fluid's velocity is:

v( y) = 1125-  0.2 ( y -75 ) ^2

where v is in mm/s and y is in mm.

In SI units it would be:

v( y) = 1.125-  200 ( y -0.075 ) ^2

To calculate the shear stress, we need to take the derivative of this expression and multiply by the fluid's viscosity:

\tau = \mu \cfrac{\partial v}{\partial y}

\tau =0.048\,   \cdot  (-400) ( y-0.075   )

for y= 0.050 \, m we have:

\tau =0.048\,   \cdot  (-400) ( 0.050 -0.075   ) = 0.48\, Pa

Which is our final result

5 0
4 years ago
Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5
pochemuha

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

Aa.Va=Ab.Vb=Q

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3

Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3

Using the volume rate:

Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s

Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s

Assuming no losses, the energy equation for fluids can be written as:

Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m

Pb = 70000\ Pa-27303\ Pa - 14715\ Pa

Pb = 27,996\ Pa = 28\ kPa

6 0
3 years ago
A master stud pattern is laid out somewhat<br> like a?
Svetradugi [14.3K]

Answer:

••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.

3 0
2 years ago
A murder in a downtown office building has been widely publicized. You’re a police detective and receive a phone call from a dig
BaLLatris [955]

Answer:

Considering the plain view doctrine, which is an exception to the warrant requirement of the Fourth Amendment, is applied by law enforcement officers and courts who can seize evidence of a crime without a warrant, if the officer observes the evidence in plain view.

Explanation:

For any digital information related to a murder case that has been seized under the plain view doctrine to be used to convict you of a crime, has to comply with three conditions:

1.  The digital evidence must be in out in the open, and easily observable by the officer, this is what "plain view" refers to.

2. The officer must have a legal right to be where he got the information related to the case.

3. The 'incriminating' character of the information must be a clear hint of the murder to fall under the plain view doctrine and the officer´s experience will help him determining whether the information is evidence or not, upon probable cause related to a crime.

7 0
3 years ago
Read 2 more answers
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