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solong [7]
3 years ago
11

What is shown in the above figure?

Engineering
2 answers:
vlabodo [156]3 years ago
3 0
A. compression ring 1
motikmotik3 years ago
3 0

Answer:

B. Oil control ring

Explanation:

I just took the test on PF

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What are the different branches of engineering involved in manufacturing a general-purpose elevator?
Anestetic [448]

Answer:

technical engineering

8 0
3 years ago
A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled
Vikentia [17]

Answer:

Force per unit plate area is 0.1344 N/m^{2}

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, \mu =1.92\times 10^{- 3} Pa-s

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

F = \tau A

where

\tau = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = (\tau ' +\tau '')A

\frac{F}{A} = \tau ' +\tau ''

Also

F = \frac{\mu v}{d}

where

\mu = dynamic coefficient of viscosity

Pressure, P = \frac{F}{A}

Therefore,

\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}

\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}

8 0
3 years ago
What are significant figures​
lesantik [10]

Answer:

rounding

Explanation:

each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit.

4 0
3 years ago
A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container
kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

delta S_{12}=\int\limits^2_1  \, \frac{dQ}{T}

In an isobaric process heat (Q) is defined as:

Q= m*Cp*dT

Replacing in the equation for entropy  

delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:  

delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}  

Solving the integral we get the expression to estimate the entropy change in the system  

delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is 0.85\frac{kJ}{Kg*K}

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

Q= m*Cp*dT

With dt=T_{2}-T_{1} clearing for T2 we get:

T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K

Now we can estimate the entropy change in the system

delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.  

b) see picture.

3 0
3 years ago
Serves as a protective barrier to prevent contact with engergized ("hot") parts<br> within the unit
erik [133]

Answer:

thanks hot hot

Explanation:

4 0
3 years ago
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