Alternating current lesson 4 exam

Describe how you would control employee exposure to excessive noise in a mining environment

Alexandra [31]

Answer:

1. Buy Quiet – select and purchase low-noise tools and machinery

2. Maintain tools and equipment routinely (such 3. as lubricate gears)

3. Reduce vibration where possible

4. Isolate the noise source in an insulated room or enclosure

5. Place a barrier between the noise source and the employee

6. Isolate the employee from the source in a room or booth (such as sound wall or window

Explanation:

Hope my answer will help u.

7 0

1 year 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

2 years ago

A rotor in a compressor stage has a mean blade radius of 0.285 m and an angular rotor velocity of 8500 RPMs. The static temperat

pantera1 [17]

Answer:

0,285 is the answer

Explanation:

6 0

2 years ago

How would an engineer know if a product design were feasible?

Masteriza [31]

- the last one ‘design meets all the criteria...’

8 0

2 years ago

Read 2 more answers

As described in "A Note About Bacterial Reproduction -- and the "Culture Bias,"" the organism Epulopisciumdoes not divide by bin

zloy xaker [14]

Answer:

A

Explanation:

The best method that will yield significantly more accurate result is to use spectrophotometer to read the turbidity of the sample and increase in turbidity is associated with increase biomass.

5 0

2 years ago