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Ivan
3 years ago
5

An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir

ed that the amplitude of motion should not exceed 0.1 mm per 1 N of unbalance force. The mass of the motor is 120 kg and the operating speed is 720 rpm Use the concept of transfer function to determine the required stiffhess coefficient of each mount.
Engineering
1 answer:
Artyom0805 [142]3 years ago
8 0

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

\frac{A}{F} =  \frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}

Formula Used:

A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

\omega = \frac{2 \pi\times N}{60}

so,

\omega = \frac{2 \pi\times 720}{60} = 75.39 rad/s

Using the given formula:

Damping is negligible, so, \zeta = 0

\frac{A}{F} will give the tranfer function

Therefore,

\frac{A}{F} = \frac{1}{\sqrt{(4K - 120\ ^{2})}}

0.1\times 10^{-3} =  \frac{1}{\sqrt{(4K - 120\ ^{2})}}

Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm

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4 years ago
A closed, rigid, 0.50 m^3 tank is filled with 12 kg of water. The initial pressure is p1 = 20 bar. The water is cooled until the
fgiga [73]

Answer:

X1= 41%

heat transfer = -3450.676 KJ

Explanation:

To get the properties for pure substance in a system we need to know at least to properties. These are usually pressure and temperature because they’re easy to measure. In this case we know the initial pressure (20 bar) which is not enough to get all the properties, but they ask to determine quality, this a property that just have meaning in the two-phase region (equilibrium) so with this information we can get the temperature of the system and all its properties.

There is another property that we can calculate from the data. This is the specific volume. This is defined as \frac{volume}{mass}. We know the mass (12 Kg) and we can assume the volume is the volume of the tank  (0.5 m^{3}) because they say that the tank was filled.  

With this we get a specific volume of  

Specific volume = \frac{0,5 m^{3}}{ 12 kg}= 0.04166667 \frac{m^{3}}{Kg}

From the thermodynamic tables we can get the data for the saturated region with a pressure of 20 bar.

Temperature of saturation = 212.385 °C  

Specific volume for the saturated steam (vg) = 0.0995805 \frac{m^{3}}/{Kg}[\tex]Specific volume for the saturated liquid (vf)= 0.00117675 [tex]\frac{m^{3}}/{Kg}[\tex] The specific volume that we calculate before 0.04166667 m^3/Kg is between 0.00117675 m^3/Kg and 0.0995805 m^3/Kg so  we can be sure that we are in two-phase region (equilibrium).The quality (X) is defined as the percentage in mass of saturated steam in a mix (Two-phase region) The relation between specific volume and quality is  [tex]v = (1-x)*v_{f} + x*v_{g}[\tex]  where  v in the specific volume in the condition (0.04166667 m^3/Kg)  vf = Specific volume for the saturated liquid (0.00117675 m^3/Kg)vg = Specific volume for the saturated steam (0.0995805 m^3/Kg)x = qualityclearing the equation we get:[tex]X = \frac{(v-v_{f})}{(v_{g}-v_{f})} 

X =\frac{(0.04166667- 0.00117675)}{ (0.0995805 – 0.00117675)} = 0.411

The quality is 41%

To calculate the heat transfer we use the next equation.  

Q = m * Cp * delta T  

Where  

Q = heat transfer (Joules, J)

m= mass of the substance (g)

Cp = specific heat (J/g*K) from tables  

Delta T = change in temperature in K for this equation.  

The mass of the substance is 12 kg or 12000 g for this equation  

Cp from tables is 4,1813 J/g*K. You can find this value for water in different states. Here we are using the value for liquid water.  

For delta T, we know the initial temperature 212.385 °C.

We also know that the system was cooled. Since we don’t have more information, we can assume that the system was cooled until a condition where all the steam condensates so now we have a saturated liquid. Since we know the pressure (4 bar), we can get the temperature of saturation for this condition from the thermodynamics tables. This is 143.613 °C, so this is the final temperature for the system.  

T(K) = T°C +273  

T1(K) = 212.385 + 273.15 = 485.535 K

T2 (K) = 143.613 +273.15= 416.763 K

Delta T (K) = (T2-T1) =416.763 K - 485.535 K = -68.772 K

Now we can calculate Q

Q = 12000g * 4,1813 J/g*K* (-68.772 K) = -3450676.36 J or -3450.676 KJ

Is negative because the heat is transfer from the water to the surroundings

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Answer:

Explanation:

Engineering Stress is defined as Load applied to the original cross-sectional area which we have taken in the start.

True stress is defined as the load divided by area of cross-section of specimen at that instant.

Engineering stress  and true stress can be expressed by relation

\sigma _T=\sigma _E\left ( 1+\epsilon _E\right )

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\sigma _T=True\ stress

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Answer:

Thomas Elva Edison

He invented light.

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3 years ago
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