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Greeley [361]
3 years ago
15

What is the damped natural frequency (in rad/s) of a second order system whose undamped natural frequency is 25 rad/s and has a

damping ratio of 0.3?
Engineering
1 answer:
TiliK225 [7]3 years ago
3 0

Answer:

damped natural frequency = 23.84 rad/s

Explanation:

given data

damping ratio = 0.3

undamped natural frequency = 25 rad/s

to find out

damped natural frequency of a second order system

solution

we know that if damping ratio is = 0

then it is undamped system

and if damping ratio is > 1

then it is overdamped system

and  and if damping ratio is ≈ 1

then it is critical damped system

so damped natural frequency of a second order system formula is

damped natural frequency = wn × \sqrt{1-r^2}

here wn is undamped natural frequency and r is damping ratio

damped natural frequency = 25 × \sqrt{1-0.3^2}

damped natural frequency = 23.84 rad/s

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Explanation:

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2 years ago
1. Which type of fit implies that a piece will never fit? a. interference fit b. construction fit c. transition fit d. impeding
VladimirAG [237]

Answer:

d

Explanation:

interference fit: the dimensions are larger and high force like heating is required to fit in the object

transition fit: the dimensions are fractionally larger and slight force like hammering is required to fit in the object

impeding fit: the dimensions are larger and the object will not fit at all

construction fit: activities making a commercial space sutiable for tenant occupation.

6 0
3 years ago
A refrigerator operating on the Carnot cycle is used to make ice. Water freezing at 32oF is the cold reservoir. Heat is rejected
Mnenie [13.5K]

Answer:

Explanation:

From the given information:

Water freezing temp. T_L = 32 ^0 \ F

Heat rejected temp T_H = 72 ^0 \ F

Recall that:

The coefficient of performance is:

COP_{ref} = \dfrac{T_L}{T_H - T_L} \\ \\  = \dfrac{32+460}{(72 +460) -(32+460)} \\ \\ =\dfrac{492}{532 -492} \\ \\ = \dfrac{492}{40} \\ \\ COP_{ref} = 12.3

Again:

The efficiency given by COP can be represented by:

COP = \dfrac{Q_L}{W} \\ \\ W = \dfrac{Q_L}{COP} \\ \\ W = \dfrac{2000 \ lbm \times 144 \ Btu/lbm}{12.3} \\ \\ W = 23414.63 \ Btu

Finally; the power input in an hour can be determined by using the formula:

Power= \dfrac{W}{t} \\ \\  Power = \dfrac{23414.63 \ Btu}{ 1 \ hr} \\ \\  Power = \dfrac{23414.634 \times 1055.056 \ J}{1 \times 3600} \\ \\  Power  = 6.86 \ kW

In hp; since 1 kW = 1.34102 hp

6.86kW will be = (6.86 × 1.34102) hp

= 9.199 hp

7 0
3 years ago
Technician a says that diesel engines can produce more power because air in fuel or not mix during the intake stroke. Technician
mariarad [96]

Answer:

Technician be says that diesel engines produce more power because they use excess air to burn feel who is correct

Explanation:

He is correct as many engines are run by diesel. It produces more power as that is how cars produce more power.

3 0
2 years ago
You have been assigned to design an open cylindrical storage tank 4 meters tall with a diameter of 8 meters to be made out of A-
Katen [24]

Answer:

The required wall thickness is 1.506 \times 10^{-3} m

Explanation:

Given:

Fluid density \rho = 1200 \frac{kg}{m^{3} }

Diameter of tank d = 8 m

Length of tank l = 4 m

F.S = 4

For A-36 steel yield stress \sigma = 250 MPa,

Allowable stress \sigma _{allow} = \frac{\sigma}{F.S}

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Pressure force is given by,

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P = 47088 Pa

Now for a vertical pipe,

\sigma _{allow} = \frac{Pd}{4t}

Where t = required thickness

 t = \frac{Pd}{4 \sigma _{allow} }

 t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }

t = 1.506 \times 10^{-3} m

Therefore, the required wall thickness is 1.506 \times 10^{-3} m

8 0
3 years ago
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