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Alex17521 [72]
2 years ago
12

Please upload a photo of a damper that you have come across in your daily life activities.Explain the type and nature of dampin

g (Viscous? Underdamped?) What do you think the damper’s function is?
Engineering
1 answer:
ella [17]2 years ago
3 0

Please upload a photo of a damper that you have come across in your daily life activities.Explain the type and nature of damping (Viscous? Underdamped?)What do you think the damper’s function is given below

Explanation:

  • A damper is a valve or plate that stops or regulates the flow of air inside a duct, chimney, VAV box, air handler, or other air-handling equipment.
  • A damper may be used to cut off central air conditioning (heating or cooling) to an unused room, or to regulate it for room-by-room temperature and climate control. Its operation can be manual or automatic.
  • Manual dampers are turned by a handle on the outside of a duct. Automatic dampers are used to regulate airflow constantly and are operated by electric or pneumatic motors, in turn controlled by a thermostat or building automation system.
  • Automatic or motorized dampers may also be controlled by a solenoid, and the degree of air-flow calibrated, perhaps according to signals from the thermostat going to the actuator of the damper in order to modulate the flow of air-conditioned air in order to effect climate control.
  • In a chimney flue, a damper closes off the flue to keep the weather (and birds and other animals) out and warm or cool air in. This is usually done in the summer, but also sometimes in the winter between uses.
  • In some cases, the damper may also be partly closed to help control the rate of combustion.
  • The damper may be accessible only by reaching up into the fireplace by hand or with a woodpoker, or sometimes by a lever or knob that sticks down or out. On a wood-burning stove or similar device, it is usually a handle on the vent duct as in an air conditioning system. Forgetting to open a damper before beginning a fire can cause serious smoke damage to the interior of a home, if not a house fire.
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The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to
skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 kN/m^{2}

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = 0.02\ m^{3}/s

(Since, 1 litre = 10^{-3} m^{3})

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, H_{loss} = 45 cm = 0.45\ m

Height, h_{roof} = 2.5\ m

Now,

The velocity in the bore is given by:

v = \frac{Q}{\pi (\frac{d}{2})^{2}}

v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s

Now, using Bernoulli's eqn:

\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k                  (1)

The velocity head is given by:

\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553

Now, by using energy conservation on the surface of water on the roof and that in the tank :

\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}

\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45

P_{tank} = 5.5\times \rho \times g

P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}

4 0
2 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
2 years ago
Instructions given by traffic police or construction flaggers _____. A. Are sometimes important to follow B. Are usually not imp
Anestetic [448]

Answer:

D. Overrule any other laws and traffic control devices.

Explanation:

Laws and traffic control devices are undoubtedly compulsory to be followed at every point in time to control traffic and other related situations. However, there are cases when certain instructions overrule these laws and traffic control devices. For example, when a traffic police is giving instructions, and though the traffic control devices too (such as traffic lights) are displaying their own preset lights to control some traffic, the instructions from the traffic police take more priority. This is because at that point in time, the instructions from the traffic control devices might not be just applicable or sufficient.

Also, in the case of instructions given by construction flaggers, these instructions have priority over those from controlling devices. This is because during construction traffic controls are redirected from the norms. Therefore, the flaggers such be given more importance.

6 0
2 years ago
A rectangular tank is filled with water to a depth of 1.5 m. Its longest side measures 2.5 m. What is the moment of the force ab
marishachu [46]

Answer:

The correct answer is option 'c': 13.8 kNm

Explanation:

We know that moment of a force equals

Moment=Force\times Arm

The hydro static force is given by Force=Pressure\times Area

We know that the hydrostatic pressure on a rectangular surface in vertical position is given by Pressure=\rho \times g\times h_{c.g}

For the given rectangular surface we have h_{c.g}=\frac{h}{2}=\frac{1.5}{2}=0.75m

Thus applying the values we get force as

Force=1000\times 9.81\times 0.75\times 1.5\times 2.5=27.59kN

This pressure will act at center of pressure of the rectangular plate whose co-ordinates is given by h/3 from base

Thus applying the calculated values we get

Moment=27.59\times \frac{1.5}{3}=13.8kN.m

8 0
2 years ago
Q1. In electronic circuits it is not unusual to encounter currents in the microampere range. Assume a 35 μA current, due to the
Anit [1.1K]

Answer:

2.9*10^14 electrons

Explanation:

An Ampere is the flow of one Coulomb per second, so 35 μA = is 35*10^-6 C per second.

An electron has a charge of 1.6*10^-19 C.

35*10^-6 / 1.6*10^-19 = 2.9*10^14 electrons

So, with a current o 35 μA you have an aevrage of 2.9*10^14 electrons flowing past a fixed reference cross section perpendicular to the direction of flow.

7 0
2 years ago
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