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julsineya [31]
3 years ago
15

The following displacement (cm), output (V) data have been recorded during a calibration of an LVDT. Displacement (cm), output (

V): 0.00 cm, 0.20 V; 0.125 cm, 0.60 V; 0.500 cm, 1.32 V; 0.769 cm, 1.95V; 1.019 cm; 2.33 V. Calculate the sensitivity of the LVDT expressed in (V/mm). (Please express your answer in V/mm and round up the answer to two (2) significant decimal digits.)

Engineering
1 answer:
nadya68 [22]3 years ago
4 0

Answer:

2.08V/cm

Explanation:

Plot the points on a graph. Draw the line of best fit. Calculate the gradietn of line of best fit.

Attached is the graph plotted on excel.

The equation of the line is

Votlage= 2.08× distance + 0.276

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3 years ago
Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a
nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer  (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

  1. Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer.  120 Vrms = 85 V and 24 Vrms = 17V = Vin
  2. Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
  3. Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

6 0
3 years ago
The insulation resistance of a motor operated by an electronic drive is to be tested using a megger. What precaution should you
EleoNora [17]
Use protective gear. Use insulated tools, Wear flame resistant clothing, safety glasses, and insulation gloves, Remove watches or other jewelry, Stand on an insulation mat. 03. Never connect the insulation tester to energized conductors or energized equipment and always follow the manufacturer's recommendations. When installing new electrical machinery or equipment, testing insulation resistance is important for two reasons. First, it ensures that the insulation is in adequate condition to begin operation. ... The test is accomplished by applying DC voltage through the de-energized circuit using an insulation tester. Insulation resistance should be approximately one megohm for each 1,000 volts of operating voltage, with a minimum value of one megohm. For example, a motor rated at 2,400 volts should have a minimum insulation resistance of 2.4 megohms.
4 0
3 years ago
You are testing a new jet engine in a test cell at sea level conditions. You measure the mass flow through the engine and find i
bulgar [2K]

Answer:

43248 newtons.

Explanation:

Force = mass x accelerations and units of force are newtons which are given in the question.

here mass = 125 of air and 2.2 of fuel, total = 125+2.2=127.5kg/s  and the velocity of the exhaust is 340m/s.

force = 340m/s * 127.5kg/s = 43248 newtons technically this is wrong (observe units) but i will expalin how i have taken acceleration as a velocity here and mass/unit time as simply mass.

see force is mass times acceleration or deceleration, here our velocity is not changing therefore it is constant 340m/s but if it were to change and become 0 in one second then there would be -340m/s^2 (note the units ) of deceleration and there would be force associated with it and that force is what i have calculated here. similarly there would be mass in flow rate of mass per second, which is also in that one second of time.

let's calculate error.

error = (actual-calculated)/actual. = (43248-60000)/43248= -38.734% less is ofcourse greater than 2%.

So the load cell is not reading correct to within 2% and it should read 43248newtons.

5 0
2 years ago
Read 2 more answers
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