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Anna71 [15]
3 years ago
8

A displacement transducer has the following specifications: Linearity error ± 0.25% reading Drift ± 0.05%/○C reading Sensitivity

error ± 0.25% reading Excitation 10 – 25 V dc Output 0 – 5 V dc Range 0 – 5 cm The transducer output is to be indicated on a voltmeter having a stated accuracy of ± 0.1% reading with a resolution of 10 μV. The system is to be used at room temperature, which can vary by ±10○C. Estimate an uncertainty in a nominal displacement of 2 cm at the design stage. Assume 95% confidence. Assume sensitivity is K = 1 V/cm.
Engineering
1 answer:
White raven [17]3 years ago
8 0

Answer:

The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

Explanation:

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

assume from specifications that k = 5v/5cm

                                                         = 1v/cm

u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2

      = 0.01225v

v = 2v * 0.001

  = 0.002v

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

= ((0.01225)^2 + (0.002)^2)^(1/2)

= 0.0124 cm

Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

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When checking the resistance of a dual voltage wye motor, there should be ____ resistance readings. 1) twelve 2) six 3) three
hjlf

Answer:

1) twelve

Explanation:

The dual voltage motors are used in day to day operations. The wye is connected with 9 lead motors. Maximum resistance can be obtained if the resistance are connected in series. To check resistance of dual voltage wye motor there must be twelve resistance readings of 1 ohm each.

8 0
3 years ago
can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and a
anyanavicka [17]

Answer:

  • branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
  • branch 2: i = 21.466∠63.435°; pf = 0.447 leading
  • total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

  X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915\times10^{-3}\approx j4.99984\,\Omega

The capacitive reactance is ...

  X_C=\dfrac{1}{j\omega C}=\dfrac{-j}{100\pi\cdot 318.31\times10^{-6}F}\approx -j10.00000\,\Omega

<u>Branch 1</u>

The impedance of branch 1 is ...

  Z1 = 8 +j4.99984 Ω

so the current is ...

  I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

<u>Branch 2</u>

The impedance of branch 2 is ...

  Z2 = 5 -j10 Ω

so the current is ...

  I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

<u>Total current</u>

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

  It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

  It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

__

The phasor diagram of the currents is attached.

_____

<em>Additional comment</em>

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

3 0
2 years ago
A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
Gekata [30.6K]

Answer:

a) \dot m_{3} = 135\,\frac{lbm}{s}, b) h_{3}=168.965\,\frac{BTU}{lbm}, c) T = 200.829\,^{\textdegree}F

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

\dot m_{1} + \dot m_{2} - \dot m_{3} = 0

The mass flow rate exiting the tank is:

\dot m_{3} = \dot m_{1} + \dot m_{2}

\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}

\dot m_{3} = 135\,\frac{lbm}{s}

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}

Properties of water are obtained from tables:

h_{1}=180.16\,\frac{BTU}{lbm}

h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)

h_{2}=29.032\,\frac{BTU}{lbm}

The specific enthalpy at outlet is:

h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }

h_{3}=168.965\,\frac{BTU}{lbm}

c) After a quick interpolation from data availables on water tables, the final temperature is:

T = 200.829\,^{\textdegree}F

8 0
3 years ago
Read 2 more answers
What is the hardest part of engineering?
Vikki [24]

ANSWER:

Aerospace Engineering. ...

Chemical Engineering. ...

Biomedical Engineering.

EXPLANATION:

This is all i know but ... I hope this helps~

7 0
1 year ago
At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15
Alenkasestr [34]

Answer:

The Poisson's Ratio of the bar is 0.247

Explanation:

The Poisson's ratio is got by using the formula

Lateral strain / longitudinal strain

Lateral strain = elongation / original width (since we are given the change in width as a result of compession)

Lateral strain = 0.15mm / 40 mm =0.00375

Please note that strain is a dimensionless quantity, hence it has no unit.

The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

Longitudinal strain = 4.1 mm / 270 mm = 0.015185

Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247

The Poisson's Ratio of the bar is 0.247

Please note also that this quantity also does not have a dimension

3 0
3 years ago
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