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vredina [299]
3 years ago
9

What is a general requirement on a sensor for the sensor to have minimal dynamic error? (The sensor is not limited to be a first

-order sensor. It is, in general, an nthorder sensor).
Engineering
1 answer:
Andrews [41]3 years ago
3 0

Answer and Explanation:

The criteria defined for the instruments that changes rapidly with time, ae called dynamic characteristics. These characteristics are namely

1. Speed of response  

2. Fidelity

3. Dynamic error

4. Measuring lag

Speed of response

It is the speed with which a measurement system responds to changes in the measured quantity.

Fidelity

It is the degree to which a measurement system indicates changes in the measured quantity without dynamic error.

Dynamic error

It is the difference between the true value of the quantity changing with time and the value indicated by the measurement system if no static error is assumed. It is also known as measurement error.

Measuring lag

It is the delay in the response of a measurement system to changes in the measured quantity. It is divided into two as follows.

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Currently, the lost time of each stage is 4 seconds, and intersection critical v/c ratio is set to be 0.75 to avoid cycle failur
KIM [24]

Find the solution in the attachments

Note: Question was incomplete, so the complete question is added in the attachments.

7 0
3 years ago
A digital filter is given by the following difference equationy[n] = x[n] − x[n − 2] −1/4y[n − 2].(a) Find the transfer function
slega [8]

Answer:

y(z) = x(z) - x(z) {z}^{ - 2}  -  \frac{1}{4} y(z) {z}^{ - 2}  \\ y(z) + \frac{1}{4} y(z) {z}^{ - 2} = x(z) - x(z) {z}^{ - 2} \\ y(z) (1 + \frac{1}{4}{z}^{ - 2}) = x(z)(1 - {z}^{ - 2}) \\  h(z) = \frac{y(z)}{x(z)}  =  \frac{(1 + \frac{1}{4}{z}^{ - 2})}{(1 - {z}^{ - 2})}

The rest is straightforward...

6 0
3 years ago
134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.
vichka [17]

Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = - 10^{\circ}C = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = 60^{\circ}C = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T =60^{\circ}C:

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at  P = 140 kPa, T =- 10^{\circ}C:

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = s'_{s} and h'_{s} = 278.06 kJ/kg.K at 700kPa

(a) Isentropic efficiency of the compressor, \eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}

\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%

(b) The temperature of the environment, T_{e} = 27^{\circ}C = 273 + 27 = 300 K

Availability at state 1, \Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg

Similarly for state 2, \Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg

Now, the efficiency of the compressor as per the second law;

\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757 = 67.57%

4 0
4 years ago
Why is the definition for second based on the Cesium 133 atom?
jarptica [38.1K]

Answer:

  The rate of vibrations of the atoms is consistent

Explanation:

A second is defined as 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.

The definition applies when the atom is in isolation at a temperature of absolute zero. Adjustments must be made for the conditions that can be maintained in practice.

Such radiation is determined by the laws of quantum mechanics, so is extremely consistent.

7 0
4 years ago
Given an unsorted array of distinct positive integers A[1..n] in the range between 1 and 10000 and an integer i in the same rang
AnnZ [28]

Answer:

Explanation:

Arbitrary means That no restrictions where placed on the number rather still each number is finite and has finite length. For the answer to the question--

Find(A,n,i)

for j =0 to 10000 do

frequency[j]=0

for j=1 to n do

frequency[A[j]]= frequency[A[j]]+1

for j =1 to n do

if i>=A[j] then

if (i-A[j])!=A[j] and frequency[i-A[j]]>0 then

return true

else if (i-A[j])==A[j] and frequency[j-A[j]]>1 then

return true

else

if (A[j]-i)!=A[j] and frequency[A[j]-i]>0 then

return true

else if (A[j]-i)==A[j] and frequency[A[j]-i]>1 then

return true

return false

7 0
3 years ago
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