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dangina [55]
3 years ago
12

10) A pressure sensor consisting of a diaphragm with strain gauges bonded to its surface has the following information in its sp

ecification: Ranges: 0 to 1400 kPa, 0 to 35 000 kPa Non-linearity error: ±0.15% of full range Hysteresis error: ±0.05% of full range What is the total error due to non-linearity and hysteresis for a reading of 1000 kPa on the 0 to 1400 kPa range?
Engineering
1 answer:
noname [10]3 years ago
8 0

The definition of absolute and percentage error allows to find the result for the total error in the pressure measurement is:

      P = ( 1000 ± 2 ) kPa

      ΔP = ±2 kPa

The magnitudes is a quantity that in addition to its value has an error due to the instruments or measurement method used, therefore the correct way to express a value is:

         Magnitude = value ± absolute error

Statistical errors can be expressed in several ways:

  • Absolute (Δx). It corresponds to the uncertainty or appreciation of the instruments or the propagation of uncertainty in the equations.
  • Relative (e_r). Indicates the fraction of the error, it is the relationship between the absolute error and the magnitude.
  • Percentage (e%). It is the relative error expressed as a percentage or relative error by 100%

In all measurements, the worst case is always taken, that is, all errors go in the same direction.

They indicate the percentage errors due to non-linearity 0.15% and the error due to hysteresis 0.05%. the measurement value is P = 1000 kPa.

Let's look for the absolute errors.

                 e_r = \frac{\Delta P}{P} = e%/100

                 \Delta P = P \ \frac{percentage \ error }{100}

             

 We apply this expression to each type of error.

Non-linearity.

                \Delta P_1 = 1000 \ \frac{0.15}{100}  

                ΔP₁ = 1.5 kPa

Hysteresis.

                \Delta P_2 = 1000 \ \frac{0.05}{100}  

                ΔP₂ = 0.5 kPa

The total absolute error in the worst case is the sum of the two errors,

              \Delta P_{total} = \Delta P_1 + \Delta P_2  

              \Delta P_{total}= 1.5 +0.5 \\\Delta P_{total} = 2 kPa

               

The measurement should be given in the following way:

            P = 1000 ± 2 kPa

In conclusion using the definition of absolute and percentage error we can find the result for the total error in the pressure measurement is:

             P = ( 1000 ± 2 ) kPa

             ΔP = ±2 kPa

Learn more here:  brainly.com/question/15114851

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BabaBlast [244]

Answer:

a) since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

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Explanation:

Given that;

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1ft is given by the equation u = 0.81 + 9.2y + (4.1 × 10³y³)

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a)

consider BG system of units

u(ft/s) = 0.81 + 9.2y + (4.1 × 10³y³)

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u = 0.81 + 9.2(0.5) + (4.1 × 10³(0.5³)

u = 0.81 + 0.46 + 0.5125

u = 1.7825 ft/s lets say this is equation 1

now consider the SI system units

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also consider y=0.05ft

1ft = 3.048×10⁻¹ (from conversion table)

so 0.05ft = 0.01524m

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u = 0.9647 m/s

1m/s = 3.281 ft per seconds ( conversion table)

so

0.9647 m/s = 0.9647(3.281)

u = 3.165 ft/s lets say this is equation 2

now since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

b)

we know that the velocity of water at the surface contact is zero

u=0

so from the equation

u = 0.81 + 9.2y + (4.1 × 10³y³)

at y = 0

u = 0.81 + 9.2(0) + (4.1 × 10³(0)³)

u = 0.81 ft/s

The velocity according to the above equation at y=0 is 0.81 ft/sec but since the fluid is flowing on flat surface which is stationary this value is wrong hence the equation is NOT CORRECT

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The uniform beam is supported by two rods AB and CD that have crosssectional areas of 10 mm2 and 15 mm2 , respectively. Determin
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Answer:

w=2.25

Explanation:

It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.  

The surface of the cross-section of the stapes was determined:  

A_ab= 10 mm^2

A-cd=  15 mm^2

The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.

σ_ab = F_ab/A_ab\leqσ_allow

σ_cd =  F_cd/A_cd\leqσ_allow

In the next step we will determine the static size: Picture b).  

We apply the conditions of equilibrium:  

∑F_x=0

∑F_y=0

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∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0

              ==> F_cd = 2*w*k*N

∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0

              ==> F_ab = w*k*N

Now we determine the load w  

<u>Sector AB:  </u>

σ_ab = F_ab/A_ab\leq σ_allow=300 KPa

         = w/10*10^-6\leq σ_allow=300 KPa

w_ab = 3*10^-3 kN/m

<u>Sector CD:  </u>

σ_cd = F_cd/A_cd\leq σ_allow=300 KPa

         = 2*w/15*10^-6\leq σ_allow=300 KPa

w_cd = 2.25*10^-3 kN/m

w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}

                                ==> w=2.25 * 10^-3 kN/m

<u>The solution is:  </u>

                                w=2.25 N/m

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