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3 years ago

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A pressure transducer has the following specifications: Input rage: 0-100 psi and the corresponding Output Range: 0-5 Volts. Lin

earity Error: 0.10% of the Reading Hysteresis Error: 0.10% of the Reading Sensitivity Error: 0.15% of the Reading Zero Drift Error: 0.20% of the Reading Its output is read via a voltmeter with instrument error of 0.10% of the reading and resolution of 0.01 V. If the applied pressure on the transducer is 65 psi, what is the design stage uncertainty of this pressure measurement system?

1 answer:

motikmotik3 years ago

6 0

Answer:

0.287

Explanation:

Design-stage uncertainty can be expressed as :

Ud = √ Uo^2 + Uc^2 ------ ( 1 )

where : Uo = 1/2( resolution value ) = 1/2 * 0.01 V = 0.005 V

Uc = √(0.10)^2 + (0.10)^2 + (0.15)^2 + (0.20)^2 = 0.287

back to equation 1

Ud = √ ( 0.005)^2 + ( 0.287 )^2 = 0.287

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A civil engineer is analyzing the compressive strength of concrete. The compressive strength is approximately normal distributed

hram777 [196]

Answer:

See explanation

Explanation:

Solution:-

- A study on compressive strength of a concrete was made. The distribution of compressive strength ( experimental testing ) was normally distributed with variance ( σ^2 ).

- A random sample of n = 12 specimens were taken and the mean compressive strength ( μ ) of 3500 psi was claimed.

- We are to test the claim made by the civil engineer regarding the mean compressive strength of the concrete. The data of compressive strength of each specimen from the sample is given below:

3273, 3229, 3256, 3272, 3201, 3247, 3267, 3237,

3286, 3210, 3265, 3273

- We will conduct the hypothesis whether the mean compressive strength of the concrete conforms to the claimed value.

Null hypothesis: μ = 3500 psi

Alternate hypothesis: μ ≠ 3500 psi

- The type of test performed on the sample data will depend on the application of Central Limit Theorem.

- The theorem states that the sample can be assumed to be normally distributed if drawn from a normally distributed population. ( We are given the population is normally distributed; hence, theorem applies )

- We will approximate the mean of the population ( μ ) with the sample mean ( x ), as per the implication specified by the theorem.

- The mean of the sample ( x ) is calculated as follows:

$x = \frac{Sum ( x_i )}{n} \\\\x = \frac{Sum ( 3273+ 3229+ 3256+ 3272+ 3201+ 3247+ 3267+ 3237+ 3286+ 3210+ 3265+3273 )}{12} \\\\x = \frac{39016}{12} \\\\x = 3251.3333$

- Since, we are testing the average compressive strength of a concrete against a claimed value. Any value that deviates significantly from the claimed value is rejected. This corroborates the use of one sample two tailed test.

- The test value may be evaluated from either z or t distribution. The conditions for z-test are given below:

The population variance is known OR sample size ( n ≥ 30 )

- The population variance is known; hence, we will use z-distribution to evaluate the testing value as follows:

$Z-test = \frac{x - u}{\sqrt{\frac{sigma^2}{n} } } \\\\Z-test = \frac{3251.333 - 3500}{\sqrt{\frac{1000^2}{12} } } \\\\Z-test = -27.24$

- The rejection region for the hypothesis is defined by the significance level ( α = 0.01 ). The Z-critical value ( limiting value for the rejection region ) is determined:

Z-critical = Z_α/2 = Z_0.005

- Use the list of correlation of significance level ( α ) and critical values of Z to determine:

Z-critical = Z_0.005 = ± 2.576

- Compare the Z-test value against the rejection region defined by the Z-critical value.

Rejection region: Z > 2.576 or Z < -2.576

- The Z-test value lies in the rejection region:

Z-test < Z-critical

-27.24 < -2.576 .... Null hypothesis rejected

Conclusion: The claim made by the civil engineer has little or no statistical evidence as per the sample data available; hence, the average compressive strength is not 3500 psi.

- To construct a confidence interval for the mean compressive strength ( μ ) we need to determine the margin of error for the population.

- The margin of error (ME) is defined by the following formula:

$ME = Z^*. \frac{sigma}{\sqrt{n} }$

Where,

- The ( Z* ) is the critical value for the defined confidence level ( CI ):

- The confidence interval and significance level are related and critical value Z* is as such:

α = 1 - CI , Z* = Z_α/2

- The critical values for ( CI = 99% & 95% ) are evaluated:

α = 1 - 0.99 = 0.01 , α = 1 - 0.95 = 0.05

Z* = Z_0.005 , Z* = Z_0.025

Z* = ± 2.58 , Z* = ± 1.96

- The formulation of Confidence interval is given by the following inequality:

[ x - ME < μ < x + ME ]

[ x - Z*√σ^2 / n < μ < x + Z*√σ^2 / n ]

- The CI of 95% yields:

[ 3251.33 - 1.96*√(1000 / 12) < μ < 3251.33 + 1.96*√(1000 / 12) ]

[ 3251.333 - 17.89227 < μ < 3251.33 + 17.89227 ]

[ 3233.44 < μ < 3269.23 ]

- The CI of 99% yields:

[ 3251.33 - 2.58*√(1000 / 12) < μ < 3251.33 + 2.58*√(1000 / 12) ]

[ 3251.333 - 23.552 < μ < 3251.33 + 23.552 ]

[ 3227.78 < μ < 3274.88 ]

- We see that the width of the confidence interval increases as the confidence level ( CI ) increases. This is due to the increase in critical value ( Z* ) associated with the significance level ( α ) increases.

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3 years ago

A 75 ohm coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 + j75 Ohm. If the diel

Hatshy [7]

Answer:

The load reflection coefficient, $\Gamma =0.62\angle 82.875^{\circ} \Omega$

Reflection coefficient at input, $\Gamma = 0.62\angle - 147.518^{\circ} \Omega$

SWR = 4.26

Given:

Characteristic impedance of the co-axial cable, $Z_{c} = 75 \Omega$

Length of the cable, L = 2.0 cm = 0.02 m

$Z_{Load} = 37.5 + j75 \Omega$

Dielectric constant, K = 2.56

frequency, f = 3.0 GHz = $3.0 \times 10^{9} Hz$

Explanation:

In order to calculate the reflection coefficient at load, we first calculate these:

The line input impedance $Z_{i}$ is given by:

$Z_{i} = Z_{c}\frac{Z_{Load} + jZ_{c} tan(\beta L)}{Z_{c} + jZ_{Load} tan (\beat L)}$ (1)

Now, we calculate the value of $\beta$ :

$\beta = \frac{2\pi}{\lambda'} = \farc{2\pi f\sqrt{K}}{c}$

(since, $\lambda' = \farc{c}{f\sqrt{K}}$ )

$\beta = \farc{2\pi f\sqrt{2.56}}{3\times 10^{8}} = 100.53$

Now, Substituting the value in eqn (1):

$Z_{i} = 75\frac{37.5 + j75 + j75 tan(100.53\times 0.02)}{75 + j(37.5 + j75) tan ( 100.53\times 0.02)} = 18.99 - j20.55 \Omega = 27.98\angle - 47.257^{\circ} \Omega$

Now, the load reflection coefficient is given by:

$\Gamma = \frac{Z_{Load} - Z_{c}}{Z_{c} + Z_{Load}}}$

Thus

$\Gamma = \frac{37.5 + j75 - 75}{75 + 37.5 + j75}} = 0.077 + j0.615 = 0.62\angle 82.875^{\circ} \Omega$

Similarly,

Reflection coefficient at input:

$\Gamma' = \frac{Z_{i} - Z_{c}}{Z_{c} + Z_{i}}}$

$\Gamma' = \frac{18.99 - j20.55 - 75}{75 + 18.99 - j20.55}} = - 0.523 - j0.334 = 0.62\angle - 147.518^{\circ} \Omega$

Now, the SWR is given by:

SWR, Standing Wave Ratio = $\frac{1 +|\Gamma|}{1 - |\Gamma|}$

SWR = $\frac{1 +|0.62|}{1 - |0.62|} = 4.26$

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3 years ago

A stream of moist air flows into an air conditioner with an initial humidity ratio of 0.6 kg(vapor)/kg(dry air), and a dry air f

BaLLatris [955]

Answer:

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Explanation:

The final humidity ratio is computed by the Principle of Mass Conservation:

Dry Air

$\dot m_{in} = \dot m_{out}$

Moist

$\dot m_{in} \cdot \omega_{in} + \dot m_{w} = \dot m_{out}\cdot \omega_{out}$

Then, the final humidity ratio is:

$\omega_{out} = \frac{\dot m_{in}\cdot \omega_{in}+\dot m_{w}}{\dot m_{out}}$

$\omega_{out} = \omega_{in} + \frac{\dot m_{w}}{\dot m_{out}}$

$\omega_{out} = 0.6\,\frac{kg\,H_{2}O}{kg\,DA} + \frac{0.4\,\frac{kg\,H_{2}O}{s} }{1.5\,\frac{kg\,DA}{s} }$

$\omega_{out} = 0.867\,\frac{kg\,H_{2}O}{kg\,DA}$

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