If \(X_{1}, X_{2^{\prime} \cdots}, X_{n}\) are independent and unbiased measurements of true values \(\mu_{1}, \mu_{2} \cdots \mu_{n}\), and \(U\left(X_{1}, X_{2^{\prime \prime}},, X_{n}\right)\) is a nonlinear function of \(X_{1}, X_{2^{\prime} \cdots}, X_{n}\), then in general \(U\left(X_{1}, X_{2^{\prime \prime}},, X_{n}\right)\).

is a biased estimate of the true value \(U\left(\mu_{1}, \mu_{2}, \ldots, \mu_{n}\right.\). A bias-corrected estimate is \(U\left(X_{1^{\prime}}, X_{2^{\prime \prime}}\right)-(1 / 2) \sum_{i=1}^{n}\left(\partial^{2} U / \partial X_{i}^{2}\right) \sigma_{X_{i}}^{2}\).When air enters a compressor at pressure \(P_{1}\) and leaves at pressure \(P_{2}\), the intermediate pressure is given by . Assume that \(P_{1}=8.1 \pm 0.1 \mathrm{MPa}\) and \(P_{2}=15.4 \pm 0.2 M P a\).

a. Estimate\(P_{3}\), and find the uncertainty in the estimate, without bias correction.

b. Compute the bias-corrected estimate of \(P_{3}\).

c. Compare the difference between the bias-corrected and non-bias-corrected estimates to the uncertainty in the non-bias-corrected estimate. Is bias correction important in this case? Explain.

Equation Transcription:

,

Text Transcription:

X_1,X_2,...,X_n

mu_1,mu_2,...mu_n

U(X_1,X_2,...,X_n)

U(X_1,, X_2,...,)-(1/2)i= sigma_i=1^n(partial^2U/partial X_i^2) X_i^2

P_1

P_2

P_3= sqrt P_1 P_2

P_1=8.10 pm .1MPa

P_2=15.4 pm 0.2MPa

P_3

Solution :

Step 1 of 3:

Let are independent and are unbiased measurement values.

Then a bias-correct estimate is

Where is non linear function .

Given,

A compressor at pressureand

Leaves at pressure .

Then the intermediate pressure is given by

.

We assume that MPa and MPa.

Our goal is :

a). We need to estimate and we have to find uncertainty in the estimate , without bias correction.

b). We need to compute the bias-corrected estimate of .

e). We need to find is bias correction important in this case? Explain.

a).

Now we have to estimate and we have to find uncertainty in the estimate , without bias correction.

Then the intermediate pressure is is

We know that and

MPa

Therefore the intermediate pressure is is 11.1687 MPa.

Given MPa

We consider and

And MPa.

We consider and .

Here we are differentiating with respect to .

Here

Then,

Therefore is 0.6894

Here we are differentiating with respect to .

Here

Then,

Therefore is 0.3626

0.10

Therefore MPa.