If X1,X2, ..., Xn are independent and unbiased

Question

Problem 23SE

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 ${P}_{3}=\sqrt{{P}_{1}{P}_{2}}$ . 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:

${X}_{1},{X}_{2},...,{X}_{n}$

${\mu{}}_{1},{\mu{}}_{2},...{\mu{}}_{n}$

$U\left({{X}_{1},{X}_{2},...,{X}_{n}}\right)$

$U({X}_{1},$ , ${X}_{2},...,)-(1/2)\sum\limits_{i=1}^{n}{(\partial{}}^{2}U/\partial{}{{X}_{i}^{2})}_{\ }^{\ } {\sigma{}}_{{X}_{i}}^{2}$

${P}_{1}$

${P}_{2}$

${P}_{3}=\sqrt{{P}_{1}{P}_{2}}$

${P}_{1}=8.1\pm{}0.1MPa$

${P}_{2}=15.4\pm{}0.2MPa$

${P}_{3}$

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

Answer 1

Step-by-Step Solution:

Solution :

Step 1 of 3:

Let ${X}_{1},{X}_{2},{X}_{3},...,{X}_{n}$ are independent and ${\mu{}}_{1},{\mu{}}_{2},{\mu{}}_{3},...,{\mu{}}_{n}$ are unbiased measurement values.

Then a bias-correct estimate is

$U({X}_{1},{X}_{2},{X}_{3},...,{X}_{n})=$ $\left({\frac{1}{2}}\right)\sum\limits_{i=1}^{n}\left({\frac{{\delta{}}^{2}U}{\delta{}{X}_{i}^{2}}}\right){\sigma{}}_{{X}_{i}}^{2}$

Where $U({X}_{1},{X}_{2},{X}_{3},...,{X}_{n})$ is non linear function ${X}_{1},{X}_{2},{X}_{3},...,{X}_{n}$ .

Given,

A compressor at pressure ${P}_{1}$ and

Leaves at pressure ${P}_{2}$ .

Then the intermediate pressure is given by

${P}_{3}=\sqrt{{P}_{1}{P}_{2}\ }$ .

We assume that ${P}_{1}=8.1\pm{}0.1$ MPa and ${P}_{2}=15.4\pm{}0.2$ MPa.

Our goal is :

a). We need to estimate ${P}_{3}$ and we have to find uncertainty in the estimate , without bias correction.

b). We need to compute the bias-corrected estimate of ${P}_{3}$ .

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

a).

Now we have to estimate ${P}_{3}$ and we have to find uncertainty in the estimate , without bias correction.

Then the intermediate pressure is ${P}_{3}$ is

${P}_{3}=\sqrt{{P}_{1}{P}_{2}}$

We know that ${P}_{1}=8.1$ and ${P}_{2}=15.4$

${P}_{3}=\sqrt{(8.1)\ (15.4)}$

${P}_{3}=\sqrt{124.74\ }$

${P}_{3}=11.1687$ MPa

Therefore the intermediate pressure is ${P}_{3}$ is 11.1687 MPa.

Given ${P}_{1}=8.1\pm{}0.1$ MPa

We consider ${P}_{1}=8.1$ and ${\sigma{}}_{{P}_{1}}=0.1$

And ${P}_{2}=15.4\pm{}0.2$ MPa.

We consider ${P}_{2}=15.4$ and ${\sigma{}}_{2}=0.2$ .

Here we are differentiating with respect to ${P}_{1}$ .

$\frac{\partial{}{P}_{3}}{\partial{}{P}_{1}}=\frac{1}{2\sqrt{P1P2\ }}\times{}{P}_{2}$

$\frac{\partial{}{P}_{3}}{\partial{}{P}_{1}}=\frac{{P}_{2}}{2\sqrt{P1P2\ }}$

Here ${P}_{3}=\sqrt{P1P2\ }$

Then,

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{1}}=\frac{15.4}{2\sqrt{(8.1)\ (15.4)\ }}$

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{1}}=\frac{15.4}{2\sqrt{124.74\ }}$

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{1}}=\frac{15.4}{2\times{}11.1687}$

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{1}}=\frac{15.4}{22.3374}$

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{1}}=0.6894$

Therefore $\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{1}}$ is 0.6894

Here we are differentiating with respect to ${P}_{1}$ .

$\frac{\partial{}{P}_{3}}{\partial{}{P}_{2}}=\frac{{P}_{1}}{2\sqrt{P1P2\ }}$

Here ${P}_{3}=\sqrt{P1P2\ }$

Then,

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{2}}=\frac{8.1}{2\sqrt{(8.1)\ (15.4)\ }}$

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{2}}=\frac{8.1}{2\sqrt{124.74\ }}$

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{2}}=\frac{8.1}{2\times{}11.1687}$

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{2}}=\frac{8.1}{22.3374}$

$\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{2}}=0.3626$

Therefore $\frac{\partial{}\sqrt{P1P2\ }}{\partial{}{P}_{2}}$ is 0.3626

${\sigma{}}_{{P}_{3}}=$ $\sqrt{{\left({\frac{\partial{}{P}_{3}}{\partial{}{P}_{1}}}\right)}^{2}{\sigma{}}_{{P}_{1}}^{2}+{\left({\frac{\partial{}{P}_{3}}{\partial{}{P}_{2}}}\right)}^{2}{\sigma{}}_{{P}_{2}}^{2}}$

${\sigma{}}_{{P}_{3}}=$ $\sqrt{{\left({\frac{\partial{}\sqrt{P1P2}}{\partial{}{P}_{1}}}\right)}^{2}{(0.1)}^{2}+{\left({\frac{\partial{}\sqrt{P1P2}}{\partial{}{P}_{2}}}\right)}^{2}{(0.2)}^{2}}$

${\sigma{}}_{{P}_{3}}=$ $\sqrt{{\left({0.6894}\right)}^{2}\times{}0.01+{\left({0.3626}\right)}^{2}\times{}0.04}$

${\sigma{}}_{{P}_{3}}=$ $\sqrt{0.4752\times{}0.01+{\left({0.13147}\right)}^{\ }\times{}0.04}$

${\sigma{}}_{{P}_{3}}=$ $\sqrt{0.004752+0.005259}$

${\sigma{}}_{{P}_{3}}=$ $\sqrt{0.0100}$

${\sigma{}}_{{P}_{3}}=$ 0.10

Therefore ${P}_{3}=11.1687\pm{}0.10$ MPa.

Step 2 of 3