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.