Question about source flow rate across line AB.

AI Thread Summary
The discussion revolves around determining the flow rate in Line AB, with two differing calculations leading to different results. One calculation yields a flow rate of 0.3012q, while the book provides a flow rate of 0.19878q, based on different values for θB. The discrepancy arises from the multi-valued nature of the arctan function, which can lead to selecting incorrect angles. Participants agree that the book's answer may be incorrect and suggest that the flow rate should be represented as Q_{AB}^{*} instead of Q_{AB}^{'} for accuracy. The conversation emphasizes the importance of selecting the correct angle when using arctan to avoid such discrepancies.
tracker890 Source h
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Homework Statement
Determine flow rate per unite width in the line
Relevant Equations
flow rate equation
Q:Please hlep me to understand which ans is correct.To determine the flow rate in Line AB.
$$\mathrm{Known}:V_A,q,r_A = constant.$$
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so/
select:## A,{B}^{\text{'}},B,A,## is control volume
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$${Q}_{AB}={Q}_{A{B}^{\text{'}}}=\iint _{A}^{}({V}_{A})dA={\int }_{{\theta }_{A}}^{{\theta }_{B}}({V}_{A}){r}_{A}d\theta $$$$\overset\rightharpoonup{V}=\triangledown \phi =<\frac{\partial \phi }{\partial r},\frac{1}{r}\frac{\partial \phi }{\partial \theta }>=<\frac{1}{r}\frac{\partial \psi }{\partial \theta },-\frac{\partial \psi }{\partial r}>=<{V}_{r},{V}_{\theta }> $$$$\therefore V_A=\frac1{r_A}\frac{\partial\psi}{\partial\theta}\;$$$$Q_{AB}=\int_{\theta_A}^{\theta_B}{(V_A)}r_Ad\theta\;=\;\int_{\theta_A}^{\theta_B}{(\frac1{r_A}\frac{\partial\psi}{\partial\theta})}r_Ad\theta=\int_{\theta_A}^{\theta_B}{(\frac{\partial\psi}{\partial\theta})}d\theta=\psi_B-\psi_A$$to find ##\psi##,
$$F(z)=\frac q{2\pi}\ln(z)=\frac q{2\pi}ln(re^{i\theta})=\frac q{2\pi}\ln r+i\frac q{2\pi}\theta=\phi+i\psi$$so $$\psi=\frac q{2\pi}\theta$$
$$Q_{AB}=\psi\left(\theta_B\right)\mathit-\psi\left(\theta_A\right)\mathit=\frac q{2\pi}(\theta_B-\theta_A)$$$$\theta_A=\tan^{-1}\left(\frac11\right)=0.7854\;rad,$$
$$\theta_B=\frac\pi2+\tan^{-1}\left(\frac1{0.5}\right)=2.6779\;rad$$
So ans by myself is
$$\therefore Q_{AB}=\frac q{2\pi}{(2.6779-0.7854)}=0.3012q............(Ans(1))$$$$////////////////////////$$
But book say:
$$\theta_A=\tan^{-1}\left(\frac yx\right)=\tan^{-1}\left(\frac{\mathit1}{\mathit1}\right)=0.7854\;rad$$$$\theta_B=\tan^{-1}\left(\frac yx\right)=\tan^{-1}\left(\frac{0.5}{-1}\right)\;=\;-0.4636\;rad$$$$Q_{AB}=\psi\left(\theta_A\right)-\psi\left(\theta_B\right)=\frac q{2\pi}{(0.7854+0.4636)}=0.19878q........(Ans(2))$$
 
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You may have noticed that the two values for θB differ by π.
The result of arctan is of course multi-valued, with the values at intervals of π. So it is always necessary to make sure that the right value is selected. Question is, which of you selected the right value? (I'm with you.)
 
haruspex said:
You may have noticed that the two values for θB differ by π.
The result of arctan is of course multi-valued, with the values at intervals of π. So it is always necessary to make sure that the right value is selected. Question is, which of you selected the right value? (I'm with you.)
I think the flow rate in book is ## {Q}_{A{B}^{*}} ## not ## {Q}_{A{B}^{\text{'}}}##.
So the book answer is not correct.
Am I right ?
1670664112315.png
 
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tracker890 Source h said:
I think the flow rate in book is ## {Q}_{A{B}^{*}} ## not ## {Q}_{A{B}^{\text{'}}}##.
So the book answer is not correct.
Am I right ?
View attachment 318535
I think so.
 
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Likes tracker890 Source h
I agree with your result. The included angle is ##\tan^{-1}2+\frac{\pi}{4}##
 
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