Chapter 19, Problem 54AP

(a)

To determine

The angle between the invar bars.

Answer to Problem 54AP

The angle between the invar bars is $\underset{\_}{\theta =2{\sin }^{-1}\left(1+{\alpha }_{Al}{T}_C/2\right)}$.

Explanation of Solution

The diagram for the three metal bars is depicted below.

Write the equation angle from trigonometry.

  $\sin \left(\theta /2\right)=\frac{L_f}{L_0}$                                                                                          (I)

Here, $\theta$ is the angle, $L_0$ is the initial length, $L_f$ is the final length.

Write the equation for the final length.

  $L_f=\left(L_0/2\right)\left[1+{\alpha }_{Al}\left(\Delta T\right)\right]$                                                                    (II)

Here, ${\alpha }_{Al}$ is the linear expansion coefficient of aluminium and $\left(\Delta T\right)$ is the change in temperature.

Conclusion:

Substitute $\left(L_0/2\right)\left[1+{\alpha }_{Al}\left(\Delta T\right)\right]$ for $L_f$ and $T_C$ for $\left(\Delta T\right)$ in equation (I) to find $\theta$.

  $\begin{array}{l}\sin \left(\theta /2\right)=\frac{\left(L_0/2\right)\left[1+{\alpha }_{Al}\left(\Delta T\right)\right]}{L_0}\\ \left(\theta /2\right)={\sin }^{-1}\left[\frac{\left[1+{\alpha }_{Al}\left(T_C\right)\right]}{2}\right]\\ \theta =2{\sin }^{-1}\left[\frac{\left[1+{\alpha }_{Al}\left(T_C\right)\right]}{2}\right]\end{array}$

Thus, the angle between the invar bars is $\underset{\_}{\theta =2{\sin }^{-1}\left(1+{\alpha }_{Al}{T}_C/2\right)}$.

(b)

To determine

The value of angle is accurate for both negative and positive temperature or not.

Answer to Problem 54AP

Yes, the value of angle is accurate for both negative and positive temperature.

Explanation of Solution

If the temperature difference positive or negative, it will not affect the value of angle.

If the temperature is positive, the bar will expand. As well as the temperature i negative, the bar will have contraction.

Thus, the value of angle is accurate for both negative and positive temperature.

(c)

To determine

The value of angle is accurate for zero degree temperature or not.

Answer to Problem 54AP

Yes, the value of angle is accurate for zero degree temperature.

Explanation of Solution

Write the expression for the angle.

$\theta =2{\sin }^{-1}\left[\frac{\left[1+{\alpha }_{Al}\left(T_C\right)\right]}{2}\right]$

At, $T_C=0$, the value of angle is

$\begin{array}{c}\theta =2{\sin }^{-1}\left(\frac{1}{2}\right)\\ =60°\end{array}$

Thus, the value of angle is accurate for zero degree temperature.

(d)

To determine

The angle between the invar bars when the invar also expand.

Answer to Problem 54AP

The angle between the invar bars when the invar also expand is $\underset{\_}{\theta =2{\sin }^{-1}\left(\frac{\left(1+{\alpha }_{Al}{T}_C\right)}{2\left(1+{\alpha }_{\text{invar}}{T}_C\right)}\right)}$.

Explanation of Solution

Refer Figure 1.

Write the equation angle from trigonometry.

  $\sin \left(\theta /2\right)=\frac{L_f}{L_{\text{in}}}$ (III)

Here, $\theta$ is the angle, $L_{\text{in}}$ is the final length of inver, $L_f$ is the final length.

Write the equation for the final length of the aluminium.

  $L_f=\left(L_0/2\right)\left[1+{\alpha }_{Al}\left(\Delta T\right)\right]$                                                             (IV)

Here, ${\alpha }_{Al}$ is the linear expansion coefficient of aluminium and $\left(\Delta T\right)$ is the change in temperature.

Write the equation for the final length of the invar.

  $L_{\text{in}}=L_0\left[1+{\alpha }_{\text{invar}}\left(\Delta T\right)\right]$                                                             (V)

Here, ${\alpha }_{\text{in}}$ is the linear expansion coefficient of inver.

Conclusion:

Substitute $\left(L_0/2\right)\left[1+{\alpha }_{Al}\left(\Delta T\right)\right]$ for $L_f$ and $T_C$ for $\left(\Delta T\right)$, $L_0\left[1+{\alpha }_{\text{invar}}\left(\Delta T\right)\right]$ for $L_{\text{in}}$ in equation (III) to find $\theta$.

  $\begin{array}{l}\sin \left(\theta /2\right)=\frac{\left[\left(L_0/2\right)\left[1+{\alpha }_{Al}\left(\Delta T\right)\right]\right]}{\left[L_0\left[1+{\alpha }_{\text{invar}}\left(\Delta T\right)\right]\right]}\\ \left(\theta /2\right)={\sin }^{-1}\left(\frac{\left[1+{\alpha }_{Al}{T}_C\right]}{\left[2\left[1+{\alpha }_{\text{invar}}{T}_C\right]\right]}\right)\\ \theta =2{\sin }^{-1}\left(\frac{\left[1+{\alpha }_{Al}{T}_C\right]}{\left[2\left[1+{\alpha }_{\text{invar}}{T}_C\right]\right]}\right)\end{array}$

Thus, the angle between the invar bars when the invar also expand is $\underset{\_}{\theta =2{\sin }^{-1}\left(\frac{\left(1+{\alpha }_{Al}{T}_C\right)}{2\left(1+{\alpha }_{\text{invar}}{T}_C\right)}\right)}$.

(e)

To determine

The greatest value of angle between the invar bars.

Answer to Problem 54AP

The greatest value of angle between the invar bars is $\underset{\_}{61.0°}$.

Explanation of Solution

Write the equation for the angle

  $\theta =2{\sin }^{-1}\left(\frac{\left[1+{\alpha }_{Al}{T}_C\right]}{\left[2\left[1+{\alpha }_{\text{invar}}{T}_C\right]\right]}\right)$                                                                  (VI)

Conclusion:

Substitute $24\times {10}^{-6}\text{/°C}$ for ${\alpha }_{Al}$ and $0.9\times {10}^{-6}\text{/°C}$ for ${\alpha }_{\text{invar}}$, $660\text{°C}$ for $T_C$ in equation (VI) to find $\theta$.

  $\begin{array}{c}\theta =2{\sin }^{-1}\left(\frac{\left[1+\left(24\times {10}^{-6}\text{/°C}\right)\left(660\text{°C}\right)\right]}{\left[2\left[1+\left(0.9\times {10}^{-6}\text{/°C}\right)\left(660\text{°C}\right)\right]\right]}\right)\\ =2\left[{\sin }^{-1}\left(0.508\right)\right]\\ =61.0°\end{array}$

Thus, the greatest value of angle between the invar bars is $\underset{\_}{61.0°}$.

(f)

To determine

The smallest value of angle between the invar bars.

Answer to Problem 54AP

The smallest value of angle between the invar bars is $\underset{\_}{59.6°}$.

Explanation of Solution

Write the equation for the angle

  $\theta =2{\sin }^{-1}\left(\frac{\left[1+{\alpha }_{Al}{T}_C\right]}{\left[2\left[1+{\alpha }_{\text{invar}}{T}_C\right]\right]}\right)$                                                                 (VI)

Conclusion:

Substitute $24\times {10}^{-6}\text{/°C}$ for ${\alpha }_{Al}$ and $0.9\times {10}^{-6}\text{/°C}$ for ${\alpha }_{\text{invar}}$, $\text{-273°C}$ for $T_C$ in equation (VI) to find $\theta$.

  $\begin{array}{c}\theta =2{\sin }^{-1}\left(\frac{\left[1+\left(24\times {10}^{-6}\text{/°C}\right)\left(-\text{273°C}\right)\right]}{\left[2\left[1+\left(0.9\times {10}^{-6}\text{/°C}\right)\left(-\text{273°C}\right)\right]\right]}\right)\\ =2\left[{\sin }^{-1}\left(0.497\right)\right]\\ =59.6°\end{array}$

Thus, The smallest value of angle between the invar bars is $\underset{\_}{59.6°}$.