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Chapter 4: Problem 55

Assume the temperature of the exhaust in an exhaust pipe can be approximated by \(T=T_{0}\left(1+a e^{-b x}\right)[1+c \cos (\omega t)],\) where \\[ T_{0}=100^{\circ} \mathrm{C}, a=3, b=0.03 \mathrm{m}^{-1}, c=0.05, \text { and } \omega=100 \mathrm{rad} / \mathrm{s} \\] If the exhaust speed is a constant \(3 \mathrm{m} / \mathrm{s}\), determine the time rate of change of temperature of the fluid particles at \(x=0\) and \(x=4 \mathrm{m}\) when \(t=0\)

Short Answer

Expert verified
The time rate of change of temperature of the fluid particles at \(x=0\) and \(x=4\) m when \(t=0\) is 0.

Step by step solution

01

Remember the Chain Rule

The Chain Rule is a fundamental tool for differentiating functions of the form \(u(v(x))\). It is stated as, \(\frac{{du}}{{dx}} = \frac{{du}}{{dv}} \cdot \frac{{dv}}{{dx}}\). In this exercise, the temperature function is made of multiple smaller functions, hence the Chain Rule will be applied extensively.
02

Time derivative of the temperature function

The supplied temperature function \(T(t, x)\) has to be differentiated with respect to time. Essentially, the time derivative of the function \(T(t, x)=T_{0}[1 + a e^{-bx}][1 + c cos (\omega t)]\) needs to be found. The time derivative, \(\frac{{dT}}{{dt}}\), is obtained by differentiating each component of the function and combining the results. This gives, \(\frac{{dT}}{{dt}} = - \omega T_0 a e^{-bx}c \sin (\omega t)\). This expression gives the rate of change of temperature with time at any particular location \(x\) along the pipe.
03

Evaluate at \(x=0\) and \(x=4\) m when \(t=0\)

The derivative function found in Step 2 must be evaluated at \(x=0\) and \(x=4\) meters. The calculation goes as follows: For \(x=0\) and \(t=0\), \(\frac{{dT}}{{dt}}_0 = - \omega T_0 a c \sin (0) = 0\). For \(x=4\) m and \(t=0\), \(\frac{{dT}}{{dt}}_4 = - \omega T_0 a e^{-b*4}c \sin (0) = 0\).

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