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Chapter 4: Problem 57

The following pressures for the air flow in Problem 4.24 were measured: $$\begin{array}{cccc} & x=0 & x=10 \mathrm{m} & x=20 \mathrm{m} \\ \hline t=0 \mathrm{s} & p=101 \mathrm{kPa} & p=101 \mathrm{kPa} & p=101 \mathrm{kPa} \\ t=1.0 \mathrm{s} & p=121 \mathrm{kPa} & p=116 \mathrm{kPa} & p=111 \mathrm{kPa} \\ t=2.0 \mathrm{s} & p=141 \mathrm{kPa} & p=131 \mathrm{kPa} & p=121 \mathrm{kPa} \\ t=3.0 \mathrm{s} & p=171 \mathrm{kPa} & p=151 \mathrm{kPa} & p=131 \mathrm{kPa} . \end{array}$$ Find the local rate of change of pressure \(\partial p / \partial t\) and the convective rate of change of pressure \(V\) on \(/ \partial x\) at \(t=2.0 \mathrm{s}\) and \(x=10 \mathrm{m}.\)

Short Answer

Expert verified
The local rate of change of pressure (\(\partial p / \partial t\)) at \(t=2.0 \mathrm{s}\) and \(x=10 \mathrm{m}\) is \(17.5 \mathrm{kPa/s}\). The convective rate of change of pressure (\(\partial p / \partial x\)) at the same point is \(-1 \mathrm{kPa/m}\).

Step by step solution

01

Calculate the local rate of change of pressure (\(\partial p / \partial t\))

The local rate of change of pressure at \(t=2.0 \mathrm{s}\) and \(x=10 \mathrm{m}\) can be calculated by taking the average of the change in pressure over the change in time between \(t=1.0 \mathrm{s}\) and \(t=3.0 \mathrm{s}\) at \(x=10 \mathrm{m}\). Accordingly, \(\partial p / \partial t = \frac{p(t=3.0 \mathrm{s}) - p(t=1.0 \mathrm{s})}{3.0 \mathrm{s} - 1.0 \mathrm{s}} = \frac{151 \mathrm{kPa} - 116 \mathrm{kPa}}{3.0 \mathrm{s} - 1.0 \mathrm{s}} = 17.5 \mathrm{kPa/s}\)
02

Calculate the convective rate of change of pressure (\(\partial p / \partial x\))

The convective rate of change of pressure at \(t=2.0 \mathrm{s}\) and \(x=10 \mathrm{m}\) can be calculated by taking the average of the change in pressure over the change in distance between \(x=0 \mathrm{m}\) and \(x=20 \mathrm{m}\) at \(t=2.0 \mathrm{s}\). Accordingly, \(\partial p / \partial x = \frac{p(x=20 \mathrm{m}) - p(x=0 \mathrm{m})}{20 \mathrm{m} - 0 \mathrm{m}} = \frac{121 \mathrm{kPa} - 141 \mathrm{kPa}}{20 \mathrm{m} - 0 \mathrm{m}} = -1 \mathrm{kPa/m}\)

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Most popular questions from this chapter

A fluid particle flowing along a stagnation streamline, as shown in Video \(V 4.9\) and Fig. \(P 4.38,\) slows down as it approaches the stagnation point. Measurements of the dye flow in the video indicate that the location of a particle starting on the stagnation streamline a distance \(s=0.6 \mathrm{ft}\) upstream of the stagnation point at \(t=0\) is given approximately by \(s=0.6 e^{-0.5 y},\) where \(t\) is in seconds and \(s\) is in feet. (a) Determine the speed of a fluid particle es a function of time, \(V_{\text {particle }}(t),\) as it flows along the streamline. (b) Determine the speed of the fluid as a function of position along the streamline, \(V=V(s)\) (c) Determine the fluid acceleration along the streamline as a function of position, \(a_{s}=a_{s}(s)\)

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