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Chapter 3: Problem 78

Helium flows through a 0.30 -m-diameter horizontal pipe with a temperature of \(20^{\circ} \mathrm{C}\) and a pressure of \(200 \mathrm{kPa}(\mathrm{abs})\) at a rate of \(0.30 \mathrm{kg} / \mathrm{s}\). If the pipe reduces to 0.25 -m-diameter, determine the pressure difference between these two sections. Assume incompressible, inviscid flow.

Short Answer

Expert verified
The pressure difference can be computed as \(delta_p = abs(p_1 - p_2)\). Specific values of \(delta_p\) will be dependent on the results from Step 1 and Step 2, and the initial conditions given in the exercise.

Step by step solution

01

Determine the velocities at the two points

The fluid velocity at the two sections of the pipe can be determined using the continuity equation, which states that the mass flow rate, \(m\), remains constant throughout. This is given by \(m= ρAv\), where A is the cross-sectional area of the pipe, ρ is the fluid density and v is the fluid velocity. Since \(m\) is constant, we have that \(A_1v_1 = A_2v_2 \). Considering that A is given by \(πd^2/4\), we find that the velocities \(v_1\) and \(v_2\) correspond to \(m/(ρπd_1^2/4)\) and \(m/(ρπd_2^2/4)\) respectively.
02

Apply Bernoulli’s equation

After determining the velocities at the two points, we then apply Bernoulli’s equation which states that the total mechanical energy at any two points in a streamline is constant. Hence, \(p_1+ 1/2 ρv_1^2 = p_2+1/2 ρv_2^2\). We then isolate \(p_2\) and get \(p_2 = p_1 + 1/2 ρ(v_1^2 - v_2^2)\). We substitute the values of \(v_1\) and \(v_2\) computed in the previous step to solve for \(p_2\).
03

Compute pressure difference

The pressure difference between the two sections of the pipe can finally be determined by computing the absolute difference between \(p_1\) and \(p_2\). This gives \(delta_p = abs(p_1 - p_2)\).

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Most popular questions from this chapter

What pressure gradient along the streamline, \(d p / d s\), is required to accelerate water upward in a vertical pipe at a rate of \(30 \mathrm{ft} / \mathrm{s}^{2} ?\) What is the the if the flow is downward?

A 40 -mph wind blowing past your house speeds up as it flows up and over the roof. If elevation effects are negligible, determine (a) the pressure at the point on the roof where the speed is 60 mph if the pressure in the free stream blowing toward your house is 14.7 psia. Would this effect tend to push the roof down against the house, or would it tend to lift the roof? (b) Determine the pressure on a window facing the wind if the window is assumed to be a stagnation point.

Consider a compressible liquid that has a constant bulk modulus. Integrate \(^{*} \mathbf{F}=m \mathbf{a}^{\prime \prime}\) along a streamline to obtain the equivalent of the Bernoulli equation for this flow. Assume steady, invicid flow.

An incompressible fluid flows steadily past a circular cylinder as shown in Fig. P3.8. The fluid velocity along the dividing streamline \((-\infty \leq x \leq-a)\) is found to be \(V=V_{0}\left(1-a^{2} / x^{2}\right),\) where \(a\) is the radius of the cylinder and \(V_{0}\) is the upstream velocity. (a) Determine the pressure gradient along this streamline. (b) If the upstream pressure is \(p_{0},\) integrate the pressure gradient to obtain the pressure \(p(x)\) for \(-\infty \leq x \leq-a,\) (c) Show from the result of part (b) that the pressure at the stagnation point \((x=-a)\) is \(p_{0}+\rho V_{0}^{2} / 2,\) as expected from the Bernoulli equation.

Air flows smoothly over the hood of your car and up past the windshield. However, a bug in the air does not follow the same path; it becomes splattered against the windshield. Explain why this is so.
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