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Chapter 3: Problem 2

Air flows steadily along a streamline from point (1) to point (2) with negligible viscous effects. The following conditions are measured: At point \((1) z_{1}=2 \mathrm{m}\) and \(p_{1}=0 \mathrm{kFa}\); at point \((2) \mathrm{z}_{2}=10 \mathrm{m}\) \(p_{2}=20 \mathrm{N} / \mathrm{m}^{2},\) and \(V_{2}=0 .\) Determine the velocity at point (1).

Short Answer

Expert verified
The speed at point (1) can be calculated by the rearranged equation from step 3: \(V_1 = \sqrt{\frac{2*(20 + 8*1.2*9.8)}{1.2}}\)

Step by step solution

01

Understanding Bernoulli's Equation

Bernoulli's equation states that for inviscid flow of a nonconducting fluid in a conservative force field, the total mechanical energy of the fluid is constant. The equation can be represented as: \(p_1 + 0.5*ρ*V_1^2 + ρ*g*z_1 = p_2 + 0.5*ρ*V_2^2 + ρ*g*z_2\), where p = pressure, ρ = density of the fluid, V = velocity of the fluid and z = height.
02

Substituting known conditions to Bernoulli's Equation

The given values are: \(p_1 = 0\), \(z_1 = 2\), \(p_2 = 20\), \(z_2 = 10\), \(V_2 = 0\). Now substituting these values in Bernoulli's equation: 0 + 0.5*ρ*V_1^2 + ρ*g*2 = 20 + 0 + ρ*g*10.
03

Simplifying to find \(V_1\)

Rearranging the equation from Step 2 to solve for \(V_1\), find: 0.5*ρ*V_1^2 = 20 + ρ*g*(10-2) = 20 + 8*ρ*g. Assuming that air is an ideal gas at normal temperature and pressure, the value for ρ (density of air) can be approximated to 1.2 kg/m^3, and g (acceleration due to gravity) is 9.8 m/s^2. Therefore: 0.5*ρ*V_1^2 = 20 + 8*1.2*9.8, which can further be solved for \(V_1\).

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Most popular questions from this chapter

Figure \(P 3.119\) shows two tall towers. Air at \(10^{\circ} \mathrm{C}\) is blowing toward the two towers at \(V_{0}=30 \mathrm{km} / \mathrm{hr}\). If the two towers are \(10 \mathrm{m}\) apart and half the air flow approaching the two towers passes between them, find the minimum air pressure between the two towers. Assume constant air density, inviscid flow, and steady-state conditions. The atmospheric pressure is \(101 \mathrm{kPa}\).

Observations show that it is not possible to blow the table tennis ball from the funnel shown in Fig. \(\mathrm{P} 3.122 a\). In fact, the ball can be kept in an inverted funnel, Fig. \(P 3.122 b,\) by blowing though it. The harder one blows through the funnel, the harder the ball is held within the funnel. Try this experiment on your own. Explain this phenomenon.

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.

Water flows in a vertical pipe of 0.15 -m diameter at a rate of \(0.2 \mathrm{m}^{3} / \mathrm{s}\) and a pressure of \(200 \mathrm{kPa}\) at an elevation of \(25 \mathrm{m}\). Determine the velocity head and pressure head at elevations of 20 end 55 m.

Water flows through a hole in the bottom of a large, open tank with a speed of \(8 \mathrm{m} / \mathrm{s}\), Determine the depth of water in the tank. Viscous effects are negligible.
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