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 \(101 \mathrm{kPa}\).

Short Answer

Expert verified
The minimum air pressure between the two towers is calculated to be approximately \(95.543 \, kPa\).

Step by step solution

01

Assumptions and Transformation of Variables

Before we begin with the calculations, some assumptions must be clarified. Firstly, the air is treated as an ideal fluid with constant density. Also, it is stated that the problem is under steady-state conditions so the properties of the air do not change over time. Then, the incoming air velocity, initially given in kilometres per hour, needs to be converted to a more practical unit (m/s). Given that \(1 \mathrm{m/s} = 3.6 \mathrm{km/hr}\), we get \(V_0 = \frac{30}{3.6} = 8.33 \mathrm{m/s}\).
02

Application of Bernoulli's Equation

Bernoulli's equation relates the pressure, velocity and height of a fluid flow. In this case, the height doesn't change so it cancels off from the equation. The principle can be expressed as \(P_{\infty} + \frac{1}{2} \rho V_{0}^{2} = P_{\text{min}} + \frac{1}{2} \rho V_{\text{min}}^{2}\). However, the velocity at the minimum pressure point between the towers (\(V_{\text{min}}\)) is double the incoming velocity, due to the conservation of mass and the assumption that exactly half the air passes between the towers. Hence, \(V_{\text{min}} = 2 \times V_0\).
03

Calculation of Minimum Air Pressure

We can insert \(V_{\text{min}}\) into Bernoulli's Equation, letting us solve for the minimum pressure, \(P_{min}\). We then get: \(P_{min} = P_{\infty} + \frac{1}{2} \rho V_{0}^{2} - \frac{1}{2} \rho V_{\text{min}}^{2}\). We use the given atmospheric pressure as \(P_{\infty} = 101 kPa\) and the air density at \(10^\circ \mathrm{C}\) as \(\rho = 1.247 \mathrm{kg/m}^3\). After calculating, we get \(P_{min} = 101000 - (0.5 \times 1.247 \times (2 \times 8.33)^{2} - 0.5 \times 1.247 \times 8.33^{2}) = 95.543 \, kPa \)

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