Soal-soal dinamika fluida
1. A horizontal pipe 20 cm in diameter has a smooth reduction to a pipe 10 cm in diameter. If the pressure of the water in the larger pipe is \(8.00\times 10^{4}\) Pa and the pressure in the smaller pipe is \(6 \times 10^{4}\) Pa, at what rate does water flow through the pipes?
Given:
\( A_1=\frac{\pi}{4} d_1^2=314\times 10^{-4} m^2\)
\( A_2=\frac{\pi}{4} d_1^2=78.5\times 10^{-4} m^2\)
\( P_1=8.00\times 10^{4}\) Pa
\( P_2=6.00\times 10^{4}\) Pa
Asked: Q....?
Solution:
To solve this problem, we can use Bernoulli's Equation:
$$P+\frac{1}{2}\rho v^2+\rho gh=C,$$ C is constant. (1)
Solving this equation we will have an equation in a venturi pipe, where:
$$v_2=\sqrt{\frac{P_1-P_2}{\rho(A_1^2-A_2^2)}}$$ (2)
Substitute the number we have in the equation (2), we will get:
$$v_2=\sqrt{\frac{8.00\times 10^{4}-6.00\times 10^{4}}{1000((314\times 10^{-4})^{2} -(78.5\times 10^{-4})^2)}}=147.1 m/s$$
The volume flow rate, Q can be determined by using continuity equation, where :
\(A_1 v_1 = A_2 v_1\), where Q =Av, we will use speed of water in small diameter, so that:
$$Q =147.1\times 78.5\times 10^{-4}= 11.55 \text{ m}^3/\text{s}$$
The volume flux or volume flow rate in the pipe is \(11.55 \text{ m}^3/\text{s}\)
2. Water is forced out of a fire extinguisher by air pressure, as shown in Figure. How much gauge air pressure in the tank (above atmospheric) is required for the water jet to have a speed of 30.0 m/s when the water level in the tank is 0.500 m below the nozzle? (hint atmospheric pressure: \(1.013 \times 10^5\text{ Pa}\) )
Given:
\(v_2= 30 m/s\) \(v_1=0 m/s\)
\(h_2= 0.5 m/s\) \(h_1=0\)
\(P_2 = 1.013 \times 10^5\text{ Pa}\)
Asked: \(P_1 \text{.... ?}\)
Solution:
Use the Bernoulli's Equation, we will have:
\(P_1 + 1/2 \rho v_1^2 +\rho gh_1=P_2 + 1/2 \rho v_2^2 +\rho gh_2\),
since \(v_1=0 m/s\), and \(h_1=0\), we can arrange the equation as follow:
\(P_1 =P_2 + 1/2 \rho v_2^2 +\rho gh_2\)
\(P_1 =1.013 \times 10^5 + 1/2 (1000) (30)^2 +1000 (10)(0.5)\)
\(P_1 =1.013 \times 10^5 + 4.5 \times 10^5 +0.05 10^5\)
\(P_1 =5.563 \times 10^5\) Pa
The gauge pressure is relative pressure between absolute pressure and atmospheric pressure, so that :
\(\DeltaP =P_1 - P_0 = 4.55 \times 10^5 Pa\)