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: $1.013 \times 10^5\text{ Pa}$ )