Question:
(a) State Bernoulli’s Principle.
(b) Figure below shows a liquid of density 1200kgm-3 flowing steadily in a tube of varying cross-sections. The cross- section at point A is 1.0 cm² and that at B is 20 mm², points A and B are in the same horizontal plane. The speed of the liquid at A is 10 cm/s. Calculate:
- the speed at B.
- the difference in pressure at A and B.
Question collected form Telegram
Group.
Solution:
(a) According to Bernoulli's principle, the total mechanical
energy of the moving fluid, which includes gravitational potential energy of
elevation, fluid pressure energy, and kinetic energy of fluid motion, remains
constant.
(b) Given:
A1=1cm2
A2 = 20mm2=$\frac{1}{5}$cm2
V1=10cm/s
(i) From Equation of Contunity:
A1 V1= A2 V2
10 × 1 =$\frac{1}{5}$ × V2
Or, V2 = 50cm/s.
Thus speed at B = 50cm/s.
(ii) The difference in pressure between points A and B can
be calculated using Bernoulli's equation:
$$p_1 + \frac{1}{2} \rho_1 v_1^2 = p_2 + \frac{1}{2} \rho_2
v_2^2$$
We know that $\rho_1 = \rho_2 = 1200 \text{ kg/m}^3$, $v_1 =
10 \text{ cm/s}$, and $v_2 = 50 \text{ cm/s}$. Substituting these values into
Bernoulli's equation, we get:
$$p_1 + \frac{1}{2} \times 1200 \text{ kg/m}^3 \times 10
\text{ cm/s}^2 = p_2 + \frac{1}{2} \times 1200 \text{ kg/m}^3 \times 50 \text{
cm/s}^2$$
Solving for $p_2$, we get:
$$p_2 = p_1 - \frac{1}{2} \times 1200 \text{ kg/m}^3 \times
(50 \text{ cm/s})^2 - \frac{1}{2} \times 1200 \text{ kg/m}^3 \times (10 \text{
cm/s})^2 = -144 \text{ Pa}$$
Therefore, the pressure at point B is 144 Pa lower than the
pressure at point A.