Question

Water flows through a horizontal tube of variable cross section. The area of cross section at A and B are 4 mm² and 2 mm² respectively. If 1 cc of water enters per second through A, find (a) the speed of water at A, (b) the speed of water at B and (c) the pressure difference P_A − P_B.

Answer 1

(a) Given:
The areas of cross-sections of the tube at A and B are 4 mm² and 2 mm², respectively.
1 cc of water enters per second through A.
Let: 

\[\rho\]= Density of the liquid
P_A = Pressure of liquid at A
P_B = Pressure of liquid at B

(a)\[\frac{Q_A}{t} = \frac{1a}{s} = \text{ Discharge }\]
\[ \Rightarrow a_A \times V_A = Q_A \]
\[\text{ Given: }\]
\[ a_A = 4 {\text{ mm }}^2 = 4 \times {10}^{- 2} {\text{cm}}^2 \]
\[ \therefore 4 \times {10}^{- 2} \times V_A = 1\text{cc/s} \]
\[ \Rightarrow V_A = 25 \text{ m/s}\]
\[(b) a_A \times V_A = a_B \times V_B \]
\[ \Rightarrow 4 \times {10}^{- 2} \times 25 = 2 \times {10}^{- 2} \times V_B \]
\[ \Rightarrow V_B = 50 \text{ cm/s}\]

(c) By Bernoulli's equation, we have:

 

\[\frac{1}{2}\rho \text{v}_A^2 + P_A = \frac{1}{2}\rho \text{ v}_B^2 + P_B \]
\[ \Rightarrow ( P_A - P_B ) = \frac{1}{2}\rho\left( v_B^2 - \text{v}_A^2 \right)\]
\[ = \frac{1}{2} \times 1 \times (2500 - 625)\]
\[ = \frac{1875}{2} = 937 . 5 \text{dyn/ {cm}}^2 \]
\[ = 93 . 75 \text{N/ m}^2\]