Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure.
The pressure difference of water between \(P\) and \(Q\) is \(15\,\text{N m}^{-2}\).
The area of cross-section at \(P\) and \(Q\) are \(40\,\text{cm}^2\) and \(20\,\text{cm}^2\), respectively.
The rate of flow of water through the pipe, in \(\text{cm}^3\text{s}^{-1}\), is:
Show Hint
Use continuity before Bernoulli equation.
Smaller area means greater speed.
For horizontal pipes, height terms cancel.
Convert units carefully.
Step 1: List what we know. Pressure difference $\Delta P = 15\,\text{N m}^{-2}$, areas $A_1 = 40\,\text{cm}^2$ at $P$ and $A_2 = 20\,\text{cm}^2$ at $Q$, density of water $\rho = 1000\,\text{kg m}^{-3}$. The pipe is horizontal. Step 2: Use continuity. The same volume of water per second crosses both sections, so $A_1 v_1 = A_2 v_2$. \[ 40 v_1 = 20 v_2 \Rightarrow v_2 = 2 v_1 \] Step 3: Apply Bernoulli's equation. Since the pipe is horizontal, the height terms cancel, leaving $P_1 + \tfrac12 \rho v_1^2 = P_2 + \tfrac12 \rho v_2^2$. \[ P_1 - P_2 = \tfrac12 \rho (v_2^2 - v_1^2) \] Step 4: Substitute the speeds. Replace $v_2$ by $2v_1$ so everything is in terms of $v_1$. \[ 15 = \tfrac12 (1000)\,(4v_1^2 - v_1^2) = 500 \times 3 v_1^2 = 1500\, v_1^2 \] Step 5: Solve for the inlet speed. \[ v_1^2 = \frac{15}{1500} = 0.01 \Rightarrow v_1 = 0.1\,\text{m s}^{-1} \] Step 6: Compute the flow rate. The volume flow rate is $Q = A_1 v_1$, with $A_1 = 40\times10^{-4}\,\text{m}^2$. \[ Q = (40\times10^{-4})(0.1) = 4\times10^{-4}\,\text{m}^3\text{s}^{-1} = 400\,\text{cm}^3\text{s}^{-1} \] \[ \boxed{400\,\text{cm}^3\text{s}^{-1}} \]
