Question

A cylindrical tank with a diameter of \(500\ \text{mm}\) contains water (density \(=1\ \text{g}\cdot\text{cm}^{-3}\)) up to a height \(h\). A \(5\ \text{mm}\) diameter round nozzle, whose center is \(1\ \text{cm}\) above the base of the tank, has its exit open to the atmosphere. The pressure above the water level in the tank is maintained at \(2\ \text{bar (abs)\). Neglect all frictional and entry/exit losses. Take \(g=10\ \text{m}\,\text{s}^{-2}\) and atmospheric pressure \(=1\ \text{bar}\). The absolute value of initial \(\dfrac{dh}{dt}\) (in \(\text{mm}\,\text{s}^{-1}\)) when \(h=51\ \text{cm}\) is ____________ (rounded off to two decimal places).}

Hint:

With pressurized tanks, add the pressure head \(\Delta p/(\rho g)\) to the static head difference between liquid level and exit.
For level change: \( \dfrac{dh}{dt}=-\dfrac{Q}{A_{\text{tank}}}\).

Answer 1

Correct Answer: 1.4

Step 1: Head available at the nozzle. Apply Bernoulli between the liquid surface (pressure \(=2\ \text{bar}\), velocity \(\approx 0\), elevation \(z=h\)) and the nozzle exit (pressure \(=1\ \text{bar}\), elevation \(z=0.01\ \text{m}\)). Pressure head difference: \[ \frac{\Delta p}{\rho g}=\frac{(2-1)\times 10^{5}}{1000\times 10}=10\ \text{m}. \] Static head difference: \[ h-z = 0.51-0.01=0.50\ \text{m}. \] Hence total head \(=10+0.50=10.5\ \text{m}\). Step 2: Exit velocity and volumetric flow rate. \[ v=\sqrt{2g(10.5)}=\sqrt{2\times10\times10.5}=\sqrt{210}=14.49\ \text{m s}^{-1}. \] Nozzle area \(A_n=\dfrac{\pi d^2}{4}=\dfrac{\pi(0.005)^2}{4}=1.963\times 10^{-5}\ \text{m}^2\). \[ Q=A_n v=(1.963\times10^{-5})(14.49)=2.84\times10^{-4}\ \text{m}^3\text{/s}. \] Step 3: Tank area and rate of level drop. Tank cross-sectional area \(A_t=\pi(0.25)^2=0.19635\ \text{m}^2\). \[ \frac{dh}{dt}=-\frac{Q}{A_t}=-\frac{2.84\times10^{-4}}{0.19635} =-1.447\times10^{-3}\ \text{m s}^{-1}=-1.447\ \text{mm s}^{-1}. \] Absolute value (initial): \[ \boxed{1.45\ \text{mm s}^{-1}}. \]