Constant Head Orifice
An orifice is an opening with a closed perimeter used to discharge or measure fluid. For constant head, discharge is based on Torricelli velocity corrected by coefficients.
$$V_t=\sqrt{2gH}$$$$Q=C_d A\sqrt{2gH}$$$$C_d=C_cC_v$$
Trajectory Method for Velocity Coefficient
If a jet travels horizontal distance $x$ while falling vertical distance $y$, the actual jet velocity can be found from projectile motion.
$$t=\sqrt{\frac{2y}{g}}$$$$V_a=\frac{x}{t}$$$$C_v=\frac{V_a}{\sqrt{2gH}}$$
Falling Head Emptying Time
For a tank of constant surface area $A_s$ emptying through an orifice area $A_o$, integrate because the head changes with time.
$$t=\frac{2A_s}{C_dA_o\sqrt{2g}}\left(\sqrt{H_1}-\sqrt{H_2}\right)$$
Orifice Discharge Under Head
A 50 mm orifice under a constant head of 3 m has $C_d=0.62$. Find $Q$.
$$A=\frac{\pi(0.05)^2}{4}=0.001963\text{ m}^2$$$$Q=0.62(0.001963)\sqrt{2(9.81)(3)}=0.00933\text{ m}^3/\text{s}$$
Answer: $Q=0.00933\text{ m}^3/\text{s}$.
Actual Discharge from Orifice Coefficients
A 40 mm diameter orifice discharges under a constant head of 2.5 m. If $C_c=0.62$ and $C_v=0.98$, determine the actual discharge.
$$C_d=C_cC_v=0.62(0.98)=0.6076$$$$A=\frac{\pi(0.04)^2}{4}=0.001257\text{ m}^2$$$$Q=C_dA\sqrt{2gH}=0.6076(0.001257)\sqrt{2(9.81)(2.5)}$$$$Q=0.00535\text{ m}^3/\text{s}$$
Answer: $Q=0.00535\text{ m}^3/\text{s}$ or about 5.35 L/s.
Falling Head Tank Emptying Time
A tank with constant plan area 3.0 m2 drains through a 30 mm orifice with $C_d=0.62$. Find the time for the head to fall from 2.0 m to 0.50 m.
$$A_o=\frac{\pi(0.03)^2}{4}=0.000707\text{ m}^2$$$$t=\frac{2A_s}{C_dA_o\sqrt{2g}}\left(\sqrt{H_1}-\sqrt{H_2}\right)$$$$t=\frac{2(3.0)}{0.62(0.000707)\sqrt{2(9.81)}}(\sqrt{2.0}-\sqrt{0.50})$$$$t=2190\text{ s}=36.5\text{ min}$$
Answer: The water level takes about 36.5 minutes to fall to 0.50 m head.
Trajectory Method Orifice Velocity
A horizontal jet from an orifice falls 0.45 m while traveling 1.80 m horizontally. The head over the orifice is 2.0 m. Determine $C_v$.
$$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(0.45)}{9.81}}=0.303\text{ s}$$$$V_a=\frac{x}{t}=\frac{1.80}{0.303}=5.94\text{ m/s}$$$$V_t=\sqrt{2gH}=\sqrt{2(9.81)(2.0)}=6.26\text{ m/s}$$$$C_v=\frac{V_a}{V_t}=\frac{5.94}{6.26}=0.949$$
Answer: $C_v=0.949$.
Problem: Submerged Orifice — Discharge Under Net Head
An orifice 75 mm in diameter connects two tanks. The upstream water level is 3.5 m above the orifice centerline and the downstream water level is 1.2 m above the centerline (both measured from the same orifice elevation). Use $C_d = 0.62$. Find the discharge through the orifice and describe how submergence affects the effective head.
For a submerged orifice, the effective head is the difference between the upstream and downstream water levels:
$$H_{eff} = H_u - H_d = 3.5 - 1.2 = 2.3 \text{ m}$$
$$A = \frac{\pi(0.075)^2}{4} = 4.418 \times 10^{-3} \text{ m}^2$$
$$Q = C_d A \sqrt{2g H_{eff}} = 0.62(4.418 \times 10^{-3})\sqrt{2(9.81)(2.3)}$$
$$Q = 0.62(4.418 \times 10^{-3})(6.717) = 0.01838 \text{ m}^3/\text{s}$$
Answer: Q = 0.01838 m³/s (18.38 L/s). Unlike a free-discharge orifice where the full upstream head drives flow, submergence reduces the effective head from 3.5 m to only 2.3 m — a 34% reduction in the head term and about 19% less discharge than if the downstream were free.
Problem: Interconnected Tanks — Time to Equalize
Two cylindrical tanks are connected at their bases by a 50 mm diameter orifice ($C_d = 0.62$). Tank 1 has a constant cross-sectional area of 4.0 m² and initial water depth of 3.0 m. Tank 2 has area 2.0 m² and initial depth of 0.50 m. Find the time in minutes for the two water levels to equalize.
Let $h$ = difference in water levels (Tank 1 higher). When levels equalize, $h = 0$. The combined effective storage area for a head change $dh$ is:
$$A_{eff} = \frac{A_1 A_2}{A_1 + A_2} = \frac{4.0(2.0)}{4.0 + 2.0} = 1.333 \text{ m}^2$$
$$h_0 = 3.0 - 0.50 = 2.50 \text{ m (initial head difference)}$$
$$h_f = 0 \text{ (equalized)}$$
$$A_{or} = \frac{\pi(0.05)^2}{4} = 1.963 \times 10^{-3} \text{ m}^2$$
$$t = \frac{2A_{eff}}{C_d A_{or}\sqrt{2g}}\left(\sqrt{h_0} - \sqrt{h_f}\right)$$
$$t = \frac{2(1.333)}{0.62(1.963 \times 10^{-3})\sqrt{2(9.81)}}(\sqrt{2.50} - 0)$$
$$t = \frac{2.666}{0.62(1.963 \times 10^{-3})(4.429)}(1.581)$$
$$t = \frac{2.666}{0.005385}(1.581) = 495.1(1.581) = 782.8 \text{ s} = 13.05 \text{ min}$$
Answer: The two tanks equalize in approximately 13.05 minutes. The effective area formula accounts for the fact that water leaving Tank 1 partially fills Tank 2, so the net head change is slower than for a single emptying tank.
Problem: Tank with Simultaneous Inflow and Orifice Outflow
A tank with a horizontal cross-sectional area of 5.0 m² has a constant inflow rate of 0.025 m³/s. It also drains through a 60 mm diameter orifice at the base ($C_d = 0.65$). Initially the water depth is 1.5 m. Determine the equilibrium depth (when inflow equals outflow) and state whether the initial depth is above or below equilibrium.
At equilibrium, $Q_{in} = Q_{out}$:
$$Q_{in} = C_d A_{or}\sqrt{2gH_{eq}}$$
$$0.025 = 0.65\frac{\pi(0.06)^2}{4}\sqrt{2(9.81)H_{eq}}$$
$$0.025 = 0.65(2.827 \times 10^{-3})(4.429)\sqrt{H_{eq}}$$
$$0.025 = 8.138 \times 10^{-3}\sqrt{H_{eq}}$$
$$\sqrt{H_{eq}} = \frac{0.025}{8.138 \times 10^{-3}} = 3.072$$
$$H_{eq} = (3.072)^2 = 9.44 \text{ m}$$
Answer: The equilibrium water depth is 9.44 m. The initial depth of 1.5 m is far below equilibrium, meaning the tank level will rise over time as inflow exceeds outflow. The tank will fill until it reaches 9.44 m if it is tall enough.