A vertical cylindrical tank 2m diameter has, at the bottom, 0.05m diameter sharp-edged orifice (Cd = 0.6). (I) If the water enters the tank at a constant rate of 0.0095 cumecs, find the depth of water above the orifice when the level in the tank becomes stable.

Cylindrical Tank with Orifice

Fluid Mechanics Problem Solution

Problem Statement

A vertical cylindrical tank 2m diameter has, at the bottom, a 0.05m diameter sharp-edged orifice (C_d = 0.6).

If water enters the tank at a constant rate of 0.0095 m³/s, find the depth of water above the orifice when the level in the tank becomes stable.
Find the time for the level to fall from 3m to 1m above the orifice when the inflow is turned off.
If water now runs into the tank at 0.02 m³/s, the orifice remaining open, find the rate of rise in water level when the level has reached a depth of 1.7m above the orifice.

Given Data

Diameter of tank (D)2 mArea of tank (A)π/4 × 2² = 3.14 m²Diameter of orifice (d)5 cm = 0.05 mArea of orifice (a)π/4 × 0.05² = 0.00196 m²Coefficient of discharge (C_d)0.6Gravitational acceleration (g)9.81 m/s²

Initial Calculations

Step 1: Calculate the constant K which characterizes the orifice discharge:

Q_o = C_d · a · √(2gh) = K · √h K = C_d · a · √(2g) = 0.6 × 0.00196 × √(2 × 9.81) = 0.0052

Step 2: Set up the general equation for tank with inflow (Q_i) and outflow (Q_o):

(Q_i - Q_o)dt = A · dh

Where:

Q_i = Inflow rate (m³/s)
Q_o = Outflow rate through orifice (m³/s)
A = Cross-sectional area of tank (m²)
h = Water depth above orifice (m)
t = Time (s)

Part I: Stable Water Level

Given: Q_i = 0.0095 m³/s

At stable condition, the water level remains constant, so dh/dt = 0, which means:

Q_i = Q_o = K · √h 0.0095 = 0.0052 · √h √h = 0.0095 ÷ 0.0052 = 1.827 h = 1.827² = 3.34 m

The stable water level is 3.34 meters above the orifice.

Part II: Time for Water Level to Fall

Given:

Initial head (H₁) = 3 m
Final head (H₂) = 1 m
Inflow is turned off, so Q_i = 0

With no inflow, the general equation becomes:

-Q_o · dt = A · dh dt = -A · dh / Q_o = -A · dh / (K · √h)

Integrating from H₁ to H₂:

t = ∫_H₁^H₂ -A · dh / (K · √h) = -2A/K · [√h]_H₁^H₂t = 2A/K · (√H₁ - √H₂) t = 2 × 3.14 / 0.0052 × (√3 - √1) t = 1207.69 × (1.732 - 1) = 1207.69 × 0.732 = 884 seconds

The time for the water level to fall from 3m to 1m is 884 seconds (14 minutes and 44 seconds).

Part III: Rate of Rise in Water Level

Given:

Inflow rate (Q_i) = 0.02 m³/s
Current head (h) = 1.7 m

From the general equation:

(Q_i - Q_o) · dt = A · dh dh/dt = (Q_i - Q_o) / A = (Q_i - K · √h) / A dh/dt = (0.02 - 0.0052 · √1.7) / 3.14 dh/dt = (0.02 - 0.0052 × 1.304) / 3.14 dh/dt = (0.02 - 0.00678) / 3.14 = 0.01322 / 3.14 = 0.00421 m/s

The rate of rise in water level is 0.00421 m/s (or 4.21 mm/s).

Summary

Part I: When water enters at 0.0095 m³/s, the stable water level is3.34 meters above the orifice.
Part II: When the inflow is turned off, it takes884 seconds for the water level to fall from 3m to 1m.
Part III: With an inflow of 0.02 m³/s and a water level of 1.7m, the water level rises at0.00421 m/s.

This problem demonstrates the application of the continuity equation and Torricelli's theorem for flow through an orifice. The solution shows how the water level in a tank responds to different inflow conditions while discharging through an orifice.