A cylindrical tank is placed with its axis vertical and is provided with a circular orifice of 4cm diameter at the bottom. A steady inflow and free discharge at the bottom of the orifice causes the depth of water in the tank to rise from 0.59m to 0.75m in 106 Sec. Further it is observed that the depth rises from 1.2m to 1.29m in 129 Sec. Determine the inflow rate and the diameter of the tank. Assume Cd = 0.62.

Cylindrical Tank with Orifice – Fluid Mechanics Solution

Cylindrical Tank with Orifice

Fluid Mechanics Problem Solution

Problem Statement

A cylindrical tank is placed with its axis vertical and is provided with a circular orifice of 4cm diameter at the bottom. A steady inflow and free discharge at the bottom of the orifice causes the depth of water in the tank to rise from 0.59m to 0.75m in 106 Sec. Further it is observed that the depth rises from 1.2m to 1.29m in 129 Sec. Determine the inflow rate and the diameter of the tank. Assume C_d = 0.62.

Given Data

Diameter of orifice (d) 4 cm = 0.04 m

Area of orifice (a) π/4 × 0.04² = 0.001257 m²

Coefficient of discharge (C_d) 0.62

Gravitational acceleration (g) 9.81 m/s²

First case – Initial depth 0.59 m

First case – Final depth 0.75 m

First case – Time interval 106 seconds

Second case – Initial depth 1.2 m

Second case – Final depth 1.29 m

Second case – Time interval 129 seconds

Initial Calculations

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

K = C_d · a · √(2g) = 0.62 × 0.001257 × √(2 × 9.81) = 0.00345 m^5/2/s

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

dh/dt = (Q_i – Q_o)/A = (Q_i – K√h)/A

First Case Analysis

Given:

Initial depth = 0.59 m
Final depth = 0.75 m
Time interval = 106 seconds

Calculating the change in depth and rate of change:

dh = 0.75 – 0.59 = 0.16 m

dt = 106 s

dh/dt = 0.16/106 = 0.001495 m/s

The average head during this interval is:

h_avg = (0.59 + 0.75)/2 = 0.67 m

Substituting into our general equation:

0.001495 = (Q_i – 0.00345√0.67)/A

0.001495A = Q_i – 0.00345√0.67

Q_i = 0.001495A + 0.00345√0.67 = 0.001495A + 0.002824

Second Case Analysis

Given:

Initial depth = 1.2 m
Final depth = 1.29 m
Time interval = 129 seconds

Calculating the change in depth and rate of change:

dh = 1.29 – 1.2 = 0.09 m

dt = 129 s

dh/dt = 0.09/129 = 0.000698 m/s

The average head during this interval is:

h_avg = (1.2 + 1.29)/2 = 1.245 m

Substituting into our general equation:

0.000698 = (Q_i – 0.00345√1.245)/A

0.000698A = Q_i – 0.00345√1.245

Q_i = 0.000698A + 0.00345√1.245 = 0.000698A + 0.003849

Solving for Unknown Variables

Step 1: We now have two equations with two unknowns (Q_i and A):

Q_i = 0.001495A + 0.002824 … (a)

Q_i = 0.000698A + 0.003849 … (b)

Equating (a) and (b):

0.001495A + 0.002824 = 0.000698A + 0.003849

0.001495A – 0.000698A = 0.003849 – 0.002824

0.000797A = 0.001025

A = 0.001025/0.000797 = 1.285 m²

Step 2: Calculate the inflow rate by substituting the value of A into equation (a):

Q_i = 0.001495 × 1.285 + 0.002824 = 0.001921 + 0.002824 = 0.0047 m³/s

Step 3: Calculate the diameter of the tank from its cross-sectional area:

A = π × D²/4

D = √(4A/π) = √(4 × 1.285/π) = √(1.635) = 1.28 m

The inflow rate (Q_i) is 0.0047 m³/s and the tank diameter (D) is 1.28 m.

Summary

By analyzing two different cases of water level rise in the tank, we established two equations relating the inflow rate and tank area.
The cross-sectional area of the tank is 1.285 m².
The inflow rate into the tank is 0.0047 m³/s (4.7 liters per second).
The diameter of the cylindrical tank is 1.28 m.

This problem demonstrates the application of conservation principles in fluid mechanics and how to solve for unknown parameters by analyzing the system behavior at different states.