A tank is in the form of frustum of a cone having top diameter of 2m, a bottom diameter of 0.8m and height 2m and is full of water. Find the time of emptying the tank through an orifice 100mm in diameter provided at the bottom. Take Cd = 0.625.

Tank with Two Orifices – Fluid Mechanics Solution

Tank with Two Orifices

Fluid Mechanics Problem Solution

Problem Statement

A tank of constant cross-sectional area of 2.8m² has two orifices each 9.3×10^-4 m² in area as shown in the figure. Calculate the time taken to lower the water level from 7.5m to 2.1m above the bottom of the tank. Assume C_d = 0.62.

Given Data

Cross-sectional area of tank (A) 2.8 m²

Area of each orifice (a) 9.3×10^-4 m²

Coefficient of discharge (C_d) 0.62

Initial water level (H₁) 7.5 m above tank bottom

Final water level (H₂) 2.1 m above tank bottom

Height of upper orifice 4.5 m above tank bottom

Height of lower orifice 0 m (at tank bottom)

Gravitational acceleration (g) 9.81 m/s²

Solution Approach

The problem consists of two phases:

Phase 1: Water level dropping from 7.5 m to 4.5 m (both orifices discharging)
Phase 2: Water level dropping from 4.5 m to 2.1 m (only lower orifice discharging)

Let’s set up the general equation for a tank with discharge through orifices:

Q · dt = -A · dh

dt = -A · dh / Q

Where:

Q = total discharge rate (m³/s)
A = cross-sectional area of tank (m²)
h = water height above lower orifice (m)
t = time (s)

Phase 1: Water Level from 7.5m to 4.5m

Step 1: Determine discharge through each orifice:

Discharge from upper orifice (Q₁) = C_d · a · √(2g(h-4.5))

Discharge from lower orifice (Q₂) = C_d · a · √(2gh)

Total discharge (Q) = Q₁ + Q₂ = C_d · a · √(2g(h-4.5)) + C_d · a · √(2gh)

Q = C_d · a · √(2g) · [√(h-4.5) + √h]

Step 2: Substitute into the time equation:

dt = -A · dh / (C_d · a · √(2g) · [√(h-4.5) + √h])

dt = -2.8 · dh / (0.62 × 9.3×10^-4 × √(2 × 9.81) · [√(h-4.5) + √h])

dt = -1096.3 · dh / [√(h-4.5) + √h]

Step 3: Integrate to find the time:

T₁ = -1096.3 ∫_4.5^7.5 dh / [√(h-4.5) + √h]

Using a substitution method and integrating:

T₁ = -1096.3 ∫_4.5^7.5 [√(h-4.5) – √h]dh / (-4.5)

T₁ = 1096.3 / 4.5 ∫_4.5^7.5 [√(h-4.5) – √h]dh

Upon evaluation of the integral:

T₁ = 942 seconds

Phase 2: Water Level from 4.5m to 2.1m

Step 1: In this phase, only the lower orifice is discharging water:

Q = Q₂ = C_d · a · √(2gh)

Step 2: For this simpler case, we can use the standard formula for time of emptying:

T₂ = 2A / (C_d · a · √(2g)) · [√H₁ – √H₂]

T₂ = 2 × 2.8 / (0.62 × 9.3×10^-4 × √(2 × 9.81)) · [√4.5 – √2.1]

T₂ = 2 × 2.8 / (0.62 × 9.3×10^-4 × √19.62) · [2.121 – 1.449]

T₂ = 5.6 / (0.62 × 9.3×10^-4 × 4.429) · [0.672]

T₂ = 5.6 / (0.00256) · 0.672 = 1474 seconds

Total Time Calculation

The total time required to lower the water level from 7.5m to 2.1m is the sum of both phases:

T_total = T₁ + T₂ = 942 + 1474 = 2416 seconds

T_total = 2416 / 60 = 40.26 minutes

The total time required to lower the water level from 7.5m to 2.1m is 2416 seconds (40.26 minutes).

Summary

The problem was divided into two phases due to the two-orifice configuration:
- Phase 1: Lowering from 7.5m to 4.5m with both orifices discharging – 942 seconds
- Phase 2: Lowering from 4.5m to 2.1m with only the lower orifice discharging – 1474 seconds
During Phase 1, the total discharge was calculated as the sum of discharge through both orifices, leading to a more complex integration.
During Phase 2, we used the simplified formula for discharge through a single orifice.
The total time required to lower the water level from 7.5m to 2.1m is 2416 seconds or 40.26 minutes.

This problem demonstrates the application of Torricelli’s theorem for flow through orifices at different heights and how to handle multi-phase emptying problems in fluid mechanics.