A tank is in the form of hemisphere of 2m diameter and having a cylindrical upper part of 2m diameter and 3m height. Find the time of emptying the tank through an orifice of 75mm diameter at its bottom if the tank is initially full of water. Take Cd = 0.62.

Hemispherical-Cylindrical Tank Emptying – Fluid Mechanics Solution

Hemispherical-Cylindrical Tank Emptying

Fluid Mechanics Problem Solution

Problem Statement

A tank is in the form of hemisphere of 2m diameter and having a cylindrical upper part of 2m diameter and 3m height. Find the time of emptying the tank through an orifice of 75mm diameter at its bottom if the tank is initially full of water. Take C_d = 0.62.

Given Data

Diameter of the tank (D) 2 m

Height of cylindrical part (H_cylinder) 3 m

Diameter of orifice (d) 75 mm = 0.075 m

Coefficient of discharge (C_d) 0.62

Radius of tank (R) 1 m

Initial water level 4 m (3m cylindrical + 1m hemispherical)

Gravitational acceleration (g) 9.81 m/s²

Solution Approach

The problem consists of two phases:

Phase 1: Water level dropping from 4 m to 1 m (emptying the cylindrical part)
Phase 2: Water level dropping from 1 m to 0 m (emptying the hemispherical part)

For each phase, we need to calculate the time separately due to the different geometries involved.

Preliminary Calculations

Step 1: Calculate the cross-sectional area of the cylindrical part:

A = π/4 × D² = π/4 × 2² = π = 3.14 m²

Step 2: Calculate the area of the orifice:

a = π/4 × d² = π/4 × 0.075² = 0.004418 m²

Phase 1: Emptying the Cylindrical Part (4m to 1m)

Step 1: For a cylindrical tank with constant cross-sectional area, we can use the standard formula:

T₁ = 2A / (C_d × a × √(2g)) × [√H₁ – √H₂]

Where:

A = cross-sectional area of tank = 3.14 m²
C_d = coefficient of discharge = 0.62
a = area of orifice = 0.004418 m²
g = gravitational acceleration = 9.81 m/s²
H₁ = initial water level = 4 m
H₂ = final water level = 1 m

Step 2: Substitute the values into the equation:

T₁ = 2 × 3.14 / (0.62 × 0.004418 × √(2 × 9.81)) × [√4 – √1]

T₁ = 2 × 3.14 / (0.62 × 0.004418 × 4.43) × [2 – 1]

T₁ = 6.28 / 0.0121 × 1 = 518 seconds

Phase 2: Emptying the Hemispherical Part (1m to 0m)

Step 1: For a hemispherical tank, we need to use a different equation due to the varying cross-sectional area:

T₂ = π / (C_d × a × √(2g)) × [4R/3 × (H₁^3/2 – H₂^3/2) – 2/5 × (H₁^5/2 – H₂^5/2)]

Where:

R = radius of hemisphere = 1 m
H₁ = initial water level in hemisphere = 1 m
H₂ = final water level in hemisphere = 0 m

Step 2: Substitute the values into the equation:

T₂ = π / (0.62 × 0.004418 × √(2 × 9.81)) × [4 × 1/3 × (1^3/2 – 0) – 2/5 × (1^5/2 – 0)]

T₂ = π / (0.62 × 0.004418 × 4.43) × [4/3 × 1 – 2/5 × 1]

T₂ = 3.14 / 0.0121 × [1.33 – 0.4]

T₂ = 3.14 / 0.0121 × 0.93 = 242 seconds

Total Time Calculation

The total time required to empty the tank is the sum of both phases:

T_total = T₁ + T₂ = 518 + 242 = 760 seconds

T_total = 760 / 60 = 12.67 minutes

The total time required to empty the tank is 760 seconds (12.67 minutes).

Summary

The problem was divided into two phases due to the composite geometry of the tank:
- Phase 1: Emptying the cylindrical part (4m to 1m) – 518 seconds
- Phase 2: Emptying the hemispherical part (1m to 0m) – 242 seconds
Different equations were used for each phase:
- For the cylindrical part, we used the standard formula for constant cross-sectional area.
- For the hemispherical part, we used a specialized formula that accounts for the varying cross-sectional area.
The total time required to empty the tank is 760 seconds or 12.67 minutes.

This problem demonstrates the application of Torricelli’s theorem for flow through an orifice and how to handle multi-phase emptying problems with different geometric configurations in fluid mechanics.