A cylindrical tank of internal diameter 0.6m, length 1.5m and axis vertical has a 5cm diameter sharp-edged orifice (Cd = 0.6) in the bottom, open to atmosphere. The tank is open at the top and empty. If water were admitted into the tank from above at a constant rate of 14lps, how long will it take to just fill the tank? How much water will escape through the orifice during that period?

Water Discharge Through a Pipe System

Problem Statement

A water reservoir supplies a 200 m long pipe system with diameter 30 cm. The pipe discharges into the atmosphere through a nozzle with diameter 10 cm. The water level in the reservoir is maintained at 40 m above the discharge point. If the friction factor for the pipe is 0.02, determine:

The discharge rate through the system in m³/s.
The velocity at the nozzle exit.
The pressure head just before the nozzle.

Given Data

Length of pipe (L)	200 m
Pipe diameter (D)	30 cm = 0.3 m
Pipe area (A_p)	π/4 × (0.3)² ≈ 0.0707 m²
Nozzle diameter (d)	10 cm = 0.1 m
Nozzle area (A_n)	π/4 × (0.1)² ≈ 0.00785 m²
Elevation difference (z)	40 m
Friction factor (f)	0.02
Gravitational acceleration (g)	9.81 m/s²

Solution Approach

To solve this problem, we’ll apply the energy equation (Bernoulli’s equation with losses) between the reservoir surface and the discharge point:

Step 1: Set up the energy equation with head loss due to friction

z + (P₁/ρg) + (V₁²/2g) = (P₂/ρg) + (V₂²/2g) + h_L

Where:

z = 40 m (elevation difference)
P₁/ρg = 0 (pressure head at water surface is atmospheric)
V₁ ≈ 0 (velocity at reservoir surface is negligible)
P₂/ρg = 0 (pressure at discharge is atmospheric)
V₂ = velocity at nozzle exit (to be determined)
h_L = head loss due to friction and contraction

Step 2: Calculate the head loss

Head loss consists of friction loss in the pipe and minor loss at the contraction:

h_L = h_f + h_m = f(L/D)(V_p²/2g) + K(V_n²/2g)

Where:

V_p = velocity in the pipe
V_n = velocity at the nozzle exit
K = minor loss coefficient for contraction ≈ 0.5

Step 3: Apply continuity equation

Q = A_p × V_p = A_n × V_n

Therefore:

V_p = V_n × (A_n/A_p) = V_n × (0.00785/0.0707) = V_n × 0.111

Step 4: Substitute into energy equation

40 = (V_n²/2g) + f(L/D)((0.111 × V_n)²/2g) + 0.5(V_n²/2g)

40 = (V_n²/2g) × [1 + f(L/D)(0.111)² + 0.5]

40 = (V_n²/2g) × [1 + 0.02(200/0.3)(0.111)² + 0.5]

40 = (V_n²/2g) × [1 + 0.02(666.67)(0.012) + 0.5]

40 = (V_n²/2g) × [1 + 0.16 + 0.5] = (V_n²/2g) × 1.66

Step 5: Solve for nozzle velocity

V_n² = (40 × 2g) / 1.66 = (40 × 2 × 9.81) / 1.66 ≈ 472.5

V_n = √472.5 ≈ 21.74 m/s

Step 6: Calculate discharge rate

Q = A_n × V_n = 0.00785 × 21.74 ≈ 0.171 m³/s

Step 7: Calculate pressure head before nozzle

Using Bernoulli’s equation between a point just before the nozzle and the exit:

(P₃/ρg) + (V_p²/2g) = (P₂/ρg) + (V_n²/2g) + h_m

(P₃/ρg) + ((0.111 × 21.74)²/2g) = 0 + (21.74²/2g) + 0.5(21.74²/2g)

(P₃/ρg) + (2.41²/2g) = (21.74²/2g) × 1.5

(P₃/ρg) = (21.74²/2g) × 1.5 – (2.41²/2g) ≈ 35.1 m

Physical Interpretation

The solution demonstrates several key fluid dynamics principles:

Energy Conservation: The total energy at the reservoir is transformed into kinetic energy at the discharge, with some energy lost due to friction and contraction.
Continuity Principle: The mass flow rate remains constant throughout the system, causing the velocity to increase as the flow area decreases at the nozzle.
Pressure-Velocity Relationship: As the fluid accelerates through the nozzle, pressure energy converts to kinetic energy, resulting in a pressure drop.
Head Loss Effects: Friction in the pipe and contraction at the nozzle reduce the available energy for conversion to velocity, affecting the final discharge rate.

Results

Discharge rate: 0.171 m³/s

Velocity at nozzle exit: 21.74 m/s

Pressure head before nozzle: 35.1 m

Problem Statement

Given Data

Solution Approach

Physical Interpretation

Results

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