Question

A fountain has to be designed to raise water to a height of 20 m from the ground level. What would be the speed of water and pressure to be applied by taking g = 10 m/s$^2$?

• A. 20 m/s and 200 kPa
• B. 10 m/s and 100 kPa
• C. 20 m/s and 100 kPa
• D. 10 m/s and 200 kPa

Hint:

For fluid problems involving height and velocity, Bernoulli's equation is fundamental. \( P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant} \). The minimum velocity to reach a height \(h\) is \( v = \sqrt{2gh} \). The hydrostatic pressure required to lift a fluid column to height \(h\) is \( P = \rho gh \). Always clarify what "speed of water" refers to (e.g., initial nozzle speed, speed at certain height).

Answer 1

The Correct Option is A

Step 1: Understand the principles involved (Bernoulli's Equation and Hydrostatics).
To raise water to a certain height, we need to overcome gravity and provide kinetic energy. This problem can be solved using Bernoulli's principle, which states that for an incompressible, inviscid fluid in steady flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline. Consider two points:
Point 1: At the ground level (pump outlet or base of the fountain), where the pressure \(P_1\) is applied and water has initial velocity \(v_1\).
Point 2: At the maximum height the water reaches (20 m), where the vertical velocity \(v_2\) becomes momentarily zero and the pressure is atmospheric (\(P_2\)).
We are given:
Height \(h = 20 \, \text{m}\)
Acceleration due to gravity \(g = 10 \, \text{m/s}^2\)
Density of water \( \rho = 1000 \, \text{kg/m}^3 \) (standard value for water).
Step 2: Calculate the minimum speed of water required.
The speed of water required refers to the initial velocity at the ground level (nozzle exit) to reach the specified height. This can be found using the principle of conservation of energy, where the initial kinetic energy is converted into potential energy at the peak height.
Using the kinematic equation \( v_f^2 = v_i^2 + 2as \), where \(v_f = 0\) (velocity at peak height), \(v_i\) is the initial speed, \(a = -g\) (acceleration due to gravity), and \(s = h\):
\[ 0^2 = v_i^2 + 2(-g)h \] \[ v_i^2 = 2gh \] \[ v_i = \sqrt{2gh} \] Substitute the given values: \[ v_i = \sqrt{2 \times 10 \, \text{m/s}^2 \times 20 \, \text{m}} \] \[ v_i = \sqrt{400 \, \text{m}^2/\text{s}^2} \] \[ v_i = 20 \, \text{m/s} \] So, the speed of water required at the ground level is 20 m/s.
Step 3: Calculate the pressure to be applied.
To find the pressure to be applied, we use Bernoulli's equation between the ground level (Point 1) and the peak height of the water jet (Point 2).
Let \(P_1\) be the gauge pressure applied at the ground level (relative to atmospheric pressure). At the peak height, the pressure \(P_2\) is atmospheric (so \(P_2 = 0\) gauge pressure). The velocity at the peak height \(v_2 = 0\). The height at the ground level \(z_1 = 0\), and at the peak \(z_2 = h = 20 \, \text{m}\).
Bernoulli's Equation: \[ \frac{P_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} + z_2 \] Substitute the values: \[ \frac{P_1}{1000 \times 10} + \frac{0^2}{2g} + 0 = \frac{0}{1000 \times 10} + \frac{0^2}{2g} + 20 \] The initial velocity \(v_1\) at the point where pressure \(P_1\) is applied (e.g., inside the pump or just before the nozzle) is considered to be very small or zero if the pressure is to generate the total head. In simpler terms, the applied pressure must provide the energy to lift the water against gravity to the required height. This is a hydrostatic pressure equivalent. \[ P_{applied} = \rho g h \] Substitute the values: \[ P_{applied} = 1000 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 20 \, \text{m} \] \[ P_{applied} = 200,000 \, \text{Pa} \] \[ P_{applied} = 200 \, \text{kPa} \] This pressure represents the static head that the pump must overcome to raise the water. Combining the results from Step 2 and Step 3, the speed of water is 20 m/s and the pressure to be applied is 200 kPa. The final answer is \( \boxed{\text{20 m/s and 200 kPa}} \).