Question

A geothermal pump is used to pump brine whose density is \(1050 \mathrm{kg} / \mathrm{m}^{3}\) at a rate of \(0.3 \mathrm{m}^{3} / \mathrm{s}\) from a depth of \(200 \mathrm{m}\). For a pump efficiency of 74 percent, determine the required power input to the pump. Disregard frictional losses in the pipes, and assume the geothermal water at \(200 \mathrm{m}\) depth to be exposed to the atmosphere.

Short answer

Answer: The required power input to the pump is approximately 834,536 Watts.

Step-by-step solution

Calculate Gravitational Potential Energy Required

First, we'll find the gravitational potential energy required to lift the water from a depth of 200 meters. The formula for gravitational potential energy is: \(PE = mgh\) where \(PE\) is potential energy, \(m\) is the mass of brine, \(g\) is the gravitational acceleration (approximately 9.81 \(\mathrm{m/s^2}\)), and \(h\) is the height (in this case, 200 meters). But we are given the flow rate (\(Q\)) in terms of volume per unit time (\(0.3\mathrm{m^3/s}\)) and density (\(\rho\)) of brine. We can relate mass and flow rate by: \(Q = \frac{dm}{dt} = \rho\frac{dV}{dt}\) Substituting this into the potential energy equation, we have: \(PE = \rho gQh\Delta t\) Now we can plug in the given values: \(PE = 1050\mathrm{kg/m^3} \times 9.81\mathrm{m/s^2} \times 0.3\mathrm{m^3/s} \times 200\mathrm{m} \times \Delta t\) Here, we are keeping the \(\Delta t\) for finding power next.

Calculate Mechanical Work Performed

As per Work-Energy principle, the work performed (\(W\)) is equal to the change in potential energy: \(W = PE = \rho gQh\Delta t\)

Calculate Power Input Required

Power (\(P\)) is the rate at which work is done: \(P = \frac{W}{\Delta t}\) Substituting the work equation, we have: \(P = \frac{\rho gQh\Delta t}{\Delta t} = \rho gQh\) Now, simply plug in the given values: \(P = 1050\mathrm{kg/m^3} \times 9.81\mathrm{m/s^2} \times 0.3\mathrm{m^3/s} \times 200\mathrm{m}\) \(P = 617757\mathrm{W}\)

Apply Efficiency to Calculate Required Power Input

Finally, we will apply the pump's efficiency (74 percent) to find the required power input: \(P_{input} = \frac{P}{\text{efficiency}}\) \(P_{input} = \frac{617757\mathrm{W}}{0.74}\) \(P_{input} = 834536\mathrm{W}\) Thus, the required power input to the pump is approximately 834,536 Watts.