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Water at \(20^{\circ} \mathrm{C}\) flows through a pump in which the inflow pipe has a diameter of \(150 \mathrm{~mm}\) and the outflow pipe has a diameter of \(75 \mathrm{~mm}\). The centerlines of the inflow and outflow pipes are aligned so that there is no difference in elevation. When the flow rate through the pump is \(30 \mathrm{~L} / \mathrm{s}\), the inflow and outflow pressures are \(150 \mathrm{kPa}\) and \(500 \mathrm{kPa}\), respectively, and measurements indicate that the temperature of the water increases by \(0.1^{\circ} \mathrm{C}\) between the inflow and the outflow. Estimate the rate at which energy is being delivered by the pump to the water. Approximately what percentage of the energy input by the pump goes toward raising the temperature?

Step by step solution

01

Calculate Mass Flow Rate

First, we calculate the mass flow rate \( \dot{m} \) using the volumetric flow rate \( Q \) and the density of water \( \rho \). At \(20^{\circ} \mathrm{C}, \rho \approx 1000 \, \text{kg/m}^3\). Given that \( Q = 30 \, \text{L/s} = 0.03 \, \text{m}^3/\text{s}\), we have:\[ \dot{m} = \rho \times Q = 1000 \, \text{kg/m}^3 \times 0.03 \, \text{m}^3/\text{s} = 30 \, \text{kg/s} \]

02

Apply Bernoulli's Equation for Energy

Using Bernoulli's equation between the inflow and the outflow of the pump:\[ \Delta P = \frac{1}{2} \rho (v_2^2 - v_1^2) + \rho g (z_2 - z_1) + W_{\text{pump}} \]Where \( \Delta P = P_2 - P_1 = 500,000 \, \text{Pa} - 150,000 \, \text{Pa} = 350,000 \, \text{Pa} \) and the change in elevation \( z_2 - z_1 = 0 \) since the pipes are aligned on the same level.

03

Calculate Velocities in the Pipes

The cross-sectional area \( A \) of the pipe is \( A = \pi (d/2)^2 \).- For the inflow pipe of diameter 150 mm: \[ A_1 = \pi (0.15 / 2)^2 = 0.017674 \, \text{m}^2 \] Velocity \( v_1 = \frac{Q}{A_1} = \frac{0.03}{0.017674} = 1.698 \, \text{m/s} \)- For the outflow pipe of diameter 75 mm: \[ A_2 = \pi (0.075 / 2)^2 = 0.004418 \, \text{m}^2 \] Velocity \( v_2 = \frac{Q}{A_2} = \frac{0.03}{0.004418} = 6.791 \, \text{m/s} \)

04

Calculate Kinetic Energy Change

Substitute the velocities into Bernoulli's equation:\[ \frac{1}{2} \rho (v_2^2 - v_1^2) = \frac{1}{2} \times 1000 \times (6.791^2 - 1.698^2) = 21328 \, \text{J/s} \]

05

Calculate Work Done by the Pump

Rearrange Bernoulli's energy equation to solve for the pump work per second, \( W_{\text{pump}} \):\[ W_{\text{pump}} = \dot{m} \cdot ( \Delta P + 21328 ) = 30 \times 350,000 + 21328 = 375,328 \, \text{W} \]

06

Calculate Energy to Increase Water Temperature

The energy needed to raise the temperature is:\[ Q_{\text{heat}} = \dot{m} \cdot c_p \cdot \Delta T\]where \( c_p \approx 4180 \, \text{J/kg}^\circ \text{C} \)\[ Q_{\text{heat}} = 30 \cdot 4180 \cdot 0.1 = 1254 \, \text{J/s} \]

07

Calculate Percentage of Energy for Heating

Calculate the percentage of energy dedicated to heating:\[ \text{Percentage} = \left(\frac{Q_{\text{heat}}}{W_{\text{pump}}}\right) \times 100 = \left(\frac{1254}{375,328}\right) \times 100 \approx 0.33\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bernoulli's equation

Bernoulli's equation is a cornerstone of fluid mechanics. It relates the pressure, velocity, and elevation changes in fluid flow to the work done in a system. This powerful equation can be used to analyze fluid flow in various scenarios, including this pump system. In our problem, the elevation change is zero because the pump's inlet and outlet are aligned horizontally. The equation simplifies to reflect the balance between pressure change, velocity change, and the work done by the pump. This simplification lets us calculate how much energy the pump adds to the fluid flow.

mass flow rate

Mass flow rate is crucial because it tells us how much mass is moving through a system per unit of time. It is usually obtained by multiplying the density of the fluid by the volumetric flow rate. In this exercise, knowing the mass flow rate allows us to calculate both the kinetic energy change in the fluid and the energy conversion due to the pump. We calculated the mass flow rate using water's density at a given temperature and the volumetric flow, arriving at a result of 30 kg/s.

energy conservation

Energy conservation principles are fundamental in fluid mechanics and help us understand how energy is transferred and transformed in systems. The energy delivered by the pump is split into kinetic energy changes and slight temperature changes in the fluid. Bernoulli's equation helps illustrate this energy conservation, demonstrating how input energy is utilized efficiently to either increase pressure, change velocity or slightly increase temperature.

pump work

Pump work represents the energy supplied by the pump to the fluid to induce flow and overcome resistance. In our exercise, we computed this work by considering both pressure changes and kinetic energy changes in the flowing water. The pump does a remarkable job in increasing the pressure, which is evident from the calculations, thereby providing the necessary flow conditions required for the system.

temperature change

The temperature change of the fluid indicates that some pump energy is used to increase the fluid's temperature. Although a 0.1°C increase seems minor, calculating the energy required for this rise gives a deeper understanding of how fluids store energy internally. By comparing this energy utilization to the total energy supplied by the pump, we found that 0.33% of the pump's energy is used for heating the water. This value offers insight into the system's thermal efficiency and showcases how even small temperature differences can be significant in high-energy systems.