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Chapter 4: Problem 53

A cutting tool uses a nozzle that generates a high-speed jet of liquid water. Assume an exit velocity of \(500 \mathrm{~m} / \mathrm{s}\) of \(20^{\circ} \mathrm{C}\) liquid water with a jet diameter of \(2 \mathrm{~mm}(0.002 \mathrm{~m})\). What is the mass-flow rate? What size (power) pump is needed to generate this from a steady supply of \(20^{\circ} \mathrm{C}\) liquid water at \(200 \mathrm{kPa}\) ?

Short Answer

Expert verified
Mass flow rate: 1.568 kg/s; Pump power needed: 196,000 W.

Step by step solution

01

Determine the Jet's Mass Flow Rate

The mass flow rate \( \dot{m} \) can be determined using the equation: \( \dot{m} = \rho \cdot A \cdot V \), where \( \rho \) is the density of water, \( A \) is the cross-sectional area of the jet, and \( V \) is the exit velocity of the water. At \( 20^{\circ} \text{C} \), the density \( \rho \) of water is approximately \( 998 \text{ kg/m}^3 \). The cross-sectional area \( A \) of the jet is \( \pi \times (d/2)^2 = \pi \times (0.002/2)^2 \text{ m}^2 \). Inserting the known values: \( \dot{m} = 998 \times \pi \times (0.002/2)^2 \times 500 \). Calculate \( \dot{m} \) to find the mass flow rate.
02

Calculate the Cross-sectional Area

Compute the cross-sectional area \( A \) of the jet: \( A = \pi \times (0.002/2)^2 \). This gives \( A \approx 3.14 \times 10^{-6} \text{ m}^2 \).
03

Calculate the Mass Flow Rate

Insert the calculated area \( A \) and the given values for \( \rho \) and \( V \) back into the mass flow rate equation: \( \dot{m} = 998 \times 3.14 \times 10^{-6} \times 500 \). Solving this gives \( \dot{m} \approx 1.568 \text{ kg/s} \).
04

Calculate Pump Power

To find the required pump power \( P \), use the formula: \( P = \dot{m} \times g \times h + \frac{1}{2} \dot{m} \times V^2 \), where \( h \) is the head corresponding to the pressure in pascals divided by \( \rho \cdot g \), and \( g \approx 9.81 \text{ m/s}^2 \). For \( 200 \text{ kPa} \) pressure, \( h = \frac{200,000}{998 \times 9.81} \), and substituting gives: \( P = 1.568 \times \left(9.81 \times \frac{200,000}{998 \times 9.81} + 0.5 \times 500^2\right) \). Simplify and calculate for \( P \).
05

Finalize the Pump Power Calculation

First calculate the head: \( h = \frac{200,000}{998 \times 9.81} \approx 20.42 \text{ m} \). Substitute \( h \) into the power formula and find \( P \approx 196,000 \text{ W} \).

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Key Concepts

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

Mass Flow Rate
The mass flow rate, denoted as \( \dot{m} \), is a fundamental measurement in fluid dynamics. It represents the mass of fluid that passes through a given area per unit of time. For any fluid flowing out of a nozzle, calculating the mass flow rate helps us understand its behavior and performance upon exiting.
  • To determine the mass flow rate, the equation \( \dot{m} = \rho \cdot A \cdot V \) is used, where \( \rho \) is the fluid density, \( A \) is the cross-sectional area the fluid flows through, and \( V \) is the velocity of the fluid.
  • For our exercise, we consider water at 20掳C, which has a density \( \rho = 998 \, \text{kg/m}^3 \).
  • The nozzle鈥檚 jet diameter was given as \(0.002 \, m\), allowing us to compute the area \( A \) using the circle area formula: \( A = \pi \times (d/2)^2 \) which gives approximately \(3.14 \times 10^{-6} \, \text{m}^2 \).
  • Substituting these values back into the mass flow rate equation, with an exit velocity of \(500 \, \text{m/s}\), results in a mass flow rate of approximately \(1.568 \, \text{kg/s}\).
Nozzle Flow
Nozzle flow focuses on how fluid behaves as it moves through and exits from a nozzle. This challenges our understanding of velocity changes under constrained pathways and continuity equation applications.
  • A nozzle constricts flow to increase velocity, crucial for applications needing high-speed streams, such as cutting tools.
  • The continuity equation forms the backbone for understanding nozzle flow: \( A_1V_1 = A_2V_2 \), indicating that the product of cross-sectional area and velocity remains constant across the nozzle.
  • This exercise highlights the exit velocity of 500 m/s. Such high velocity requires precisely calculated nozzle dimensions, as seen in the mass flow calculation.
  • Understanding nozzle flow involves not just mechanics, but thinking about energy conservation and optimal equipment design.
Pump Power
Pump power is the required power to drive fluid through a system, ensuring it reaches the desired pressures and velocities. Calculating the power required for a pump is critical for selecting and designing pumps adequate to application needs.
  • Pump power is calculated using the formula: \( P = \dot{m} \times g \times h + \frac{1}{2} \dot{m} \times V^2 \).
  • This equation considers both the gravitational potential energy (\(g\) for gravity, and \(h\) for the hydraulic head) and kinetic energy components (using the exit velocity \(V\)).
  • In this exercise, it's crucial to calculate the head \(h\), which is derived from hydraulic losses: \(h = \frac{200,000}{998 \times 9.81} \approx 20.42 \text{m}\).
  • The final calculated pump power rounded off ends at approximately \(196,000 \text{W}\), showcasing the energy demands of accelerating water to high speeds.
Thermodynamics Concepts
Thermodynamics connects intimately with fluid dynamics, especially in systems like pumps and nozzles, where energy transformations are at play. Understanding this interplay ensures effective design and operation of hydraulic systems.
  • The First Law of Thermodynamics, often related to the conservation of energy, is applicable when determining energy exchanges in pump and nozzle operations.
  • In our scenario, the energy conservation principle helps calculate energy added by the pump to overcome pressure differences and increase kinetic energy in the water jet.
  • The Bernoulli equation, another thermodynamic principle, helps correlate pressure, velocity, and height in fluid flow scenarios, aiding in head loss calculations.
  • Recognizing the role of thermal dynamics not only in heat engines but also hydraulically driven systems, enlarges the toolkit for engineers.

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Most popular questions from this chapter

Carbon dioxide gas enters a steady-state, steady-flow heater at \(45 \mathrm{lbf} / \mathrm{in} .^{2}, 60 \mathrm{~F}\) and exits at \(40 \mathrm{lbf} / \mathrm{in} .^{2}, 1800 \mathrm{~F}\). Changes in kinetic and potential energies are negligible. Calculate the required heat transfer per lbm of carbon dioxide flowing through the heater.

A boiler receives a constant flow of \(5000 \mathrm{~kg} / \mathrm{h}\) liquid water at \(5 \mathrm{MPa}\) and \(20^{\circ} \mathrm{C}\), and it heats the flow such that the exit state is \(450^{\circ} \mathrm{C}\) with a pressure of \(4 \mathrm{MPa}\). Determine the necessary minimum pipe flow area in both the inlet and exit pipe(s) if there should be no velocities larger than \(20 \mathrm{~m} / \mathrm{s}\).

A superheater takes \(3 \mathrm{~kg} / \mathrm{s}\) saturated water vapor in and heats it at \(2000 \mathrm{kPa}\) to \(350^{\circ} \mathrm{C}\). Find the specific heat transfer in the process.

Methane at \(1 \mathrm{MPa}, 250 \mathrm{~K}\) is throttled through a valve to \(100 \mathrm{kPa}\). Assume no change in the kinetic energy. What is the exit temperature?

An irrigation pump takes water from a river at \(10^{\circ} \mathrm{C}, 100 \mathrm{kPa}\) and pumps it up to an open canal, where it flows out \(100 \mathrm{~m}\) higher at \(10^{\circ} \mathrm{C}\). The pipe diameter in and out of the pump is \(0.1 \mathrm{~m}\), and the motor driving the unit is \(5 \mathrm{hp}\). What is the flow rate, neglecting kinetic energy and losses?
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