/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 A pump steadily draws water thro... [FREE SOLUTION] | 91Ó°ÊÓ

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A pump steadily draws water through a pipe from a reservoir at a volumetric flow rate of \(1.3 \mathrm{~L} / \mathrm{s}\). At the pipe inlet, the pressure is \(101.4 \mathrm{kPa}\), the temperature is \(18^{\circ} \mathrm{C}\), and the velocity is \(3 \mathrm{~m} / \mathrm{s}\). At the pump exit, the pressure is \(240 \mathrm{kPa}\), the temperature is \(18^{\circ} \mathrm{C}\), and the velocity is \(12 \mathrm{~m} / \mathrm{s}\). The pump exit is located \(12 \mathrm{~m}\) above the pipe inlet. Ignoring heat transfer, determine the power required by the pump, in \(\mathrm{kJ} / \mathrm{s}\) and \(\mathrm{kW}\). The local acceleration of gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\).

Short Answer

Expert verified
The power required by the pump is 323.694 kW.

Step by step solution

01

- Convert units if necessary

Ensure all units are consistent. Given flow rate is in \(\text{L/s}\). Convert it to \( \text{m}^3/\text{s} \): \(1.3 \text{ L/s} = 1.3 \times 10^{-3} \text{ m}^3/\text{s} \).
02

- Write down the Bernoulli’s Equation

Using the Bernoulli's equation in the extended form with the addition of the pump work term, we have: \[\frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 + \frac{W_{pump}}{\rho g} = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2\].
03

- Solve for hydrodynamic power of the pump

First, rearrange the Bernoulli's equation to solve for the pump work term (\( W_{pump} \)): \( W_{pump} = \rho g \frac{ \big( \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2 \big) - \big( \frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 \big) } \). Here, \(W_{pump} \) is the power required.
04

- Insert given values

Given: \(P_1 = 101.4 \text{ kPa} = 101,400 \text{ Pa}\), \(V_1 = 3 \text{ m/s}\), \(P_2 = 240 \text{ kPa} = 240,000 \text{ Pa}\), \(V_2 = 12 \text{ m/s}\), \( z_1 = 0 \text{ m} \), \( z_2 = 12 \text{ m} \), \( g = 9.8 \text{ m/s}^2 \), and \( \rho = 1000 \text{ kg/m}^3 \) (assuming water).
05

- Calculate each term

1. \(\frac{P_1}{\rho g} = \frac{101,400}{1000 \times 9.8} = 10.35 \text{ m}\).2. \(\frac{V_1^2}{2g} = \frac{3^2}{2 \times 9.8} = 0.46 \text{ m}\).3. \(\frac{P_2}{\rho g} = \frac{240,000}{1000 \times 9.8} = 24.49 \text{ m}\).4. \(\frac{V_2^2}{2g} = \frac{12^2}{2 \times 9.8} = 7.35 \text{ m}\).
06

- Plug in the calculated values

Using these values: \(W_{pump} = \rho g \big(24.49 + 7.35 + 12 - (10.35 + 0.46)\big)\) \(W_{pump} = 1000 \times 9.8 \times 33.03 \) \(W_{pump} = 323,694 \text{ W} \) or \(W_{pump} = 323.694 \text{ kW} \).

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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 describes the conservation of energy in a fluid flow system. It relates the pressure, velocity, and elevation at two points in a fluid under steady flow conditions. The complete form of Bernoulli's equation, including the pump work, is given by: In this exercise, Bernoulli's equation helps us analyze the energy changes between the inlet and outlet of the pump. By examining the fluid's properties at these two points, we can determine the required pump power.
Hydrodynamic Power
Hydrodynamic power is the power required to move a fluid through a system, overcoming resistance and lifting it to a higher elevation. To find the hydrodynamic power of the pump, we rearrange Bernoulli's equation to isolate the pump work term: The power required by the pump is directly influenced by changes in pressure, velocity, and elevation of the fluid. By plugging in the given values, we can calculate the work done by the pump and thereby determine the hydrodynamic power.
Energy Conversion in Fluids
Energy conversion in fluids involves transforming one form of energy into another, such as converting kinetic energy to pressure energy or gravitational potential energy. In the context of the pump problem, the pump converts electrical or mechanical energy into hydrodynamic energy, which results in changes in the fluid's pressure, velocity, and elevation. Understanding this conversion is crucial for determining the power requirements and efficiency of pumps and other fluid-handling equipment.
Engineering Problem Solving
Engineering problem solving often involves breaking a complex problem into manageable steps. This methodical approach helps ensure all relevant factors are considered and integrated into the solution. For the pump power calculation, we:
  • Convert units to consistent measurements
  • Apply Bernoulli's equation to the system
  • Isolate and solve for the desired term, accommodating real-world conditions such as elevation changes and fluid properties

This step-by-step process helps in systematically arriving at an accurate and reliable solution, demonstrating the importance of a structured approach in engineering tasks.

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Most popular questions from this chapter

Air enters a compressor operating at steady state with a pressure of 1 bar, a temperature of \(20^{\circ} \mathrm{C}\), and a volumetric flow rate of \(0.25 \mathrm{~m}^{3} / \mathrm{s}\). The air velocity in the exit pipe is 210 \(\mathrm{m} / \mathrm{s}\) and the exit pressure is \(1 \mathrm{MPa}\). If each unit mass of air passing from inlet to exit undergoes a process described by \(p v^{1.34}=\) constant, determine the exit temperature, in \({ }^{\circ} \mathrm{C}\), and the diameter of the exit pipe, in \(\mathrm{mm}\).

A large pipe carries steam as a two-phase liquid-vapor mixture at \(1.0 \mathrm{MPa}\). A small quantity is withdrawn through a throttling calorimeter, where it undergoes a throttling process to an exit pressure of \(0.1 \mathrm{MPa}\). For what range of exit temperatures, in \({ }^{\circ} \mathrm{C}\), can the calorimeter be used to determine the quality of the steam in the pipe? What is the corresponding range of steam quality values?

Air enters a one-inlet, one-exit control volume at 8 bar, \(600 \mathrm{~K}\), and \(40 \mathrm{~m} / \mathrm{s}\) through a flow area of \(20 \mathrm{~cm}^{2}\). At the exit, the pressure is 2 bar, the temperature is \(400 \mathrm{~K}\), and the velocity is \(350 \mathrm{~m} / \mathrm{s}\). The air behaves as an ideal gas. For steady-state operation, determine (a) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\). (b) the exit flow area, in \(\mathrm{cm}^{2}\).

A water heater operating under steady flow conditions receives water at the rate of \(5 \mathrm{~kg} / \mathrm{s}\) at \(80^{\circ} \mathrm{C}\) temperature with specific enthalpy of \(320.5 \mathrm{~kJ} / \mathrm{kg}\). Water is heated by mixing steam at temperature \(100.5^{\circ} \mathrm{C}\) and specific enthalpy of 2650 \(\mathrm{kJ} / \mathrm{kg}\). The mixture of water and steam leaves the heater in the form of liquid water at temperature \(100^{\circ} \mathrm{C}\) with specific enthalpy of \(421 \mathrm{~kJ} / \mathrm{kg}\). Calculate the required steam flow rate to the heater per hour.

Steam at \(11032 \mathrm{kPa}, 538^{\circ} \mathrm{C}\), and a velocity of \(0.6 \mathrm{~m} / \mathrm{s}\) enters a turbine operating at steady state. As shown in Fig. P4.43, \(22 \%\) of the entering mass flow is extracted at \(1103 \mathrm{kPa}, 232^{\circ} \mathrm{C}\), with a velocity of \(3 \mathrm{~m} / \mathrm{s}\). The rest of the steam exits as a two-phase liquid-vapor mixture at \(7 \mathrm{kPa}\), with a quality of \(85 \%\) and a velocity of \(45 \mathrm{~m} / \mathrm{s}\). The turbine develops a power output of \(2.6 \times 10^{5} \mathrm{~kW}\). Neglecting potential energy effects and heat transfer between the turbine and its surroundings, determine (a) the mass flow rate of the steam entering the turbine, in \(\mathrm{kg} / \mathrm{h}\). (b) the diameter of the extraction duct, in \(\mathrm{m}\).

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