/*! 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 94 Liquid water enters a valve at \... [FREE SOLUTION] | 91Ó°ÊÓ

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Liquid water enters a valve at \(300 \mathrm{kPa}\) and exits at \(275 \mathrm{kPa}\). As water flows through the valve, the change in its temperature, stray heat transfer with the surroundings, and potential energy effects are negligible. Operation is at steady state. Modeling the water as incompressible with constant \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), determine the change in kinetic energy per unit mass of water flowing through the valve, in \(\mathrm{kJ} / \mathrm{kg}\).

Short Answer

Expert verified
0.025 kJ/kg

Step by step solution

01

- Identify Given Data

Given data from the problem are:1. Inlet pressure, \(P_1 = 300\) kPa2. Outlet pressure, \(P_2 = 275\) kPa3. Density, \(\rho = 1000\) kg/m³4. Negligible temperature change, heat transfer, and potential energy effects.
02

- Apply Energy Balance

Since heat transfer, work, and potential energy effects are negligible, the energy balance reduces to the change in kinetic energy. At steady state, the energy balance for an incompressible fluid can be simplified to:\[\Delta \frac{v^2}{2} = \frac{P_1 - P_2}{\rho}\]
03

- Plug in Known Values

Substitute \(P_1\), \(P_2\), and \(\rho\) into the simplified energy balance equation:\[\Delta \frac{v^2}{2} = \frac{300 \text{ kPa} - 275 \text{ kPa}}{1000 \text{ kg/m}^3}\]Convert the pressures from kPa to Pa:\[300 \text{ kPa} = 300,000 \text{ Pa}\]\[275 \text{ kPa} = 275,000 \text{ Pa}\]So the equation becomes:\[\Delta \frac{v^2}{2} = \frac{300,000 \text{ Pa} - 275,000 \text{ Pa}}{1000 \text{ kg/m}^3}\]
04

- Final Calculation

Calculate the change in kinetic energy:\[\Delta \frac{v^2}{2} = \frac{25,000 \text{ Pa}}{1000 \text{ kg/m}^3} = 25 \text{ m}^2/\text{s}^2\]Since 1 m²/s² = 0.001 kJ/kg, the change in kinetic energy per unit mass in kJ/kg is:\[\Delta \frac{v^2}{2} = 25 \text{ m}^2/\text{s}^2 * 0.001 \text{ kJ/kg per m}^2/\text{s}^2 = 0.025 \text{ kJ/kg}\]

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Key Concepts

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

Energy Balance Equations
Energy balance equations are essential tools in engineering thermodynamics to account for the energy entering and leaving a system. In this specific scenario, the fluid flows through a valve, and the problem specifies that heat transfer, work, and potential energy effects are negligible. As a result, the energy balance equation simplifies to focus on the kinetic energy change. For an incompressible fluid, the reduced form of the energy balance is given by \[ \frac{\text{d}\frac{v^2}{2}}{\text{d}m} = \frac{P_1 - P_2}{\rho} \], where \(\frac{v^2}{2}\) is the specific kinetic energy, \(P_1\) and \(P_2\) are the pressures at the inlet and outlet, and \( \rho \) is the density of the fluid. This simplified equation helps us focus on the most significant contributor to changes in the system's energy, making it easier to solve real-world engineering problems.
Incompressible Fluid
An incompressible fluid has a constant density regardless of pressure changes. This simplification is valuable in fluid mechanics because it allows us to ignore the effects of compression and expansion, thus making calculations more straightforward. In the given problem, water is treated as an incompressible fluid with a constant density of 1000 kg/m³. This assumption is valid for liquids like water because their density does not significantly change with pressure in most practical engineering applications. Understanding incompressible fluids is crucial in settings where pressure changes are involved but where density remains effectively constant, simplifying analysis and design.
Kinetic Energy Change
The kinetic energy of a fluid is associated with its velocity. In the given problem, we need to determine the change in kinetic energy per unit mass of the water flowing through the valve. Using the simplified energy balance equation, the change in specific kinetic energy is given by \[ \frac{v^2}{2} = \frac{P_1 - P_2}{\rho} \]. With the pressures provided in kPa (300 kPa at the inlet and 275 kPa at the outlet) and assuming a constant density, the change in kinetic energy can be computed straightforwardly. By converting the pressures to Pascal and applying the formula, we find that the specific kinetic energy change is 0.025 kJ/kg. This calculation confirms how pressure differentials can influence the kinetic energy of an incompressible fluid.
Fluid Mechanics
Fluid mechanics is the study of fluids (liquids and gases) and the forces acting on them. It encompasses various sub-disciplines, including fluid statics, fluid kinematics, and fluid dynamics. In the context of the given problem, fluid mechanics principles help us understand how pressure variations within a valve affect the fluid's velocity and kinetic energy. By modeling water as an incompressible fluid and using energy balance equations, we can predict the changes in kinetic energy with remarkable precision. This understanding is fundamental in designing and optimizing devices like valves, pumps, and turbines, where fluid movement is a critical component.
Pressure Differentials
Pressure differentials drive the movement of fluids in pipelines and various engineering systems. In the problem, water flows through a valve from a higher pressure (300 kPa) to a lower pressure (275 kPa). This pressure difference results in a change in kinetic energy, which was computed using the energy balance equations for an incompressible fluid. Understanding pressure differentials is crucial not only for predicting fluid behavior but also for designing systems that control fluid flow effectively. By grasping how pressure differentials impact velocity and kinetic energy, engineers can optimize systems to ensure efficient and safe operation.

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Most popular questions from this chapter

Air expands through a turbine from 8 bar, \(960 \mathrm{~K}\) to 1 bar, \(450 \mathrm{~K}\). The inlet velocity is small compared to the exit velocity of \(90 \mathrm{~m} / \mathrm{s}\). The turbine operates at steady state and develops a power output of \(2500 \mathrm{~kW}\). Heat transfer between the turbine and its surroundings and potential energy effects are negligible. Modeling air as an ideal gas, calculate the mass flow rate of air, in \(\mathrm{kg} / \mathrm{s}\), and the exit area, in \(\mathrm{m}^{2}\).

During cardiac surgery, a heart-lung machine achieves extracorporeal circulation of the patient's blood using a pump operating at steady state. Blood enters the wellinsulated pump at a rate of 5 liters/min. The temperature change of the blood is negligible as it flows through the pump. The pump requires \(20 \mathrm{~W}\) of power input. Modeling the blood as an incompressible substance with negligible kinetic and potential energy effects, determine the pressure change, in \(\mathrm{kPa}\), of the blood as it flows through the pump.

A tank of volume \(1.2 \mathrm{~m}^{3}\) initially contains steam at \(8 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\). Steam is withdrawn slowly from the tank until the pressure drops to \(p\). Heat transfer to the tank contents maintains the temperature constant at \(400^{\circ} \mathrm{C}\). Neglecting all kinetic and potential energy effects and assuming specific enthalpy of the exiting steam is linear with the mass in the tank. (a) determine the heat transfer, in \(\mathrm{kJ}\), if \(p=2 \mathrm{MPa}\). (b) plot the heat transfer, in \(\mathrm{kJ}\), versus \(p\) ranging from \(0.5\) to \(8 \mathrm{MPa}\).

Refrigerant \(134 \mathrm{a}\) flows at steady state through a horizontal tube having an inside diameter of \(0.05 \mathrm{~m}\). The refrigerant enters the tube with a quality of \(0.1\), temperature of \(36^{\circ} \mathrm{C}\), and velocity of \(10 \mathrm{~m} / \mathrm{s}\). The refrigerant exits the tube at 9 bar as a saturated liquid. Determine (a) the mass flow rate of the refrigerant, in \(\mathrm{kg} / \mathrm{s}\). (b) the velocity of the refrigerant at the exit, in \(\mathrm{m} / \mathrm{s}\). (c) the rate of heat transfer, in \(\mathrm{kW}\), and its associated direction with respect to the refrigerant.

Air with a mass flow rate of \(2.3 \mathrm{~kg} / \mathrm{s}\) enters a horizontal nozzle operating at steady state at \(450 \mathrm{~K}, 350 \mathrm{kPa}\), and velocity of \(3 \mathrm{~m} / \mathrm{s}\). At the exit, the temperature is \(300 \mathrm{~K}\) and the velocity is \(460 \mathrm{~m} / \mathrm{s}\). Using the ideal gas model for air with constant \(c_{p}=1.011 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the area at the inlet, in \(\mathrm{m}^{2}\). (b) the heat transfer between the nozzle at its surroundings, in kW. Specify whether the heat transfer is to or from the air.

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