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Does the amount of mass entering a control volume have to be equal to the amount of mass leaving during an unsteady-flow process?

Short Answer

Expert verified
Answer: No, during an unsteady-flow process, the amount of mass entering a control volume does not have to be equal to the amount of mass leaving it. This can be explained using the principle of conservation of mass and the concept of net mass flow rate into and out of the control volume.

Step by step solution

01

Understand the principle of conservation of mass

According to the principle of conservation of mass, the total mass within a system remains constant over time unless mass is added or removed from the system. In other words, the mass of a system can neither be created nor destroyed.
02

Define control volume and unsteady-flow process

A control volume is a defined region in space through which fluid (or gas) can flow. The control volume is usually chosen to simplify the analysis of fluid behavior, and its boundaries can be fixed, moving, or deformable. An unsteady-flow process is a fluid flow process in which the mass, volume, or other fluid properties within the control volume change with time.
03

Relate the conservation of mass to unsteady-flow process

For an unsteady-flow process, the mass within the control volume changes with time. According to the conservation of mass, the change in mass within the control volume must be equal to the difference between the mass entering the control volume and the mass leaving it. This can be written mathematically as: \[ \frac{dm_{CV}}{dt} = m_{in} - m_{out} \]
04

Determine if mass entering is equal to mass leaving

In an unsteady-flow process, the amount of mass entering the control volume does not necessarily have to be equal to the amount of mass leaving it. If the net mass flow rate into the control volume is positive, the mass within the control volume will increase (i.e., accumulating), and if the net mass flow rate is negative, the mass within the control volume will decrease (i.e., depleting). This can be illustrated using the mathematical equation from step 3. In conclusion, during an unsteady-flow process, the amount of mass entering a control volume does not have to be equal to the amount of mass leaving it. This can be explained using the principle of conservation of mass and the concept of net mass flow rate into and out of the control volume.

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Key Concepts

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

Conservation of Mass
Understanding the conservation of mass is crucial when studying fluid dynamics, especially in scenarios involving unsteady-flow processes. The core of this principle is that mass isn't magically created or annihilated; it's steadfast. It just gets redistributed or transformed from one form to another. Imagine a balloon: as you inflate it, the balloon's size changes, but the amount of air (mass) you've pushed into it has simply moved from your lungs to the balloon. The overall mass of the air hasn't changed.

In a more formal sense, this principle tells us that for any control volume - think of an imaginary box through which fluid flows - the rate at which mass changes inside this box (\( \frac{dm_{CV}}{dt} \) - where 'm' stands for mass and 't' stands for time) is precisely the imbalance between how much mass enters and how much leaves over time. It's a simple yet powerful concept that lays the foundation for more complex fluid mechanics calculations.
Control Volume
A control volume is a virtual or imaginary box that engineers and scientists draw around a part of space to analyze fluid flow. This 'box' can come in any shape or size and can be as flexible or as rigid as the problem requires. It might be a section of a pipe, a room, or even a section of the ocean! The boundaries could be fixed, like the walls of a container, or they could move and deform, like the surface of a bubble. What's important is choosing the right control volume to simplify the analysis of the problem.

When we talk about an unsteady-flow process, we're looking at scenarios where the flow properties inside our chosen control volume change with time. Maybe more water is being poured into our container, or gas is being pumped out of it. These changes are the heart of the unsteady-flow process and provide insights into how systems evolve over time. By strategically selecting the control volume, engineers can isolate and study these changes effectively.
Net Mass Flow Rate
Net mass flow rate is all about the balance - or imbalance - of mass coming in and out of our control volume. Think of it like a checkout line at the grocery store; sometimes more people are joining the line than leaving, and other times it's the other way around. The net mass flow rate tells us which way that 'line' is trending. It's calculated by subtracting the mass flow rate out (mass leaving per unit time, or \( m_{out} \) ) from the mass flow rate in (mass entering per unit time, or \( m_{in} \) ).

The relationship is pretty straightforward: if more mass flows into the control volume than out, we say the net mass flow rate is positive, indicating accumulation. If less mass flows in than out, the net mass flow rate is negative, signifying depletion. During unsteady-flow processes, these rates frequently fluctuate, which is why the mass inside our control volume doesn’t always remain constant and doesn’t necessarily equal the mass leaving, reflecting the dynamic nature of fluid systems.

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Most popular questions from this chapter

In a steam heating system, air is heated by being passed over some tubes through which steam flows steadily. Steam enters the heat exchanger at 30 psia and \(400^{\circ} \mathrm{F}\) at a rate of 15 lbm/min and leaves at 25 psia and \(212^{\circ} \mathrm{F}\). Air enters at 14.7 psia and \(80^{\circ} \mathrm{F}\) and leaves at \(130^{\circ} \mathrm{F}\). Determine the volume flow rate of air at the inlet.

A heat exchanger is used to heat cold water at \(15^{\circ} \mathrm{C}\) entering at a rate of \(2 \mathrm{kg} / \mathrm{s}\) by hot air at \(85^{\circ} \mathrm{C}\) entering at a rate of \(3 \mathrm{kg} / \mathrm{s}\). The heat exchanger is not insulated and is losing heat at a rate of \(25 \mathrm{kJ} / \mathrm{s}\). If the exit temperature of hot air is \(20^{\circ} \mathrm{C},\) the exit temperature of cold water is \((a) 28^{\circ} \mathrm{C}\) (b) \(35^{\circ} \mathrm{C}\) \((c) 38^{\circ} \mathrm{C}\) \((d) 78^{\circ} \mathrm{C}\) \((e) 90^{\circ} \mathrm{C}\)

Hot combustion gases (assumed to have the properties of air at room temperature) enter a gas turbine at \(1 \mathrm{MPa}\) and \(1500 \mathrm{K}\) at a rate of \(0.1 \mathrm{kg} / \mathrm{s}\), and exit at \(0.2 \mathrm{MPa}\) and \(900 \mathrm{K} .\) If heat is lost from the turbine to the surroundings at a rate of \(15 \mathrm{kJ} / \mathrm{s}\), the power output of the gas turbine is \((a) 15 \mathrm{kW}\) (b) \(30 \mathrm{kW}\) \((c) 45 \mathrm{kW}\) \((d) 60 \mathrm{kW}\) \((e) 75 \mathrm{kW}\)

A constant-pressure \(R-134\) a vapor separation unit separates the liquid and vapor portions of a saturated mixture into two separate outlet streams. Determine the flow power needed to pass \(6 \mathrm{L} / \mathrm{s}\) of \(\mathrm{R}-134 \mathrm{a}\) at \(320 \mathrm{kPa}\) and 55 percent quality through this unit. What is the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), of the two outlet streams?

Consider two identical buildings: one in Los Angeles, California, where the atmospheric pressure is \(101 \mathrm{kPa}\) and the other in Denver, Colorado, where the atmospheric pressure is 83 kPa. Both buildings are maintained at \(21^{\circ} \mathrm{C}\), and the infiltration rate for both buildings is 1.2 air changes per hour (ACH). That is, the entire air in the building is replaced completely by the outdoor air 1.2 times per hour on a day when the outdoor temperature at both locations is \(10^{\circ} \mathrm{C}\). Disregarding latent heat, determine the ratio of the heat losses by infiltration at the two cities.

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