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Chapter 4: Problem 2

A control volume has one inlet and one exit. The mass flow rates in and out are, respectively, \(\dot{m}_{i}=1.5\) and \(\dot{m}_{\mathrm{e}}=\) \(1.5\left(1-e^{-0.002}\right.\) ), where \(t\) is in seconds and \(\dot{m}\) is in \(\mathrm{kg} / \mathrm{s}\). Plot the time rate of change of mass, in \(\mathrm{kg} / \mathrm{s}\), and the net change in the amount of mass, in \(\mathrm{kg}\), in the control volume versus time, in s, ranging from 0 to \(3600 \mathrm{~s}\).

Short Answer

Expert verified
Integrate and plot \(1.5e^{-0.002t}\) for mass rate change and \(1.5 \left(1 - e^{-0.002 t}\right)\) for net mass change over time.

Step by step solution

01

Write down the given information

Given information: - Inlet mass flow rate, \(\dot{m}_{i}=1.5 \, \text{kg/s}\)- Exit mass flow rate, \(\dot{m}_{e}=1.5\left(1-e^{-0.002 t}\right) \, \text{kg/s}\)- Time range is from 0 to 3600 seconds
02

Write the mass balance equation

Using the mass balance equation for a control volume, we have: \[ \frac{d m_\text{cv}}{d t} = \dot{m}_{i} - \dot{m}_{e}\]where \(m_\text{cv}\) is the mass in the control volume.Substitute the given mass flow rates:\[ \frac{d m_\text{cv}}{d t} = 1.5 - 1.5\left(1 - e^{-0.002 t}\right)\]
03

Simplify the expression

Simplify the expression for the rate of change of mass:\[ \frac{d m_\text{cv}}{d t} = 1.5 - 1.5 + 1.5 e^{-0.002 t} = 1.5 e^{-0.002 t}\]This is the time rate of change of mass in the control volume.
04

Calculate the net change in the amount of mass

Integrate the rate of change of mass to find the net change in mass, \(\Delta m\), over time:\[ \Delta m = \int_0^t \frac{d m_\text{cv}}{d t} \, dt = \int_0^t 1.5 e^{-0.002 t} \, dt\]\(u = -0.002 t\), then \(\frac{du}{dt} = -0.002\)So we need to change the limits accordingly and integrate:\[ \Delta m = 1.5 \int_0^t e^{-0.002 t} \, dt = -750 \left[e^{-0.002 t}\right]_0^t = 1.5 \left[1 - e^{-0.002 t}\right]\]
05

Plot the results

Using a computational tool like MATLAB or Python, plot the following:1. Time rate of change of mass, \(\frac{d m_\text{cv}}{d t} = 1.5 e^{-0.002 t}\), vs time (0 to 3600 seconds).2. Net change in the amount of mass, \(\Delta m = 1.5 \left[1 - e^{-0.002 t}\right]\), vs time (0 to 3600 seconds).

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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 balance equation is essential to understand the changes within a control volume. The general form is:
\[ \frac{d m_{cv}}{d t} = \dot{m}_{i} - \dot{m}_{e} \]
In this exercise, it translates to:
\[ \dot{m}_{cv} = 1.5 - 1.5(1 - e^{-0.002 t}) \]
Mass balance ensures the difference between inlet and outlet flow rates equals the rate of mass change within the control volume. Simplifying, we have:
\[ \frac{d m_{cv}}{d t} = 1.5e^{-0.002 t} \]
Rate of Change of Mass
In thermodynamics, analyzing the rate of change of mass helps us understand how mass within the control volume adjusts over time. Using the simplified equation:
\[ \frac{d m_{cv}}{d t} = 1.5e^{-0.002 t} \]
This function shows that as time increases, the rate of mass change decays exponentially. Initially, the rate is at its peak, then gradually diminishes. When implementing this in a graph, expect a downward trend starting from a maximum value of 1.5kg/s, tapering off as time progresses.
Integration in Thermodynamics
Integration is key to finding the total change or 'net change' in the amount of mass over a period. From the rate of change expression, we integrate:
\[ \Delta m = \int_0^t1.5e^{-0.002 t}dt \]
Using substitution and integration process:
\[ \Delta m = 1.5\big[ 1 - e^{-0.002 t}\big] \]
This integral encompasses the sum of all infinitesimal changes over time. It provides the cumulative mass added or removed from the control volume.
Time-Dependent Analysis
Time-dependent analysis helps us visualize how a system evolves. In this problem, both the exit mass flow rate and the rate of mass change are time-dependent. They rely on the exponential term (e^{-0.002 t}). Such time-based fluctuations are common in real-world systems, where inputs and outputs rarely remain constant.
Tools like MATLAB or Python are ideal for plotting these variations. The plots illustrate the system dynamics:
- Rate of change versus time: Shows the exponential decay
- Net change versus time: Displays how cumulative mass shifts over the period of 3600 seconds. These visualizations provide powerful insights into the long-term behavior of the control volume.

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Most popular questions from this chapter

A rigid tank of volume \(0.75 \mathrm{~m}^{3}\) is initially evacuated. \(\mathrm{A}\) hole develops in the wall, and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) flows in until the pressure in the tank reaches \(1 \mathrm{bar}\). Heat transfer between the contents of the tank and the surroundings is negligible. Determine the final temperature in the tank, in \({ }^{\circ} \mathrm{C}\).

Ammonia enters a heat exchanger operating at steady state as a superheated vapor at 14 bar, \(60^{\circ} \mathrm{C}\), where it is cooled and condensed to saturated liquid at 14 bar. The mass flow rate of the refrigerant is \(450 \mathrm{~kg} / \mathrm{h}\). A separate stream of air enters the heat exchanger at \(17^{\circ} \mathrm{C}, 1\) bar and exits at \(42^{\circ} \mathrm{C}, 1\) bar. Ignoring heat transfer from the outside of the heat exchanger and neglecting kinetic and potential energy effects, determine the mass flow rate of the air, in \(\mathrm{kg} / \mathrm{min}\).

Pumped-hydraulic storage power plants use relatively inexpensive off-peak baseload electricity to pump water from a lower reservoir to a higher reservoir. During periods of peak demand, electricity is produced by discharging water from the upper to the lower reservoir through a hydraulic turbinegenerator. A single device normally plays the role of the pump during upper-reservoir charging and the turbine-generator during discharging. The ratio of the power developed during discharging to the power required for charging is typically much less than \(100 \%\). Write a report describing the features of the pump-turbines used for such applications and their size and cost. Include in your report a discussion of the economic feasibility of pumped-hydraulic storage power plants,

Carbon dioxide gas is heated as it flows steadily through a 2.5-cm-diameter pipe. At the inlet, the pressure is 2 bar, the temperature is \(300 \mathrm{~K}\), and the velocity is \(100 \mathrm{~m} / \mathrm{s}\). At the exit, the pressure and velocity are \(0.9413\) bar and \(400 \mathrm{~m} / \mathrm{s}\), respectively. The gas can be treated as an ideal gas with constant specific heat \(c_{p}=0.94 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Neglecting potential energy effects, determine the rate of heat transfer to the carbon dioxide, in \(\mathrm{kW}\).

The mass flow rate at the inlet of a one-inlet, one-exit control volume varies with time according to \(\dot{m}_{\mathrm{i}}=100\left(1-e^{-2 l}\right)\), where \(\dot{m}_{1}\) has units of \(\mathrm{kg} / \mathrm{h}\) and \(t\) is in \(\mathrm{h}\). At the exit, the mass flow rate is constant at \(100 \mathrm{~kg} / \mathrm{h}\). The initial mass in the control volume is \(50 \mathrm{~kg}\). (a) Plot the inlet and exit mass flow rates, the instantaneous rate of change of mass, and the amount of mass contained in the control volume as functions of time, for \(t\) ranging from 0 to \(3 \mathrm{~h}\). (b) Estimate the time, in \(\mathrm{h}\), when the tank is nearly empty.
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