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A storage battery develops a power output of $$ \dot{W}=1.2 \exp (-t / 60) $$ where \(\dot{W}\) is power, in \(\mathrm{kW}\), and \(t\) is time, in s. Ignoring heat transfer (a) plot the power output, in \(\mathrm{kW}\), and the change in energy of the battery, in \(\mathrm{kJ}\), each as a function of time. (b) What are the limiting values for the power output and the change in energy of the battery as \(t \rightarrow \infty\) ? Discuss.

Short Answer

Expert verified
1.2 \(\text{\( kW \)}\) decreases to 0 kW; energy increases to 72 kJ.

Step by step solution

01

Define and Understand Given Functions

The power output \(\text{\( \dot{W} \)}\) is given by \(\text{\( \dot{W} = 1.2 \exp \left( - \frac{t}{60} \right) \)}\) in kW. Here, \(\text{\(t \)}\) is time in seconds and \(\text{\( \dot{W} \)}\) is the output power in kW.
02

Plot Power Output as a Function of Time

To plot \(\text{\( \dot{W} \)}\) as a function of \(\text{\( t \)}\), evaluate \(\text{\( \dot{W}(t) \)}\) for various values of \(\text{\( t \)}\) including \(\text{\( t = 0, 60, 120, 180, 240, 300 \)}\) to understand how power output changes over time. Use a graphing tool to create the plot.
03

Integrate Power to Find Energy Change

The change in energy \(\text{\( \Delta E \)}\) of the battery is found by integrating the power output with respect to time. Integrate \(\text{\( \int_0^t \dot{W}(t') dt' \)}\) to find \(\text{\( \Delta E(t) \)}\).
04

Compute Energy Change Function

Solve the integral \(\text{\( \int_0^t 1.2 \exp \left( -\frac{t'}{60} \right) dt' \)}\). This evaluates to \(\text{\( \Delta E(t) = 1.2 \times -60 \left( -\exp \left( -\frac{t}{60} \right) + 1 \right) \)}\). Simplify to: \(\text{\( \Delta E(t) = 72 (1 - \exp \left( -\frac{t}{60} \right)) \)}\) in kJ.
05

Plot Change in Energy as Function of Time

Plot \(\text{\( \Delta E \)}\) as a function of time \(\text{\( t \)}\). Use a graphing tool to create the plot with key time points such as t = 0, 60, 120, 180, 240, 300 seconds.
06

Determine Limiting Values as t Approaches Infinity

Evaluate the limiting values as \(\text{\( t \rightarrow \infty \)}\): \(\text{\( \dot{W} = 1.2 \exp \left( -\frac{t}{60} \right) \rightarrow 0 \)}\) and \(\text{\( \Delta E = 72 (1 - \exp \left( -\frac{t}{60} \right)) \rightarrow 72 \)}\) kJ.

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Key Concepts

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

Integration in Thermodynamics
In thermodynamics, integration helps us determine energy changes over time. For our problem, we start with a given power output \( \text{\( \dot{W} = 1.2 \exp\left( -\frac{t}{60} \right) \)} \) where \( t \) is time in seconds and \( \dot{W} \) is power in kW.
This power function describes how power decreases exponentially over time.
To find the total energy change in the battery, we integrate the power function with respect to time.
We set up our integral from 0 to some time \( t \). The integral looks like this: \[ \Delta E(t) = \int_0^t 1.2 \exp\left( -\frac{t'}{60} \right) dt' \]
Here, \( t' \) is a dummy integration variable. By solving it, we simplify to get \[ \Delta E(t) = 72 (1 - \exp\left( -\frac{t}{60} \right)) \]
This result gives us the energy change \( \Delta E(t) \) in kJ over time.
Let's look deeper at this as the next key concept involves analyzing energy change over time.
Energy Change Over Time
The concept of energy change over time provides insight into how the battery's stored energy evolves.
In our case, we determined that: \[ \Delta E(t) = 72 (1 - \exp\left( -\frac{t}{60} \right)) \]
This function means the change in energy starts from 0 kJ and increases over time, approaching 72 kJ as time goes to infinity.
Initially, the rate of increase in energy is high.
However, as time progresses, this rate slows down markedly.
This behavior can be seen when graphing the function at various points (t = 0, 60, 120, 180, 240, 300 seconds).
Understanding this change is vital because it reveals how efficient the battery is in converting power-output over time into stored energy.
Next, we will dive into an analysis to understand the battery discharge process.
Battery Discharge Analysis
Analyzing battery discharge involves examining the power output and energy change over time.
The power output function \( \text{\( \dot{W}(t) = 1.2 \exp\left( -\frac{t}{60} \right) \)} \) models how power decreases as the battery discharges.
Initially, at t = 0 seconds, power output is highest at 1.2 kW.
As time progresses, power output decreases exponentially.
We observe this by plotting \( \text{\( \dot{W}(t) \)} \) for various time values.
As t approaches infinity, the power output approaches 0 kW, effectively meaning that the battery is completely discharged.
On the other hand, by integrating this power output, we found the energy change function.
This energy curve, \(\text{\( \Delta E(t) = 72 (1 - \exp\left( -\frac{t}{60} \right)) \)} \) shows how the battery accumulates energy over time.
As t goes to infinity, \( \text{\( \Delta E(t) \)} \) reaches its maximum value of 72 kJ, indicating full energy storage.
This analysis helps in understanding the effectiveness of energy storage and discharge, crucial for designing and optimizing battery systems.

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Most popular questions from this chapter

A closed system undergoes a process during which there is energy transfer from the system by heat at a constant rate of \(10 \mathrm{~kW}\), and the power varies with time according to $$ \dot{W}= \begin{cases}-8 t & 01 \mathrm{~h}\end{cases} $$ where \(t\) is time, in \(\mathrm{h}\), and \(\dot{W}\) is in \(\mathrm{kW}\). (a) What is the time rate of change of system energy at \(t=\) \(0.6 \mathrm{~h}\), in \(\mathrm{kW}\) ? (b) Determine the change in system energy after \(2 \mathrm{~h}\), in \(\mathrm{kJ}\).

A major force opposing the motion of a vehicle is the rolling resistance of the tires, \(F_{r}\), given by $$ F_{\mathrm{r}}=f^{\circ} W $$ where \(f\) is a constant called the rolling resistance coefficient and \(W\) is the vehicle weight. Determine the power, in \(\mathrm{kW}\), required to overcome rolling resistance for a truck weighing \(322.5 \mathrm{kN}\) that is moving at \(110 \mathrm{~km} / \mathrm{h}\). Let \(f=0.0069\).

List examples of heat transfer by conduction, radiation, and convection you might find in a kitchen.

Air undergoes two processes in series: Process 1-2: polytropic compression, with \(n=1.3\), from \(p_{1}=\) \(100 \mathrm{kPa}, v_{1}=0.04 \mathrm{~m}^{3} / \mathrm{kg}\) to \(v_{2}=0.02 \mathrm{~m}^{3} / \mathrm{kg}\) Process 2-3: constant-pressure process to \(v_{3}=v_{1}\) Sketch the processes on a \(p-v\) diagram and determine the work per unit mass of air, in \(\mathrm{kJ} / \mathrm{kg}\).

An electric generator coupled to a windmill produces an average electric power output of \(15 \mathrm{~kW}\). The power is used to charge a storage battery. Heat transfer from the battery to the surroundings occurs at a constant rate of \(1.8 \mathrm{~kW}\). Determine, for \(8 \mathrm{~h}\) of operation (a) the total amount of energy stored in the battery, in \(\mathrm{kJ}\). (b) the value of the stored energy, in \(\$$ if electricity is valued at \)\$ 0.08\( per \)\mathrm{kW} \cdot \mathrm{h}$.

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