Question 3.11: Using the Energy Balance and Software One kmol of carbon dio...

Known One kmol of CO _{2} undergoes a constant-pressure process in a piston–cylinder assembly. The initial temperature, T_{1}, and the pressure are known.

Find Plot the heat transfer versus the final temperature, T_{2}. Use the ideal gas model and evaluate \Delta \bar{u} using (a) \bar{u} data from IT, (b) constant \bar{c}_{v} evaluated at T_{1} from IT.

Schematic and Given Data:

Engineering Model

1. The carbon dioxide is a closed system.

2. The piston is the only work mode, and the process occurs at constant pressure.

3. The carbon dioxide behaves as an ideal gas.

4. Kinetic and potential energy effects are negligible.

Analysis

The heat transfer is found using the closed system energy balance, which reduces to

Using Eq. 2.17 at constant pressure (assumption 2)

W=\int_{V_{1}}^{V_{2}} p d V (2.17)

Then, with \Delta U=n\left(\bar{u}_{2}-\bar{u}_{1}\right), the energy balance becomes

Solving for Q

1 Q=n\left[\left(\bar{u}_{2}-\bar{u}_{1}\right)+p\left(\bar{v}_{2}-\bar{v}_{1}\right)\right]

With p \bar{v}=\bar{R} T, this becomes

The object is to plot Q versus T_{2} for each of the following cases: (a) values for \bar{u}_{1} \text { and } \bar{u}_{2} \text { at } T_{1} \text { and } T_{2}, respectively, are provided by IT, (b) Eq. 3.50 is used on a molar basis, namely,

u\left(T_{2}\right)-u\left(T_{1}\right)=c_{v}\left(T_{2}-T_{1}\right) (3.50)

where the value of \bar{c}_{v} is evaluated at T_{1} using IT.

The IT program follows, where Rbar denotes \bar{R}, cvb denotes \bar{c}_{v}, and ubar1 and ubar2 denote \bar{u}_{1} \text { and } \bar{u}_{2}, respectively.

Use the Solve button to obtain the solution for the sample case of T_{2}=1500 K. For part (a), the program returns Q_{ a }=6.16 \times 10^{4} kJ. The solution can be checked using CO _{2} data from Table A-23, as follows:

Thus, the result obtained using CO _{2} data from Table A-23 is in close agreement with the computer solution for the sample case. For part (b), IT returns \bar{c}_{v}=28.95 kJ / kmol \cdot K \text { at } T_{1} \text { , giving } Q_{ b }=4.472 \times 10^{4} kJ \text { when } T_{2}=1500 K. This value agrees with the result obtained using the specific heat c_{v} at 300 K from Table A-20, as can be verified.

Now that the computer program has been verified, use the Explore button to vary T_{2} from 300 to 1500 K in steps of 10. Construct the following graph using the Graph button:

As expected, the heat transfer is seen to increase as the final temperature increases. From the plots, we also see that using constant \bar{c}_{v} \text { evaluated at } T_{1} \text { for calculating } \Delta \bar{u}, and hence Q, can lead to considerable error when compared to using \bar{u} data. The two solutions compare favorably up to about 500 K, but differ by approximately 27% when heating to a temperature of 1500 K.

1 Alternatively, this expression for Q can be written as

Introducing \bar{h}=\bar{u}+p \bar{v}, the expression for Q becomes

Skills Developed

Ability to…

• define a closed system and identify interactions on its boundary.

• apply the energy balance using the ideal gas model.

• use IT to retrieve property data for CO _{2} as an ideal gas and plot calculated data.

Quick Quiz

Repeat part (b) using \bar{c}_{v} \text { evaluated at } T_{\text {average }}=\left(T_{1}+T_{2}\right) / 2. Which approach gives better agreement with the results of part (a): evaluating \bar{c}_{v} \text { at } T_{1} \text { or at } T_{\text {average }} ? \text { Ans. At } T_{\text {average }}