Question 5.6: Consider a piston–cylinder device, as shown in Figure 5.8, w...
Consider a piston–cylinder device, as shown in Figure 5.8, which initially contains air at a pressure of 400 kPa and a temperature of 17 °C. The air in the piston–cylinder device receives 50 kJ of work per each kg of air in the device through the rotating paddle. The air in the device receives thermal energy (heat) from a source with a temperature of 600 °C. The piston–cylinder device operates at a constant pressure of approximately 400 kPa, which makes the process an isobaric process. The air volume in the device increases to three times its original volume. (a) Write the mass, energy, entropy, and exergy balance equations for this system, (b) find the boundary work produced by the mechanism, (c) find the heat input from the source to the system, (d) calculate the entropy generation, and (e) calculate the exergy destruction.
a) Write the mass, energy, entropy, and exergy balance equations.
Analyze the piston–cylinder device using the balance equations, which can be written as:
\text { MBE }: m_{1}=m_{2}=\text { constant }EBE : m_{1} u_{1}+Q_{i n}+W_{i n}=m_{2} u_{2}+W_{out}
Here, W_{out} is the boundary movement work as produced by the system.
EnBE : m_{1} S_{1}+Q_{i n} / T_{s}+S_{\text {gen }}=m_{2} S_{2}\operatorname{ExBE}: m_{1} e x_{1}+\left(1-\left(T_{o} / T_{s}\right)\right) Q_{i n}+W_{i n}=m_{2} e x_{2}+W_{o u t}+E x_{d}
b) Calculate the boundary work produced by the system.
Note that the temperature of the air inside the device remains inconstant through the process and the pressure remains constant due to the piston motion and production of boundary work. One can write the ideal gas equation for this as
P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}and find T_2 =870 K
In addition, the EBE is reduced to
u_{1}+q_{i n}+w_{i n}=u_{2}+w_{out}Since the system operates at a constant pressure (so-called: isobaric process), one can then use the following equation to calculate the boundary work produced by the system:
w_{out}=w_{b}=P\left(v_{2}-v_{1}\right)=166.5 kJ / kgc) The heat input to the system is calculated as follows, by using the EBE in specific terms:
207.1 kJ / kg +q_{i n}+50 kJ / kg =166.5 kJ / kg +650 kJ / kgq_{i n}=559.3 kJ / kg
d) The entropy generation is calculated based on the entropy balance equation written at the beginning of the solution as the first step in solving energy systems:
m_{1} S_{1}+Q_{i n} / T_{s}+S_{\text {gen }}=m_{2} S_{2}By dividing the above equation by the mass of the system, which is constant since the system is a closed energy system, the division result is:
s_{1}+q_{i n} / T_{S}+s_{g e n}=s_{2}5.274+559.3 /(600+273.15)+s_{\text {gen }}=6.417
s_{g e n}=0.502 kJ /( kg K )
e) The exergy destruction is calculated using the ExBE as follows:
m_{1} e x_{1}+\left(1-\left(T_{o} / T_{s}\right)\right) Q_{i n}+W_{i n}=m_{2} e x_{2}+W_{o u t}+E x_{d}E x_{d}=m\left(e x_{1}-e x_{2}\right)+\left(1-\left(T_{o} / T_{s}\right)\right) Q_{i n}+W_{i n}-W_{o u t}
E x_{d}=m\left(u_{1}-u_{2}-T_{0}\left(s_{1}-s_{2}\right)\right)+\left(1-\left(T_{o} / T_{s}\right)\right) Q_{\text {in }}+W_{\text {in }}-W_{\text {out }}
E x_{d}=\left((207.1-650) k J / k g-298 K(5.274-6.417) \frac{k J}{k g K}\right)+\left(1-\left(\frac{298}{873.15}\right)\right) 559.3 \frac{ kJ }{ kg }+50 \frac{ kJ }{ kg }-166.5 \frac{ kJ }{ kg }
E x_{d}=149.6 \frac{k J}{k g}