The aim of this problem is to find the **entropy generation** value of an **adiabatic process** in which **nitrogen** is compressed at a given **temperature** and **pressure.** The concept required to solve this problem is related to **thermodynamics,** which includes **the entropy generation formula.**

In **general** terms, **entropy** is described as a standard of **randomness** or **disruption** of a **system.** In the **thermodynamics** point of view, **entropy** is used to explain the **behavior** of a **system** in spans of **thermodynamic** characteristics such as **pressure, temperature,** and **heat capacity.**

If a process undergoes an **entropy change** $(\bigtriangleup S)$, it is described as the **quantity** of **heat** $(q)$ **radiated** or **soaked isothermally** and **reversibly separated** by the absolute **temperature** $(T)$. Its **formula** is given as:

\[\bigtriangleup S=\dfrac{q_{rev,iso}}{T}\]

The total **entropy change** can be found using:

\[\bigtriangleup S_{total}=\bigtriangleup S_{surroundings} + \bigtriangleup S_{system}\]

If the system **radiates heat** $(q)$ at a **temperature** $(T_1)$, which is acquired by surroundings at a **temperature** $(T_2)$, $ \bigtriangleup S_{total}$ becomes:

\[\bigtriangleup S_{total}=-\dfrac{q}{T_1} + \dfrac{q}{T_2} \]

One more important **concept** regarding this problem is **entropy change** for **isothermal expansion** of **gas:**

\[\bigtriangleup S_{total}=nR\ln (\dfrac{V_2}{V_1}) \]

## Expert Answer

Given **information:**

**Initial pressure,** $P_1=100kPa$,

**Initial temperature,** $T_1=25^{\circ}$,

**Final pressure,** $P_2=600kPa$,

**Final temperature,** $T_1=290^{\circ}$.

The properties of **nitrogen** at the given **temperature** are:

**Specific heat capacity,** $c_p=1047\space J/kgK$ and,

**Universal** **gas constant,** $R=296.8$.

Now apply the total **entropy equation** on the **compressor:**

\[S_{in} – S_{out} + S_{gen}=\bigtriangleup S_{system} \]

\[S_{1-2} + S_{gen} = 0\]

\[q_m\cdot (s_{1} – s_2)+S_{gen} = 0 \]

\[S_{gen} = q_m\cdot (s_2 – s_1)\]

Since the **amount** of **heat exchange** between the **system** and the **surroundings** is **negligible,** the **induced entropy** rate is just the difference between the **entropy** at **discharge** and the **inlet.**

The formula to **calculate** the **entropy change** is derived from the **expression** $s = s(T,p)$:

\[\dfrac{S_{gen}}{q_m} = s_{gen} = s_2 – s_1 \]

Using the **isothermal expansion** equations to **simplify:**

\[=c_p\ln (\dfrac{T_2}{T_1}) – R\ln (\dfrac{P_2}{P_1})\]

\[=1047\ln (\dfrac{290+273}{25+273}) – 296.8\ln (\dfrac{600\cdot 10^3}{100\cdot 10^3}) \]

\[s_{gen}= 134 J/kgK \]

## Numerical Result

The **entropy generation** for this **process** is $s_{gen}= 134 J/kgK$.

## Example

Find the **minimum work input** when nitrogen is condensed in an **adiabatic compressor.**

The **thermodynamic properties** of **nitrogen** at an expected intermediate **temperature** of $400 K$ are $c_p = 1.044 kJ/kg·K$ and $k = 1.397$.

Since there is only **one channel in** and **one exit,** thus $s_1 = s_2 = s$. Let’s take the **compressor** as the **system,** then the **energy balance** for this **system** can be yielded as:

\[E_{in} – E_{out} = \bigtriangleup E_{system} = 0\]

**Rearranging,**

\[E_{in} = E_{out} \]

\[mh_1 + W_{in} = mh_2 \]

\[ W_{in} = m(h_2 – h_1) \]

For **minimum work,** the **process** should be **reversible** and **adiabatic** as given in the **statement,** so the exit **temperature** will be:

\[ T_2 = T_1 \{\dfrac{P_2}{P_1}\}^{(k-1)/k} \]

\[ T_2 = 303\{\dfrac{600 K}{120 K}\}^{(0.397)/1.397} = 479 K \]

**Substituting** into the **energy equation** gives us:

\[ W_{in}= m(h_2 – h_1) \]

\[ W_{in} = c_p (T_2 – T_1) \]

\[ W_{in} = 1.044(479- 303) \]

\[ W_{in}= 184 kJ/kg \]