Q.4.34 - An Introduction to Thermal Physics

Question

Consider an ideal Hampson-Linde cycle in which no heat is lost to the environment.

(a) Argue that the combination of the throttling valve and the heat exchanger is a constant-enthalpy device, so that the total enthalpy of the fluid coming out of this combination is the same as the enthalpy of the fluid going in.

(b) Let $x$ be the fraction of the fluid that liquefies on each pass through the cycle. Show that

$x=\frac{H_{\text{out }}-H_{\text{in }}}{H_{\text{out }}-H_{\text{liq }}},$

where $H_{\text{in }}$ is the enthalpy of each mole of compressed gas that goes into the heat exchanger, $H_{\text{out }}$ is the enthalpy of each mole of low-pressure gas that comes out of the heat exchanger, and $H_{\text{liq }}$ is the enthalpy of each mole of liquid produced.

(c) Use the data in Table $4.5$ to calculate the fraction of nitrogen liquefied on each pass through a Hampson-Linde cycle operating between 1 bar and 100 bars, with an input temperature of $300 \mathrm{K}$. Assume that the heat exchanger works perfectly, so the temperature of the low-pressure gas coming out of it is the same as the temperature of the high-pressure gas going in. Repeat the calculation for an input temperature of $200 \mathrm{K}$.

Step-by-Step Solution

Answer

a) As a result, the throttling valve and heat exchanger work together to provide a constant enthalpy device.

b) As a result, the formula for the fraction of fluid x that liquefies on each cycle pass is verified to be.$x=\frac{H_{\text{oul }}-H_{\text{in }}}{H_{\text{oul }}-H_{\text{liq }}}$

c) Thus, the fraction of nitrogen liquefied on each pass through a Hampson-Linde cycle with a $300 \mathrm{K}$ input temperature is $0.045$, while the fraction of nitrogen liquefied on each pass through a Hampson-Linde cycle with a $200 \mathrm{K}$ input temperature is $0.147.$

Part (a) - Step 1: To find

The combination of throttling valve and heat exchanger is a constant enthalpy device, according to the above argument.

Part (a) - Step 2: Explanation

Given: There is no heat loss to the environment in an ideal Hampson-Linde cycle.

Explanation:Throttling and a heat exchanger are used in the Hampson Linde cycle. In the throttling process, the enthalpy is conserved. When the heat exchanger becomes an ideal process, absorbing heat equal to the quantity lost in this cycle, the composite process conserves enthalpy.

Write the expression of the energy of system

$E=q-p v$

Here, $E$ is energy,

          $q$ is heat absorbed

         $p$ is pressure and

        $v$ is volume of system.

The difference between the system's final and initial energies:

$$\begin{gathered} E_{\mathrm{F}}-E_{\mathrm{I}}=(q-pv{)}_{\mathrm{F}}-(q-pv{)}_{\mathrm{I}} \\ =p_{\mathrm{I}}{v}_{\mathrm{I}}-p_{\mathrm{F}}{v}_{\mathrm{F}} \end{gathered}$$

Here, the subscript F denotes final value and 

          the subscript I denotes initial value of variable.

Now,Rearrange the above equation

$E_{\mathrm{F}}+p_{\mathrm{F}}{v}_{\mathrm{F}}=E_{\mathrm{I}}+p_{\mathrm{I}}{v}_{\mathrm{I}}\dots \text{ (1) }$

Write the system's enthalpy $H$ expression.

$H=E+pv\dots \text{ (2) }$

Equation (1) can be simplified by using equation (2)

$H_{\mathrm{F}}=H_{\mathrm{I}}$

As a result, following the cycle, the initial and final values of enthalpy are the same.

Part(b) - Step 3: To find

The fraction of fluid that liquefies on each run through the cycle, expressed as an expression.

Part (b) - Step 4: Explanation

Throttling and a heat exchanger are used in the Hampson-Linde cycle to liquefy gases. An ideal Hampson-Linde cycle conserves the enthalpy.

After one pass through the cycle, write the expression for final enthalpy.

$H_{\mathrm{F}}=(1-x)H_{\text{out }}+x{H}_{\text{liq }}$

Here, $x$ is fraction of fluid that liquefies on each pass,

         $H_{\text{out }}$ is enthalpy of each mole of low pressure gas and

          $H_{\mathrm{liq}}$ is enthalpy of each mole of liquid produced.

Since the enthalpy is conserved during the cycle,

$H_{\mathrm{F}}=H_1\dots \text{ (3) }$

Since the initial enthalpy per mole is the incoming enthalpy, therefore

$H_{\mathrm{I}}=H_{\text{in }}$

Here, $H_{\text{in }}$is enthalpy of each mole of compressed gas going into heat exchanger.

$$\begin{gathered} \text{ Substitute }{H}_{\mathrm{F}}:(1-x)H_{\text{out }}+x{H}_{\text{liq }} \\ H_{\mathrm{I}}=H_{\text{in }}\text{ for in equation (3) } \end{gathered}$$

$(1-x)H_{\text{out }}+x{H}_{\text{liq }}=H_{\text{in }}$

Rearrange the above expression for $x$

$x=\frac{H_{\text{oul }}-H_{\text{in }}}{H_{\text{oul }}-H_{\text{liq }}}$

Hence the formula for the fraction of fluid x that liquefies on each cycle pass is verified to be $x=\frac{H_{\text{oul }}-H_{\text{in }}}{H_{\text{oul }}-H_{\text{liq }}}$

Part (c) - Step 5: To find

On each pass through a Hampson-Linde cycle with an input temperature of $300 \mathrm{K}$, the proportion of nitrogen liquefied. On each pass through a Hampson-Linde cycle with an input temperature of $200 \mathrm{K}$ the proportion of nitrogen liquefied.

Part(c) - Step 6: Explanation

Given: 

The cycle operates at pressures ranging from 1 to 100 bar.

The heat exchanger is designed to keep the temperature of the outgoing low-pressure gas and the incoming high-pressure gas the same.

Formula:

Write an equation for the fraction of liquid$x$that liquefies during each cycle pass.$X=\frac{H_{\text{out }}-H_{\text{in }}}{H_{\text{cut }}-H_{\text{liq }}}\dots .\left(4\right)$

• The liquid has an enthalpy $H_{\mathrm{liq}}$ of $-3407 \mathrm{J}$.Table (4.5)
• The value of incoming enthalpy $H_{\text{in }}$at temperature $300 \mathrm{K}$ and

 pressure 100 bar is $8174 \mathrm{J}$.

• The value of outgoing enthalpy $H_{\text{out }}$ at temperature $300 \mathrm{K}$ and pressure 1 bar is $8717 \mathrm{J}$.
• The value of incoming enthalpy $H_{\text{in }}$at temperature $200 \mathrm{K}$ and

pressure$100\mathrm{bar}$ is $4442 \mathrm{J}$. 

• The value of outgoing enthalpy $H_{\text{out }}$ at temperature $200 \mathrm{K}$ and pressure $1\mathrm{bar}$ is $5800 \mathrm{J}$.  

Part(c)- Step 7: Calculation part

Calculation:

$$\begin{gathered} \text{ Substitute} H_{\text{out }}=8717 \mathrm{J} \\ H_{\text{in }}=8174 \mathrm{J} \\ H_{\text{liq }}=-3407 \mathrm{J}\text{ at }300 \mathrm{K}\text{ in equation (4) } \end{gathered}$$

$$\begin{gathered} x=\frac{(8717 \mathrm{J})-(8174 \mathrm{J})}{(8717 \mathrm{J})-(-3407 \mathrm{J})} \\ =0.045 \end{gathered}$$

$$\begin{gathered} \text{ Substitute }{H}_{\text{out }}=5800 \mathrm{J} \\ H_{\text{in }}=4442 \mathrm{J} \\ H_{\text{liq }}=-3407 \mathrm{J}\text{ at }200 \mathrm{K}\text{ in equation (4) } \end{gathered}$$

$$\begin{gathered} x=\frac{(5800 \mathrm{J})-(4442 \mathrm{J})}{(5800 \mathrm{J})-(-3407 \mathrm{J})} \\ =0.147 \end{gathered}$$

Hence Thus, the fraction of nitrogen liquefied on each pass through a Hampson-Linde cycle with a $300 \mathrm{K}$ input temperature is $0.045$, while the fraction of nitrogen liquefied on each pass through a Hampson-Linde cycle with a $200 \mathrm{K}$ input temperature is $0.147.$