91Ó°ÊÓ

The final temperature

Given:A nitrogen throttling process operates between pressures 1 bar and 100 bars. The initial temperature is $300\mathrm{K}$.

Formula:The value of the initial enthalpy at temperature $300\mathrm{K}$and pressure 100 bars is $8174\mathbf{J}$, the enthalpy at temperature $300\mathrm{K}$and pressure 1 bar is $8717\mathrm{J}$, and the enthalpy at temperature $200\mathrm{K}$and pressure 1 bar is $5800\mathrm{J}$from table 4.5

The liquid nitrogen fraction x has the following value:

$$\begin{gathered} x=\frac{(8174\mathrm{J})-(8717\mathrm{J})}{(5800\mathrm{J})-(8717\mathrm{J})} \\ =0.186 \end{gathered}$$

$\text{The expression of the final temperature}{T}_{\mathrm{F}}$

$T_{\mathrm{F}}=(200\mathrm{K})x+(300\mathrm{K})(1-x)…\left(1\right)$

Calculation:Substitute $x=0.186inequation\left(1\right)$

$$\begin{gathered} T_{\mathrm{F}}=(200\mathrm{K})(0.186)+(300\mathrm{K})(1-0.186) \\ =281.4k \end{gathered}$$

Hence the final temperature is$281.4k$

The final temperature

Given:

A nitrogen throttling process operates between pressures 1 bar and 100 bars. The initial temperature is $200\mathrm{K}$.

Formula : The value of the initial enthalpy at temperature $200\mathrm{K}$and pressure 100 bars is $4442\mathrm{J}$, the enthalpy at temperature $200\mathrm{K}$and pressure $1\mathrm{bar}$is $5800\mathrm{J}$, and the enthalpy at temperature $100\mathrm{K}$and pressure 1 bar is $2856\mathrm{J}$from table 4.5.

The fraction of nitrogen x that becomes liquefied has the following value:

$$\begin{gathered} x=\frac{(4442\mathrm{J})-(5800\mathrm{J})}{(2856\mathrm{J})-(5800\mathrm{J})} \\ =0.461 \end{gathered}$$

The expression of the final temperature $T_{\mathrm{F}}$

$T_{\mathrm{F}}=(100\mathrm{K})x+(200\mathrm{K})(1-x)….\left(2\right)$

Substitute X= 0.461 in equation (2)

$$\begin{gathered} T_{\mathrm{F}}=(100\mathrm{K})(0.461)+(200\mathrm{K})(1-0.461) \\ =153.9\mathrm{K} \end{gathered}$$

Hence the final temperature is 153.9K

The final temperature. and the fraction of nitrogen that becomes liquefied.

Given:

A nitrogen throttling process operates between pressures 1 bar and 100 bars. The initial temperature is $100\mathrm{K}$.

Formula: In order to conserve the initial enthalpy, the temperature has to decrease. Since the decreased temperature lies between the enthalpies for gaseous and liquid nitrogen at pressure 1 bar and temperature $77\mathrm{K}$, therefore, the temperature will be$77\mathrm{K}$after the throttling process.

$$\begin{gathered} \text{The expression of the initial enthalpy}H\text{at temperature}100\mathrm{K} \\ \text{and pressure}100\text{bars} \end{gathered}$$

$H=H_{\mathrm{liq}}x+H_{\mathrm{gas}}(1-x)$

Here, x is fraction of liquefaction, $H_{\text{liq}}$is the enthalpy of liquefied nitrogen and $H_{\text{gas}}$is the enthalpy of gaseous nitrogen at pressure$1\mathrm{bar}$and temperature $77\mathrm{K}$.

Rearrange the above expression for $x$

$x=\frac{H-H_{gE}}{H_{\mathrm{fiq}}-H_{gas}}…\text{(3)}$

Calculation:Substitute H= $-1946\mathrm{J}$$H_{\text{liq}}$=$-3407J$androle="math" localid="1648599827773" $H_{\text{gas}}=2161\mathrm{J}$in equation (3)

$$\begin{gathered} x=\frac{(-1946\mathrm{J})-(2161\mathrm{J})}{(-3407\mathrm{J})-(2161\mathrm{J})} \\ =0.7376 \end{gathered}$$

Thus, the final temperature after the throttling process is $77\mathrm{K}$.

Thus, the fraction of nitrogen that becomes liquefied is $\underset{¯}{0.7376}$.

The highest initial temperature at which some liquefaction takes place.

Given:$$\begin{gathered} \text{A nitrogen throttling process operates between pressures}1\text{bar and} \\ 100\text{bars.} \end{gathered}$$

Formula: The temperature at which liquefaction occurs is when the enthalpy of the liquid equals the enthalpy of the gas. It's somewhere between$100\mathrm{K}$and $200\mathrm{K}$in temperature.

The value of the enthalpy of gas is $2161\mathrm{J}$, the enthalpy at temperature $100\mathrm{K}$is $-1946\mathrm{J}$, and the enthalpy at temperature$200\mathrm{K}$is $4442\mathrm{J}$from table 4.5.

$\text{The value of the fraction of nitrogen}x\text{that becomes liquefied is:}$

$$\begin{gathered} x=\frac{(2161\mathrm{J})-(4442\mathrm{J})}{(-1946\mathrm{J})-(4442\mathrm{J})} \\ =0.357 \end{gathered}$$

The expression of the highest temperature $T_{\mathrm{H}}$

$T_{\mathrm{H}}=(100\mathrm{K})x+(200\mathrm{K})(1-x)…\left(4\right)$

Calculation:Substitute x= 0.357 in equation (4)

$$\begin{gathered} T_{\mathrm{F}}=(100\mathrm{K})(0.357)+(200\mathrm{K})(1-0.357) \\ =164.3\mathrm{K} \end{gathered}$$

$$\begin{gathered} \text{Hencethe highest initial temperature at which some liquefaction}\text{t} \\ \text{akes place is}164.3\mathrm{K} \end{gathered}$$

When the initial temperature is $600\mathrm{K}$the following is the result.

The nitrogen throttling technique works at pressures ranging from 1 bar to 100 bar.

Because the enthalpy at a pressure of 100 bars is greater than the enthalpy at a pressure of 1 bar when the initial temperature is$600\mathrm{K}$the temperature after the throttling operation will rise rather than decrease, in contrast to earlier parts.

$$\begin{gathered} \text{Hencethe final temperature increases when the initial temperatureis} \\ 600\mathrm{K}. \end{gathered}$$