Chapter 17.7, Problem 37P

(a)

To determine

The critical temperature of air.

The critical pressure of air.

The critical density of air.

Answer to Problem 37P

The critical temperature of air is $\underset{\_}{355\text{K}}$.

The critical pressure of air is $\underset{\_}{\text{167}\text{kPa}}$.

The critical density of air is $\underset{\_}{1.65\text{kg}/{\text{m}}^{\text{3}}}$.

Explanation of Solution

Determine the stagnation temperature of ideal gas.

$T_0=T+\frac{V^2}{2c_p}$ (I)

Here, the static temperature of ideal gas is $T$, the specific heat of pressure for ideal gas is $c_p$, and the velocity of the ideal gas flow is $V$.

Determine the stagnation pressure of ideal gas.

$P_0=P{\left(\frac{T_0}{T}\right)}^{k/\left(k-1\right)}$ (II)

Here, the static pressure of ideal gas is $P$, the specific heat ratio of ideal gas is $k$.

Determine the density of the ideal gas.

${\rho }_0=\frac{P_0}{R{T}_0}$ (III)

Here, the pressure of the ideal gas is $P$.

Determine the critical temperature at the throat of nozzle.

$T^{\ast }=T_0\left(\frac{2}{k+1}\right)$ (IV)

Here, the stagnation temperature of ideal gas is $T_0$.

Determine the critical pressure at the throat of nozzle.

$P^{\ast }=P_0{\left(\frac{2}{k+1}\right)}^{k/\left(k-1\right)}$ (V)

Here, the stagnation pressure of ideal gas is $P_0$.

Determine the critical density at the throat of nozzle.

${\rho }^{\ast }={\rho }_0{\left(\frac{2}{k+1}\right)}^{1/\left(k-1\right)}$ (VI)

Here, the stagnation density of ideal gas is ${\rho }_0$.

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of universal gas constant, specific heat of pressure, and the specific heat ratio of air at $300\text{K}$ temperature as $0.287\text{kJ}/\text{kg}\cdot \text{K}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ and $1.400$.

Substitute $100°\text{C}$ for $T$, $325\text{m}/\text{s}$ for $V$, and $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (I).

$\begin{array}{c}{T}_0=\left(100°\text{C}\right)+\frac{{\left(325\text{m}/\text{s}\right)}^2}{2\times \left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =\left(100°\text{C}\right)+\frac{{\left(325\text{m}/\text{s}\right)}^2}{2\times \left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ =\left(100°\text{C}\right)+\frac{\left(105625{\text{m}}^2/{\text{s}}^2\right)\times \left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)}{\left(2.01\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ =152.5°\text{C}\end{array}$

    $\begin{array}{l}=152.5°\text{C}+273.2\\ =425.7\text{K}\end{array}$

Substitute 200 kPa for $P$, $100°\text{C}$ for $T$, $425.7\text{K}$ for $T_0$, and 1.400 for k in Equation (II).

$\begin{array}{c}{P}_0=\left(200\text{kPa}\right)\times {\left(\frac{425.7\text{K}}{100°\text{C}}\right)}^{1.400/\left(1.400-1\right)}\\ =\left(200\text{kPa}\right)\times {\left(\frac{425.7\text{K}}{100°\text{C+273}\text{.2}}\right)}^{1.400/\left(1.400-1\right)}\\ =\left(200\text{kPa}\right)\times \left(1.585138\right)\\ =317.0\text{kPa}\end{array}$

Substitute 317.0 kPa for $P_0$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, and 425.7 K for $T_0$ in Equation (III).

$\begin{array}{c}{\rho }_0=\frac{317.0\text{kPa}}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(425.7\text{K}\right)}\\ =\frac{317.0\text{kPa}}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)\left(425.7\text{K}\right)}\\ =\frac{317.0\text{kPa}}{\left(122.1759\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)}\\ =2.59462\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $425.7\text{K}$ for $T_0$ and 1.400 for k in Equation (IV).

$\begin{array}{c}{T}^{\ast }=\left(425.7\text{K}\right)\times \left(\frac{2}{1.400+1}\right)\\ =\left(425.7\text{K}\right)\times \left(0.8333\right)\\ =354.75\text{K}\\ \cong \text{355}\text{K}\end{array}$

Thus, the critical temperature of air is $\underset{\_}{355\text{K}}$.

Substitute $317.0\text{kPa}$ for $P_0$ and 1.400 for k in Equation (V).

$\begin{array}{c}{P}^{\ast }=\left(317.0\text{kPa}\right)\times {\left(\frac{2}{1.400+1}\right)}^{1.400/\left(1.400-1\right)}\\ =\left(317.0\text{kPa}\right)\times \left(0.528282\right)\\ =167.4\text{kPa}\\ \cong 167\text{kPa}\end{array}$

Substitute $2.595\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_0$, and 1.400 for k in Equation (VI).

$\begin{array}{c}{\rho }^{\ast }=\left(2.595\text{kg}/{\text{m}}^{\text{3}}\right){\left(\frac{2}{1.4+1}\right)}^{1/\left(1.4-1\right)}\\ =\left(2.595\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.633938\right)\\ =1.6450\text{kg}/{\text{m}}^{\text{3}}\\ \cong 1.65\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Thus, the critical density of air is $\underset{\_}{1.65\text{kg}/{\text{m}}^{\text{3}}}$.

(b)

To determine

The critical temperature of helium.

The critical pressure of helium.

The critical density of helium.

Answer to Problem 37P

The critical temperature of helium is $\underset{\_}{256\text{K}}$.

The critical pressure of helium is $\underset{\_}{\text{104}\text{kPa}}$.

The critical density of helium is $\underset{\_}{0.195\text{kg}/{\text{m}}^{\text{3}}}$.

Explanation of Solution

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of universal gas constant, specific heat of pressure, and the specific heat ratio of helium at $300\text{K}$ temperature as $2.0769\text{kJ}/\text{kg}\cdot \text{K}$, $5.1926\text{kJ}/\text{kg}\cdot \text{K}$ and $1.667$.

Substitute $60°\text{C}$ for $T$, $300\text{m}/\text{s}$ for $V$, and $5.1926\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$ in Equation (I).

$\begin{array}{c}{T}_0=\left(60°\text{C}\right)+\frac{{\left(300\text{m}/\text{s}\right)}^2}{2\times \left(5.1926\text{kJ}/\text{kg}\cdot \text{K}\right)}\\ =\left(60°\text{C}\right)+\frac{{\left(300\text{m}/\text{s}\right)}^2}{2\times \left(5.1926\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ =\left(60°\text{C}\right)+\frac{\left(90000{\text{m}}^2/{\text{s}}^2\right)\times \left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)}{\left(10.3852\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ =68.66°\text{C}\end{array}$

    $\begin{array}{l}=68.66°\text{C}+273.2\\ =341.866\text{K}\\ \cong \text{341}\text{.9}\text{K}\end{array}$

Substitute 200 kPa for $P$, $60°\text{C}$ for $T$, $\text{341}\text{.9}\text{K}$ for $T_0$, and 1.667 for k in Equation (II).

$\begin{array}{c}{P}_0=\left(200\text{kPa}\right)\times {\left(\frac{\text{341}\text{.9}\text{K}}{60°\text{C}}\right)}^{1.667/\left(1.667-1\right)}\\ =\left(200\text{kPa}\right)\times {\left(\frac{341.9\text{K}}{333.2\text{K}}\right)}^{1.667/\left(1.667-1\right)}\\ =\left(200\text{kPa}\right)\times \left(1.066539\right)\\ =213.3\text{kPa}\end{array}$

Substitute 213.3 kPa for $P_0$, $2.0769\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, and 341.9 K for $T_0$ in Equation (III).

$\begin{array}{c}{\rho }_0=\frac{213.3\text{kPa}}{\left(2.0769\text{kJ}/\text{kg}\cdot \text{K}\right)\left(341.9\text{K}\right)}\\ =\frac{213.3\text{kPa}}{\left(2.0769\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)\left(341.9\text{K}\right)}\\ =\frac{213.3\text{kPa}}{\left(710.0921\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)}\\ =0.30038\text{kg}/{\text{m}}^{\text{3}}\end{array}$

     $\cong 0.3004\text{kg}/{\text{m}}^{\text{3}}$

Substitute $341.9\text{K}$ for $T_0$ and 1.667 for k in Equation (IV).

$\begin{array}{c}{T}^{\ast }=\left(341.9\text{K}\right)\times \left(\frac{2}{1.667+1}\right)\\ =\left(341.9\text{K}\right)\times \left(0.749906\right)\\ =256.39\text{K}\\ \cong \text{256}\text{K}\end{array}$

Thus, the critical temperature of helium is $\underset{\_}{256\text{K}}$.

Substitute $213.3\text{kPa}$ for $P_0$ and 1.667 for k in Equation (V).

$\begin{array}{c}{P}^{\ast }=\left(213.3\text{kPa}\right)\times {\left(\frac{2}{1.667+1}\right)}^{1.667/\left(1.667-1\right)}\\ =\left(213.3\text{kPa}\right)\times \left(0.487092\right)\\ =103.89\text{kPa}\\ \cong 104\text{kPa}\end{array}$

Thus, the critical pressure of helium is $\underset{\_}{104\text{kPa}}$

Substitute $0.3004\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_0$, and 1.667 for k in Equation (VI).

$\begin{array}{c}{\rho }^{\ast }=\left(0.3004\text{kg}/{\text{m}}^{\text{3}}\right){\left(\frac{2}{1.667+1}\right)}^{1/\left(1.667-1\right)}\\ =\left(0.3004\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.649537\right)\\ =0.195\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Thus, the critical density of helium is $\underset{\_}{0.195\text{kg}/{\text{m}}^{\text{3}}}$.