Chapter 17.7, Problem 110P

Air is heated as it flows through a 6 in × 6 in square duct with negligible friction. At the inlet, air is at T₁ = 700 R, P₁ = 80 psia, and V₁ = 260 ft/s. Determine the rate at which heat must be transferred to the air to choke the flow at the duct exit, and the entropy change of air during this process.

To determine

The rate of heat transfer in the duct.

The entropy change in the duct.

Answer to Problem 110P

The rate of heat transfer in the duct is $\underset{\_}{16,090\text{Btu}/\text{s}}$.

The entropy change in the duct is $\underset{\_}{0.434\text{Btu}/\text{lbm}\cdot \text{R}}$.

Explanation of Solution

Determine the inlet density of air.

${\rho }_1=\frac{P_1}{R{T}_1}$ (I)

Here, the inlet pressure of air is $P_1$, the universal gas constant of air is $R$, and the inlet temperature of air is $T_1$.

Determine the mass flow rate of the duct.

${\dot{m}}_{\text{air}}={\rho }_1{A}_{c1}V$ (II)

Here, the inlet velocity of air is $V_1$ and the density of the air is ${\rho }_1$.

Determine the inlet stagnation temperature of air.

$T_{01}=T_1+\frac{V_1^2}{2c_p}$ (III)

Here, the inlet static temperature of ideal gas is $T_3$, the specific heat of pressure for ideal gas is $c_p$, and the inlet velocity of the ideal gas flow is $V_1$.

Determine the relation of ideal gas speed of sound at the inlet.

$c_1=\sqrt{kR{T}_1}$ (IV)

Here, the specific heat ratio of air is $k$, the gas constant of the air is $R$, and the inlet temperature of the air is $T_1$.

Determine the speed of sound at the inlet.

${\text{Ma}}_1=\frac{V_1}{c_1}$ (V)

The inlet velocity of the air flow in the device is $V_1$ and the inlet Mach number is ${\text{Ma}}_1$.

Determine the static temperature in the duct.

$\frac{T_2}{T_1}=\frac{T_2/T^{\ast }}{T_1/T^{\ast }}$ (VI)

Here, the ratio of Rayleigh flow for inlet temperature is $T_1/T^{\ast }$ and the ratio of Rayleigh flow for outlet temperature is $T_2/T^{\ast }$.

Determine the static pressure in the duct.

$\frac{P_2}{P_1}=\frac{P_2/P^{\ast }}{P_1/P^{\ast }}$ (VII)

Here, the ratio of Rayleigh flow for inlet pressure is $P_1/P^{\ast }$ and the ratio of Rayleigh flow for outlet pressure is $P_2/P^{\ast }$.

Determine the stagnation temperature in the duct.

$\frac{T_{02}}{T_{01}}=\frac{T_{02}/T^{\ast }}{T_{01}/T^{\ast }}$ (VIII)

Here, the ratio of Rayleigh flow for exit stagnation temperature is $T_{01}/T^{\ast }$ and the ratio of Rayleigh flow for outlet stagnation temperature is $T_{02}/T^{\ast }$.

Determine the rate of heat transfer of the duct.

$\dot{Q}={\dot{m}}_{\text{air}}{c}_p\left(T_{02}-T_{01}\right)$ (IX)

Determine the entropy change of the duct.

$\Delta s=c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}$ (X)

Conclusion:

From the Table A-2E, “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 $80°\text{F}$ temperature as $0.06855\text{Btu}/\text{lbm}\cdot \text{R}$, $0.2400\text{Btu}/\text{lbm}\cdot \text{R}$ and $1.4$.

Substitute $80\text{psia}$ for $P_1$, $0.06855\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$, and $700\text{R}$ for $T_1$ in Equation (I).

$\begin{array}{c}{\rho }_1=\frac{80\text{psia}}{\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(700\text{R}\right)}\\ =\frac{80\text{psia}}{\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\times \left(\frac{5.40395\text{psia}\cdot {\text{ft}}^{\text{3}}}{\text{1}\text{Btu}}\right)\right)\left(700\text{R}\right)}\\ =\frac{80\text{psia}}{\left(0.3704\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(700\text{R}\right)}\\ =0.3085\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

Substitute $260\text{ft}/\text{s}$ for $V_1$, $0.3085\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_1$, and $6\text{in}\times 6\text{in}$ for $A_{c1}$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{\text{air}}=\left(0.3085\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(6\text{in}\times 6\text{in}\right)\left(260\text{ft}/\text{s}\right)\\ =\left(0.3085\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(6\text{in}\times 6\text{in}\right)\times \left(\frac{1{\text{ft}}^2}{144\text{in}}\right)\left(260\text{ft}/\text{s}\right)\\ =20.052\text{lbm}/\text{s}\end{array}$

Substitute $700\text{R}$ for $T_1$, $260\text{ft}/\text{s}$ for $V_1$, and $0.2400\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ in Equation (III).

$\begin{array}{c}{T}_{01}=\left(700\text{R}\right)+\frac{{\left(260\text{ft}/\text{s}\right)}^2}{2\times \left(0.2400\text{Btu}/\text{lbm}\cdot \text{R}\right)}\\ =\left(700\text{R}\right)+\frac{\left(67600{\text{ft}}^2/{\text{s}}^2\right)}{\left(0.48\text{Btu}/\text{lbm}\cdot \text{R}\right)}\\ =\left(700\text{R}\right)+\frac{\left(67600{\text{ft}}^2/{\text{s}}^2\right)\times \left(\frac{1\text{Btu}/\text{lbm}}{25,037{\text{ft}}^2/{\text{s}}^2}\right)}{\left(0.48\text{Btu}/\text{lbm}\cdot \text{R}\right)}\\ =\left(700\text{R}\right)+\left(5.625\text{R}\right)\end{array}$

$=705.6\text{R}$

Substitute 1.4 for k, $0.06855\text{Btu}/\text{lbm}\cdot \text{R}$ for R, and $700\text{R}$ for $T_1$ in Equation (IV).

$\begin{array}{c}{c}_1=\sqrt{\left(1.4\right)\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \left(700\text{R}\right)}\\ =\sqrt{\left(1.4\right)\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\times \left(\frac{25,037{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{Btu}/\text{lbm}}\right)\times \left(700\text{R}\right)}\\ =\sqrt{1681960.6{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =1296.9\text{ft}/\text{s}\end{array}$

   $=1297\text{ft}/\text{s}$

Substitute $260\text{ft}/\text{s}$ for $c_{\text{1}}$ and $1297\text{ft}/\text{s}$ for $V_1$ in Equation (V).

$\begin{array}{c}{\text{Ma}}_1=\frac{\left(260\text{ft}/\text{s}\right)}{\left(1297\text{ft}/\text{s}\right)}\\ =0.20046\\ \cong 0.2005\end{array}$

Refer to Table A-34, “Rayleigh flow function for an ideal gas with k=1.4”, to obtain the value ratio of static temperature, pressure, and stagnation temperature at $0.2005$ inlet Mach number using interpolation method of two variables.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (XI)

Here, the variables denote by x and y is ratio of stagnation temperature and Mach number.

Show the Mach number at $0.2$ and $0.3$ as in Table (1).

S. No	Mach number $\left(x\right)$	ratio of stagnation temperature $\left(y\right)$
1	$0.2$	$0.1736$
2	$0.2005$	$y_2=?$
3	$0.3$	$0.3469$

Calculate ratio of static temperature, pressure, and stagnation temperature at $0.2005$ inlet Mach number using interpolation method.

Substitute $0.2$ for $x_1$, $0.2005$ for $x_2$, $0.3$ for $x_3$, $0.1736$ for $y_1$, and $0.3469$ for $y_3$ in Equation (XI).

$\begin{array}{c}{y}_2=\frac{\left(0.2005-0.2\right)\left(0.3469-0.1736\right)}{\left(0.3-0.2\right)}+0.1736\\ =0.1744\end{array}$

From above calculation the ratio of stagnation temperature at $0.2005$ is inlet Mach number $0.1744$.

Repeat the Equation (XII), to obtain the value of inlet ratio of static temperature and pressure at $0.2005$ inlet Mach number as:

$\begin{array}{l}{T}_1/T^{\ast }=0.2075\\ P_1/P^{\ast }=2.272\end{array}$

From the Table A-34, “Rayleigh flow function for an ideal gas with k=1.4”, to obtain the value of the outlet ratio of temperature, pressure, and velocity at 1 outlet Mach number as:

$\begin{array}{l}{T}_2/T^{\ast }=1\\ P_2/P^{\ast }=1\\ V_2/V^{\ast }=1\end{array}$

Substitute $700\text{R}$ for $T_1$, 1 for $T_2/T^{\ast }$, and $0.2075$ for $T_1/T^{\ast }$ in Equation (VI).

$\begin{array}{l}\frac{T_2}{\left(700\text{R}\right)}=\frac{1}{0.2075}\\ T_2=\frac{\left(700\text{R}\right)}{0.2075}\\ T_2=3373.49\text{R}\end{array}$

Substitute 80 psia for $P_1$, 1 for $P_2/P^{\ast }$, and $2.272$ for $P_1/P^{\ast }$ in Equation (VII).

$\begin{array}{l}\frac{P_2}{\left(80\text{psia}\right)}=\frac{1}{2.272}\\ P_2=\frac{\left(80\text{psia}\right)}{2.272}\\ P_2=35.21\text{psia}\end{array}$

Substitute $705.6\text{R}$ for $T_{01}$, 1 for $T_{02}/T^{\ast }$, and $0.1743$ for $T_{01}/T^{\ast }$ in Equation (VIII).

$\begin{array}{l}\frac{T_{02}}{\left(705.6\text{R}\right)}=\frac{1}{0.1743}\\ T_{02}=\frac{\left(705.6\text{R}\right)}{0.1743}\\ T_{02}=4048\text{R}\end{array}$

Substitute $20.052\text{lbm}/\text{s}$ for ${\dot{m}}_{\text{air}}$, $0.2400\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $4048\text{R}$ for $T_{02}$ and $705.6\text{R}$ for $T_{01}$ in Equation (IX).

$\begin{array}{c}\dot{Q}=\left(20.052\text{lbm}/\text{s}\right)\left(0.2400\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(4048-705.6\right)\text{R}\\ =\left(20.052\text{lbm}/\text{s}\right)\left(0.2400\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(3342.4\text{R}\right)\\ =16085\text{Btu}/\text{s}\\ \cong 16090\text{Btu}/\text{s}\end{array}$

Thus, the rate of heat transfer in the duct is $\underset{\_}{16,090\text{Btu}/\text{s}}$.

Substitute $0.2400\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $3373.49\text{R}$ for $T_2$, $700\text{R}$ for $T_1$, $0.06855\text{Btu}/\text{lbm}\cdot \text{R}$ for R, $35.21\text{psia}$ for $P_2$,and $80\text{psia}$ for $P_1$ in Equation (X).

$\begin{array}{c}\Delta s=\left[\begin{array}{l}\left(0.2400\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(\ln \frac{\left(3373.49\text{R}\right)}{\left(700\text{R}\right)}\right)\\ -\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(\ln \frac{\left(35.21\text{psia}\right)}{\left(80\text{psia}\right)}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(0.2400\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(1.572\right)\\ -\left(0.06855\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(-0.8207\right)\end{array}\right]\\ =0.4336\text{Btu}/\text{lbm}\cdot \text{R}\\ \cong 0.434\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Thus, the entropy change in the duct is $\underset{\_}{0.434\text{Btu}/\text{lbm}\cdot \text{R}}$.