Chapter 12, Problem 95P

To determine

The rate of fuel burn.

The exit temperature of air.

Answer to Problem 95P

The rate of fuel burn is $0.493\text{kg}/\text{s}$.

The exit temperature of air is $5041.37K$.

Explanation of Solution

Given information:

The diameter of combustion chamber is $16\text{cm}$, inlet temperature of air is $450\text{K}$, inlet pressure of air is $380\text{kPa}$, heating value of fuel is $39000\text{kJ}/\text{kg}$, inlet velocity of air is $55\text{m}/\text{s}$ and exit Mach number is $0.8$.

Write the expression for inlet cross- section area of combustion chamber.

   $A=\frac{\pi }{4}{D}^2$     ...... (I)

Here, diameter of combustion chamber is $D$.

Write the expression for inlet density of air.

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

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

Write the expression for inlet mass flow rate.

   $\dot{m}={\rho }_{1}A{V}_1$     ...... (III)

Here, velocity at inlet is $V_1$.

Write the expression for stagnation temperature.

   $T_0=T+\frac{V^2}{2c_P}$     ...... (IV)

Here, initial temperature is $T$, velocity is $V$ and specific heat at constant pressure is $c_P$.

Write the expression for Mach number at inlet.

   ${\text{Ma}}_1=\frac{V_1}{\sqrt{kR{T}_1}}$     ...... (V)

Here, specific heat ratio is $k$.

Write the expression for heat transfer.

   $\dot{Q}=\dot{m}{c}_P\left(T_{02}-T_{01}\right)$     ...... (VI)

Here, stagnation temperature at inlet is $T_{01}$ and stagnation pressure at outlet is $T_{02}$.

Write the expression for mass flow rate.

   ${\dot{m}}_f=\frac{\dot{Q}}{HV}$     ...... (VII)

Here, heating value is $HV$.

Calculation:

Substitute $16\text{cm}$ for $D$ in Equation (I).

   $\begin{array}{c}A=\frac{\pi }{4}{\left(16\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(16\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =\frac{\pi }{4}{\left(0.16\text{m}\right)}^2\\ =0.0201{\text{m}}^{\text{2}}\end{array}$

Refer to Table-A-1 “Molar mass, gas constant, and ideal gas specific heat of some substances” to obtain gas constant of air as $287\text{J}/\text{kg}\cdot \text{K}$, specific heat ratio as $1.4$ and specific heat of air at constant pressure as $1005\text{J}/\text{kg}\cdot \text{K}$

Substitute $380\text{kPa}$ for $P_1$, $287\text{J}/\text{kg}\cdot \text{K}$ for $R$ and $450\text{K}$ for $T_1$ in Equation (II).

   $\begin{array}{c}{\rho }_1=\frac{380\text{kPa}}{\left(287\text{J}/\text{kg}\cdot \text{K}\right)\left(450\text{K}\right)}\\ =\frac{380\text{kPa}\times \frac{1000\text{Pa}}{1\text{kPa}}}{\left(287\text{J}/\text{kg}\cdot \text{K}\right)\left(450\text{K}\right)}\\ =\frac{380000\text{Pa}\times \frac{\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}}{129150\text{J}/\text{kg}\times \frac{\text{1}\text{Nm}/\text{kg}}{1\text{J}/\text{kg}}}\\ =2.942\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $2.942\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_1$, $0.0201{\text{m}}^{\text{2}}$ for $A$ and $55\text{m}/\text{s}$ for $V_1$ in Equation (III).

   $\begin{array}{c}\dot{m}=\left(2.942\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.0201{\text{m}}^{\text{2}}\right)\left(55\text{m}/\text{s}\right)\\ =\left(0.0591342\text{kg}/\text{m}\right)\left(55\text{m}/\text{s}\right)\\ =3.252\text{kg}/\text{s}\end{array}$

Substitute $1005\text{J}/\text{kg}\cdot \text{K}$ for $c_P$, $55\text{m}/\text{s}$ for $V_1$, $T_{01}$ for $T_0$ and $450\text{K}$ for $T_1$ in Equation (IV).

   $\begin{array}{c}{T}_{01}=450\text{K}+\frac{{\left(55\text{m}/\text{s}\right)}^2}{2\left(1005\text{J}/\text{kg}\cdot \text{K}\right)}\\ =450\text{K}+\frac{{\left(55\text{m}/\text{s}\right)}^2}{2\left(1005\text{J}/\text{kg}\cdot \text{K}\times \frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{K}}{1\text{J}/\text{kg}\cdot \text{K}}\right)}\\ =450\text{K}+\frac{{\left(55\text{m}/\text{s}\right)}^2}{2\left(1005{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{K}\times \frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{K}}{1\text{J}/\text{kg}\cdot \text{K}}\right)}\\ =451.5\text{K}\end{array}$

Substitute $1.4$ for $k$, $55\text{m}/\text{s}$ for $V_1$, $287\text{J}/\text{kg}\cdot \text{K}$ for $R$ and $450\text{K}$ for $T_1$ in Equation (V).

   $\begin{array}{c}{\text{Ma}}_1=\frac{55\text{m}/\text{s}}{\sqrt{1.4\left(287\text{J}/\text{kg}\cdot \text{K}\right)\left(450\text{K}\right)}}\\ =\frac{55\text{m}/\text{s}}{\sqrt{180810\text{J}/\text{kg}\times \frac{1{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{J}/\text{kg}}}}\\ =\frac{55\text{m}/\text{s}}{425.2175\text{m}/\text{s}}\\ =0.129\end{array}$

Refer to Table-A-15 “Rayleigh flow function for an ideal gas with $k=1.4$ ” at Mach number $0.129$ to obtain ratio of pressure temperature and velocity.

Write the relation of temperature at initial stagnation state and sonic state.

   $\frac{T_1}{T^{*}}=0.0915373$     ...... (VIII)

Here, temperature inlet sonic state is $T^{*}$.

Write the relation of velocity at initial stagnation state and sonic state.

   $\frac{V_1}{V^{*}}=0.0915373$     ...... (IX)

Here, inlet velocity at sonic state is $V^{*}$.

Refer to Table-A-15 “Rayleigh flow function for an ideal gas with $k=1.4$ ” at Mach number $0.8$ to obtain ratio of temperature and velocity.

Write the relation of temperature at initial stagnation state and sonic state.

   $\frac{T_2}{T^{*}}=1.0255$     ...... (X)

Here, temperature inlet sonic state is $T^{*}$.

Write the relation of velocity at initial stagnation state and sonic state.

   $\frac{V_2}{V^{*}}=0.8101$     ...... (XI)

Here, inlet velocity at sonic state is $V^{*}$.

Substitute $450\text{K}$ for $T_1$ in Equation (VIII).

   $\begin{array}{l}\frac{450\text{K}}{T^{*}}=0.0915373\\ T^{*}=\frac{450\text{K}}{0.0915373}\\ T^{*}=4916.02\text{K}\end{array}$

Substitute $55\text{m}/\text{s}$ for $V_1$ in Equation (IX).

   $\begin{array}{l}\frac{55\text{m}/\text{s}}{V^{*}}=0.0915373\\ V^{*}=\frac{55\text{m}/\text{s}}{0.0915373}\\ V^{*}=1409.2\text{m}/\text{s}\end{array}$

Substitute $4916.02\text{K}$ for $T^{*}$ in Equation (X).

   $\begin{array}{l}\frac{T_2}{4916.02\text{K}}=1.0255\\ T_2=4916.02\text{K}\times 1.0255\\ T_2=5041.37K\end{array}$

Substitute $1409.2\text{m}/\text{s}$ for $V^{*}$ in Equation (XI).

   $\begin{array}{l}\frac{V_2}{1409.2\text{m}/\text{s}}=0.8101\\ V_2=1409.2\text{m}/\text{s}\times 0.8101\\ V_2=1141.45\text{m}/\text{s}\end{array}$

Substitute $1141.45\text{m}/\text{s}$ for $V$, $5041.37K$ for $T$, $T_{02}$ for $T$ and $1005\text{J}/\text{kg}\cdot \text{K}$ for $c_P$ in Equation (IV).

   $\begin{array}{c}{T}_{02}=5041.37\text{K}+\frac{{\left(1141.45\text{m}/\text{s}\right)}^2}{2\left(1005\text{J}/\text{kg}\cdot \text{K}\right)}\\ =5041.37\text{K}+\frac{{\left(1141.45\text{m}/\text{s}\right)}^2}{2\left(1005\text{J}/\text{kg}\cdot \text{K}\times \frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{K}}{1\text{J}/\text{kg}\cdot \text{K}}\right)}\\ =5041.37\text{K}+\frac{{\left(1141.45\text{m}/\text{s}\right)}^2}{2\left(1005{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{K}\times \frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\cdot \text{K}}{1\text{J}/\text{kg}\cdot \text{K}}\right)}\\ =6337.7\text{K}\end{array}$

Substitute $6337.7\text{K}$ for $T_{02}$, $451.5\text{K}$ for $T_{01}$, $3.252\text{kg}/\text{s}$ for $\dot{m}$ and $1005\text{J}/\text{kg}\cdot \text{K}$ for $c_P$ in Equation (VI).

   $\begin{array}{c}\dot{Q}=\left(3.252\text{kg}/\text{s}\right)1005\text{J}/\text{kg}\cdot \text{K}\left(6337.7\text{K}-451.5\text{K}\right)\\ =\left(3.252\text{kg}/\text{s}\right)\left(5915631\text{J}/\text{kg}\right)\\ =19237.632\times {10}^3\text{J}/\text{s}\end{array}$

Substitute $19237.632\times {10}^3\text{J}/\text{s}$ for $\dot{Q}$ and $39000\text{kJ}/\text{kg}$ for $HV$ in Equation (VII).

   $\begin{array}{c}{\dot{m}}_f=\frac{19237.632\times {10}^3\text{J}/\text{s}}{39000\text{kJ}/\text{kg}\times \frac{1000\text{J}/\text{kg}}{1\text{kJ}/\text{kg}}}\\ =\frac{19237.632\times {10}^3\text{J}/\text{s}}{39000\times {10}^3\text{J}/\text{kg}}\\ =0.493\text{kg}/\text{s}\end{array}$

Conclusion:

The rate of fuel burn is $0.493\text{kg}/\text{s}$.

The exit temperature of air is $5041.37K$.