Question

13.37 Octane gas \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) at \(25^{\circ} \mathrm{C}\) enters a jet engine and burns completely with \(300 \%\) of theoretical air entering at \(25^{\circ} \mathrm{C}\), 1 atm with a volumetric flow rate of \(42 \mathrm{~m}^{3} / \mathrm{s}\). Products of combustion exit at \(990 \mathrm{~K}, 1 \mathrm{~atm}\). If the fuel and air enter with negligible velocities, determine the thrust produced by the engine in \(\mathrm{kN}\).

Short answer

Calculate the air-fuel ratio, mass flow rates, and exhaust velocity to find the thrust.

Step-by-step solution

Determine the stoichiometric air-fuel ratio

For the complete combustion of octane \(\text{C}_{8} \text{H}_{18}\), the balanced chemical equation is: \[ \text{C}_{8} \text{H}_{18} + 12.5\text{O}_2 + 3.76(12.5\text{N}_2) \rightarrow 8\text{CO}_2 + 9\text{H}_2O + 3.76(12.5\text{N}_2) \] The stoichiometric air-fuel ratio by mass is given by the formula: \[ \frac{\text{mass of air}}{\text{mass of fuel}} = \frac{12.5(32 + 3.76 \times 28.97)}{114} \] Calculate this ratio.

Calculate theoretical and actual air flow rates

Given that actual air is 300% of theoretical, we can write: \[ \text{Actual air flow} = 3 \times \text{Theoretical air flow} \] The volumetric flow rate of air at standard conditions (25^{\text{C}}, 1 ATM) is provided as 42 m^3/s. Use the ideal gas law to convert this to mass flow rate: \[ \text{mass flow rate of air, } \text{ṁ}_{\text{air}} = \frac{\text{volumetric flow rate} \times \text{density}}{ \text{molar mass} } \]where density at standard conditions can be found using \[ \text{density} = \frac{\text{P}}{\text{R} \times \text{T}} \]

Find mass flow rate of octane

Determine the mass flow rate of fuel using the air to fuel ratio from Step 1 and the mass flow rate of air from Step 2: \[ \text{ṁ}_{\text{fuel}} = \frac{\text{ṁ}_{\text{air}}}{\text{A/F ratio}} \]

Calculate the velocity of expelled gases

Using the mass of products and the ideal gas law, find the velocity of exhaust gases exiting at 990 K and 1 ATM. The exhaust velocity, \[ V_e = \frac{\text{volumetric flow rate}}{\text{cross-sectional area}} \] Determine the volumetric flow rate of the exhaust gases by considering the complete combustion process.

Calculate thrust produced

Using the momentum equation for thrust calculation: \[ F = \text{ṁ}_{\text{fuel}} \times V_e \] Simplify this with the values obtained from previous steps to determine the thrust force.

Key concepts

stoichiometric air-fuel ratio

The stoichiometric air-fuel ratio is critical in ensuring complete combustion of a fuel. In our case, we are working with octane (\text{C}_8 \text{H}_{18}). The balanced chemical equation for the complete combustion is: \[ \text{C}_8 \text{H}_{18} + 12.5\text{O}_2 + 3.76(12.5\text{N}_2) \rightarrow 8\text{CO}_2 + 9\text{H}_2O + 3.76(12.5\text{N}_2) \] Here, we need to calculate the mass air-fuel ratio. This ratio is calculated using the formula: \[ \frac{\text{mass of air}}{\text{mass of fuel}} = \frac{12.5(32 + 3.76 \times 28.97)}{114} \] This ratio helps us know how much air is needed to fully burn 1 kg of fuel under ideal conditions. If we have too much fuel or too little air, combustion will be incomplete, wasting fuel and producing more pollution.

mass flow rate

The mass flow rate, \text{ṁ}, refers to the amount of mass traveling through a cross-section per unit time. It’s typically measured in kilograms per second (kg/s). For our jet engine problem, we have the following steps to calculate the mass flow rates:

• We start with the volumetric flow rate of air which is given as 42 m^3/s at standard conditions.
• Using the ideal gas law, we convert this volumetric flow rate to mass flow rate.

The mass flow rate of air can be determined using the equation: \[ \text{ṁ}_{\text{air}} = \frac{\text{volumetric flow rate} \times \text{density}}{ \text{molar mass} } \] Density at standard conditions can be calculated using: \[ \text{density} = \frac{\text{P}}{\text{R} \times \text{T}} \]

ideal gas law

The ideal gas law is a fundamental equation in thermodynamics, describing the behavior of ideal gases. It is written as: \[ \text{PV} = \text{nRT} \] where P is pressure, V is volume, n is number of moles, R is the ideal gas constant (8.314 J·mol^-1·K^-1), and T is temperature. In our problem, the ideal gas law helps us convert the volumetric flow rate of the air into a mass flow rate. Using rearrangement, we get: \[ \text{n} = \frac{\text{P} \times \text{V}}{\text{R} \times \text{T}} \] By evaluating n, the number of moles, we can then find the mass using the molar mass relation: \[ \text{mass} = \text{n} \times \text{molar mass} \]

combustion process

Combustion is a chemical process where fuel reacts with an oxidant, releasing heat and producing products like CO2, H2O, etc. For our octane (C8H18) in the jet engine, combustion can be defined by the balanced equation: \[ \text{C}_8 \text{H}_{18} + 12.5\text{O}_2 + 3.76(12.5\text{N}_2) \rightarrow 8\text{CO}_2 + 9\text{H}_2O + 3.76(12.5\text{N}_2) \] To ensure complete combustion, sufficient oxygen is essential. We use a stoichiometric ratio to determine the precise amounts of fuel and air needed for complete combustion. This process converts chemical energy in the fuel into thermal energy, which in turn, gets used for propulsion by the jet engine.

momentum equation

Thrust in a jet engine is a result of the momentum change of gases. The thrust can be determined using the momentum equation: \[ F = \text{ṁ}_{\text{fuel}} \times V_e \] Where F is the thrust, \text{ṁ}_{\text{fuel}} is the mass flow rate of the fuel, and V_e is the exhaust velocity. In our problem, we calculate the velocity of the expelled gases at the nozzle exit. Given that we know the mass flow rates and exhaust conditions, we can determine thrust effectively. This equation essentially measures how fast the exhaust gases are pushed out of the engine, propelling the jet forward.