State the differences between Gaseous fuels and solid fuels. A gaseous fuel has the following composition: CH 4 = 38%, H 2 = 26%, N 2 = 6%, CO = 26% and CO 2 = 4%. Find the stoichiometric air/fuel ratio by volume.

Assume 100 volumes of fuel.

• For $CH_4$ (38 volumes): $$CH_4 + 2O_2 \to CO_2 + 2H_2O$$ Oxygen required: $38 \times 2 = 76$ volumes of $O_2$.
• For $H_2$ (26 volumes): $$H_2 + \frac{1}{2}O_2 \to H_2O$$ Oxygen required: $26 \times \frac{1}{2} = 13$ volumes of $O_2$.
• For $CO$ (26 volumes): $$CO + \frac{1}{2}O_2 \to CO_2$$ Oxygen required: $26 \times \frac{1}{2} = 13$ volumes of $O_2$.
• $N_2$ (6 volumes) and $CO_2$ (4 volumes) are inert and do not require oxygen for combustion.

Total oxygen required per 100 volumes of fuel = $76 + 13 + 13 = 102$ volumes of $O_2$.

Air is approximately 21% oxygen by volume.
Volume of air required = $\frac{\text{Total oxygen required}}{\text{Fraction of oxygen in air}}$

Air/fuel ratio by volume = $\frac{\text{Volume of air required}}{\text{Volume of fuel}}$

The stoichiometric air/fuel ratio by volume is $\boxed{\text{4.857}}$.

3. a) State the steady flow energy equation.

The Steady Flow Energy Equation (SFEE) for a single inlet and single outlet system is given by:

where:

• $\dot{Q}$ is the rate of heat transfer to the system.
• $\dot{W}$ is the rate of work done by the system.
• $\dot{m}$ is the mass flow rate.
• $h$ is the specific enthalpy ($h = u + Pv$).
• $V$ is the velocity.
• $g$ is the acceleration due to gravity.
• $z$ is the elevation.
• Subscripts 1 and 2 denote inlet and outlet conditions, respectively.

3. b) In a steady flow system, a substance flows at a rate of $5.2 \text{ kg/sec}$. It enters at a pressure of $620 \text{ kN/m}^2$, a velocity of $298 \text{ m/sec}$, internal energy $2100 \text{ kJ/kg}$, and specific volume $0.37 \text{ m}^3\text{/kg}$. It leaves at a pressure of $130 \text{ kN/m}^2$, a velocity of $145 \text{ m/sec}$, internal energy $1500 \text{ kJ/kg}$, and specific volume $1.2 \text{ m}^3\text{/kg}$. During its passage through the system, the substance rejects $120 \text{ kJ/sec}$ of heat. Determine the work done per kg of the substance.

Given data:

• Mass flow rate $\dot{m} = 5.2 \text{ kg/s}$
• Inlet (State 1): $P_1 = 620 \text{ kPa}$, $V_1 = 298 \text{ m/s}$, $u_1 = 2100 \text{ kJ/kg}$, $v_1 = 0.37 \text{ m}^3\text{/kg}$
• Exit (State 2): $P_2 = 130 \text{ kPa}$, $V_2 = 145 \text{ m/s}$, $u_2 = 1500 \text{ kJ/kg}$, $v_2 = 1.2 \text{ m}^3\text{/kg}$
• Heat rejected $\dot{Q}_{out} = 120 \text{ kJ/s}$. Therefore, heat added to the system $\dot{Q} = -120 \text{ kJ/s}$.
• Assume negligible change in potential energy ($g(z_2 - z_1) = 0$).

The SFEE per unit mass is:

We need to find $w$, the work done per kg. So, $w = q - (h_2 - h_1) - \frac{V_2^2 - V_1^2}{2}$.

Specific enthalpy $h = u + Pv$.

The work done per kg of the substance is $\boxed{\text{684.21 kJ/kg}}$.