A simplified schematic drawing of the carburetor of a gasoline \((S=0.75)\) engine is shown in Fig. \(P 5.96 .\) The throat trea is 0.5 in. \(^{2}\) The running engine draws air dowaward through the carburetor Venturi and maintains a throat pressure of 14.3 psia. The low throat pressure draws fuel from the float chamber and into the airstream. The chergy losses in the 0.07 -in.-diameter fuel metering line and valve are given by \\[ g h_{L}=\frac{K V^{2}}{2 g} \\] where \(\mathrm{K}=6.0\) and \(V\) is the fuel velocty in the metering line. Assume that the air is an ideal fluid having a constant density \(\rho_{A}\) \(=0.075 \mathrm{lbm} / \mathrm{ft}^{3} .\) The atmospheric pressure is 14.7 psia. Calculate the air-to-fuel ratio \(\left(\dot{m}_{2} / \dot{m}_{1}\right)\)

Step by step solution

01

Calculating the Throat Velocity

Since we are assuming air to be an ideal fluid, we can make use of Bernoulli's Principle for pressure and speed in fluid flow. From Bernoulli's equation, we can calculate the velocity of air at the throat. Bernoulli's equation is given as \[ P_A + 0.5 * \rho_A * v_A^2 = P_{throat} + 0.5 * \rho_A * v_{throat}^2 \] where \( P_A \) is the atmospheric pressure and \( v_A \) is air velocity which is regarded as zero since we're considering a standstill outside the carburetor. Therefore, the velocity at the throat, \( v_{throat} \), can be calculated as \[ v_{throat} = \sqrt{ ((P_A - P_{throat})*2)/\rho_A } \], substituting the given value.

02

Calculating the Mass Flow Rate of Air

The equation for mass flow rate is given as \( \dot{m} = \rho * A * v \), where \( \dot{m} \) is the mass flow rate. In our case, \( \dot{m}_{air} = \rho_A * A_{throat} * v_{throat} \). Substituting the values obtained from step 1 for velocity, and other given values, calculate \( \dot{m}_{air} \).

03

Calculating the Fuel Velocity

Knowing that the energy losses of fuel in the fuel metering line and valve can be equated to the height difference in pressure, and with the Gillespie's equation (given in problem: \( g h_{L}=\frac{K V^{2}}{2 g} \)), we can calculate fuel velocity \( v_{fuel} \) as \( v_{fuel} = \sqrt{(2*g*h_L/K)} \).

04

Calculating the Mass Flow Rate of Fuel

The density of fuel \( \rho_{fuel} \) can be found by using given specific gravity (S), through the equation: \( \rho_{fuel} = S * \rho_{water} \), where density of water \( \rho_{water} \) is a constant and is known. Then, using the formula from step 2, we find the mass flow rate of fuel \( \dot{m}_{fuel} = \rho_{fuel} * A_{fuel} * v_{fuel} \). Here, area of fuel pipe \( A_{fuel} = \pi * D_{fuel}^2 / 4 \), where \( D_{fuel} \) is the diameter of fuel line. Substituting these values we can find \( \dot{m}_{fuel} \).

05

Calculating the Air-to-Fuel Ratio

From the mass flow rates calculated in steps 2 and 4, we can find the air-to-fuel ratio \( \dot{m}_{2} / \dot{m}_{1} \) as \( \dot{m}_{air} / \dot{m}_{fuel} \).

06

Final Result

The calculated ratio gives the air-to-fuel ratio.

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