Solve the following numerical on four cylinder four stroke engine and calculate diameter of venture and nozzle.

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written 8.2 years ago by

ak.amitkhare.ak • 560

• modified 8.2 years ago

A four cylinder four stroke engine has a cubic capacity of 1490 cc. it develops maximum power at 4200 rpm and air fuel ratio is 13:1. The air speed at venture is limited to 90 m/s. the volumetric efficiency of engine is 70%. Nozzle lip is 6 mm and atmospheric pressure and temperature are 1.013 bar and 293 K. An allowance is to be made for emulsion tube whose diameter should be taken as 1 /2.5 of venture diameter. Taking following data, calculate the diameter of venture and nozzle. Cda-0.85,Cdf=0.66 and density of fuel=740Kg/m3

Mumbai University > Mechanical Engineering > Sem 5 > IC Engines

Marks: 10M

Year: Dec 2015

ic engines

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1 Answer

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written 8.2 years ago by

ak.amitkhare.ak • 560

A single carburetor supplies air-fuel mixture to the engine. The area through which air is flowing is

${A_a} = \frac{\pi }{4}({D^2} - {d^2})$

$= \frac{\pi }{4}({D^2} - {(0.4D)^2})$ as $d=0.4D$

$= \frac{\pi }{4}({D^2} - 0.16D)$

$= \frac{\pi }{4} \times 0.84{D^2} = 0.66{D^2}$

The volume of air supplied to the engine at atmospheric conditions per second

${\text{ = swept volume}} \times \frac{{RPM}}{{2 \times 60}}{\eta _v} = 1490 \times \frac{{4200 \times 0.7}}{{2 \times 60}}$

$= 0.036505{m^3}/\sec$

The mass of air is given by

${m_a} = \frac{{\rho V}}{{RT}} = \frac{{1.013 \times {{10}^5} \times 0.0360505}}{{287 \times 293}} = 0.0434{\text{ }}kg/\sec$

the air velocity through venturi is given by

${V_a} = \sqrt {2{C_v} \times {T_1}\left[ {1 - ({{\frac{{{P_2}}}{{{P_1}}}}^{\frac{{\gamma - 1}}{\gamma }}})} \right]}$ where

${{C_v}}$ (coefficient ofvelocity)

Where, the compression of air through venture is considered isentropic.

$90 = \sqrt {2 \times 1000 \times 293\left[ {1 - ({{\frac{{{P_2}}}{{{P_1}}}}^{0.286}})} \right]}$

$= 1 - ({\frac{{{P_2}}}{{{P_1}}}^{0.286}})$

$= \frac{{90 \times 90}}{{2000 \times 293}}$

$= 0.1382(\frac{{{P_2}}}{{{P_1}}})$

$= {(1 - 0.01382)^{3.5}}$

${p_2} = 0.9524bar$

Considering the compression of air passing through venture is isentropic

${p_1}v_1^\gamma = {p_2}v_2^\gamma$

${v_2} = {v_1}{\left( {\frac{{{p_1}}}{{{p_2}}}} \right)^{\frac{1}{\gamma }}}$

$= 1450 \times {10^{ - 6}}{\left( {\frac{1}{{0.9524}}} \right)^{\frac{1}{\gamma }}}$

$= 0.00150{\text{ }}{{\text{m}}^3}/\sec$

The volume of air passing through the carburetor is given by

${v_2} = {A_a}{V_a}{C_{da}}$

$0.00150 = 0.66{D^2} \times 90 \times 0.85 \times {10^{ - 2}}$

$D = 0.0545{\text{ }}m$

The fuel mass flow is given by

${m_f} = \frac{{{m_a}}}{{13}} = \frac{{0.00150}}{{13}} = 0.0001153$

${m_f} = \frac{{{m_a}}}{{13}} = \frac{{0.00150}}{{13}} = 0.0001153{\text{ Kg/sec}}$

The fuel mass flow through the carburetor is given by

${m_f} = {A_f}{C_{df}}\sqrt {2{\rho _f}\left( {\Delta p - Zg{\rho _f}} \right)}$ where

${\Delta _p} = {p_1} - {p_2} = 1.013 - 0.9524 = 0.0606{\text{ }}bar$

Substituting the values in the above equation

$0.0001153 = \frac{\pi }{4}d_f^2 \times 0.66\sqrt {2 \times 740(0.0606 \times {{10}^5} - 6 \times {{10}^{ - 3}} \times 9.81 \times 740)}$

${d_f} = 0.000273{\text{ m}}$

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