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Solve the following numaricle.

During trail on single cylinder, four stroke oil engine the following observations were recorded:

Bore and stroke: 300mmX450mm,

Duration of trial: 60 min,

Engine speed: 220 RPM,

Fuel consumption: 7kg,

Calorific value of fuel: 45000KJ/kg,

Indicated mean effective pressure: 5.867 bar,

Brake drum diameter: 1650 mm,

Total weight of jacketed cooling water: 500kg,

Temperature rise of jacket cooling water: 40 degree centigrade,

Temperature of exhaust gases: 300 degree centigrade

Air consumption: 300kg,

Specific heat of exhaust gases: 1.004 KJ/KJK

Room temperature: 25℃,

Determine: Mechanical efficiency, indicated and brake thermal efficiency and heat balance sheet on minute and percentage basis.

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Mechanical Efficiency:

$$\eta = \frac{{B.P.}}{{I.P.}}$$

$$B.P. = \frac{{2\pi NT}}{{1000}} = \frac{{2\pi N(mgr)}}{{1000}} = \frac{{2\pi \times 220 \times 130 \times 9.81 \times 825}}{{60 \times 1000}}$$

$$B.P. = 24.24{\text{ KW}}$$

$$IP = \frac{{100 \times {P_m} \times LAn}}{1}$$

$$= \frac{{100 \times {P_m} \times LAn \times no.{\text{ of cyllinders}}}}{2}$$

$$= \frac{{100 \times 5.867 \times 0.45 \times \pi \times {{0.3}^2} \times 220}}{{60 \times 2 \times 4}}$$

$$IP = 34.214KW$$

$$\eta = \frac{{B.P.}}{{I.P.}} = \frac{{24.14}}{{34.214}} = 0.7084 = 70.84\%$$

Indicated and brake thermal efficiency:

$${\eta _{it}} = \frac{{3600}}{{{C_i} \times CV}}{C_i} = \frac{{fuel{\text{ consumption}}}}{{IP}} = \frac{7}{{34.24}} = 0.2044$$

$$= \frac{{3600}}{{0.2044 \times 45000}} = 0.3914$$

$${\eta _{it}} = 39.14\% {\text{ indicated thermal efficiency}}$$

Heat balance sheet:

Heat supplied per minute

$$= \frac{{7 \times 45000}}{{60}} = 5250KJ/Min$$

Heat in BP:

$$= 24.24 \times 60 = 1454.4KJ/\min$$

Heat in cooling water:

$$= \frac{{500 \times 40 \times 4.2}}{{60}} = 1400KJ/\min$$

Heat in dry exhaust gases:

Qg: Heat taken by gases leaving the calorimeter

$$= {C_{pg}}{m_g}\left( {{T_g} - {T_a}} \right)$$

$${m_g} = {m_a} + {m_f} = {m_f}\left( {\frac{{{m_a}}}{{{m_f}}} + 1} \right) = \frac{{10}}{{60}}\left( {\frac{{300}}{{10}} + 1} \right) = 5.17Kg/\min$$

$$= 1.004 \times 5.17 \times \left( {300 - 25} \right)$$

$${Q_g} = 1427.437KJ/\min$$

Unaccounted Heat:

$$= {Q_s} - \left( {{Q_w} + {Q_{BP}} + {Q_g}} \right) = 5250 - \left( {1400 + 1454 + 1427.437} \right)$$

$$= 968.563Kj/\min$$

Percentage for heat lost in BP: $$= \frac{{1454.4}}{{5250}} \times 100 = 27.70\%$$ Percentage for heat lost by cooling water: $$= \frac{{1400}}{{5250}} \times 100 = 26.67\%$$ Percentage for heat lost by exhaust gases: $$= \frac{{1427.437}}{{5250}} \times 100 = 27.2\%$$ Percentage of heat unaccounted for: $$= \frac{{968.563}}{{5250}} \times 100 = 18.44\%$$