A diesel engine having compression ratio of 16 uses the fuel $C_{7}H_{16}$ . The compression follows the law $pv^{1.4}$ = constant. If the engine uses 64% excess air and the temperature at the beginning of compression is 325 K, find the percentage of stroke at which combustion is completed.
Assume $c_{p} = 1.0 + 21 × 10^{–5}$ T kJ/kg K and T is in degrees’ kelvin. C.V. = 44000 kJ/kg.
Assume that air contains 23% by weight of $O_{2}$ .

Question

A diesel engine having compression ratio of 16 uses the fuel [latex] C_{7}H_{16} [/latex]. The compression follows the law [latex] pv^{1.4}[/latex] = constant. If the engine uses 64% excess air and the temperature at the beginning of compression is 325 K, find the percentage of stroke at which combustion is completed.
Assume [latex] c_{p} = 1.0 + 21 × 10^{–5} [/latex] T kJ/kg K and T is in degrees’ kelvin. C.V. = 44000 kJ/kg.
Assume that air contains 23% by weight of [latex] O_{2} [/latex].

Accepted Answer

Given : r = 16 ; Excess air used = 64%;
[latex]T_{1}[/latex] = 325 K ; C.V. = 44000 kJ/kg.
Fig. 44 shows the cycle of operation. The minimum amount of air required to burn 1 kg of fuel can be calculated by using the following chemical equation :

[latex] C_{7}H_{16} + 11O_{2} = 7 CO_{2} + 8 H_{2}O [/latex]

100 + 352 = 308 + 144

∴ 1 kg of fuel requires [latex] \frac{352}{100} kg of O_{2}[/latex] and

[latex] \frac{352}{100} × \frac{100}{23} kg[/latex] of air = 15.3 kg

Actual air supplied = 15.3 × 1.64 = 25.092 kg/kg of fuel.

% cut-off = [latex] \frac{v_{3} - v_{2}}{v_{1} - v_{2}} = \frac{v_{3} - v_{2}}{16 v_{2}} = \frac{1}{16} (\frac{v_{3} }{v_{2} } - 1) [/latex] ...(i)

From compression process 1–2, we have ( ∵ [latex] \frac{v_{1}}{ v_{2}} [/latex] = r = 16)

[latex] \frac{T_{2}}{T_{1}} = (\frac{v_{1}}{v_{2}})^{γ - 1} = (16)^{1.4 - 1} , or , T_{2} = 325 × (16)^{0.4}[/latex] = 985 K

The heat supplied per kg of fuel = Heat given to the gases during the process 2–3.

∴ 44000 = [latex]\int_{ T_{2}}^{ T_{3}}{(\frac{m_{a} }{m_{f}} + 1) c_{p} dT} [/latex]

44000 = [latex] \int_{ T_{2}}^{ T_{3}} {( 25.092 + 1) (1.0 + 21 × 10^{–5} T) dT}[/latex]

or, [latex] \frac{44000}{ 26.092} = \int_{ 985}^{ T_{3}} {(1.0 + 21 × 10^{–5} T) dT}[/latex]

or, 1686.34 =[latex] (T_{3} – 985) + \frac{21 × 10^{–5}}{2} [T_{3}² - (985)²][/latex]

or, 1686.34 = [latex] T_{3} – 985 + 10.5 × 10^{–5} T_{3}² [/latex] – 101.87

or, 10.5 ×[latex] 10^{–5} T_{3}² + T_{3} [/latex] – 2762.7 = 0

or, [latex] T_{3} = \frac{- 1 ± \sqrt{1² + 4 × 10.5 × 10^{–5} × 2762.7} }{2 × 10.5 × 10^{–5} }[/latex]

= [latex] \frac{- 1 + 1.47}{2 × 10.5 × 10^{–5}}[/latex] = 2238 K.

From constant pressure process 2–3, we have

[latex] \frac{v_{2}}{T_{2}} = \frac{v_{3}}{T_{3}} , or , \frac{v_{3}}{v_{2}} = \frac{T_{3}}{T_{2}} = \frac{2238}{985} [/latex] = 2.272

Substituting this in eqn. (i), we have

∴ % cut-off = [latex] \frac{1}{16}[/latex] (2.272 – 1 ) = 0.0795 or 7.95%.

Question 8.42: A diesel engine having compression ratio of 16 uses the fuel...

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