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Air in the cylinder of a Diesel Engine is at $30^o\ $$C$ and $138\text{ kPa}$ in compression. If it is further compressed to 1/18th of its original volume. Which of the following most nearly equals the work done in compression of the displacement volume of the cylinder is $14.2$ Liters

Answer is $11\ \text{kJ}$

$$T_1 = 30^o\ C + 273.15 = 303.15\ K$$ $$P_1 = 138\ kPa$$ $$V_d = 14.2\ L*(1\ m^3/1000L) = 0.0142\ m^3$$

Attempt 1: Using isentropic relation

$$r_k = 18 = V_1/V_2$$ $$V_1 = \frac{mRT}{P_2} = \frac{1*0.287*303.15}{138} = 0.63046\ m^3$$ $$V2 = 0.03502578502\ m^3$$ $$P2 = \frac{mRT}{V2}=\frac{287*303.15}{0.0350257} = 2484\ Pa$$ for air: $$k = 1.4$$ $$W = \frac{(P_2V_2 - P_1V_1)}{k - 1} = \text{(A very very small value)}$$


Attempt 2: Using $r_k$ and displacement volume:

$$r_k = \frac{(V_d + V_c)}{V_c}$$ $$V_d = 14.2L = 0.0142\ m^3$$ $$V_c = 8.3529411*10^{-4} = V_2$$ $$V_1 = V_c + V_d = 0.01583529\ m^3$$ $$P_1V_1^k = P_2V_2^k $$ $$\frac{138*0.01583529^{1.4}}{(8.3529411*10^{-4})^{1.4}} = P2 = 8487.512\ kPa$$

Plug in everything into the work function $$W = \frac{P_2V_2 - P_1V_1}{k-1} = 12.1 \ kJ$$

Ok, I'm getting close but I don't know what I'm doing wrong... Any hints? And why doesnt' method 1 work?

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Your second attempt is correct, you just had two mistakes:

  1. If you re-calculated $V_1$ you'll find that it equals $0.01503$ not $0.0158$ $m^3$.

  2. same issue with $P_2$. $$P_2 = \frac{138*10^{3}*0.01503^{1.4}}{(8.3529411*10^{-4})^{1.4}}= 7.889*10^6\ Pa$$

Substituting in isentropic work equation: $$W = \frac{P_2V_2 - P_1V_1}{k-1} = \frac{(7.867*10^6 * 8.3529411*10^{-4}) - (138*10^3*0.01503)}{1.4-1}=11.2 \ kJ$$

Which is very close to your problem set answer, I think it was a multiple choice question as he is asking about which answer nearly equals.

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Considering that the following is already sufficient to solve the "intensive" problem:

Air in the cylinder of a Diesel Engine is at $30\,{}^o\mathrm{C}$ and $138\,\mathrm{kPa}$. If it is further compressed to $1/18\,$th of its original volume, calculate the work done on the system.

the compression ratio is $\varrho=v_1/v_2=18$, and the barometric ratio $\beta=P_2/P_1$ is rapidly derived from the isoentropic relation (given that $\gamma=1.4$ is temperature independent): $$ P_1v_1^\gamma=P_2v_2^\gamma\to \frac{P_2}{P_1}=\left(\frac{v_1}{v_2}\right)^\gamma\to \beta=\varrho^\gamma $$ this can be directly used inside the work equation. The latter is derived from the integral of $P\,\mathrm{d}v$ on Clapeyron's Diagram (isoentropic shaft work on a closed system), using the intensive relation, as demonstrated below: \begin{align} w=-\int_{v_1}^{v_2}P\,\mathrm{d}v&=-\int_{v_1}^{v_2}\frac{P_1v_1^{\gamma}}{v^\gamma}\,\mathrm{d}v\\ &=P_1v_1^{\gamma}\int_{v_1}^{v_2}-v^{-\gamma}\,\mathrm{d}v\\ &=P_1v_1^{\gamma}\frac{1}{\gamma-1}\left(v_2^{1-\gamma}-v_1^{1-\gamma}\right)\\ &=RT_1\frac{1}{\gamma-1}\left(v_2^{1-\gamma}/v_1^{1-\gamma}-1\right)\\ &=RT_1\frac{1}{\gamma-1}\left(\varrho^{\gamma-1}-1\right) \end{align}

Since the result will describe an intensive quantity, the extensive result will depend on the chemical substance contained (and not consumed) inside the system. That is derived from the displaced volume and the compression ratio, namely rearranging the terms to highlight $\varrho$:

the compression work done, given the displacement volume of the cylinder as $\Delta V=14.2\,\mathrm{L}$

$$ \Delta V=V_1-V_2\to\frac{\Delta V}{V_2}=\frac{V_1-V_2}{V_2}=\varrho-1\to V_2=\frac{\Delta V}{\varrho-1} $$ from here the initial volume is known (substituting back the $V_2$ term): $$ V_1=V_2+\Delta V=\frac{\Delta V}{\varrho-1}+\Delta V=\frac{14.2\,\mathrm{L}}{17}+14.2\,\mathrm{L}=15.03\,\mathrm{L}=15.03\times10^{-3}\,\mathrm{m^3} $$ substituting all the first intensive parameters inside the Ideal Gas EoS, the constant molar quantity of air is known: $$ n=\frac{P_1V_1}{RT_1}=\frac{15.03\times10^{-3}\,\mathrm{m^3}\cdot138\times10^{3}\,\mathrm{Pa}}{8.314\,(\mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}})\cdot303.15\,\mathrm{K}}=0.823\,\mathrm{mol} $$ Knowing that the extensive parameter $W$ is expressed as $nw$, and from the previous equation $w=13.72\,\mathrm{kJ\,mol^{-1}}$, the requested value is $W=0.823\,\mathrm{mol}\cdot 13.72\,\mathrm{kJ\,mol^{-1}}=11.29\,\mathrm{kJ}$.

At this point, even if the result is different from the first given, there can be one explanation. Since in your first method you considered an initial mass of air equal to $1\,\mathrm{kg}$, this is contradictory with respect to the Equation of State. Because the initial (and constant) molar quantity of air depends to the initial extensive state $(T_1,P_1,V_1)$, then the only variable left $V_1$ is derived if the displacement volume and compression ratio are known.

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    $\begingroup$ Great work Andrea, but why did you calculate intensive mechanical work through $w=\int_{P_1}^{P_2}v\,\mathrm{d}P$ as isentropic steady-flow shaft work (we have a closed system here) and not $w=\int_{v_1}^{v_2}P\,\mathrm{d}v$ as boundary work? $\endgroup$ – Algo Jul 12 '15 at 17:46
  • $\begingroup$ @Algo My reasoning went with the assumption that inside a non constant control volume the intensive differential relation $T\mathrm{d}s=\mathrm{d}h-v\mathrm{d}P$ would hold, because here I evaluated $w$ as the enthalpy variation. As if the compression stage (within existance of inlet flow, inside a real engine) would be equivalent to an Operating Machine $\endgroup$ – TheVal Jul 12 '15 at 19:51
  • $\begingroup$ But this can't apply to our case, not to mention that it's not a complete diesel cycle this is just a simple isentropic compression process. However, when dealing with idealized simple power cycle we usually assume that all expansion and compression processes take place in a quasi-equilibrium manner. $\endgroup$ – Algo Jul 12 '15 at 22:26
  • $\begingroup$ Anyway, $p\,dv$ is never neglected physics.stackexchange.com/questions/52001/… $\endgroup$ – Algo Jul 12 '15 at 22:29
  • $\begingroup$ @Algo Here I've modified it and thus corrected it. Thanks again for noticing the mistake ;) $\endgroup$ – TheVal Sep 8 '15 at 19:57

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