Engine Turbochargers and Intercoolers. The power output of an automobile engine is directly proportional to the mass of air that can be forced into the volume of the engine’s cylinders to react chemically with gasoline. Many cars have a turbocharger, which compresses the air before it enters the engine, giving a greater mass of air per volume. This rapid, essentially adiabatic compression also heats the air. To compress it further, the air then passes through an intercooler in which the air exchanges heat with its surroundings at essentially constant pressure. The air is then drawn into the cylinders. In a typical installation, air is taken into the turbocharger at atmospheric pressure 11.01 * 105 Pa2, density r = 1.23 kg>m3, and temperature 15.0°C. It is compressed adiabatically to 1.45 * 105 Pa. In the intercooler, the air is cooled to the original temperature of 15.0°C at a constant pressure of 1.45 * 105 Pa. (a) Draw a pV-diagram for this sequence of processes. (b) If the volume of one of the engine’s cylinders is 575 cm3 , what mass of air exiting from the intercooler will fill the cylinder at 1.45 * 105 Pa? Compared to the power output of an engine that takes in air at 1.01 * 105 Pa at 15.0°C, what percentage increase in power is obtained by using the turbocharger and intercooler? (c) If the intercooler is not used, what mass of air exiting from the turbocharger will fill the cylinder at 1.45 * 105 Pa? Compared to the power output of an engine that takes in air at 1.01 * 105 Pa at 15.0°C, what percentage increase in power is obtained by using the turbocharger alone?

Step by step solution

01

Step 1:

pressure =11.01 * 105 Pa

density r = 1.23 kg/m3

temperature =15.0°C

adiabatically compressed pressure=1.45 * 105 Pa

temperature =15.0°C

constant pressure = 1.45 * 105 Pa

02

step 2:

(a)p-v diagram

$$\begin{gathered} {\mathrm{P}}_1=1.01\times {10}^5\mathrm{Pa} \\ {\mathrm{\rho }}_1=1.23\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Are the gas and pressure before adiabatic compression.

${\mathrm{P}}_2=1.45\times {10}^5\mathrm{Pa}$

Is pressure after compression gas attains pressure.

03

Step 3:

(b)

$$\begin{gathered} {\mathrm{P}}_1=1.01\times {10}^5\mathrm{Pa} \\ {\mathrm{T}}_1=15{.0}^0\mathrm{C}=288.15\mathrm{K} \\ {\mathrm{\rho }}_1=1.23\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Are pressure, temperature, density before adiabatic compression

$$\begin{gathered} {\mathrm{P}}_2=1.45\times {10}^5\mathrm{Pa} \\ {\mathrm{T}}_2=15{.0}^0\mathrm{C}=288.15\mathrm{K} \\ {\mathrm{V}}_2=575{\mathrm{cm}}^3=5.75\times {10}^{-4}{\mathrm{m}}^3 \end{gathered}$$

Are pressure, temperature, volume of engine-cylinder,

$$\begin{gathered} \frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2} \\ {\mathrm{V}}_1=\frac{{\mathrm{P}}_2{\mathrm{V}}_2{\mathrm{T}}_1}{{\mathrm{T}}_2} \\ {\mathrm{V}}_1=\frac{\left(1.45\times {10}^5\mathrm{Pa}\right)\left(5.75\times {10}^{-4}{\mathrm{m}}^3\right)(288.15\mathrm{K})}{(288.15\mathrm{K})\left(1.01\times {10}^5\mathrm{Pa}\right)} \\ {\mathrm{V}}_1=8.255\times {10}^{-4}{\mathrm{m}}^3 \end{gathered}$$

Mass of the gas,

$$\begin{gathered} {\mathrm{m}}_{\mathrm{c}}={\mathrm{\rho V}}_1 \\ {\mathrm{m}}_{\mathrm{c}}=1.23\mathrm{kg}/{\mathrm{m}}^3\left(8.255\times {10}^{-4}{\mathrm{m}}^3\right) \\ {\mathrm{m}}_{\mathrm{c}}=10.154\times {10}^{-3}\mathrm{kg} \\ {\mathrm{m}}_{\mathrm{c}}=10.154\mathrm{gm} \end{gathered}$$

Mass of air,

$$\begin{gathered} {\mathrm{m}}_{\mathrm{a}}={\mathrm{\rho V}}_2 \\ {\mathrm{m}}_{\mathrm{a}}=1.23\mathrm{kg}/{\mathrm{m}}^3\left(5.75\times {10}^{-4}{\mathrm{m}}^3\right) \\ {\mathrm{m}}_{\mathrm{a}}=7.073\times {10}^{-3}\mathrm{kg} \\ {\mathrm{m}}_{\mathrm{a}}=7.0723\mathrm{gm} \end{gathered}$$

Percentage increase in engine power,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{\%}}=1.01\times {10}^5\mathrm{Pa} \\ {\mathrm{P}}_{\mathrm{\%}}=\left(\frac{(10.154\mathrm{gm})-(7.073\mathrm{gm})}{(7.073\mathrm{gm})}\right)\times 100 \\ {\mathrm{P}}_{\mathrm{\%}}=43.57\mathrm{\%} \end{gathered}$$

04

Step 4:

( c)

Gas, pressure, temperature before adiabatic compression,

$$\begin{gathered} {\mathrm{P}}_1=1.01\times {10}^5\mathrm{Pa} \\ {\mathrm{T}}_1=15{.0}^{\circ }\mathrm{C}=288.15\mathrm{K} \\ {\mathrm{\rho }}_1=1.23\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

After turbo charger,

$$\begin{gathered} {\mathrm{P}}_2=1.45\times {10}^5\mathrm{Pa} \\ {\mathrm{V}}_2=575\mathrm{cm}3=5.75\times {10}^{-4}{\mathrm{m}}^3 \end{gathered}$$

Specific heat ratio,

$$\begin{gathered} \gamma =1.40 \\ {\mathrm{P}}_1^{1-Y_T^Y}{T}_1^{\mathrm{Y}}={\mathrm{P}}_2^{1-\mathrm{Y}}{\mathrm{T}}_2^{\mathrm{Y}} \\ {\mathrm{T}}_2^{\mathrm{Y}}=\frac{{\mathrm{P}}_1^{1-\mathrm{Y}}{\mathrm{T}}_1^{\mathrm{Y}}}{{\mathrm{P}}_2^{1-\mathrm{Y}}} \\ {\mathrm{T}}_2={\mathrm{T}}_1{\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)}^{\frac{\mathrm{Y}-1}{\mathrm{Y}}} \\ {\mathrm{T}}_2=\left(288.15\mathrm{K}\right){\left(\frac{\left(1.45\times {10}^5\mathrm{Pa}\right)}{\left(1.01\times {10}^5\mathrm{Pa}\right)}\right)}^{\frac{1.4-1}{1.4}} \\ {\mathrm{T}}_2=\left(288.15\mathrm{K}\right)(1.4356{)}^{2/7} \\ {\mathrm{T}}_2=319.51\mathrm{K} \\ {\mathrm{T}}_2={46.363}^{\mathrm{o}}\mathrm{C} \end{gathered}$$

Volume of gas,

$$\begin{gathered} \frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2} \\ {\mathrm{V}}_1=\frac{{\mathrm{P}}_2{\mathrm{V}}_2{\mathrm{T}}_1}{{\mathrm{T}}_2} \\ {\mathrm{V}}_1=\frac{\left(1.45\times {10}^5\mathrm{Pa}\right)\left(5.75\times {10}^{-4}{\mathrm{m}}^3\right)(288.15\mathrm{K})}{(319.51\mathrm{K})\left(1.01\times {10}^5\mathrm{Pa}\right)} \\ {\mathrm{V}}_1=7.445\times {10}^{-4}{\mathrm{m}}^3 \end{gathered}$$

Mass of gas,

$$\begin{gathered} {\mathrm{m}}_{\mathrm{c}}=\rho {\mathrm{V}}_1 \\ {\mathrm{m}}_{\mathrm{c}}=1.23\mathrm{kg}/{\mathrm{m}}^3\times \left(7.445\times {10}^{-4}{\mathrm{m}}^3\right) \\ {\mathrm{m}}_{\mathrm{c}}=9.157\times {10}^{-3}\mathrm{kg} \\ {\mathrm{m}}_{\mathrm{c}}=9.157\mathrm{gm} \\ {\mathrm{m}}_1=9.157\mathrm{gm} \end{gathered}$$

Mass of air,

$$\begin{gathered} {\mathrm{m}}_{\mathrm{a}}={\mathrm{\rho V}}_2 \\ {\mathrm{m}}_{\mathrm{a}}=1.23\mathrm{kg}/{\mathrm{m}}^3\left(5.75\times {10}^{-4}{\mathrm{m}}^3\right) \\ {\mathrm{m}}_{\mathrm{a}}=7.073\times {10}^{-3}\mathrm{kg} \end{gathered}$$

Power output in engine,

$$\begin{gathered} {\mathrm{m}}_{\mathrm{a}}=7.0723\mathrm{gm} \\ {\mathrm{P}}_{\mathrm{\%}}=\frac{{\mathrm{m}}_{\mathrm{c}}-{\mathrm{m}}_{\mathrm{a}}}{{\mathrm{m}}_{\mathrm{a}}}\times 100 \\ {\mathrm{P}}_{\mathrm{\%}}=\left(\frac{\left(9.157\mathrm{gm}\right)-\left(7.073\mathrm{gm}\right)}{\left(7.073\mathrm{gm}\right)}\right)\times 100 \\ {\mathrm{P}}_{\mathrm{\%}}=29.47\mathrm{\%} \end{gathered}$$

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