Chapter 5.5, Problem 188RP

The turbocharger of an internal combustion engine consists of a turbine and a compressor. Hot exhaust gases flow through the turbine to produce work, and the work output from the turbine is used as the work input to the compressor. The pressure of ambient air is increased as it flows through the compressor before it enters the engine cylinders. Thus, the purpose of a turbocharger is to increase the pressure of air so that more air gets into the cylinder. Consequently, more fuel can be burned and more power can be produced by the engine.

In a turbocharger, exhaust gases enter the turbine at 400°C and 120 kPa at a rate of 0.02 kg/s and leave at 350°C. Air enters the compressor at 50°C and 100 kPa and leaves at 130 kPa at a rate of 0.018 kg/s. The compressor increases the air pressure with a side effect: It also increases the air temperature, which increases the possibility that a gasoline engine will experience an engine knock. To avoid this, an aftercooler is placed after the compressor to cool the warm air with cold ambient air before it enters the engine cylinders. It is estimated that the aftercooler must decrease the air temperature below 80°C if knock is to be avoided. The cold ambient air enters the aftercooler at 30°C and leaves at 40°C. Disregarding any frictional losses in the turbine and the compressor and treating the exhaust gases as air, determine (a) the temperature of the air at the compressor outlet and (b) the minimum volume flow rate of ambient air required to avoid knock.

FIGURE P5–188

To determine

The temperature of the air at the compressor outlet.

Answer to Problem 188RP

The temperature of the air at the compressor outlet is $108.6°\text{C}$.

Explanation of Solution

Draw the schematic diagram of the given turbo charger of the engine as shown in

Figure 1.

Write the general energy rate balance equation.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}\\ \left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}\end{array}$ (I)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Refer Figure 1 and Refer Equation (I).

For turbine:

Consider the turbine is adiabatic, neglect the heat transfer. Also neglect the kinetic and potential energy changes. The work done is by the system (turbine) and the work done on the system is zero i.e. ${\dot{W}}_1=0$.

The Equations (I) reduced as follows to obtain the work output of compressor.

$\begin{array}{l}\left[0+0+{\dot{m}}_{\text{exh}}\left(h_{\text{exh,1}}+0+0\right)\right]-\left[0+{\dot{W}}_{\text{T,2}}+{\dot{m}}_{\text{exh}}\left(h_{\text{exh,2}}+0+0\right)\right]=0\\ {\dot{m}}_{\text{exh}}{h}_{\text{exh,1}}-\left({\dot{W}}_{\text{T,2}}+{\dot{m}}_{\text{exh}}{h}_{\text{exh,2}}\right)=0\\ {\dot{W}}_{\text{T,2}}={\dot{m}}_{\text{exh}}{h}_{\text{exh,1}}-{\dot{m}}_{\text{exh}}{h}_{\text{exh,2}}\\ {\dot{W}}_{\text{T,2}}={\dot{m}}_{\text{exh}}\left(h_{\text{exh,1}}-h_{\text{exh,2}}\right)\end{array}$ (II)

The change in enthalpy is expressed as follow.

$h_{\text{exh,1}}-h_{\text{exh,2}}=c_{p,\text{exh}}\left(T_{\text{exh,1}}-T_{\text{exh,2}}\right)$

Here, the specific heat of exhaust gas is $c_{p,\text{}\text{exh}}$, the exit temperature of exhaust gas is $T_{\text{exh,}2}$ and the inlet temperature of exhaust gas is $T_{\text{exh,}1}$.

Substitute $c_{p,\text{exh}}\left(T_{\text{exh,1}}-T_{\text{exh,2}}\right)$ for $h_{\text{exh,1}}-h_{\text{exh,2}}$ in Equation (II).

${\dot{W}}_{\text{T,2}}={\dot{m}}_{\text{exh}}{c}_{p,\text{exh}}\left(T_{\text{exh,1}}-T_{\text{exh,2}}\right)$ (III)

For compressor:

Consider the compressor is adiabatic, neglect the heat transfer. Also neglect the kinetic and potential energy changes. The work done is on the system (compressor) and the work done by the system is zero i.e. ${\dot{W}}_2=0$.

The Equations (I) reduced as follows to obtain the work input of compressor.

$\begin{array}{l}\left[0+{\dot{W}}_{\text{C,1}}+{\dot{m}}_a\left(h_{a\text{,1}}+0+0\right)\right]-\left[0+0+{\dot{m}}_{\text{a}}\left(h_{a\text{,2}}+0+0\right)\right]=0\\ {\dot{W}}_{\text{C,1}}+{\dot{m}}_a{h}_{a\text{,1}}-{\dot{m}}_a{h}_{a\text{,2}}=0\\ {\dot{W}}_{\text{C,1}}={\dot{m}}_a{h}_{a\text{,2}}-{\dot{m}}_a{h}_{a\text{,1}}\\ {\dot{W}}_{\text{C,1}}={\dot{m}}_a\left(h_{a\text{,2}}-h_{a\text{,1}}\right)\end{array}$

${\dot{W}}_{\text{C,1}}={\dot{m}}_a{c}_{p,a}\left(T_{a\text{,2}}-T_{a\text{,1}}\right)$ (IV)

Here, the mass flow rate is $\dot{m}$, the enthalpy is $h$, the temperature is $T$, the subscripts 1 and 2 indicates the inlet and outlet states, and the subscript $a$ indicates the air, the subscript $\text{exh}$ indicates exhaust gas.

Refer Table A-2(b), “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure $\left(c_{p,\text{exh}}\right)$ of exhaust gas corresponding to its temperature is $1.063\text{kJ/kg}\cdot \text{°C}$.

The specific heat at constant pressure $\left(c_{p,a}\right)$ of warm air corresponding to its temperature is $1.008\text{kJ/kg}\cdot \text{°C}$.

The specific heat at constant pressure $\left(c_{p,ca}\right)$ of cold ambient air corresponding to its temperature is $1.005\text{kJ/kg}\cdot \text{°C}$.

Conclusion:

Substitute $0.02\text{kg/s}$ for ${\dot{m}}_{\text{exh}}$, $1.063\text{kJ/kg}\cdot \text{°C}$ for $c_{p,\text{exh}}$, $400°\text{C}$ for $T_{\text{exh,1}}$, and $350°\text{C}$ for $T_{\text{exh,2}}$ in Equation (III).

$\begin{array}{c}{\dot{W}}_{\text{T,2}}=\left(0.02\text{kg/s}\right)\left(1.063\text{kJ/kg}\cdot \text{°C}\right)\left(400°\text{C}-350°\text{C}\right)\\ =1.063\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =1.063\text{kW}\end{array}$

Here, ${\dot{W}}_{\text{T,2}}={\dot{W}}_{\text{C,1}}=1.063\text{kW}$

Substitute $1.063\text{kW}$ for ${\dot{W}}_{\text{C,1}}$, $0.018\text{kg/s}$ for ${\dot{m}}_a$, $1.008\text{kJ/kg}\cdot \text{°C}$ for $c_{p,a}$, and $50°\text{C}$ for $T_{a\text{,1}}$ in Equation (IV).

$\begin{array}{l}1.063\text{kJ/s}=\left(0.018\text{kg/s}\right)\left(1.008\text{kJ/kg}\cdot \text{°C}\right)\left(T_{a,2}-50°\text{C}\right)\\ T_{a,2}-50°\text{C}=\frac{1.063\text{kJ/s}}{0.018144\text{kJ/s}\cdot \text{°C}}\\ T_{a,2}=58.5867°\text{C}+50°\text{C}\\ T_{a,2}=108.5867°\text{C}\end{array}$

$T_{a,2}\simeq 108.6°\text{C}$

Thus, the temperature of the air at the compressor outlet is $108.6°\text{C}$.

To determine

The minimum volume flow rate of ambient air required to avoid knock.

Answer to Problem 188RP

The minimum volume flow rate of ambient air required to avoid knock is $44.9\text{L/s}$.

Explanation of Solution

Refer Figure 1 and Refer Equation (I).

For aftercooler:

The after cooler is the two inlet and two outlet system.

Refer the Equation (I) Express the energy rate balance equation for aftercooler as follows

$\begin{array}{l}{\dot{m}}_a\left(h_{a\text{,3}}-h_{a\text{,2}}\right)={\dot{m}}_{ca}\left(h_{ca\text{,2}}-h_{ca\text{,1}}\right)\\ {\dot{m}}_a{c}_{p,a}\left(T_{a\text{,3}}-T_{a\text{,2}}\right)={\dot{m}}_{ca}{c}_{p,ca}\left(T_{ca\text{,2}}-T_{ca\text{,1}}\right)\end{array}$ (V)

Here, subscript $ca$ indicates the cold air.

Write the formula for volume flow rate of cold air.

${\dot{V}}_{ca}=\frac{{\dot{m}}_{ca}R{T}_{ca,1}}{P_{ca,1}}$ (VI)

Here, the gas constant of air is $R$, the temperature is $T$ and the pressure is $P$.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant $\left(R\right)$ of air is $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$.

Conclusion:

Substitute, $0.018\text{kg/s}$ for ${\dot{m}}_a$, $1.008\text{kJ/kg}\cdot \text{°C}$ for $c_{p,a}$, $108.6°\text{C}$ for $T_{a,2}$, $80°\text{C}$ for $T_{a\text{,2}}$, $1.005\text{kJ/kg}\cdot \text{°C}$ for $c_{p,ca}$, $40°\text{C}$ for $T_{ca\text{,2}}$, and $30°\text{C}$ for $T_{ca\text{,1}}$ in Equation (V).

$\begin{array}{l}\left[\begin{array}{l}\left(0.018\text{kg/s}\right)\left(1.008\text{kJ/kg}\cdot \text{°C}\right)\\ \left(108.6°\text{C}-80°\text{C}\right)\end{array}\right]={\dot{m}}_{ca}\left(1.005\text{kJ/kg}\cdot \text{°C}\right)\left(40°\text{C}-30°\text{C}\right)\\ 0.5189\text{kJ/s}={\dot{m}}_{ca}\left(10.05\text{kJ/kg}\right)\\ {\dot{m}}_{ca}=\frac{0.5189\text{kJ/s}}{10.05\text{kJ/kg}}\\ {\dot{m}}_{ca}=0.05163\text{kg/s}\end{array}$

Substitute $0.05163\text{kg/s}$ for ${\dot{m}}_{ca}$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, $30°\text{C}$ for $T_{ca\text{,1}}$, and $100\text{kPa}$ for $P_{ca,1}$ in Equation (VI).

$\begin{array}{c}{\dot{V}}_{ca}=\frac{\left(0.05163\text{kg/s}\right)\left(0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\right)\left(30°\text{C}\right)}{100\text{kPa}}\\ =\frac{\left(0.05163\text{kg/s}\right)\left(0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\right)\left(30+273\right)\text{K}}{100\text{kPa}}\\ =\frac{4.4901\text{kPa}\cdot {\text{m}}^3\text{/s}}{100\text{kPa}}\\ =0.0449{\text{m}}^3\text{/s}\times \frac{1000\text{}\text{L}}{1{\text{m}}^3}\end{array}$

$=44.9\text{L/s}$

Thus, the minimum volume flow rate of ambient air required to avoid knock is $44.9\text{L/s}$.