Chapter 9.12, Problem 167RP

(a)

To determine

The temperature at the end of expansion process.

Answer to Problem 167RP

The temperature at the end of expansion process is $\underset{\_}{1098\text{K}}$.

Explanation of Solution

Determine the state 2 temperature in the polytropic compression process 1-2.

$T_2=T_1{\left(\frac{v_1}{v_2}\right)}^{n-1}$ (I)

Here, the state 1 temperature is $T_1$, the specific volume at state 1 is $v_1$, the specific volume at state 2 is $v_2$, and the polytropic constant is $n$.

Determine the state 2 pressure in the polytropic compression process 1-2.

$P_2=P_1{\left(\frac{v_1}{v_2}\right)}^n$ (II)

Here, the state 1 pressure is $P_1$, the specific volume at state 1 is $v_1$, and the specific volume at state 2 is $v_2$.

Determine the work per unit mass in the polytropic compression process 1-2.

$w_{12}=\frac{R\left(T_2-T_1\right)}{1-n}$ (III)

Here, the universal gas constant is $R$.

Determine the state 3 temperature in the constant volume heat addition process 2-3.

$T_3=T_2\left(\frac{P_3}{P_2}\right)$ (IV)

Here, the state 2 temperature is $T_2$ and the state 3 pressure is $P_3$.

Determine the heat transfer per unit mass in the constant volume heat addition process 2-3.

$\begin{array}{c}{q}_{\text{in}}=u_3-u_2\\ =c_v\left(T_3-T_2\right)\end{array}$ (V)

Here, the specific heat of constant volume is $c_v$.

Determine the state 4 temperature in the polytropic expansion process 3-4.

$T_4=T_3{\left(\frac{v_3}{v_4}\right)}^{n-1}$ (VI)

Here, the specific volume at state 3 is $v_3$ and the specific volume at state 4 is $v_4$.

Determine the state 4 pressure in the polytropic expansion process 3-4.

$P_4=P_3{\left(\frac{v_3}{v_4}\right)}^n$ (VII)

Here, the state 3 pressure is $P_3$.

Determine the work per unit mass in the polytropic compression process 3-4.

$w_{34}=\frac{R\left(T_4-T_3\right)}{1-n}$ (VIII)

Here, the universal gas constant is $R$.

Conclusion:

From the Table A-2 (a), “Ideal-gas specific heats of various common gases”, obtain the value of universal gas constant of air is $0.287\text{kJ}/\text{kg}\cdot \text{K}$.

Refer to Table A-2 (b), “Ideal-gas specific heats of various common gases”, obtain the below properties at the average temperature of 850 K using interpolation method of two variables.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (I)

Here, the variables denote by x and y are temperature and specific heat of constant pressure.

Show the temperature at 800 K and 900 K as in Table (1).

S. No	Temperature, K $\left(x\right)$	specific heat of constant pressure, $\text{kJ}/\text{kg}\cdot \text{K}$ $\left(y\right)$
1	800 K	1.099
2	850 K	$y_2=?$
3	900 K	1.121

Calculate specific heat of constant pressure at an average temperature of 850 K for liquid phase using interpolation method.

Substitute 800 K for $x_1$, 850 K for $x_2$, 900 K for $x_3$, $1.099\text{kJ}/\text{kg}\cdot \text{K}$ for $y_1$, and $1.121\text{kJ}/\text{kg}\cdot \text{K}$ for $y_3$ in Equation (I).

$\begin{array}{c}{y}_2=\frac{\left(850\text{K}-800\text{K}\right)\left(1.121\text{kJ}/\text{kg}\cdot \text{K}-1.099\text{kJ}/\text{kg}\cdot \text{K}\right)}{\left(900\text{K}-800\text{K}\right)}+1.099\text{kJ}/\text{kg}\cdot \text{K}\\ =\frac{\left(50\text{K}\right)\left(0.022\text{kJ}/\text{kg}\cdot \text{K}\right)}{\left(100\text{K}\right)}+1.099\text{kJ}/\text{kg}\cdot \text{K}\\ =\text{1}\text{.110}\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From above calculation the specific heat of constant pressure is $\text{1}\text{.110}\text{kJ}/\text{kg}\cdot \text{K}$ at a temperature of 850 K.

Similarly repeat the interpolation method for specific heat of constant volume and ratio of specific heat as:

$c_v=0.823\text{kJ}/\text{kg}\cdot \text{K}$.

$k=1.349$.

Substitute $37°\text{C}$ for $T_1$, $11$ for $v_1/v_2$, and $1.3$ for $n$ in Equation (I).

$\begin{array}{c}{T}_2=\left(37°\text{C}\right)\times {\left(11\right)}^{1.3-1}\\ =\left(37°\text{C}\text{+}\text{273}\right)\times {\left(11\right)}^{0.3}\\ =\left(310\text{K}\right)\times \left(2.053136\right)\\ =636.47\text{K}\end{array}$

    $\cong 636.5\text{K}$

Substitute $100\text{kPa}$ for $P_1$, $11$ for $v_1/v_2$, and $1.3$ for $n$ in Equation (II).

$\begin{array}{c}{P}_2=\left(100\text{kPa}\right)\times {\left(11\right)}^{1.3}\\ =\left(100\text{kPa}\right)\times \left(22.5845\right)\\ =2258\text{K}\end{array}$

Substitute $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $636.5\text{K}$ for $T_2$, $37°\text{C}$ for $T_1$, and $1.3$ for $n$ in Equation (III).

$\begin{array}{c}{w}_{12}=\frac{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(636.5\text{K}-37°\text{C}\right)}{1-1.3}\\ =\frac{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(636.5\text{K}-\left(37°\text{C}+\text{273}\right)\right)}{1-1.3}\\ =\frac{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(636.5\text{K}-310\text{K}\right)}{1-1.3}\\ =\frac{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(326.5\text{K}\right)}{-0.3}\end{array}$

      $=-312.3\text{kJ}/\text{kg}$

Substitute $636.5\text{K}$ for $T_2$, 8 MPa for $P_3$, and $2258\text{kPa}$ for $P_2$ in Equation (IV).

$\begin{array}{c}{T}_3=\left(636.5\text{K}\right)\times \left(\frac{8\text{MPa}}{2258\text{kPa}}\right)\\ =\left(636.5\text{K}\right)\times \left(\frac{8\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)}{2258\text{kPa}}\right)\\ =\left(636.5\text{K}\right)\times \left(\frac{8000\text{kPa}}{2258\text{kPa}}\right)\\ =2255\text{K}\end{array}$

Substitute $0.823\text{kJ}/\text{kg}\cdot \text{K}$ for $c_v$, 2255 K for $T_3$, and 636.5 K for $T_2$ in Equation (V).

$\begin{array}{c}{q}_{\text{in}}=\left(0.823\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(2255-636.5\right)\text{K}\\ =\left(0.823\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(1618.5\text{K}\right)\\ =1332\text{kJ}/\text{kg}\end{array}$

Substitute $2255\text{K}$ for $T_3$, $1/11$ for $v_3/v_4$, and $1.3$ for $n$ in Equation (VI).

$\begin{array}{c}{T}_4=\left(2255\text{K}\right)\times {\left(\frac{1}{11}\right)}^{1.3-1}\\ =\left(2255\text{K}\right)\times {\left(\frac{1}{11}\right)}^{0.3}\\ =\left(2255\text{K}\right)\times \left(0.48706\right)\\ =1098\text{K}\end{array}$

Thus, the temperature at the end of expansion process is $\underset{\_}{1098\text{K}}$.

Substitute $8\text{MPa}$ for $P_3$, $1/11$ for $v_2/v_1$, and $1.3$ for $n$ in Equation (VII).

$\begin{array}{c}{P}_4=\left(8\text{MPa}\right)\times {\left(\frac{1}{11}\right)}^{1.3}\\ =\left(8\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)\right)\times \left(0.044278\right)\\ =354.2\text{kPa}\end{array}$

Substitute $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $1098\text{K}$ for $T_4$, $2255\text{K}$ for $T_3$, and $1.3$ for $n$ in Equation (VIII).

$\begin{array}{c}{w}_{34}=\frac{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1098\text{K}-2255\text{K}\right)}{1-1.3}\\ =\frac{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(-1157\text{K}\right)}{-0.3}\\ =1106\text{kJ}/\text{kg}\end{array}$

(b)

To determine

The net-work output at the constant volume heat rejection.

The thermal efficiency at the constant volume heat rejection.

Answer to Problem 167RP

The net-work output at the constant volume heat rejection is $\underset{\_}{794\text{kJ}/\text{kg}}$.

The thermal efficiency at the constant volume heat rejection is $\underset{\_}{59.6\%}$.

Explanation of Solution

Determine the net-work output at the constant volume heat rejection.

$w_{\text{net,out}}=w_{34}-w_{12}$ (IX)

Determine the thermal efficiency at the constant volume heat rejection.

${\eta }_{\text{th}}=\frac{w_{\text{net,out}}}{q_{\text{in}}}$ (X)

Conclusion:

Substitute $1106\text{kJ}/\text{kg}$ for $w_{34}$ and $312.3\text{kJ}/\text{kg}$ for $w_{12}$ in Equation (IX).

$\begin{array}{c}{w}_{\text{net,out}}=\left(1106\text{kJ}/\text{kg}\right)-\left(312.3\text{kJ}/\text{kg}\right)\\ =793.7\text{kJ}/\text{kg}\\ \cong 794\text{kJ}/\text{kg}\end{array}$

Thus, the net-work output at the constant volume heat rejection is $\underset{\_}{794\text{kJ}/\text{kg}}$.

Substitute $794\text{kJ}/\text{kg}$ for $w_{\text{net,out}}$ and $1332\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (X).

$\begin{array}{c}{\eta }_{\text{th}}=\frac{794\text{kJ}/\text{kg}}{1332\text{kJ}/\text{kg}}\\ =0.596\\ =0.596\times 100\\ \cong 59.6\%\end{array}$

Thus, the thermal efficiency at the constant volume heat rejection is $\underset{\_}{59.6\%}$.

(c)

To determine

The mean effective pressure at the constant volume heat rejection.

Answer to Problem 167RP

The mean effective pressure at the constant volume heat rejection is $\underset{\_}{982\text{kPa}}$.

Explanation of Solution

Determine the initial volume at the constant volume heat rejection.

${\nu }_1=\frac{R{T}_1}{P_1}$ (XI)

Determine the mean effective pressure at the constant volume heat rejection.

$\begin{array}{c}\text{MEP}=\frac{w_{\text{net,out}}}{{\nu }_1-{\nu }_2}\\ =\frac{w_{\text{net,out}}}{{\nu }_1\left(1-1/r\right)}\end{array}$ (XII)

Here, the compression ratio is $r$.

Note: ${\nu }_{\min }={\nu }_2=\frac{{\nu }_{\max }}{r}$.

Conclusion:

Substitute $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $100\text{kPa}$ for $P_1$, $37°\text{C}$ for $T_1$ in Equation (XI).

$\begin{array}{c}{\nu }_1=\frac{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(37°\text{C}\right)}{\left(100\text{kPa}\right)}\\ =\frac{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)\times \left(37°\text{C}+\text{273}\right)}{\left(100\text{kPa}\right)}\\ =\frac{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\times \left(310\text{K}\right)}{\left(100\text{kPa}\right)}\\ =0.8897{\text{m}}^3/\text{kg}\end{array}$

Substitute $794\text{kJ}/\text{kg}$ for $w_{\text{net,out}}$, 11 for $r$, and $0.8897{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (XII).

$\begin{array}{c}\text{MEP}=\frac{794\text{kJ}/\text{kg}}{\left(0.8897{\text{m}}^{\text{3}}/\text{kg}\right)\left(1-1/11\right)}\\ =\frac{794\text{kJ}/\text{kg}}{\left(0.8897{\text{m}}^{\text{3}}/\text{kg}\right)\left(0.909091\right)}\\ =\frac{794\text{kJ}/\text{kg}}{\left(0.808818{\text{m}}^{\text{3}}/\text{kg}\right)}\times \left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)\\ =981.6\text{kPa}\end{array}$

        $\cong 982\text{kPa}$

Thus, the mean effective pressure at the constant volume heat rejection is $\underset{\_}{982\text{kPa}}$.

(d)

To determine

The engine speed for a given net power.

Answer to Problem 167RP

The engine speed for a given net power is $\underset{\_}{3820\text{rev}/\text{min}}$.

Explanation of Solution

Determine the clearance volume at the beginning of compression process.

$r=\frac{{\nu }_c+v_d}{{\nu }_c}$ (XIII)

Here, the volume of the gasoline engine is ${\nu }_d$.

Determine the initial volume.

${\nu }_1={\nu }_c+{\nu }_d$ (XIV)

Determine the total mass contained in the cylinder.

$m_t=\frac{P_1{\nu }_1}{R{T}_1}$ (XV)

Determine the engine speed for a net power output of 50 kW.

$\dot{n}=2\times \frac{{\dot{W}}_{\text{net}}}{m_t{w}_{\text{net}}}$ (XVI)

Note: the two revolutions in one cycle in four-stroke engines.

Conclusion:

Substitute $1.6\text{L}$ for ${\nu }_d$ and 11 for $r$ in Equation (XIII).

$\begin{array}{l}11=\frac{{\nu }_c+1.6\text{L}}{{\nu }_c}\\ 11=\frac{{\nu }_c+1.6\text{L}\times \left(\frac{1{\text{m}}^{\text{3}}}{1000\text{L}}\right)}{{\nu }_c}\\ 11=\frac{{\nu }_c+0.0016{\text{m}}^{\text{3}}}{{\nu }_c}\\ {\nu }_c=0.00016{\text{m}}^{\text{3}}\end{array}$

Substitute $0.00016{\text{m}}^{\text{3}}$ for ${\nu }_c$ and $1.6\text{L}$ for ${\nu }_d$ in Equation (XIV).

$\begin{array}{c}{\nu }_1=\left(0.00016{\text{m}}^{\text{3}}\right)\times \left(1.6\text{L}\right)\\ =\left(0.00016{\text{m}}^{\text{3}}\right)\times \left(1.6\text{L}\times \left(\frac{1{\text{m}}^{\text{3}}}{1000\text{L}}\right)\right)\\ =0.00176{\text{m}}^{\text{3}}\end{array}$

Substitute $100\text{kPa}$ for $P_1$, $0.00176{\text{m}}^{\text{3}}$ for ${\nu }_1$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, $37°\text{C}$ for $T_1$ in Equation (XV).

$\begin{array}{c}{m}_t=\frac{\left(100\text{kPa}\right)\left(0.00176{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(37°\text{C}\right)}\\ =\frac{\left(100\text{kPa}\right)\left(0.00176{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)\left(37°\text{C}+\text{273}\right)}\\ =\frac{\left(100\text{kPa}\right)\left(0.00176{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(310\text{K}\right)}\\ =0.001978\text{kg}\end{array}$

Substitute $50\text{kW}$ for ${\dot{W}}_{\text{net}}$, $0.001978\text{kg}$ for $m_t$, $749\text{kJ}/\text{kg}$ in Equation (XVI).

$\begin{array}{c}\dot{n}=2\left(\text{rev}/\text{cycle}\right)\times \frac{\left(50\text{kW}\right)}{\left(0.001978\text{kg}\right)\left(749\text{kJ}/\text{kg}\cdot \text{cycle}\right)}\\ =2\left(\text{rev}/\text{cycle}\right)\times \frac{\left(50\text{kW}\times \left(\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right)\right)}{\left(0.001978\text{kg}\right)\left(749\text{kJ}/\text{kg}\cdot \text{cycle}\right)}\\ =2\left(\text{rev}/\text{cycle}\right)\times \frac{50\text{kJ}/\text{s}}{\left(0.001978\text{kg}\right)\left(749\text{kJ}/\text{kg}\cdot \text{cycle}\right)}\times \left(\frac{60\text{s}}{1\min }\right)\\ =3820\text{rev}/\text{min}\end{array}$

Thus, the engine speed for a given net power is $\underset{\_}{3820\text{rev}/\text{min}}$.

(e)

To determine

The specific fuel consumption.

Answer to Problem 167RP

The specific fuel consumption is $\underset{\_}{267\text{g}/\text{kWh}}$.

Explanation of Solution

Determine the mass of fuel burned during one cycle.

$\begin{array}{c}\text{AF}=\frac{m_a}{m_f}\\ =\frac{m_t-m_f}{m_f}\end{array}$ (XVII)

Here, the air-fuel ratio is $\text{AF}$.

Determine the specific fuel consumption.

$\text{sfc}=\frac{m_f}{m_t{w}_{\text{net}}}$ (XVIII)

Conclusion:

Substitute 16 for AF and $0.001978\text{kg}$ for $m_t$ in Equation (XVII).

$\begin{array}{l}16=\frac{\left(0.001978\text{kg}\right)-m_f}{m_f}\\ m_f=0.0001164\text{kg}\end{array}$

Substitute $0.0001164\text{kg}$ for $m_f$, $0.001978\text{kg}$ for $m_t$, and $794\text{kJ}/\text{kg}$ for $w_{\text{net}}$ in Equation (XVIII).

$\begin{array}{c}\text{sfc}=\frac{0.0001164\text{kg}}{\left(0.001978\text{kg}\right)\left(794\text{kJ}/\text{kg}\right)}\\ =\frac{0.0001164\text{kg}}{\left(1.570532\text{kJ}\right)}\times \left(\frac{1000\text{g}}{1\text{kg}}\right)\left(\frac{3600\text{kJ}}{1\text{kWh}}\right)\\ =2668\text{g}/\text{kWh}\\ \cong \text{267}\text{g}/\text{kWh}\end{array}$

Thus, the specific fuel consumption is $\underset{\_}{267\text{g}/\text{kWh}}$.