Question 13.16: A four-cylinder, 3.50 liter automobile engine operates on an...

For each cylinder, the displacement volume is \sout{V}_d = 3.50/4 = 0.875   \text{L} = 8.75 × 10^{–4}  \text{m}^3 . The clearance volume, \sout{V}_c, is calculated from the expansion ratio as \text{ER} = (\sout{V}_c + \sout{V}_d)/\sout{V}_c, or

\sout{V}_c = \sout{V}_d/(\text{ER} − 1) = (8.75 × 10^{−4})/(10.0−1) = 9.72 × 10^{−5} \text{m}^3

Then, from Figure 13.52b,

\sout{V}_c = \sout{V}_1 = \sout{V}_{7s} = \sout{V}_4 = 9.72 × 10^{-5}  \text{m}^3

Also

\sout{V}_{6s} = \sout{V}_{2s} = \sout{V}_3 = \sout{V}_d +\sout{V}_c = 8.75 × 10^{−4}  \text{m}^3 + 9.72 × 10^{−5}  \text{m}^3 = 9.72 × 10^{−4}  \text{m}^3

and

\sout{V}_5 = \sout{V}_{7s} × \text{CR} = (9.72 × 10^{−5}  \text{m}^3) × 8.00 = 7.78 × 10^{−4}  \text{m}^3

Now, we can compute the temperature and pressure at each point in the cycle by starting at state 5 in Figure 13.52b, the closing of the intake valve, where we know the pressure and temperature.

State 5 in Figure 13.52b

This is where the intake valve closes and the following information is given in the problem statement: p_5 = 200  \text{kPa} and T_5 = 40.0º\text{C} = 313  \text{K}.

State 6 in Figure 13.52b

The process from states 5 to 6 is isentropic, so

p_{6s} = p_5 (\frac{\sout{V}_5}{\sout{V}_{6s}} )^k = (200  \text{kPa}) (\frac{7.78 × 10^{−4}  \text{m}^3}{9.72 × 10^{−4}  \text{m}^3} ) = 148  \text{kPa}

and

T_{6s} = T_5 (\frac{\sout{V}_5}{\sout{V}_{6s}} )^{k-1} = (313 \text{K}) (\frac{7.78 × 10^{−4}  \text{m}^3}{9.72 × 10^{−4}  \text{m}^3} ) = 648  \text{K}

State 7s in Figure 13.52b

The process from states 5 to 7s is also isentropic, so

p_{7s} = p_5(\sout{V}_5/\sout{V}_7s)^k = p_5 (\text{CR})^k = (200.  \text{kPa})(8.00)^{1.35} = 3310  \text{kPa}

and

T_{7s} = T_5(\sout{V}_5/\sout{V}_7s)^{k-1} = T_5 (\text{CR})^{k-1} = (313 \text{K})(8.00)^{0.35} = 648  \text{K}

State 1 in Figure 13.52b

The mass of air in the cylinder is

m_\text{air} = \frac{p_{6s} \sout{V}_{6s}}{RT_{6s}} = \frac{(148  \text{kPa})(9.72 × 10^{4}  \text{m}^3}{(0.287  \text{kJ/kg.K})(648  \text{K})} = 1.73 × 10^{−3}  \text{kg}

Since we have an air–fuel ratio of 15.0, the mass of fuel in the cylinder is m_\text{fuel} = \frac{m_\text{air}}{\text{AF} +1} = \frac{1.73 × 10^{−3}  \text{kg}}{15.0 + 1 } = 1.08 × 10^{−4}  \text{kg}

The heat produced by the combustion process is then

Q_\text{comb}=m_\text{fuel} × \text{Fuel heating value} = (1.08 × 10^{−4}  \text{kg})(43300  \text{ kJ/kg-fuel}) = 4.69  \text{kJ}.

Also, Q_\text{comb} = m_\text{air}c_\text{v-air}(T_1 – T_{7s}) = (1.73 × 10^{–3}  \text{kg})(T_1 – 648  \text{K}) = 4.69  \text{kJ}. Solving for T_1 gives T_1 = 3940  \text{K}. Then, since the process from state 7s to 1 is a constant volume process, p_1 = p_{7s}(T_1/T_{7s}) = (3310  \text{kPa})(3940  \text{K}/624  \text{K}) = 20.2  \text{MPa} .

State 2s in Figure 13.52b

The process from 1 to 2s is isentropic, so

p_{2s} = p_1 (\frac{\sout{V}_1}{\sout{V}_{2s}} )^k = (20.2 × 10^{3}\text{kPa}) (\frac{9.72 × 10^{−5}  \text{m}^3}{9.72 × 10^{−4}  \text{m}^3} )^{1.35} = 901  \text{kPa}

and

T_{2s} = T_1 (\frac{\sout{V}_1}{\sout{V}_{2s}} )^{k-1} = (3920  \text{K}) (\frac{9.72 × 10^{−5}  \text{m}^3}{9.72 × 10^{−4}  \text{m}^3} ) = 1760  \text{K}

State 3 in Figure 13.52b

p_3 = p_\text{exhaust} = 101  \text{kPa}, and since the process from 2s to 3 is a constant volume process, we have T_3 = T_{2s}(p_3/p_{2s}) = (1760  \text{K}) (101  \text{kPa}/901  \text{kPa}) = 196  \text{K}.

State 4 in Figure 13.52b

p_4 = p_3 = 101  \text{kPa} and T_4 = \text{atmospheric temperature}.