Question

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 750 kJ/kg of heat is transferred to air during the constant-volume heat-addition process. Taking into account the variation of specific heats with temperature, determine (a) the pressure and temperature at the end of the heat-addition process, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle.

Answer 1

Answer:

Part A) 3899 kPa  

Part B) 392.33 kJ/kg  

Part C) 0.523

Part D) 495 kPa

Explanation:

Part A

First from the temperature at state 1 the relative specific volume and the internal energy at that state are determined from:

$u_{1}$ = 214.07 kJ/kg  

$\alpha$$r_{1}$ = 621.2  

The relative specific volume at state 2 is obtained from the compression ratio:  

$\alpha$$r_{2}$ = $\frac{\alpha r_{1} }{r}$

     =621.2/ 8

    = 77.65  

From this the temperature and internal energy at state 2 can be determined using interpolation with data from A-17(table):  

$T_{2}$ = 673 K

$u_{2}$ = 491.2 kJ/kg  

The pressure at state 2 can be determined by manipulating the ideal gas relations at state 1 and 2:  

$P_{2}$ =  $P_{1} r\frac{T_{2} }{T_{1} }$

       = 95*8*673/300

      = 1705 kPa  

Now from the energy balance for stage 2-3 the internal energy at state 3 can be obtained:  

$deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}$

     = 1241.2 kJ/kg

From this the temperature and relative specific volume at state 3 can be determined by interpolation with data from A-17(table):  

$T_{3}$ = 1539 K  

$\alpha r_{3}$ = 6.588  

The pressure at state 3 can be obtained by manipulating the ideal gas relations for state 2 and 3:  

$P_{3} =P_{2} \frac{T_{3} }{T_{2} }$

     = 3899 kPa  

Part B

The relative specific volume at state 4 is obtained from the compression ratio:  

$\alpha r_{4}= r\alpha r_{3}$

      = 52.7

From this the temperature and internal energy at state 4 can be determined by interpolation with data from A-17:  

$T_{4}$=775 K

$u_{4}$= 571.74 kJ/kg  

The net work output is the difference of the heat input and heat rejection where the heat rejection is determined from the decrease in internal energy in stage 4-1:  

$w=q_{in}-q_{out}\\q_{in}-(u_{4} -u_{1} )\\=392.33 kJ/kg$

Part C  

The thermal efficiency is obtained from the work and the heat input:  

η=$\frac{w}{q_{in} }$

=0.523

Part D  

The mean effective pressure is determined from its standard relation:  

MEP=$\frac{w}{\alpha_{1}- \alpha_{2} }$

      =$\frac{w}{\alpha_{1}(1- \frac{1}{r} }$

      =$\frac{rwP_{1} }{RT_{1} (r-1) }$

      =495 kPa