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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.

Respuesta :

Hashirriaz830

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:

[tex]u_{1}[/tex] = 214.07 kJ/kg  

[tex]\alpha[/tex][tex]r_{1}[/tex] = 621.2  

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

[tex]\alpha[/tex][tex]r_{2}[/tex] = [tex]\frac{\alpha r_{1} }{r}[/tex]

     =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):  

[tex]T_{2}[/tex] = 673 K

[tex]u_{2}[/tex] = 491.2 kJ/kg  

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

[tex]P_{2}[/tex] =  [tex]P_{1} r\frac{T_{2} }{T_{1} }[/tex]

       = 95*8*673/300

      = 1705 kPa  

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

[tex]deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}[/tex]

     = 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):  

[tex]T_{3}[/tex] = 1539 K  

[tex]\alpha r_{3}[/tex] = 6.588  

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

[tex]P_{3} =P_{2} \frac{T_{3} }{T_{2} }[/tex]

     = 3899 kPa  

Part B

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

[tex]\alpha r_{4}= r\alpha r_{3}[/tex]

      = 52.7

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

[tex]T_{4}[/tex]=775 K

[tex]u_{4}[/tex]= 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:  

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

Part C  

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

η=[tex]\frac{w}{q_{in} }[/tex]

=0.523

Part D  

The mean effective pressure is determined from its standard relation:  

MEP=[tex]\frac{w}{\alpha_{1}- \alpha_{2} }[/tex]

      =[tex]\frac{w}{\alpha_{1}(1- \frac{1}{r} }[/tex]

      =[tex]\frac{rwP_{1} }{RT_{1} (r-1) }[/tex]

      =495 kPa

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