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
