Respuesta :
Answer:
Net work, W = 842.52 kJ/kg
Power, P = 32.82 kW
Given:
compression ratio, r = 10:1
Total volume displaced, [tex]V_{d} = 2.3 l = 2.3\times 10^{- 3} m^{3}[/tex]
[tex]T_{inlet}[/tex] = 280 K
[tex]P_{inlet}[/tex] = 70 kPa
Speed, N = 2100 rpm = 35 rps
Heat energy added, Q = 1400 kJ/kg
Solution:
Now, calculation of the overall efficiency:
[tex]\eta = 1- r^(1 - k)[/tex]
where
k = 1.4
[tex]\eta = 1- 10^(0.4) = 0.6018[/tex]
Also,
Net work done, W = [tex]\eta Q = 0.6018\times 1400 = 842.52 kJ/kg[/tex]
Now, specific volume is given by:
[tex]P_{inlet}\vartheta = RT_{inlet}[/tex]
[tex]70\times 10^{3}\vartheta = 0.287\times 280[/tex]
[tex]\vartheta = 1.148 m^{3}/kg[/tex]
Now, mean effective pressure, P' can be calculated as:
[tex]W = P'(\vartheta - \vartheta')[/tex]
[tex]\frac{W}{(\vartheta - \vartheta')} = P'[/tex]
[tex]\frac{842.52}{\vartheta(1 - \frac{\vartheta'}{\vartheta})} = P'[/tex]
(Since, r = [tex]\frac{\vartheta}{\vartheta'}[/tex])
[tex]\frac{842.52}{\vartheta(1 - \frac{1}{r})} = P' = \frac{842.52}{1.148(1 - \frac{1}{10})}[/tex]
P' = 815.45 kPa
Now, for power produced, P:
P = [tex]\frac{1}{2}P'V_{d} N[/tex]
P = [tex]\frac{1}{2}\times 815.45\times 2.3\time 10^{- 3}\times 35[/tex]
P = 32.82 kW
