Answer:
(a) W/m = 49.334 Btu/lb
(b) [tex]\frac{E_{d} }{m}[/tex] = 22.12 Btu/lb
Explanation:
For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.
(a) Using the thermodynamic table for air.
At the temperature ([tex]T_{1}[/tex])of 800 ºR and pressure ([tex]P_{1}[/tex]) of 75 Ibf/in.^2, we can deduce that:
Specific enthalpy ([tex]h_{1}[/tex]) = 191.81 BTu/lb
Specific entropy ([tex]s_{1}[/tex]) = 0.6956 Btu/(lb.ºR)
At the temperature ([tex]T_{2}[/tex])of 600 ºR and pressure ([tex]P_{2}[/tex]) of 15 Ibf/in.^2, we can deduce that:
Specific enthalpy ([tex]h_{2}[/tex]) = 143.47 BTu/lb
Specific entropy ([tex]s_{2}[/tex]) = 0.6261 Btu/(lb.ºR)
The work done can be calculated using energy rate equation:
[tex]\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}[/tex]
Q/m = heat transfer = -2 Btu/lb
[tex]V_{1}[/tex] = 400 ft/s
[tex]V_{2}[/tex] = 100 ft/s
[tex]\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}[/tex][/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb
(b) To calculate the exergy destruction, we will use the equation for exergy rate:
[tex]\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}][/tex]
The equation above is further simplified to:
[tex]\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }][/tex]
Using a reference temperature (To) = 500 °R
Average surface temperature (Tb = 620°R
[tex]\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}][/tex]
[tex]\frac{E_{d} }{m}[/tex] = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb
