Chapter 8 - Entropy Accounting
	Updated --- 8 November 2015
#	Hint	Answer	Update Date
8.1
8.2
8.3
8.4
8.5	About as basic as it gets.	0.0291 W/K
8.6	This is a long one!	a) intermediate torque: 261.25 ftlb, output torque: 310.23 ft-lb b) Box 1: 1.837 Btu/(hr R), Box 2: 1.745 Btu/(hr R), total = 3.583 Btu(hr R)
8.7
8.8	Standard COE and EA question here, nothing tricky. If you do not remember what a reversible process is, look it up; it has something to do with entropy generation.	a) COP = 4 b) 6.36 Btu/(hr R)
8.9	Standard COE and EA.	a) 13.1 HP, 66,667 Btu/hr, 51.9 Btu/(hr R) b) 12.67, 7895 Btu/hr
8.10	For part b) and c), use EA rate form and integrate over the cycle to get the finite form. Since this is a cycle, the entropy at the initial state and final state must be the same; what does that imply will happen to the left-hand side of the equation?	a) 500 kJ, 1 kJ/K b) 1100 kJ, -1 kJ/K, 0.55 c) 800 kJ, 0.4
8.11
8.12
8.13
8.14
8.15	Plenty of hints in the problem statement.	a) s2-s1=0, 30kW b) 0.07 kJ/K, 50.9 kW
8.16	Prove to yourself the nozzle operates isentropically; use this result along with a substance model to find a relationship between pressure and temperature.	a) 475 ^oC, 686 m/s b) 0.2 kW/K, 606 m/s
8.17	For Part b) first consider the resistor alone as the system: how does the electrical work relate to the heat transfer out of the resistor? Then consider the water alone as the system: the heat transfer into this system is solely from the resistor across the boundary with an average temperature of 97^oC.	a) 17,500 kJ b) 8.88 kJ/K c) 56.2 kJ/K
8.18	How can you compute the air density if you know the inlet temperature and pressure? You can use EA to solve for the outlet temperature if you use the hint provided in the problem statement.	279 kW
8.19
8.20
8.21
8.22
8.23