Modern Engineering Thermodynamics

106 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms
type of work or heat transport present. However, you must always remember that W, _W, Q, and _Q are also net
terms and represent a summation of all the different types of work and heat transports of energy present. This is
illustrated in the following example.
EXAMPLE 4.2
Determine the energy transport rate for the system shown in Figure 4.4.
Fuel flow
E fuel = 15,000 Btu/min
Top heat loss
180,000 Btu/h
Exhaust flow
E exhaust = 500. Btu/min
System
boundary
Engine−generator set
200. hp
50.0 hp
Electrical workout
Bottom heat loss
54,000 Btu/h
FIGURE 4.4
Example 4.2
Solution
From Eq. (4.14), the total energy transport rate is
_E T = _Q − _W +∑ _Emass
flow
where
and
_Q = net heat transfer into the system
= − 180: × 10 3 Btu/h − 54:0 × 10 3 Btu/h = − 234 × 10 3 Btu/h
_W = net work rate out of the system = 200: hp + 50:0hp= 250: hp
while
So
∑ _Emass = net mass flow of energy into the system
flow
= 15:0 × 10 3 Btu/min − 500:Btu/min = 14:5 × 10 3 Btu/min
_E T = ð‒234 × 103 Btu/hÞ½1 h/ ð60 minÞŠ− ð250: hpÞ½42:4 Btu/ ðhp.minÞŠ+ 14:5 × 103 Btu/min = 0:00 Btu/min
Exercises
4. Determine the energy transport rate that occurs in Example 4.2 when the work mode directions are reversed.
Answer: _E T = 21:2 × 10 3 Btu/min:
5. Determine the net rate of energy gain of a closed system that receives heat at a rate of 4500. kJ/s and produces work at a
rate of 1500. kJ/s. Answer: _E G = 3000: KJ/s:
6. An insulated open system has a net gain of 700. Btu of energy while producing 500. Btu of work. Determine the mass
flow energy transport. Answer: E mass flow = 1.20 × 10 3 Btu.
The system of Example 4.2 has no net energy transport rate, even though it has six energy transport rates. Note
that the energy rate balance (Eq. (4.2)) for this system is _E G = _E T ; therefore, this system also has no net gain of
energy. That is, the total energy E of this system is constant in time.