Modern Engineering Thermodynamics

9.4 Open System Entropy Balance Equations 283
EXAMPLE 9.1
You are now an attorney in a patent office. An inventor wants to patent a new domestic hot water heater. The inventor
claims to have a secret process that heats liquid water from 15.0ºC to 50.0ºC using only 100. W of electrical energy. Evaluate
the inventor’s claim and decide whether or not to issue a patent.
Solution
First, draw a sketch of the system (Figure 9.1).
Since the water heater is a steady state open system with an isothermal boundary,
we can provide a preliminary evaluation of the inventor’s claim with Eq. (9.10).
If this equation produces a positive entropy production rate, then the claim is at
least possible. If it produces a negative entropy production rate, then we can say
without equivocation that the claim is impossible to achieve.
Assumptions: (1) all the input electrical energy is converted into heating the water,
(2) the system boundary temperature is the same as the ambient temperature, or 20ºC.
Since the inventor does not provide us with the water flow rate, we can calculate it from the modified energy rate balance as
_m = _Q /ðh 2 − h 1 Þ, where for an incompressible liquid, h 2 – h 1 = c(T 2 – T 1 )+v(p 2 – p 1 ). Neglecting the pressure drop across
the water heater, we can now compute the expected water flow rate as
_m =
_Q
0:100 kJ/s
=
= 0:000686 kg/s
h 2 − h 1 ð4:186 kJ/kg.KÞ½50 + 273:15 − ð15 + 273:15Þ KŠ
Not a very fast water flow rate. For liquid water, s out – s in = c ln(T out /T in ), so
50:0 + 273:15
s out − s in = c lnðT out /T in Þ = ð4:186 kJ=kg.KÞ ln = 0:480 kJ/kg.K
15:0 + 273:15
Then, Eq. (9.10) gives
15°C Domestic
water
heater
Room temp. = 20°C
Q
_S P = _m ðs out − s in Þ− _ 0:100 kJ/s
= ð0:000686 kg/sÞð0:480 kJ/kg.KÞ −
T b
20:0 + 273:15 K = −1:36 × 10−5 kJ/s.K
Since the entropy production rate is negative, this water heater cannot possibly meet the claims of the inventor, so we
should reject the patent application.
S P
0?
FIGURE 9.1
Example 9.1.
50°C
Q = 100. W
HOW CAN YOU HEAT THE WATER IN EXAMPLE 9.1 IF THE SYSTEM
BOUNDARY TEMPERATURE IS ONLY 20ºC?
Doesn’t the boundary have to be higher than the fluid temperature for heat to go into the water? The boundary temperature in
Eq. (9.10) can be the “average” boundary temperature, but if 20ºC is the “average” boundary temperature, then 20ºC =(T max +
15ºC)/2, and the maximum boundary temperature is T max = 2(20) – 15=25ºC, and that is not hot enough to heat the water to 50ºC.
This is what happens when you make incorrect assumptions. The average boundary temperature cannot be 20ºC, it has to
be higher. Suppose the inventor now tells us that it is 150ºC. Then, we get
_S P = ð0:000686 kg/sÞð0:480 kJ/kg.KÞ −
0:100 kJ/s
150:0 + 273:15 K = 9:12 × 10− 5 kJ/s.K
The patent application is now alright, since the entropy production rate is positive. But the water flow rate here is only 6.86 ×
10 –4 kg/s = 0.686 grams per second, or 42.2 grams per minute, or 2.47 kg per hour. Not a very effective water heater.
Exercises
1. Suppose the inventor in Example 9.1 corrected his patent claim and said the heater was 1000. W instead of 100. W and
the heat transfer boundary temperature was 150ºC instead of 20ºC. What would be the water flow rate end entropy
production rate under these conditions? Answer: _m = 6:83 grams per second and _S P = 9:12 × 10 −4 kJ/s.K.
2. OK, so now the inventor says the heater is 10.0 kW, the heater boundary temperature is 150ºC and the water flow rate
is 0.500 kg/s. What are the water outlet temperature and the unit’s entropy production rate? Answer: T out = 19.8ºC and
_S P = 0:0108 kJ/s.K.
3. Alright, now the inventor hires an engineer to determine the heat transfer rate and entropy production rate needed to
heat 0.500 kg/s from 15ºC to50ºC. You are the engineer, so what are the answers? Hint: Is _Q (a) 20.9 kW, (b) 73.3 kW,
or (c) 103. kW? Is _S P (a) 0.0220 kJ/s · K, (b) 0.137 kJ/s · K, or (c) 0.0669 kJ/s · K?