Modern Engineering Thermodynamics

15.5 Standard Reference State 603
If moisture enters the combustion process as humidity in the inlet air, this moisture is carried through the
reaction as an inert element and adds to the combustion water in the products. This has the net effect of raising
the dew point temperature. This is illustrated in the next example.
EXAMPLE 15.5
During the automobile engine fuel combustion test discussed in Example 15.3, the dry bulb and wet bulb temperatures of
the inlet air were measured to be 90.0°F and 75.0°F, respectively. Determine (a) the amount of water carried into the engine
in the form of inlet humidity and (b) the new dew point temperature of the exhaust products. Assume the exhaust is at a
total pressure of 14.7 psia.
Solution
From the psychrometric chart, Figure D.6a of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find
that, for T DB = 90°F andT WB = 76°F, the relative humidity ϕ = 50% and the humidity ratio, ω = (105 grains of H 2 O per lbm of
dry air) × (1 lbm/7000. grains) = 0.0150 lbm H 2 O/lbm dry air. On a molar basis, the humidity ratio is
ω = ð0:0150 lbm H 2 O/lbm dry airÞ = 0:0241 lbmole H 2 O/lbmole dry air
28:97 lbm dry air/lbmole dry air
18:016 lbm H 2 O/lbmole H 2 O
From the balanced reaction equation of part a of Example 15.3, we find that the amount of dry air used per mole of fuel is
21.9(1 + 3.76) = 104 moles, and this now carries with it 0.0241(104) = 2.51 moles of water. Assuming this water passes
through the reaction unchanged, the total amount of water now in the exhaust is 9:00 + 2:51 = 11:5 moles per mole of fuel.
Consequently, the total moles of product are 111.5, the mole fraction of water vapor in the exhaust is now
χ H2O = n H2O
= 11:5/111:5 = 0:103
n total
and Eq. (12.23) gives the partial pressure of the water vapor in the exhaust as
p H2O = 0:103ð14:7Þ = 1:52 psia
Again, interpolating in Table C.1a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find
T sat ð1:52 psiaÞ = T DP = 116°F = 46:5°C
Exercises
13. What happens to the water vapor in the engine’s exhaust in Example 15.5 if the surrounding air temperature is 20.0°C?
Answer: It condenses into liquid water, since the surrounding air temperature is less than the dew point temperature of
the water vapor.
14. If the inlet air in Example 15.5 contains 3.214 moles of water vapor per mole of fuel burned (instead of 2.50 moles of
H 2 O per mole of fuel), determine the new dew point temperature. Answer: T DP = 118°F.
15. If the inlet air in Example 15.5 has a relative humidity of 100.% and a dry bulb temperature of 90.0°F, what is the new
exhaust dew point temperature? Answer: T DP = 123°F.
By comparing the results of Examples 15.4 and 15.5, we see that combustion air with 50.0% relative humidity
has a dew point temperature 7.6°F (4.2°C) higher than that of dry combustion air.
15.5 STANDARD REFERENCE STATE
Because we deal with a variety of elements and compounds in combustion reactions, it is necessary to define
a common thermodynamic reference state for all these substances. Recall that, in developing the steam tables,
we chose the triple point of water as the reference state and arbitrarily set the specific internal energy of
liquid water equal to zero at that point. Therefore, the values of u and h in the steam tables are not the actual
specific internal energies and enthalpies of steam, they are only relative values. This is sufficient, since most of
our formulae use u 2 − u 1 or h 2 − h 1 for changes occurring within a system and the effect of the reference
state cancels out in the subtraction process. In the case of the gas tables, we take 0 K and 1 atm as the
thermodynamic reference state and arbitrarily set the specific internal energy equal to zero at this state. However,
in the case of combustion processes, a more pragmatic thermodynamic reference state of 25.0°C and
0.100 MPa (approximately 1 atm) is chosen. This is called the standard reference state (SRS) for combustion
reactions. However, since most of the calorimeters used to study combustion processes are steady state, steady
flow, open systems, it is more convenient to set the specific enthalpy rather than the specific internal energy of
the elements equal to zero at this state.