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MEM 310: Thermodynamic Analysis I
Fall 2010
Homework 5 Solutions
Wednesday, December 1, 2010
- The Clapeyron equation is: (^) ( dP dT
sat
sfg vfg
hfg T vfg
The derivative (dP/dT )sat can be computed using values from the water saturation table A-4: ( dP dT
sat, 140 ◦C
Psat, 145 ◦C − Psat, 135 ◦C 145 ◦C − 135 ◦C
415 .68 kPa − 313 .22 kPa 145 ◦C − 135 ◦C = 10.25 kPa/K.
From this table, we also have:
vfg, 140 ◦C = vg, 140 ◦C − vf, 140 ◦C = 0.50850 m^3 /kg − 0 .001080 m^3 /kg = 0.50742 m^3 /kg.
For our computations we will also need the temperature in Kelvins:
T = 140◦C + 273.15 K = 413.15 K.
Using these results gives:
hfg, 140 ◦C = T vfg, 140 ◦C
dP dT
sat, 140 ◦C
= (413.15 K)(0.50742 m^3 /kg)(10.25 kPa/K) = 2148.98 kJ/kg,
sfg, 140 ◦C = hfg,^140 ◦C T =^2148 .98 kJ/kg 413 .15 K = 5.1990 kJ/(kg · K).
These results match the tabulated values to three decimal places. The tabulated values are:
hfg, 140 ◦C = 2144.3 kJ/kg, sfg, 140 ◦C = 5.1901 kJ/(kg · K).
- The molar masses and gas constants for the two species are (from table A-1):
MN 2 = 28.013 kg/kmol, MO 2 = 31.999 kg/kmol, RN 2 = 0.2968 kJ/(kg · K), RO 2 = 0.2598 kJ/(kg · K). (a) The number of moles of each component is:
nN 2 = mN 2 MN 2 =^
0 .4 kg 28 .013 kg/kmol = 0.014279 kmol, nO 2 = mO 2 MO 2 =^
0 .6 kg 31 .999 kg/kmol = 0.018751 kmol. So the total number of moles of gas is:
n = nN 2 + nO 2 = 0.014279 kmol + 0.018751 kmol = 0.033030 kmol,
while the total mass of the gas is:
m = mN 2 + mO 2 = 0.4 kg + 0.6 kg = 1.0 kg. The average molar mass of this mixture is therefore:
M = m n = 1 .0 kg 0 .033030 kmol = 30.276 kg/kmol.
(b) The average gas constant can be computed from the average molar mass:
R =
Ru M
8 .31447 kJ/(kmol · K) 30 .276 kg/kmol = 0.27462 kJ/(kg · K).
(c) The partial pressure of each gas can be computed by ignoring the presence of the other gas and simply applying the ideal gas law:
PN 2 =
mN 2 RN 2 T V =
(0.4 kg)(0.2968 kJ/(kg · K))(300 K) 0 .5 m^3 = 71.232 kPa, PO 2 = mO 2 RO 2 T V
(0.6 kg)(0.2598 kJ/(kg · K))(300 K) 0 .5 m^3 = 93.528 kPa.
(d) Ideal gas case Under the ideal gas assumptions, the total pressure is simply the sum of the individual partial pressures of the gases: Pideal = PN 2 + PO 2 = 71.232 kPa + 93.528 kPa = 164.76 kPa. Kay’s rule case To do the Kay’s rule computation, we need the critical properties of the two gases:
Tcr,N 2 = 126.2 K, Tcr,O 2 = 154.8 K, Pcr,N 2 = 3.39 MPa, Pcr,O 2 = 5.08 MPa,
The averaged parameters are:
¯a =
yN 2 ¯a^1 N/ 22 + yO 2 ¯a^1 O/ 22
(0.43200)(0.13702 Pa · m^6 /mol^2 )^1 /^2 + (0.56770)(0.13757 Pa · m^6 /mol^2 )^1 /^2
= 0.13725 Pa · m^6 /mol^2 , ¯b = yN 2 ¯bN 2 + yO 2 ¯bO 2 = (0.43200)(3. 8690 · 10 −^5 m^3 /mol) + (0.56770)(3. 1670 · 10 −^5 m^3 /mol) = 3. 4693 · 10 −^5 m^3 /mol. Using these values, we can solve for the pressure in the van der Waals equation of state:
PvdW = RuT v ¯ − ¯b − ¯a ¯v^2 = (8.31447 J/(mol · K))(300 K) 0 .015138 m^3 /mol − 3. 4693 · 10 −^5 m^3 /mol
0 .13725 Pa · m^6 /mol^2 (0.015138 m^3 /mol)^2 = 165.14 kPa. This closely matches the other results, as expected for this comparatively low-density gas.
(e) Under the assumption of constant specific heats, and using the fact that the volume of the gas cannot change in its rigid container, we can see that the heat added in warming the gas from 300 K to 330 K must be: Q = mcv,300K∆T. The average value of the constant-volume specific heat may be computed by averaging the specific heats of the individual gases:
cv,300K = yN 2 cv,N 2 ,300K + yO 2 cv,O 2 ,300K = (0.43200)(0.743 kJ/(kg · K)) + (0.56770)(0.657 kJ/(kg · K)) = 0.694 kJ/(kg · K), where the cv values for the individual gases are derived from table A-2. Using this result, we have:
Q = mcv,300K∆T = (1.0 kg)(0.694 kJ/(kg · K))(330 K − 300 K) = 20.8 kJ.
The change in pressure is:
∆P =
nRu∆T V
(0.033030 kmol)(8.31447 kJ/(kmol · K))(330 K − 300 K) 0 .5 m^3 = 16.48 kPa.
- (a) The given pressure, dry bulb, and wet bulb temperatures are:
P = 110 kPa, Tdb = 20◦C, Twb = 15◦C.
Since we are operating near atmospheric pressure, the specific humidity may be determined from the equation: ωs = cp,air,db(Twb − Tdb) + ω′hfg,water,wb hg,water,db − hf,water,wb^ , where: ω′^ =
- 622 Psat,water,wb P − Psat,water,wb
The constant pressure specific heat of air at the dry-bulb temperature (here 293 K) may be obtained from table A-2: cp,air,293K ≈ cp,air,300K = 1.005 kJ/(kg · K). The saturation pressure and specific enthalpy values for saturated water at various temperatures may be obtained from table A-4:
Psat,water,wb = 1.2281 kPa +^1 .7057 kPa^ −^1 .2281 kPa 15 ◦C − 10 ◦C (14◦C − 10 ◦C) = 1.6102 kPa, hg,water,db = 2537.4 kJ/kg,
hfg,water,wb = 2477.2 kJ/kg + 2465 .4 kJ/kg − 2477 .2 kJ/kg 15 ◦C − 10 ◦C (14◦C − 10 ◦C) = 2467.8 kJ/kg,
hf,water,wb = 42.022 kJ/kg + 62 .982 kJ/kg − 42 .022 kJ/kg 15 ◦C − 10 ◦C (14◦C − 10 ◦C) = 58.790 kJ/kg.
Using these values gives:
ω′^ = 0 .622(1.6102 kPa) 110 kPa − 1 .6102 kPa = 0.0092402 (kg H^2 O)/(kg dry air), and inserting this into the equation for the specific humidity gives:
ωs = (1.005 kJ/(kg · K))(15◦C − 20 ◦C) + (0.0092402 (kg H 2 O)/(kg dry air))(2467.8 kJ/kg) 2537 .4 kJ/kg − 58 .790 kJ/kg = 0.011227 (kg H 2 O)/(kg dry air).
(b) To compute the relative humidity we also need the saturation pressure at the dry bulb temperature:
Psat,water,db = 2.3392 kPa. The relative humidity is:
φ = ωsP (0.622 + ωs)Psat,water,db = (0.011227 (kg H 2 O)/(kg dry air))(110 kPa) (0.622 + 0.011227 (kg H 2 O)/(kg dry air))(2.3392 kPa) = 0. 83376.
(c) The vapor pressure of water under the conditions described above is:
Pv = φPsat,water,db = (0.83376)(2.3392 kPa) = 1.9503 kPa.
