The air in a room has a dry-bulb temperature of 26°C and a wet-bulb temperature of 21°C. Assuming a pressure of 100 kPa, determine (a) the specific humidity, (b) the relative humidity, and (c) the dew-point temperature.
(a)
The specific humidity.
Answer to Problem 29P
The specific humidity is 0.0138 kg H2O/kg dry air.
Explanation of Solution
Express the specific humidity.
ω1=cp(T2−T1)+ω2hfg2hg1−hf2=cp(T2−T1)+ω2hfg@21°Chg@26°C−hf@21°C (I)
Here, specific heat at constant pressure of air is cp, initial and final temperature is T1 and T2 respectively, specific humidity at exit is ω2, initial specific enthalpy at saturated vapor is hg1, final specific enthalpy at saturated liquid is hf2 and final specific enthalpy at evaporation is hfg2.
Express specific humidity at exit.
ω2=0.622Pg2P2−Pg2=0.622Psat@21°CP2−Psat@21°C (II)
Here, final pressure is P2, final vapor pressure is Pg2 and saturation pressure at temperature of 21°C is Psat@21°C.
Conclusion:
Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure at temperature of 21°C using an interpolation method.
Write the formula of interpolation method of two variables.
y2=(x2−x1)(y3−y1)(x3−x1)+y1 (III)
Here, the variables denote by x and y is temperature and saturation pressure respectively.
Show thesaturation pressure corresponding to temperature as in Table (1).
Temperature T(°C) |
Saturation pressure Psat(kPa) |
20 (x1) | 2.3392 (y1) |
21 (x2) | (y2=?) |
25 (x3) | 3.1698 (y3) |
Substitute 20°C,21°C, and 25°C for x1,x2 and x3 respectively, and 2.3392 kPa for y1 and 3.1698 kPa for y3 in Equation (III).
y2=(21°C−20°C)(3.1698 kPa−2.3392 kPa)(25°C−20°C)+2.3392 kPa=2.488 kPa=Psat@21°C
Thus, the saturation pressure at temperature of 21°C is,
Psat@21°C=2.488 kPa
Substitute 2.488 kPa for Psat@21°C and 100 kPa for P2 in Equation (II).
ω2=0.622(2.488 kPa)100 kPa−2.488 kPa=0.01587 kg H2O/kg dry air
Refer Table A-4, “saturated water-temperature table”, and write final specific enthalpy evaporation at temperature of 21°C using an interpolation method.
Show thefinal specific enthalpy evaporationcorresponding to temperature as in Table(2).
Temperature T(°C) |
Final specific enthalpy evaporation hfg2(kJ/kg) |
20 (x1) | 2453.5 (y1) |
21 (x2) | (y2=?) |
25 (x3) | 2441.7 (y3) |
Use excels and tabulates the values from Table (2) in Equation (III) to get,
hfg2=2451.2 kJ/kg
Refer Table A-4, “saturated water-temperature table”, and write initial specific enthalpy saturated vapor at temperature of 26°C using an interpolation method.
Show theinitial specific enthalpy saturated vapor corresponding to temperature as in Table (3).
Temperature T(°C) |
Initial specific enthalpy saturated vapor hg1(kJ/kg) |
25(x1) | 2546.5(y1) |
26(x2) | (y2=?) |
30(x3) | 2555.6(y3) |
Use excels and tabulates the values from Table (3) in Equation (III) to get,
hg1=2548.3 kJ/kg
Refer Table A-4, “saturated water-temperature table”, and write final specific enthalpy saturated liquid at temperature of 21°C using an interpolation method.
Show thefinal specific enthalpy evaporationcorresponding to temperature as in Table(4).
Temperature T(°C) |
Final specific enthalpy saturated liquid hf2(kJ/kg) |
20 (x1) | 83.915 (y1) |
21 (x2) | (y2=?) |
25 (x3) | 104.83 (y3) |
Use excels and tabulates the values from Table (4) in Equation (III) to get,
hf2=88.10 kJ/kg
Refer Table A-2 (a), “ideal gas specific heats of various common gases”, and write specific heat at constant pressure of dry air.
cp=1.005 kJ/kg°C
Substitute 1.005 kJ/kg°C for cp, 21°C for T2 and 26°C for T1, 0.01587 for ω2, 2451.2 kJ/kg for hfg2, 2548.3 kJ/kg for hg1 and 88.10 kJ/kg for hf2 in Equation (I).
ω1=(1.005 kJ/kg°C)(21−26)°C+(0.01587)(2451.2 kJ/kg)(2548.3−88.10) kJ/kg=0.0138 kg H2O/kg dry air
Hence, the specific humidity is 0.0138 kg H2O/kg dry air.
(b)
The relative humidity.
Answer to Problem 29P
The relative humidity is 64.4%.
Explanation of Solution
Express saturation pressure of water at temperature of 30°C.
ϕ1=ω1P1(0.622+ω1)Pg1=ω1P1(0.622+ω1)Psat@26°C (IV)
Here, initial pressure is P1 and saturation pressure at temperature of 26°C is Psat@26°C.
Conclusion:
Refer Table A-4, “saturated water-temperature table”, and write the saturation pressure at temperature of 26°C using an interpolation method.
Show thesaturation pressure corresponding to temperature as in Table (5).
Temperature T(°C) |
Saturation pressure Psat(kPa) |
25 (x1) | 3.1698 (y1) |
26 (x2) | (y2=?) |
30 (x3) | 4.2469 (y3) |
Use excels and tabulates the values from Table (5) in Equation (III) to get,
Psat@26°C=3.3638 kPa
Substitute 0.01377 for ω1, 100 kPa for P1 and 3.3638 kPa for Psat@26°C in Equation (IV).
ϕ1=(0.01377)(100 kPa)(0.622+0.01377)(3.3638 kPa)=0.644=64.4%
Hence, the relative humidity is 64.4%.
(c)
The dew point temperature.
Answer to Problem 29P
The dew point temperature is 18.8°C.
Explanation of Solution
Express initial partial pressure of water vapor.
Pν1=ϕ1Pg1=ϕ1Psat@26°C (V)
Express the dew point temperature
Tdp=Tsat@Pν (VI)
Conclusion:
Substitute 0.644 for ϕ1 and 3.3638 kPa for Psat@26°C in Equation (V).
Pν1=(0.644)(3.3638 kPa)=2.166 kPa
Substitute 2.166 kPa for Pν1 in Equation (VI).
Tdp=Tsat@2.166 kPa (VII)
Here, saturation pressure at pressure of 2.166 kPa is Tsat@2.166 kPa.
Refer Table A-4, “saturated water-temperature table”, and write temperature at saturation pressure of 2.166 kPa using an interpolation method.
Show thetemperature corresponding to saturation pressure as in Table (6).
Saturation pressure Psat(kPa) |
Temperature T(°C) |
1.7057 (x1) | 15 (y1) |
2.166 (x2) | (y2=?) |
2.3392 (x3) | 20 (y3) |
Use excels and tabulates the values from Table (6) in Equation (III) to get,
Tsat@2.166 kPa=18.8°C
Substitute 18.8°C for Tsat@2.166 kPa in Equation (VII).
Tdp=18.8°C
Hence, the dew point temperature is 18.8°C.
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Chapter 14 Solutions
Thermodynamics: An Engineering Approach
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