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Q12.4 Outdoor air of 5 C, e = 0.8 and p = 77.3 cm Hg is heated isobarically to 20 C. What is
Q12.4 Outdoor air of 5 ° C, e = 0.8 and p = 77.3 cm Hg is heated isobarically to 20 ° C.
What is the humidity level then?
How much water must evaporate per m³ of cold air to bring e back to 0.8?
Maximum pressure, density for water and saturated water vapor
T | pmax | ρL | ρverz |
(°C) | (bar) | (kg/m3) | (kg/m3) |
0 | 0,00609 | 999,8 | 0,00485 |
5 | 0,00866 | 1000 | 0,00680 |
10 | 0,01226 | 999,6 | 0,00940 |
15 | 0,0169 | 999 | 0,0128 |
20 | 0,0233 | 998 | 0,0173 |
25 | 0,0317 | 997 | 0,0230 |
30 | 0,0424 | 996 | 0,0304 |
35 | 0,0561 | 994 | 0,0396 |
40 | 0,0737 | 992 | 0,0511 |
45 | 0,0957 | 990 | 0,0654 |
50 | 0,1232 | 988 | 0,0830 |
55 | 0,157 | 986 | 0,104 |
60 | 0,200 | 983 | 0,130 |
65 | 0,250 | 980 | 0,161 |
70 | 0,312 | 978 | 0,198 |
75 | 0,385 | 975 | 0,242 |
80 | 0,473 | 972 | 0,293 |
85 | 0,577 | 969 | 0,353 |
90 | 0,700 | 965 | 0,424 |
95 | 0,844 | 962 | 0,505 |
100 | 1,013 | 958 | 0,595 |
105 | 1,207 | 955 | 0,705 |
110 | 1,432 | 951 | 0,827 |
115 | 1,688 | 947 | 0,965 |
120 | 1,984 | 943 | 1,122 |
125 | 2,319 | 939 | 1,30 |
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From ideal gas equation we can write nRTPV P773 cm Hg1017atm Humiditydensity of dry airdensity ...Get Instant Access to Expert-Tailored Solutions
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