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If the at mospheric conditions are 20°C, 1.013 bar and specific humidity of 0.0095 kg/kg of dry air, the partial pressure of vapour will be nearly (a) 0.076 bar (b) 0.056 bar (c) 0.036 bar(d) 0.016 bar |
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Answer» Correct option is (d) 0.016 bar DBT = 20°C = 293 K Pt = 1.013 bar w = 0.622 x \(\cfrac{P_v}{p_t-p_v}\) 0.0095 = 0.622 x \(\cfrac{p_v}{1.013-p_v}\) 0.0152 = \(\cfrac{p_v}{1.013-p_v}\) 0.0153 – 0.0152Pv = Pv 0.0153 = 1.0152Pv Pv = 0.016 bar |
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