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calculate: 1) partial pressure of water vapor and dry air; 2) Dew point temp; 3) specific humidity of air; 4) Enthalpy of air per kg of dry air.


A sample of moist air has a dry bulb temperature of 25°C and a relative humidity of 50 percent. The barometric pressure is 740 mm of Hg. Without using Psychometric chart, calculate: 1) partial pressure of water vapor and dry air; 2) Dew point temp; 3) specific humidity of air; 4) Enthalpy of air per kg of dry air.

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1) Partial pressure of water vapour and dry air: Let $p_{v}=$ partial pressure of water vapour. We know that relative humidity $\phi:$ $$ \phi=\frac{p_{v}}{p_{s}} $$ From steam table, we find that the saturation pressure of vapour corresponding to dry bulb temperature of $25^{\circ} \mathrm{C}$ is $$ \begin{array}{c} p_{s}=0.031698 b a r=3169.8 \mathrm{~N} / \mathrm{m}^{2} \\ 0.5=\frac{p_{v}}{3169.8} \\ p_{v}=1584.9 \mathrm{~N} / \mathrm{m}^{2} \end{array} $$ 2) Dew point temperature: Since the DPT is the saturation temperature corresponding to the partial pressure of water vapour $p_{v}$ therefore from steam tables, we find that corresponding to a pressure of $1584.9 \mathrm{~N} / \mathrm{m}^{2}$ the dew point temperature is, $$ \begin{array}{c} T_{d p}-10 \\ \frac{1.5849-1.2281}{T_{d p}}=13.735 \end{array} $$ 3) Specific humidity: $$ \begin{aligned} W=\frac{0.622 p_{v}}{p_{b}-p_{v}} &=\frac{0.622 \times 1584.9}{98658.6-1584.9} \\ W &=0.010 \end{aligned} $$ 4) Enthalpy of air per kg of dry air: From steam tables, latent heat of vaporization of water corresponding to a dew point temperature of $13.735^{\circ} \mathrm{C}$, $$ \begin{array}{c} \frac{h_{f g d p}-2477.2}{13.735-10}=\frac{2465.4-2477.2}{15-10} \\ h_{f g d p}=2468.38 \mathrm{~kJ} / \mathrm{kg} \end{array} $$ We know that specific enthalpy, $$ \begin{array}{c} h=1.022 t_{d}+W\left(h_{f g d p}+2.3 t_{d p}\right)=1.022 \times 13.735+0.01(2468.38+2.3 \times 13.735) \\ h=39.036 \frac{\mathrm{kJ}}{\mathrm{Kg}} \text { of } d r y \text { air } \end{array} $$

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