written 2.0 years ago by
Sharad
• 90


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,
$$
\begin{array}{c}
T_{d p}10 \\
\frac{1.58491.2281}{T_{d p}}=13.735
\end{array}
$$
3) Specific humidity:
$$
\begin{aligned}
W=\frac{0.622 p_{v}}{p_{b}p_{v}} &=\frac{0.622 \times 1584.9}{98658.61584.9} \\
W &=0.010
\end{aligned}
$$
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 $13.735^{\circ} \mathrm{C}$,
$$
\begin{array}{c}
\frac{h_{f g d p}2477.2}{13.73510}=\frac{2465.42477.2}{1510} \\
h_{f g d p}=2468.38 \mathrm{~kJ} / \mathrm{kg}
\end{array}
$$
We know that specific enthalpy,
$$
\begin{array}{c}
h=1.022 t_{d}+W\left(h_{f g d p}+2.3 t_{d p}\right)=1.022 \times 13.735+0.01(2468.38+2.3 \times 13.735) \\
h=39.036 \frac{\mathrm{kJ}}{\mathrm{Kg}} \text { of } d r y \text { air }
\end{array}
$$