Answer to Question #196808 in Mechanical Engineering for muhammed mudhasir

Question

A sling psychrometery in a laboratory test recorded the following readings. Dry bulb temperature 35°C

Wet bulb temperature = 25°C Calculate the following

(i) Specific humidity, (ii) Relative humidity

(iii) Vapour density in air dew point depression.

(iv) Dew point temperature and (v) Enthalpy of mixture per kg of dry air

Take atmospheric pressure = 1.0132 bar

Answer 1

$(i)$ Specific humidity,

W$=\frac{0.622p_v}{p_t-p_v}=\frac{0.622\times0.0252}{1.0312-0.0252}=0.01586$ kg/kg of dry air

$(ii)$ Relative humidity, $\phi=\frac{p_v}{p_{vs}}=\frac{0.0252}{0.0563}$

$(p_{vs=0.0563}$ bar corresponding to 35$^0C,$ from steam tables)

$=0.447$ or $44.7$ %

$(iii)$ Vapour density:

using characteristic gas equation , we have

$p_vV_v=m_vR_vT_v$

$p_v=\frac{m_v}{V_v}R_vT_v=\rho_vR_vT_v$

where ,$\rho_v=$ Vapuor density, $R_v=$ $\frac{Universal \space gas \space content}{Molecular\space weight of H_20}=\frac{8314.3}{18}$

$0.0252\times 10^5=\rho_v\times\frac{8314.3}{18}\times(273+35)$

$\rho_v=\frac{0.0252\times10^5\times18}{8314.3\times308}=0.0177kg/m^3$

$(iv)$Dew point temperature;$t_{dp}:$ Corresponding to 0.0252 bar, from steam tables(by interpolation)

$t_{dp}=21+(22-21)\times\frac{0.0252-0.0249}{0.0264-0.0249}$

$=21.2^0C$

$(v)$ Enthalpy of mixture per kg of dry air,h:

h$=1.005t_{db}+W(2500+1.88t_{db})$

$=1.005\times35+0.01586[2500+1.88\times35]$

$=35.175+40.69$

$=75.86kJ/kg$ of dry air