Question

A room (10 m x 5 m x 5 m) contains moist air at 38°C /1 bar with relative humidity 70 %. Determine the: (i) Specific humidity. (ii) Dew point temperature.(ii) Mass of dry air and water vapour with the sir in the room. (tv) Determine the quantity of water vapour that will condense out when the roo temperature is brought down to 10°C.

Answer 1

i) Dry bulb temperature , DBT =$38^oC$

Total pressure, $p_t=1 bar=100kPa$

From the steam table, the saturation pressure of water vapor corresponding to DBT 38^o C is $P_{vs}=6.62kPa$

As relative humidity, RH or $\phi = \frac{p_v}{p_{vs}}$

$0.70= \frac{p_v}{6.62} \implies 6.62 \times 0.70 = 4.634 kPa$

Specific humidity, w is also known also absolute humidity or humidity ratio is given by

$w=\frac{m_v}{m_a}=0.622 \frac{p_v}{p_t-p_u}=0.622 \frac{6.62 \times 0.70}{1-6.62 \times 0.70}=0.0326$ kg of water vapour per kg of dry air.

ii) Dew point temperature is the temperature at which water vapor in moist can start condensing. This is the saturation temperature corresponding to the partial pressure of water vapor.

Therefore, $p_v=4.634 kPa$

From the steam table, the saturation temperature corresponding to interpolation of $\frac{5-4.5}{32.9-31}=\frac{5-4.634}{32.9-t} \implies t= 31.50^o C$

iii) Volume of moist air =$10 \times 5\times5=250 m^3$

For dry air, $P_a*V_a =m_aRaT_a \implies m_a=\frac{P_aV_a}{R_aT_a}=\frac{(P_t-P_v)*V}{R_aT_a}=\frac{(100-4.634)*V}{\frac{8.314}{29} \times311}=267.399kg$

$m_v=W.m_a=0.0326 \times 267.399 =8.717 kg$

iv) New DBT =10⁰C, so $p_{vs}=1.2276 kPa$

$\frac{pv}{1.2276}=0.7 \implies pv=0.85932 kPa$

$w=\frac{m_v}{m_a}=0.622 \frac{p_v}{p_t-p_u}=0.622 \frac{0.85932}{1-0.85932}=0.00539$ kg of water vapour per kg of dry air.

$m_v=W.m_a=0.00539 \times 267.399 =1.44 kg$

So the amount of water vapour condensed, $8.717-1.44=7.277 kg$