HUMID SPECIFIC HEAT

Eq. (20.16) can be written as,

h= (c_pa + wc_pv)t_d + w(h_fg)_0°C

is termed as the humid specific heat. It is the specific heal of moist air (1 + w) kg per kg of dry air.

At low temperatures of air-conditioning the value of w is very small. An approximate value of c_p of 1.0216 kJ/(kg d.a.).K may be taken for nil practical purposes

Example 20.1

A mixture of dry air and water vapour is at a temperature of 20°C under a total pressure of 735 mm Hg. The dew point temperature is 15°C. Calculate the following:

1. Partial pressure of water vapour,
2. Relative humidity,
3. Specific humidity,
4. Specific enthalpy of water vapour
5. Enthalpy of air per kg of dry air, and
6. Specific volume of air per kg of dry air.

Solution

1. From steam tables, the partial pressure of water vapour at t_dp = 15°Cp_v =12.79mm HgNow  760mm Hg = 1.01325×10⁵ N/m²1mm Hg = 133.32 N/m²∴   p_v =12.79×133.32=1705.16N/m²
2. Saturation pressure of water vapour at 20°C DBTp_s =17.54mm Hg = 17.54 × 133.32 = 2338.4 N/m²Relative humidity, 
3. Specific humidity, w = 
4. Latent heat of vaporization of water at t_d = 20°C, (h_fg)_20°C = 2454.1kJ/kgLatent heat of vaporization of water at t_dp =15°C, (h_fg)_15°C = 2465.9 kJ/kgSpecific enthalpy of water vapour,h_v =[4.1868t_dp + (h_fg)_dp +1.88(t_d − t_dp)]=[4.1868×15+ 2465.9+1.88(20 −15)]= 2538.1 kJ/kg w.v
5. Enthalpy of air per kg d.a., h = 1.005 t_d + wh_v= 1.005 × 20 + 0.011 × 2538.1= 48.02 kJ/kg d.a.
6. Specific volume of air per kg of d.a. = 0.873 m³/kg d.a

Example 20.2

A sling psychrometer gave the following readings:

DBT = 30°C, WBT = 20°C, Barometer reading = 740 mm of Hg.

Determine: (a) dew point temperature, (b) relative humidity, (c) specific humidity (d) degree of saturation, (e) vapour density, and (f) enthalpy of mixture per kg of dry air

Solution

Given: t_d = 30°C, t_w = 20°C, p_b = 740 mm Hg

1. Corresponding to t_w = 20°C, from steam tables, saturation pressure, p_w = 0.023385 barBarometric pressure, p_b = 740 × 133.32 × 10⁻⁵ = 0.986568 barPartial pressure of water vapourCorresponding to p_v = 0.01703 bar pressure, from steam tables, dew point temperaturet_dp = 15°C
2. Corresponding to t_d = 30°C, from steam tables, saturation pressure of water vapour,p_s = 0.04246 bar
3. Specific humidity, = 0.010925 kg/kg d.a
4. Specific humidity of saturated air,  Degree of saturation,
5. Vapour density,= 0.01218 kg/m³ of d.a.
6. From steam tables, (h_fg)_dp = _15°C = 2465.9 kJ/kgEnthalpy of mixture per kg d.a.,h = 1.0216 t_d + w[(h_fg)_dp + 2.3068 t_dp]= 1.0216 × 30 + 0.010925 [2465.9 + 2.3068 × 15] = 57.966 kJ/kg d.a.

Example 20.3

A room 5m × 4m × 3m contains moist air at 40°C. The barometric pressure is 1 bar and relative humidity is 70%. Determine: (a) humidity ratio, (b) dew point temperature, (c) mass of dry air, and (d) mass of water vapour.

If the moist air is further cooled at constant pressure until the temperature is 15°C, find the amount of water vapour condensed.

Solution

Given: V = 5 × 4 × 3 = 60 m³, t_d = 40°C, p_b = 1 bar, ϕ = 0.70.

1. Saturation pressure of vapour corresponding to t_d = 40°C, from steam tables is,p_s = 7.384 kPa or 0.07384 barp_v = 0.7× 0.07384 = 0.05169 barHumidity ratio,= 0.0339 kg/kg d.a or 33.9 g/kg d.a
2. DPT, t_dp = saturation temperature corresponding to p_v = 0.05169 bar or 5.169 kPa= 33.46°C from steam tables
3. p_a = p_b − p_v = 1.0 − 0.05169 = 0.94831 barMass of dry air, 
4. Now w = Mass of water vapour, m_v = 0.0339 × 63.34 = 2.147 kgSaturation pressure corresponding to 15°C, from steam tables is,p_s = p_v = 1.707 kPa or 0.01707 barNow  p_a = p_b − p_v = 1 − 0.01707 = 0.98293 bar or 98.293 kPaMass of dry air, Mass of water vapour, m_v = wm_a = 0.0108 × 71.35 = 0.77 kgAmount of water vapour condensed = 2.147 − 0.77 = 1.377 kg