E3.13 Cooling of Moist Air

Example 3.13: Moist air at 90 °F dry-bulb temperature and 60% rh (relative humidity) enters a cooling coil at 5000 cfm and is processed to final saturation conditions at 45 °F. Find the tons of refrigeration required.

Solution Air conditioning

Locate state 1 as the intersection of dry-bulb temperature, t = 90 °F, and relative humidity φ = 60%.

Humidity ratio, W₁ = 0.0183 lb_w/lb_a

Humid air volume, υ = 14.262 ft³/lb_a

Humid enthalpy, h₁ = 41.818 Btu/lb_a

Locate state 2 on the saturation curve at 45 °F

Humidity ratio, W₂ = 0.006334 lb_w/lb_a

h₂ = 17.653 Btu/lb_a

Enthalpy of water at t₂ = h_w2 =13.09 Btu/lb_w

Mass flow rate of dry air, m_a = 5000/14.262 = 350.58 lb_a/min

Heat withdrawal rate, q₂ = m_a[(h₁ - h₂) - (W₁ - W₂)h_w2] = 8417 Btu/min

One ton of refrigeration is equal to heat withdrawal rate of 200 Btu/min.

Refrigeration capacity = 42.08 ton.