Thermodynamics HW Solutions 966

Thermodynamics HW Solutions 966 - Assumptions All the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 12 Radiation Heat Transfer 12-82E An average person produces 0.50 lbm of moisture while taking a shower. The contribution of showers of a family of four to the latent heat load of the air-conditioner per day is to be determined. Moisture 0.5 lbm Assumptions All the water vapor from the shower is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 1050 Btu/lbm. Analysis The amount of moisture produced per day is m vapor = (Moisture produced per person)(No. of persons) = (0.5 lbm/person)(4 persons/day) = 2 lbm/day Then the latent heat load due to showers becomes Q latent = m vapor h fg = (2 lbm/day)(1050 Btu/lbm) = 2100 Btu/day 12-83 There are 100 chickens in a breeding room. The rate of total heat generation and the rate of moisture production in the room are to be determined.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Assumptions All the moisture from the chickens is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 2430 kJ/kg. The average metabolic rate of chicken during normal activity is 10.2 W (3.78 W sensible and 6.42 W latent). Analysis The total rate of heat generation of the chickens in the breeding room is W 1020 = chickens) (100 W/chicken) 2 . 10 ( chickens) of (No. total gen, total gen, = = q Q & & 100 Chickens 10.2 W The latent heat generated by the chicken and the rate of moisture production are kW 0.642 = W 642 = chickens) )(100 W/chicken 42 . 6 ( chickens) of (No. latent gen, latent gen, = = q Q & & g/s 0.264 kg/s 000264 . kJ/kg 2430 kJ/s 642 . fg latent gen, moisture = = = = h Q m & & 12-63...
View Full Document

Ask a homework question - tutors are online