Question

1) In a low-temperature drying operation, air at 60°C and 10% RH is being passed over diced carrots at the rate of 20 kg dry air/s. The constant rate of evaporation of water from the carrots was measured as 0.16 kg moisture/s. Estimate the temperature and RH of the air leaving the dryer. (8 pts) 2) 100 kg of bulghur with an initial moisture content of 80% on wet basis and a surface area of 12 m² is dried in a tray dryer. Assuming constant-rate drying and air inlet temperatures of 120°C dry bulb and 50°C wet bulb, estimate the time needed to dry to 50% moisture content on wet basis. Under the conditions in the dryer, measurements indicate the mass-transfer coefficient (kg) from the bulghur surface to the air to be 0.018 kg/m²s. (12 pts) Equations: NE WdX A do N = k,(Y. - Y) where Y, is the saturated humidity at saturation T, and y, is the humidity of the air at the prevailing conditions of air

Answer 1

Andr (A) flow rate of air 20 kg/s change is weight of canorts = 0.16 kg/s Humidity gain by air = ool 0.16 20 = 0.008 kg Kadoyair T = 60c from psychrometricchart. humidity of air is = 0.0125 kg /kg dry air = 0.0125 +0.008 Humidity of air leaving doyer. = 0.205 kg kgaryair from psychometric chart. T=416 42% RH (2) kg z orol8 kg/m² weight of bulgher (w): 60 (1-08) = 20 zo Initiae moisture content content X final water Yo = 08/(1-08) - ukg/kg |(1-0,5) = 1kg/kg : 0.5

from Psychrometric chart YS=0.087 kg/kg Ya = 0.055 kg/kg t w(% -%) ng "(-ya) to 2014-1 0.018x12x (0.087-61055) 8680 sec Time required ip 2.41 hr. please give the thumbs up for this Answerd. If you are any doutes put in comments. thank you.