A large-scale drying oven operates at steady-state-steady-flow as part of a tras

Question

A large-scale drying oven operates at steady-state-steady-flow as part of a
trash-to-energy facility. Damp paper and cardboardat 20 C is fed into the
dryer at a total rate of 18,000 kg/hr. The initialcomposition of the damp trash
is 80% solid matter and 20% water (by mass). The temperature of the exiting
material is 80 C. The drying process reduces the moisture contentof the
material to 5% by mass (and 95% solid matter) at the exit. Dry atmospheric
air at 20 C (with negligible moisture content) also enters the oven. The dry
air along with the evaporated water vapor flows outof the oven exhaust (also
at 80 C) as a humid air mixture, saturated at 100% relative humidity.
Changes in kinetic and potential energy and external heat transfer from the
drying oven to the atmosphere are negligible. The absolute pressure is
constant at standard atmospheric pressure of 101.3 kPa.
NOTE: The humid air mixture pressure = pvapor + pDA = pATM
with the vapor pressure = psat,H2Ox relative humidity
The mol fractions & volume fractions of the dry airand vapor within the humid air mixture are:
NDA/Nmix= VDA/Vmix= pDA/pATMand Nvapor/Nmix= Vvapor/Vmix= pvapor/pATM
The humidity ratio (by mass) of the humid air mixture is then:
mvapor/mDA= (MH2O Nvapor) / (MDANDA) = (Mvapor/MDA) (Nvapor/NDA) = (Mvapor/MDA)(pvapor)/pDA)
The specific heat of the solid matter, Csolid = 1.30 kJ/kg-K
For dry air: M
DA= 28.97 kg/kmol, k = c
p/cv= 1.4, R = Ru /M, cv= R/(k-1), cp= R k /(k-1)
For the H2O: Mvapor= 18.02 kg/kmol, see the steam tables for p
satand the enthalpy of liquid and vapor
Find the evaporation rate,   m &
vapor,out = _____ kg/s and the required dry air mass flowrate,   m &
DA,IN= _____ kg/s.
Find the required oven heat transfer rate input, Q&=___________ kJ/s

Explanation / Answer

intial mass of water in damp cardboard = 18000 kg / hr * 0.20 = 3600 kg/hr

mass of solid = 0.8 * 18000 = 14,400 kg/hr

In the final cardboard solid is 95 %

so.. 0.95 * total mass = 14,400

so total mass = 14400/0.95 = 15,157.895 kg/hr

final mass of water = 0.05 * 15,157.895 = 757.895 kg/hr

so mass of water evaporated = 3600 - 757.895 = 2,842.105 kg/hr

so evaporation rate = 2,842.105 kg / hr = 2,842.105 / 3600 kg/sec = 0.7895 kg/sec

psat@ 80 celcius = 47.416 KPa

mvapour / mdry air = (Mvapor/MDA)(pvapor)/pDA)

pvapour = 47.416 KPa

patm = 101.325 Kpa

so PDA = patm - pvap = 101.325 - 47.416 = 53.909 Kpa

so mvapour / mdry air = ( 18.02 / 28.97 ) * ( 47.416 /53.909) = 0.5471

mvapour = 0.7895 kg / sec

so mDA = 0.7895 /0.5471 = 1.443053 kg/sec

so enthalpy of liquid water at 20 Celcius = h1 = 83.915 KJ/kg

enthalpy of saturated water vapour at 80 C = h2 = 2643 KJ/kg

so heat transfer rate = ( h2 - h1 ) * mwatervap

= (2643 - 83.915) * 0.7895 = 2,020.3976075 KJ/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.