Question

A stream of air at 650.0 °C and 835 torr with a dew point of 50.0 °C flowing at a rate of 1915 L/s is to be cooled in a spray cooler.

A fine mist of liquid water at 15.0 °C is sprayed into the hot air at a rate of 110.0 g/s and evaporates completely.

The cooled air emerges at 1 atm.

For the flow diagram below, use two material balances and three constraints to determine all unknowns except for the outlet temperature T₃.

Problem 8.38 Spray Cooler Stepped submission A stream of air at 650.0 °C and 835 torr with a dew point of 50.0 °C flowing at a rate of 1915 L/s is to be cooled in a spray cooler. A fine mist of liquid water at 15.0 °C is sprayed into the hot air at a rate of 110.0 g/s and evaporates completely. The cooled air emerges at 1 atm. Physical Property Tables Material Balances x Check significant figures throughout the calculation. It is convenient to use the Antoine equation. Make sure that you have used kg or moles consistently. For the flow diagram below, use two material balances and three constraints to determine all unknowns except for the outlet temperature T. g HąO(l)/s n (mol H2O(l)/s) 960 1991/ ni (mol/s) X1 (mol H2O (v) /mol) (1-x)(mol DA/mol) p. 05 Tdp = 0.0 ng (mol/s) X3 (mol H2O(v)/mol) (1-x) (mol DA/mol) T3 (°C), 1 atm mol/s Z movs mol/s mol H2O(v)/mol mol H20(V)/mol

Answer 1

Given Values:

Dry bulb temperature of inlet air,Tdb = 650.0 0C

Pressure,P = 835 torr

Dew point of inlet air, Tdp = 50.0 0C

Volumetric flow rate of inlet air, V` = 1915 L/s

Mass flow rate of liquid water = 110.0 g/s

Temperature of liquid water, Tw = 15.0 0C

Calculations of molar flow rates:

$1586766396552_blob.png$

Assuming air to be an ideal gas.

PV = nRT

Where R is universal gas constant, R = 62.364 L.torr/mol.K

ni =

Plugging in values,

835 * 1915 62.364* (650+ 273) mol/s " 1

ni = 27.78 mol/

$1586766396871_blob.png$

Molecular weight of water = 18 g /mol

Molar flowrate of water, $1586766396810_blob.png$ = Mass flow rate/Molecular weight = 110/18 mol/s

niz = 6.11 mol/

$1586766396830_blob.png$

Since all the water is evaporated, the molar flowrate out will be the sum of the two flowrates.

ng = nii + niz

n3 = (27.78 +6.11) mol/s

ng = 33.81 mol/s