Question

A dryer, working continuously, is being fed with a wet material that has 60% solids and 40% water. 1000 kg / h of product with 10% humidity come out of the dryer. The dryer receives dry air (79% mol of N2 and 21% mol of oxygen) through a 70 cm internal diameter duct. Air enters at a speed of 60 km / h at a pressure of 585 mmHg of absolute pressure and 100 ° C.

Calculate:
a) The amount of wet material that enters the drying in Kg / h.
b) Evaporated water in Kg / h.
c) The amount of humid air that comes out of the dryer in Kg / h.
d) Air humidity in Kg of water / Kg of dry air

Answer 1

Given that,

A dryer, working continuously, is being fed with a wet material that has 60% solids and 40% water.

Let the wet feed be

$x_{F}$ be the fraction of feed that is water

CF = 0.4

The amount of product is given as $P=1000kg/h$ which has a humidity of 10%

$x_{P}$ be the fraction of product that is water

Тр 0.1

The dryer has an internal diameter of

D = 70cm = 0.7m

The dryer recieves dry air.

The air enters at a velocity of $v=60km/h=16.67m/s$

and at a pressure of $p=585mmHg=77993.6Pa$ and temperature of $T=100^{0}C=373K$

Now we will find the solutions of the questions,

Doing balance on the dry solid coming inside and going outside we get,

$(1-x_{F})F=(1-x_{p})P$

$\Rightarrow (1-0.4)F=(1-0.1)*1000$

$\Rightarrow F=1500kg/h$

So

amount of wet material that enters the dryer is

b).Taking the overall mass balance on the dryer we get,

So,

$\Rightarrow water\: evaporated=500kg/h$

So, the water evaporated will be at the rate of

c)

The volumetric flow rate of the air entering can be calculated as,

$q=vA$

Where is the area of the cross section of the dryer, $A=\frac{\pi }{4}D^{2}=\frac{\pi }{4}0.7^{2}=0.385m^{2}$

So,

We should multiply the volumetric flow rate with it's density to find the mass flow rate of the dry air entering,

The average molecular weight of air can be calculated as,

$M=0.79*M_{N_{2}}+0.21*M_{O_{2}}=0.79*28+0.21*32=28.84g/mol=28.84*10^{-3}kg/mol$

Considering the air to be ideal gas, density is given as,

$\rho =\frac{pM}{RT}=\frac{77993.6*28.84*10^{-3}}{8.314*373}=0.725kg/m^{3}$

So, mass flow rate is,

$m=q*\rho=23090.75*0.725=16740.8\: kg/h$

Now we have to calculate the flow rate of humid air that comes out of the dryer which is,

$\Rightarrow humid \: air=500+16740.8=17240.8kg/h$

d) So now the humidity will be,

$Y=\frac{water}{dry\: air}=\frac{500}{16740.8}=0.02987\; kg\; water/kg\; dry\; air$