I am studying drying theory by myself, using this book:

http://www.nzifst.org.nz/unitoperations/drying1.htm

and I have a question. In the problem that I copy below, I am not sure if the total heat supplied to the product is independent of the way it is supplied. I mean, is it the same to:

a) Evaporate all water at 24 ºC and then heating the steam in air to 71 ºC?

b) Heating all water up to 71 ºC and then evaporating?

c) Or some intermediate path?

I mean, is the total heat supplied dependent only on the starting and final states or does the path matter?

**EXAMPLE
7.4. Efficiency of a potato dryer**

A dryer reduces the moisture content of 100 kg of a potato product from
80% to 10% moisture. 250 kg of steam at 70 kPa gauge is used to heat 49,800
m^{3} of air to 80°C, and the air is cooled to 71°C in
passing through the dryer. Calculate the efficiency of the dryer. The
specific heat of potato is 3.43 kJ kg^{-1} °C^{-1}.
Assume potato enters at 24°C, which is also the ambient air temperature,
and leaves at the same temperature as the exit air.

*In 100 kg of raw
material there is 80% moisture, that is 80 kg water and 20 kg dry material,*

total weight of dry product = 20 x (10/9)

= 22.2 kg

weight of water = (22.2 - 20)

= 2.2 kg.

water removed = (80 - 2.2)

= 77.8 kg.

*Heat supplied to
potato product *

= sensible heat to raise potato product temperature from 24°C to 71°C
+ latent heat of vaporization.

*Now, the latent
heat of vaporization corresponding to a saturation temperature of 71°C
is 2331 kJ kg*^{-1}

Heat (minimum) supplied/100 kg potato

= 100 x (71 - 24) x 3.43 + 77.8 x 2331

= 16 x 10^{3} + 181 x 10^{3}

= 1.97 x 10^{5} kJ.

Heat to evaporate water only = 77.8 x 2331

= 1.81 x 10^{5} kJ

*The specific heat
of air is 1.0 J kg*^{-1 }°C^{-1} and the density of
air is 1.06 kg m^{-3 (Appendix 3)}

Heat given up by air/100 kg
potato

= 1.0 x (80 - 71) x 49,800 x 1.06

= 4.75 x 10^{5} kJ.

*The latent heat of
steam at 70 kPa gauge is 2283 kJ kg*^{-1}

Heat in steam = 250 x 2283

= 5.71 x 10^{5} kJ.

*Therefore (a) efficiency
based on latent heat of vaporisation only:*

= (1.81 x 10^{5})/ (5.71 x 10^{5})

= 32%

(b)
efficiency assuming sensible heat remaining in food after drying is unavailable

= (1.97 x 10^{5})/ (5.71 x 10^{5})

= 36%

(c) efficiency based heat input and output, in drying air

= (80 - 71)/ (80 - 24)

= __16%__

*Whichever
of these is chosen depends on the objective for considering efficiency.
For example in a spray dryer, the efficiency calculated on the air temperatures
shows clearly and emphatically the advantages gained by operating at the
highest feasible air inlet temperature and the lowest air outlet temperatures
that can be employed in the dryer.*

## Comments rated to be "almost" Good Answers:

Re: Question About Drying Theory" by Silverthorn on 12/03/2012 5:31 PM (score 1)