FOOD ENGINEERING

Module 3. Food dehydration

Lesson 27
ESTIMATION OF DRYING TIME FOR FOOD PRODUCTS

27.1 Introduction

In order to determine the time required to achieve the desired reduction in product moisture content, the rate of moisture removal or drying rate must be predicted. The rate of drying depends on properties of drying air (the dry bulb temperature, RH, and velocity of air and the surface heat transfer coefficient), the properties of food (moisture content, surface to volume ratio and the surface temperature) and rate of moisture loss. The size of the pieces has an important effect on the drying rate in both the constant and falling rate periods. In the constant rate period, smaller pieces have a larger surface area available for evaporation where as in falling rate period smaller pieces have a shorter distance for moisture to travel through the food. Other factors which influence the rate of drying include:

1. The fat content of the food (higher fat contents generally results in slower drying, as water is trapped with in the food).

2. The method of preparation of food (cut pieces lose moisture more quickly than losses through skin.

3. The amount of food placed in a dryer in relation to its size (in a given dryer faster drying is achieved with smaller quantities of food).

For constant rate drying period the following general expression would apply:

R_c = d_w / d_t = w_o – w_c / t_c --------------(1)

Where,

w₀ = Initial Moisture Content (kg water / kg dry solid )

w_c = Critical moisture content (kg water / kg dry solid ) and

t_c = Time for constant rate drying.

During falling rate drying, the following analysis would apply.

- d_w/dt = R_c / w_c (w) or

Where the limits of integration are between critical moisture content w_c of end of constant rate drying, t_c and some desired final moisture content, w.

On integration:

t –t_c = w_c/R_c ln(w_c/w) or time for falling rate becomes

t_f = w_c/R_c x ln (w_c/w) ------------(3) and

The total drying times becomes

t = (w_o - w_c) /R_c + w_c/R_c X ln (w_c/w) ----------(4)

The above equation indicates that the time for complete drying from some initial moisture content ‘w_o’ to a desirable final moisture content w depends on knowledge of critical moisture content ‘w_c’, the time for constant rate drying t_c, and the rate for constant drying R_c.

Example 27.2

A tunnel dryer is being designed for drying apple halves from initial moisture content of 70% (wet basis) to final moisture content of 5% (wet basis). An experimental drying curve for the product indicates that the critical moisture content is 25% (wet basis) and the time for constant drying is 5 min. Based on the information provided, estimate the total drying time for product.

Solution

Initial product moisture content w_o­ = 0.7 / 0.3 = 2.33 kg H₂O / kg solids

Critical moisture content w_c = 0.25 / 0.75 = 0.333 kg H₂O / kg solids.

Final moisture content w = 0.05 / 0.95 = 0.0526 kg H₂O / kg solids.

Time for constant rate drying t_c = 5 min

Required:

Total drying time

Solution:

R_c = (w_o – w_c) / t_c = (2.33 – 0.33) / 5 min = 0.4 kg H₂O / kg solids min

Final drying time t_F = w_c/R_c X ln(w_c/w)

= 0.333/0.4 X ln (0.333 / 0.0526)

= 1.54 min

Total drying time becomes t = 5 + 1.54 = 6.54 min.

Example 27.3

Estimate the drying rate and time needed to reduce the moisture content of a 100 m m diameter spherically shaped droplet, falling in a spray dryer from 60 to 35 %. The initial density of the droplet is 900 kg/m³. The droplet is in an air stream such that T _a = 200^oC, p = 101.3 k Pa, h = 200 w/m^2oC and T_wb = 60^oC. Assume that constant rate drying applies over the total drying process and droplet doesn’t change in size.

Solution

Diameter of the particle = 100 m m = 0.0001 m

Density of the droplet = 900 kg / m³

Dry bulb temperature of air = 60^oC

Convective heat transfer coefficient h = 200 w/m^2oC

Pressure in the chamber is 101.3 kPa.

The surface of the droplet will be assumed to be at the wet bulb temperature of the air, thus T_s = T_wb = 60^oC.