Question 5 of 6: Drying-Rate Curve from Tray-Drying Data

Question B2: Drying-Rate Curve from Tray-Drying Data (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The drying-rate curve — drying rate $-dW/dt$ as a function of moisture content — and its constant-rate and falling-rate regions.

Approach. Because the sample dries to zero moisture, the tabulated values are the free water; differentiate the moisture–time data numerically to get the rate, then plot rate against the mean moisture over each interval and read off the constant-rate value and critical moisture.

1. Compute the drying rate. Over each interval the rate is $$R=-\frac{\Delta W}{\Delta t}.$$ For the first interval, $R=-(4.29-4.57)/0.25=1.12$ lb/hr; the early intervals scatter around a steady value. The interval rates, listed against the mean moisture $\bar W$ of each interval, are:

   $t$ (hr)	$\bar W$ (lb)	$R$ (lb/hr)	$t$ (hr)	$\bar W$ (lb)	$R$ (lb/hr)
   0–0.25	4.43	1.12	3.75–4.00	1.07	0.92
   0.25–0.50	4.17	0.96	4.00–4.25	0.87	0.68
   0.50–1.00	3.95, 3.72	0.84, 0.96	4.25–4.50	0.69	0.72
   1.00–2.00	3.49–2.80	0.92, 1.00, 0.84, 0.92	4.50–5.00	0.54, 0.42	0.48, 0.48
   2.00–3.00	2.58–1.91	0.84, 0.92, 0.88, 0.92	5.00–6.00	0.31, 0.20	0.20, 0.24
   3.00–3.75	1.68–1.29	0.92, 0.68, 0.84	6.00–7.50	0.11, 0.05, 0.01	0.14, 0.10, 0.04

2. Constant-rate period. Averaging the flat early region ($0\le t\le4.0$ hr, $W:4.57\rightarrow0.95$) gives $$R_c=\frac{4.57-0.95}{4.0}\approx\boxed{0.90\ \mathrm{lb/hr}}$$ and a least-squares line through the first 17 points gives the same 0.90. The isolated 0.68 at 3.25–3.50 hr is scatter, because the next two intervals return 0.84 and 0.92. Here surface moisture evaporates as fast as it reaches the surface, so the rate is set by external heat/mass transfer, not the solid.
3. Critical moisture and falling-rate period. The last constant-rate interval ends at $t=4.0$ hr ($W=0.95$ lb; 0.92 lb/hr at $\bar W=1.07$). After it the rate drops to about 0.70 and keeps falling, so the surface can no longer stay wetted below $W_c\approx1.0$ lb. The rate then falls — roughly 0.70→0.48→0.22→0.12→0.04 lb/hr — approaching zero as the last internally-diffusing water leaves. $$\boxed{W_c\approx1.0\ \mathrm{lb}}$$
4. Assemble the curve. Plotting $R$ against the mean moisture $\bar W$ of each interval (Figure B2) gives the classic drying-rate curve: a horizontal constant-rate plateau at 0.90 lb/hr to the right of $W_c$, then a descending falling-rate branch to the origin. Total water removed is 4.57 lb.