Question

T(SEC)	f(t)
0	1
9	0.91
18	0.834
60	0.69
69	0.636
78	0.587
87	0.536
120	0.492
129	0.451
138	0.415
147	0.391
180	0.347
189	0.319
198	0.291
207	0.27
240	0.246
249	0.226
258	0.207
267	0.19
300	0.175

5.3 Determine the time constant by differentiation using the following method: (a) At any point on the graph of f(t) vs t, the slope line (tangent) will always reach zero in a time t= t. Given t-to f(t)= f(t)e ? it can be shown that (do not prove) At t= f(t)- co?f(t.)-f(t. + At) (b) Apply the above equation for t to each of the data points to the first 3 minutes and average the values of t to obtain the time constant.

Answer 1

t/ sec	f(t)	Δt/ sec	τ(t)/ sec
0	1	9	100.0
9	0.91	9	107.8
18	0.834	42	243.3
60	0.69	9	115.0
69	0.636	9	116.8
78	0.587	9	103.6
87	0.536	33	402.0
120	0.492	9	108.0
129	0.451	9	112.8
138	0.415	9	155.6
147	0.391	33	293.3
180	0.347	9	111.5
189	0.319	-	-

Δt gives the step time and τ in the table is the time constant at each data point.

The average of τ over the first 3 minutes (180 seconds) is 164.1 seconds.