Reservoir Inflow Over an Eight-Hour Window
A municipal water reservoir is filled by a pump whose inflow rate is monitored over an 8-hour maintenance window. From t=0 to t=4 hours, the rate is r(t)=3+2t kilolitres per hour. At t=4 the pump is switched to a steady setting, and from t=4 to t=8 the rate is r(t)=11 kL/h. The control room records only the rate function, not the total volume. The supervisor needs the total volume added during the 8 hours, the average inflow rate over the window, and the time at which the instantaneous inflow rate equals that average. A trainee suggests averaging the two endpoint rates, 3 and 11, and multiplying by 8; the supervisor is unsure because the rate is not linear over the whole interval. The reservoir already holds 40 kL, and the overflow alarm is set at 120 kL.
1. Compute the total volume added over the 8-hour window.
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Split at t=4 because the rule changes. Volume = ∫_0^4 (3+2t)dt + ∫_4^8 11 dt. First integral: [3t+t^2]_0^4 = (12+16)-0=28 kL. Second integral: 11×(8-4)=44 kL. Total = 28+44=72 kL. The rate is continuous at t=4 because 3+2(4)=11, but continuity alone does not justify a single linear formula; splitting is required.
2. Find the average inflow rate over 8 hours and the time when the instantaneous rate equals this average.
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Average rate = total volume / total time =72/8=9 kL/h. On 0≤t≤4, solve 3+2t=9, so 2t=6 and t=3 h. This lies in the interval. On 4<t≤8, r(t)=11, which never equals 9. Hence the only time is t=3 h. The average is not simply the average of the endpoint rates; it is a time-weighted mean.
3. Evaluate the trainee's endpoint-average method and decide whether the overflow alarm will sound.
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The trainee computes (3+11)/2=7 kL/h and predicts 7×8=56 kL. This assumes r(t) is linear over the whole 8 hours. In fact r rises linearly for 4 hours and then stays at 11, so the true average is 9 kL/h and the true volume is 72 kL. Final content = initial 40+72=112 kL. The alarm threshold is 120 kL, so the alarm will not sound; the margin is 8 kL. The trainee's method underestimates the volume by 16 kL and would give a falsely safe margin of 24 kL.