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Problem 6 The sketch shows one trough of an air humidifier system. The vertical cross - section of the trough is triangular with depth D

Problem 6
The sketch shows one trough of an air humidifier system. The vertical cross-section of the trough is triangular with depth D and half-angle 45, while the horizontal cross-section is rectangular with length L=30cm. Initially, the trough is empty. At time t=0, the valve is opened, and water enters at a constant flowrate of Qin=1.5Lh. Water evaporates from the liquid surface into the air at a rate Qevap=0.0015As, where As is the liquid surface area (wL) in cm2 and Qevap is in cm3s. The aim is to predict the depth h of water in the trough versus time t, and the minimum depth D required for the trough to prevent overflow.
Your analysis should be carried out as follows:
(i) derive the differential equation for h(t).
(ii) solve the differential equation and the appropriate initial conditions to find h(t).
(iii) calculate the time required for the trough to fill to a depth h=2cm.
(iv) find D.
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