[ASK] Debit: Determining Flow Rates In/Out Of Container

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SUMMARY

The discussion focuses on calculating the flow rates of three water channels (Channel I, Channel II, and Channel III) used to fill and empty a water tub. The combined flow rates are established through equations derived from the time taken to fill and empty the tub. Specifically, Channel I and II together fill the tub in 1 hour and 12 minutes, while Channel III empties it. The flow rates are determined as follows: Channel I (C1) = 1/120 tubs per minute, Channel II (C2) = 1/180 tubs per minute, and Channel III (C3) = 1/720 tubs per minute.

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Monoxdifly
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A water tub has 3 water channels. Channel I and II for filling the tub and channel III to empty it. Those 3 channels together fulfill the tub for 1 hour and 20 minutes. Channel I and II together fulfill the tub for 1 hour and 12 minutes from the empty condition. Channel II and III together fulfill the tub for 4 hours from the empty condition. The duration of each channel I and channel II fulfill the tub and channel III empties the tub are...

Well, I don't know where to start. But if channel I and II together (without channel III being open) fulfill the tub in 1 hour and 12 minutes, does it mean that channel III can empty the tub just within 8 minutes (because 01.20 - 01.12)?
 
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Re: [ASK] Debit

This question was cross-posted on another site, where I responded:

I would begin by defining $C_1,\,C_2,\,C_3$ as the flow rates of the 3 channels, in tubs per minute.

From the given information, we know:

$$C_1+C_2-C_3=\frac{1}{80}\tag{1}$$

$$C_1+C_2=\frac{1}{72}\tag{2}$$

$$C_2-C_3=\frac{1}{240}\tag{3}$$

Suppose you subtract (3) from (1)...what do you get?

The OP correctly deduced:

$$C_1=\frac{1}{120}$$

$$C_2=\frac{1}{180}$$

$$C_3=\frac{1}{720}$$

I wish I had seen the thread here first. :p
 

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