When Will the Cistern Become Empty with Pipes X, Y, and Z Operating?

[22990atinesh]

Homework Statement

Q. A cistern has 3 pipes X, Y, Z. X and Y are filler pipes used in filling the tank in 4 & 5 hours respectively. Z is an exhaust pipe which empties the tank in 2 hours. If the pipes are opened in order at 4 AM, 5 AM & 6 AM respectively. Then When will the cistern become empty ?

Homework Equations

The Attempt at a Solution

$R_X$ Rate of pipe X = $\frac {1}{4}$
$R_Y$ Rate of pipe Y = $\frac {1}{5}$
$R_Z$ Rate of pipe Z = $\frac {1}{2}$

X --> 4 AM --> t
Y --> 5 AM --> t+1
Z --> 6 AM --> t+2

"Cistern become Empty" represents 0 work

$t*R_X$ + $(t+1)*R_Y$ + $(t+2)*R_Z$ = 0

$t=14 hours => 8 PM$

Is it the correct answer and does my approach is correct. If anybody has a more simpler approach then please share...

 

[DrClaude]

Your answer is incorrect. There are many signs that are wrong: should all the rates have the same sign? And what about time: is the clock time at which a pipe is opened the correct value to consider?

 

[Mark44]

Homework Statement

Q. A cistern has 3 pipes X, Y, Z. X and Y are filler pipes used in filling the tank in 4 & 5 hours respectively. Z is an exhaust pipe which empties the tank in 2 hours. If the pipes are opened in order at 4 AM, 5 AM & 6 AM respectively. Then When will the cistern become empty ?

Homework Equations

The Attempt at a Solution

$R_X$ Rate of pipe X = $\frac {1}{4}$
$R_Y$ Rate of pipe Y = $\frac {1}{5}$
$R_Z$ Rate of pipe Z = $\frac {1}{2}$

X --> 4 AM --> t
Y --> 5 AM --> t+1
Z --> 6 AM --> t+2

"Cistern become Empty" represents 0 work

$t*R_X$ + $(t+1)*R_Y$ + $(t+2)*R_Z$ = 0

$t=14 hours => 8 PM$

If I substitute the flow rates into your equation above, I get
t(1/4) + (t + 1)(1/5) + (t + 2)(1/2) = 0
=> t/4 + t/5 + 1/5 + t/2 + 1 = 0
=> 19t/20 = -24/20
=> t = -24/19 (hours)

That's a long way from your 14 hours.

Is it the correct answer and does my approach is correct. If anybody has a more simpler approach then please share...

No, your answer is incorrect, and the equation your started with is incorrect. Also, the work you did in solving your equation doesn't result in 14 hours, so you must have made a mistake there, as well.

There are several things wrong with your analysis.
1. The flow rates (not work rates) can't all be positive. Each flow rate is in units of tank/hour. Water is coming into the tank through two pipes, and is going out of the tank through the third pipe. The flow rates need to take this into account, which isn't happening in your equation.
2. If water starts coming into the tank through pipe X at 4AM and water comes in through pipe Y at 5AM, then the time that water comes in via pipe X is longer than for pipe Y. You have

X --> 4 AM --> t
Y --> 5 AM --> t+1
Z --> 6 AM --> t+2

What you have implies that pipe Z is running two hours more than pipe X , and pipe Y is running one hour longer than pipe X. Neither of these is true.
3. I don't know how you got t = 14 hours from your equation.

 

[jk22]

I don't know if this problem is solvable with those datas, does the state of the cistern at 4AM not influence the result ?

 

[DrClaude]

I don't know if this problem is solvable with those datas, does the state of the cistern at 4AM not influence the result ?

My reading of the problem is that you should take the cistern to be initially empty.

 

[22990atinesh]

I really don't have clue how I got t=14 hours, May be in hurry is mis-solved it. But Whats the correct answer of this question.

 

[Mark44]

Use the hints given in this thread to write the correct equation, and then solve the equation.

It is against the rules of this forum to give the answers. This is from the forum rules (https://www.physicsforums.com/showthread.php?t=414380).

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