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chap0090

# chap0090 - Pipe Networks M.S Ghidaoui Spring 2000 7 6 5 1 3...

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Node 1 Q Q 1 Q 2 Q 3 Pipe Networks M.S. Ghidaoui, Spring 2000 3 1 2 4 5 6 7 There are 7 nodes in the above network: 1 ,2, 3, 4, 5, 6 and 7. Continuity Equation at a Node: inflows to node i = outflows from nodes i Example: Continuity at Node 1: Q = Q 1 + Q 2 + Q 3 or : Q 1 + Q 2 + Q 3 – Q = 0 Q 1 + Q 2 + Q 3 + (-Q) = 0 Therefore, continuity states that:

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de 7: Q4 Q1 Q5 Sum of the outflows at node 1 –the sum of the inflows into node 1=0 Continuity at Node 7: Q4=flow from node 3 towards node 7 and Q5=flow from node 7 towards node 6. That is, Q4 is an inflow into node 7 but Q5 is an outflow from node 7. Q 1 + Q 4 = Q 5 or -Q 1 – Q 4 +Q 5 = 0 Therefore, continuity states that: Sum of the outflows at node 7 – the sum of the inflows into node 7=0 Continuity at Node i: Let the total number of inflows and outflows at node i be N. The inflows are numbered from p = 1 to j and the outflows are numbered form p= j+1 to N. Therefore, continuity (i.e., mass conservation) dictates that: 0 1 1 = - = + = j p p N j p p Q Q Energy equation for loops: Reminder; Energy is scalar Energy is unique at a point Energy is path independent but work and heat are path dependent
inflow outflow (2) (1) line 1 line2 Apply the energy equation form 1 to 2 along line 1 ( 29 + + - + + = + + + = + + g V Z P g V Z P h h g V Z P g V Z P Line L Line losses 2 2 ) ( 2 2 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 γ γ γ γ Apply the energy equation between, and but following line 2: ( 29 + + - + + = g V Z P g V Z P h Line L 2 2 2 2 2 2 2 1 1 1 2 γ γ Conclusion: ( 29 ( 29 ( 29 ( 29 0 2 1 2 1 = - = Line L Line L Line L Line L h h or h h The sum of all head losses around a loop=0. Mathematically, this energy condition is as follows: ( h losses ) arounded a loop = 0 Sign Convention: Often, the summation around the loop is taken in the clockwise direction. Of course, there is no problem with making the summation in anti-

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