ME33_Lecture_22

ME33_Lecture_22 - ME 33 Fall 2005 Lecture 22 10-24-05...

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Example Problem – Major and Minor Losses in a Piping System Given : Water ( ρ = 998. kg/m 3 , µ = 1.00 × 10 -3 kg/m s) flows by gravity alone from one large tank to another, as sketched. The elevation difference between the two surfaces is H = 35.0 m. The pipe is 2.5 cm I.D. with an average roughness of 0.010 cm. The total pipe length is 20.0 m. The entrance and exit are sharp. There are two regular threaded 90-degree elbows, and one fully open threaded globe valve. D H 2 1 Control volume To do : Calculate the volume flow rate through this piping system. Solution : First we draw a control volume, as shown by the dashed line. We cut through the surface of both reservoirs (inlet 1 and outlet 2), where we know that the velocity is nearly zero and the pressure is atmospheric. The rest of the control volume simply surrounds the piping system. We apply the head form of the energy equation from the inlet (1) to the outlet (2):
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ME33_Lecture_22 - ME 33 Fall 2005 Lecture 22 10-24-05...

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