Explanation Given information: The diameter of the pipe = 2.0 cm = 0.02 m The loss coefficient at sharp inlet valve = 0.50 Loss coefficient for each valve = 2.4 Loss coefficient for the elbow = 0.90 The contraction at the exit reduces the diameter by a factor = 0.60 Minor loss coefficient of the contraction = 0.15 Total length of the pipe = 8.75 m. Elevation difference (z 1 − z 2 ) = 4.6 m Calculation: When free delivery conditions are given means that the net head across the pump will be zero. With the net head being zero, there will be neither gain of pressure or loss of pressure across the pump. So, for the calculation, the pressure gain, or loss can easily be neglected. Let us write energy equation in the head form between two locations. Let us consider the location 1 and location 2. H r e q = h p u m p = 0 = P 2 − P 1 ρ g + α 2 V 2 2 − V 1 2 2 g + ( z 2 − z 1 ) + h t u r b i n e + h l .......(1) The free surface at location 1 and the free surface at location 2 are both open to atmosphere, hence, the pressure head in the energy equation will become zero. The velocity of flow at the inlet is very small, so the term V 1 = 0. Thus, equation (1) will become: ( z 1 − z 2 ) = α 2 V 2 2 2 g + h l ........ (2) The irreversible head loss in equation (2) includes both type of losses i.e. major losses and the minor losses. Now, the minor losses considered here are of two types. First, the minor losses due to flow by the mean velocity V and second, the minor losses due to the contraction. The losses due to contraction can be calculated based on exit velocity V 2 . ( z 1 − z 2 ) = α 2 V 2 2 2 g + V 2 2 g ( f L D + ∑ p i p e K L ) + V 2 2 2 g K L , c o n t r a c t i o n ........ (3) Where K L , c o n t r a c t i o n is the summation of factors of losses. ∑ p i p e K L = 0.50 + 2 × ( 2.4 ) + 3 × ( 0

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Applied Fluid Mechanics

A fluid is a substance that is continuously deformed by the application of shear stress. The rate of deformation of a fluid depends on its viscosity. The fluid tries to flow when it interacts with some other system.

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Chapter 14, Problem 25P

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The volume flow rate through this piping system.

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Chapter 14, Problem 24P

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Select duct sizes for the rectangular duct system shown in the Figure, using the equal friction method. The total pressure available for the duct system is 0.12 in. wg (30 Pa), and the loss in total pressure for each diffuser at the specified flow rate is 0.02 in. wg (5 Pa). Assume the velocity in mean duct 600 fpm. 150 che 25 Boo ae und torectang a5On e S. 150 ct 0.095 mi 071 15m

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The volume flow rate through this piping system.

Solution Summary: The author explains that the volume flow rate through this piping system is Q=36Lpm.

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Chapter 14 Solutions