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Fluid Mechanics

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Component Head Loss The Minor Loss Coefficient, K:-

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**Combined Head Loss Equation**

(Total head loss)=(Pipe head Loss)+(Component head loss) βπΏ= πππππ πβ πΏ π· β( π£ 2 2βπ ) + πππππππππ‘π πΎβ( π£ 2 2βπ ) = π£ 2 2βπ β πππππ πβ( πΏ π· ) + πππππππππ‘π πΎ

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Example If oil (Ξ½ = 4 Γ 10-5 m2/s; S = 0.9) flows from the upper to the lower reservoir at a rate of m3/s in the 15 cm smooth pipe, what is the elevation of the oil surface in the upper reservoir?

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**Solution Minor head loss coefficients:- entrance = Ke = 0.5**

bend = Kb = 0.19 outlet = KE = 1.0 π΄π= Ξ 4 β ππ 2 = β = π 2 π£= π π΄π = =1.58π/π π
π= πβπ· π£ = 1.58β0.15 4β 10 β5 =5925

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The flow is turbulent Assume e=0 π= ( log 10 ( π 3.7βπ· π
π 0.9 ) ) 2 = 0.25 ( log 10 ( 0 +( )))^2 =0.036 Head Loss:- βπΏ= π£ 2 2βπ β πππππ πβ( πΏ π· ) + πππππππππ‘π πΎ

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βπΏ= β9.80 β 0.036β β βπΏ=6.255π π1 Ι£ +π§1+ π£1 2 2βπ +βπ= π2 Ι£ +π§2+ π£2 2 2βπ +βπ‘+βπΏ Where Hp=ht=0 V1=v2=0 P1=P2=Patm=0 π§1=π§2+βπΏ= =136.25π

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Non-Round Conduits If the conduit is not a pipe it is a square, triangle or any other shapes we replace the diameter by the hydraulic diameter. π·β= 4βπΆπππ π π΄πππ πππ‘π‘ππ πππππππ‘ππ For Example and rectangular tube with L(Length) and W(wide) the hydraulic diameter will be:- π·β= 4βπΏβπ 2(πΏ+π) = 2βπΏβπ (πΏ+π€)

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Example Air (T = 20Β°C and p = 101 kPa absolute) flows at a rate of 2.5 m3/s in a horizontal, commercial steel, HVAC duct. (Note that HVAC is an acronym for heating, ventilating, and air conditioning.) What is the pressure drop in inches of water per 50 m of duct?

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**Solution π΄=πΏβπ=0.6β0.3=0.18 π 2 π£= π π΄ = 2.5 0.18 = 13.89π π **

π·β= 4βπΏβπ 2β(πΏ+π) = 4β0.6β0.3 2β( ) =0.4π π
π= πβπ·β π£ = 13.89β β 10 β6 =368000 The flow is turbulent π= π

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π= ( log 10 ( π 3.7βπ·β π
π 0.9 ) ) 2 = 0.25 ( log 10 ( ( β0.4 )+( )))^2 =0.015 βπΏ=πβ πΏ π·β β π£ 2 2βπ = β β9.81 =18.6π

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ππππ π =ΟβπββπΏ=1.2β9.814β18.6=220ππ 1000β9.81ββπ€=220 βπ€=0.0224π=2.24ππ=0.883πππβ

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Pumps a centrifugal pump is a machine that uses a rotating set of blades situated within a housing to add energy to a flowing fluid.

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Pump Curve

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Example A pump is to be used to transfer crude oil (! = 2 Γ 10!4 lbf-s/ft2, 1 = 0*86) from the lower tank to the upper tank at a flow rate of 100 gpm. The loss coefficient for the check valve is 5.0. The loss coefficients for the elbow and the inlet are 0.9 and 0.5, respectively. The 2-in. pipe is made from commercial steel (e = 0*002in.) and is 40 ft long. The elevation distance between the liquid surfaces in the tanks is 10 ft. The pump efficiency is 80%. Find the power required to operate the pump.

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Solution

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Pipes in Parallel The Losses is equal in both branches as the energy different in point (1) and point (2) is equal according to Energy equation.

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βπΏ1=βπΏ2 π1β πΏ1 π·1 β π£1 2 2βπ =π2β πΏ2 π·2 β π£ βπ ( π1 π2 ) 2 = π2βπΏ2βπ·1 π1βπΏ1βπ·2 π1 π2 = ( π2βπΏ2βπ·1 π1βπΏ1βπ·2 ) 1/2

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Example A piping system consists of parallel pipes as shown in the following diagram. One pipe has an internal diameter of 0.5 m and is 1000 m long. The other pipe has an internal diameter of 1 m and is 1500 m long. Both pipes are made of cast iron (e = 0.26 mm). The pipes are transporting water at 20C (Ο= 1000 kg/m3, v= 10^-6 m2/s). The total flow rate is 4 m3/s. Find the flow rate in each pipe and the pressure drop in the system. There is no elevation change. Neglect minor losses.

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Solution

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**Take π1=0.017 πππ π2=0.0145 ππ ππππ‘ππ ππ’ππ π **

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