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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) ℎ𝐿= 𝑝𝑖𝑝𝑒𝑠 𝑓∗ 𝐿 𝐷 ∗( 𝑣 2 2∗𝑔 ) + 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝐾∗( 𝑣 2 2∗𝑔 ) = 𝑣 2 2∗𝑔 ∗ 𝑝𝑖𝑝𝑒𝑠 𝑓∗( 𝐿 𝐷 ) + 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝐾

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Example If oil (ν = 4 × 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 𝐴𝑝= Π 4 ∗ 𝑑𝑝 2 = ∗ = 𝑚 2 𝑣= 𝑄 𝐴𝑝 = =1.58𝑚/𝑠 𝑅𝑒= 𝑉∗𝐷 𝑣 = 1.58∗0.15 4∗ 10 −5 =5925

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The flow is turbulent Assume e=0 𝑓= ( log 10 ( 𝑒 3.7∗𝐷 𝑅𝑒 0.9 ) ) 2 = 0.25 ( log 10 ( 0 +( )))^2 =0.036 Head Loss:- ℎ𝐿= 𝑣 2 2∗𝑔 ∗ 𝑝𝑖𝑝𝑒𝑠 𝑓∗( 𝐿 𝐷 ) + 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝐾

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ℎ𝐿= ∗9.80 ∗ 0.036∗ ∗ ℎ𝐿=6.255𝑚 𝑃1 ɣ +𝑧1+ 𝑣1 2 2∗𝑔 +ℎ𝑝= 𝑃2 ɣ +𝑧2+ 𝑣2 2 2∗𝑔 +ℎ𝑡+ℎ𝐿 Where Hp=ht=0 V1=v2=0 P1=P2=Patm=0 𝑧1=𝑧2+ℎ𝐿= =136.25𝑚

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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. 𝐷ℎ= 4∗𝐶𝑟𝑜𝑠𝑠 𝐴𝑟𝑒𝑎 𝑊𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 For Example and rectangular tube with L(Length) and W(wide) the hydraulic diameter will be:- 𝐷ℎ= 4∗𝐿∗𝑊 2(𝐿+𝑊) = 2∗𝐿∗𝑊 (𝐿+𝑤)

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Example Air (T = 20°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 𝐴=𝐿∗𝑊=0.6∗0.3=0.18 𝑚 2 𝑣= 𝑄 𝐴 = 2.5 0.18 = 13.89𝑚 𝑠**

𝐷ℎ= 4∗𝐿∗𝑊 2∗(𝐿+𝑊) = 4∗0.6∗0.3 2∗( ) =0.4𝑚 𝑅𝑒= 𝑉∗𝐷ℎ 𝑣 = 13.89∗ ∗ 10 −6 =368000 The flow is turbulent 𝑒= 𝑚

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𝑓= ( log 10 ( 𝑒 3.7∗𝐷ℎ 𝑅𝑒 0.9 ) ) 2 = 0.25 ( log 10 ( ( ∗0.4 )+( )))^2 =0.015 ℎ𝐿=𝑓∗ 𝐿 𝐷ℎ ∗ 𝑣 2 2∗𝑔 = ∗ ∗9.81 =18.6𝑚

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𝑃𝑙𝑜𝑠𝑠=ρ∗𝑔∗ℎ𝐿=1.2∗9.814∗18.6=220𝑃𝑎 1000∗9.81∗ℎ𝑤=220 ℎ𝑤=0.0224𝑚=2.24𝑐𝑚=0.883𝑖𝑛𝑐ℎ

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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 × 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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ℎ𝐿1=ℎ𝐿2 𝑓1∗ 𝐿1 𝐷1 ∗ 𝑣1 2 2∗𝑔 =𝑓2∗ 𝐿2 𝐷2 ∗ 𝑣 ∗𝑔 ( 𝑉1 𝑉2 ) 2 = 𝑓2∗𝐿2∗𝐷1 𝑓1∗𝐿1∗𝐷2 𝑉1 𝑉2 = ( 𝑓2∗𝐿2∗𝐷1 𝑓1∗𝐿1∗𝐷2 ) 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 (ρ= 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 𝑓1=0.017 𝑎𝑛𝑑 𝑓2=0.0145 𝑎𝑠 𝑖𝑛𝑖𝑡𝑎𝑙 𝑔𝑢𝑒𝑠𝑠**

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Hydraulic Engineering

Hydraulic Engineering

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