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Unformatted text preview: ChE 374—Lecture 13—Mechanical Energy Analysis of Steady Flow 0 Kinetic Energy Correction Factor — g term appears in energy equations
— Usually assume uniform v.
— In general 11 is nonuniform (e.g., pipes have v20 at the walls). * For mass ﬂow7 just use average velocity, then can treat as uniform. * For K.E. this does not work since '52 # 11—2
— Correct with a fudge factor, 0:: K.E. : @172.
— a 2 2 for laminar flow and a = 1.04 — 1.11 for turbulent ﬂow.
— Often ignored: * Most flows are turbulent * ICE. is small compared to AP, or A2. >I< Small error compared to other assumptions or unknowns. 0 Friction and Mechanical Energy Losses — SS Energy Equation: . . AP A2
Q+W5=mAu+m<T+Tv+9Az).
W Aemech — We’ve been ignoring Q, W5, Au.
— Now, account for W5 and “Losses”.
— Observations:
1 Take 2 W5 = Au 2 0 —> Aemec;Z = 0 —> emech is conserved.
2 Consider a heated pipe, no friction, then mAu :  So that heat a Au
3 Friction converts mechanical energy to internal energy.
 Au not from heat transfer is from friction losses. Rearrange Energy Eqn:  F is +.
~ F decreases emech.  Book calls this Emech,loss~ 4 Head Form:
AP A212 ——+—+Az=hw—hL
pg 29
 with hL = F/Thg, hw = W's/nag.
 Normally F is friction in pipes, with losses in pumps accounted with efﬁciency. o Examples: e Example 1: Raise a liquid: hm : A2 + hL.
— Example 2: Pump a liquid: hw 2 Ap—f + hL. . T _ AP _ A 2
_ Example 3. Nozzle. —E A 2”; + hL. — In each case, hL is wasted: Az, AP, and Av are less than without loss.
~ Text 594. 0 Note: one equation, one unknown: (m, AP, A112, Az, W3, 77, — Set all but one and solve for that one. CONSIDER THIS IN BOOK EXAMPLES, HW. Laban (K — ['Vlfctaa’s/CCJ‘ Eat/k
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 Fall '12
 DavidLignell

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