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Unformatted text preview: PGE 322K — TRANSPORT PHENOMENA
Fall 2008
EXAM 1
Sept 24, 2008
Except where noted, do all calculations in SI units
BEWARE OF UNNECESSARY INFORMATION.
DO NOT SPEND TOO LONG ON ANY ONE PROBLEM.
DO NOT LEAVE ANY PROBLEM BLANK!
YOU CAN START ANWERS FROM EQUATIONS IN BSL, JUST GIVE THE
EQUATION NUMBER
Total:100 pts NAME Fall 2008 1. (30 pts) Below are plots of either the shear force per unit area (17) versus the shear rate (y), or the effective viscosity (u) versus the shear rate. Each of the plots is for a particular ﬂuid. Below each plot, label the ﬂuid as either Newtonian, power law shear thinning, power law shear thickening, or Bingham plastic. ’17 T ‘C
r t v
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i t i
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Q) “datum Fall 2008 P = 2.00 x105 Pa P = 2.07 x105 Pa 2) (40 pts) Your ﬁrst assignment as a new engineer is to design a pipe that will conduct a
hydraulic ﬂuid at 0.3 ml/min (5 x 10'9 m3/sec) at a certain pressure gradient. The pipe is
angled with the top of the pipe being 2 m higher than the bottom of the pipe. The total
length of the pipe is 3 m. The pressure at the top of the pipe is held at 2.00 x 105 Pa, and
the pressure at the bottom of the pipe is at 2.07 x 105 Pa. The hydraulic ﬂuid has a
viscosity of 10 cP (10'2 Pas), and its density is 1100 kg/m3. Assume the ﬂow is laminar. a) Which way would the ﬂuid ﬂow in the pipe? To the right or to the left? MO LET {—20 oh! (BOT «Mr C CALCQUX‘TQ kgpx’ M0 HL : DNOOMQSPQ. 4’ “(\xt03\k\°)(9\3 “I, Quarxxtog ?e. g
H >1¢¥¥w§ ?k ‘lr O = 1.9:}x“) 9‘»
R . QWNT Mgr) I Lo»: 05 To NC. (L \ 0%T Fall 2008 b) What is the total head gradient (AH/L) driving the ﬂow? Ad 3.11mi; '110’3‘XH7< LS‘X(O\‘ pk.
L ’ V r — OM 0) Calculate the diameter of the pipe that is needed so the ﬂuid will ﬂow at the desired ﬂow rate (5 x 10'9 m3/sec). If you are uncertain about the head gradient in part b)
just use AH/L = 103 Pa/m. "1
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3“ Fall 2008 d) Is the assumption of laminar ﬂow valid for this size pipe? €£94V>
QC: N a... 1:
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AV): ’9‘ z 3x“? M/m' ~. 3.9XW "Vs
no} (ownmf’nl
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WCJM Lb OK Fall 2008 velocity proﬁle. We assume that VZ = VZ(X), and that P = P(x) (no z—or y dependence in
either). a) You do not need to do a shell balance over again, as this problem should look familiar. This problem is slightly different than that in BSL. Write down the last equation
in ESL that is correct for this problem. ‘E \M‘L OQ (lﬁmc Q05
aw : 950‘ Q txg€1 6.039 X b) Now write down the correct boundary conditions for this problem. Fall 2008 c) Now solve for v2 as a function of x. Fall 2008 ...
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 Spring '08
 dicarlo
 Fluid Dynamics, Shear Stress, power law shear

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