Quiz 2 – 2013.11.27
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Transcript of Quiz 2 – 2013.11.27
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Quiz 2 – 2013.11.27
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Questions
1. What is the Reynolds number?2. Differentiate the flow patterns observed in laminar
flow from those in turbulent flow.3. How does temperature affect the dynamic viscosity
of a fluid?
TIME IS UP!!!
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Overall Balances
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Outline
1.Mass Balance
2.Energy Balance
3.Momentum Balance
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Mass Balance
For an overall mass balance, no mass is being generated. Why?
rate of mass output rate of mass input from control volume to control volume
rate of accumulation 0of mass within control volume
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Mass Balance
Imagine the control volume as having infinitesimal surfaces dA. We need to find the net outflow of mass across the control surface.
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Mass Balance
For every dA element, a streamline of velocity vector v passes through it.
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Mass Balance
For every dA element, a unit normal vector n exists.
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Mass Balance
The component of velocity vector v in the direction of the unit normal vector n is given by: cosv n vn
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Mass Balance
The rate of mass efflux through dA:
vn A v n A ( )d cos d
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Mass Balance
What do we get when we integrate over the entire control surface?
vn A d
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Mass Balance
POSITIVE: net outflow of massNEGATIVE: net inflow of massZERO: ?
vn A d
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Mass Balance
Rate of mass outflow across control surface (and control volume):
Rate of mass accumulation in control volume:
vn A d
dVt
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Mass Balance
rate of mass output rate of mass input from control volume to control volume
rate of accumulation 0of mass within control volume
vn A Vt
d d 0
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Overall Mass Balance
vn A Vt
d d 0
Mmt
d 0d
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A well-stirred storage vessel contains 10000 kg of dilute methanol solution (xMetOH = 0.05). A constant flow of 500 kg/min of pure water is suddenly introduced into the tank and a constant rate of withdrawal of 500 kg/min of solution is started. These two flows are continued and remain constant. Assuming that the densities of the solutions are the same and that the total contents of the tank remain constant at 10,000 kg of solution, calculate the time for the alcohol content to drop to 1.0 wt.%.
Overall Mass Balance
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Outline
1.Mass Balance
2.Energy Balance
3.Momentum Balance
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• Possessed/Carried by fluid– Internal Energy– Potential Energy– Kinetic Energy– PV-work
• Transferred between system and surroundings– Heat– Shaft work
Forms of Energy
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• Intrinsic property of the fluid• Molecules in random motion
Internal Energy (U)
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• Position of the fluid with respect to an arbitrary reference plane
cggz
Potential Energy (mgz)
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• Due to fluid motion• Correction factor, a– To account for velocity distribution– Ranges from 0.5 (laminar) to 1.0 (turbulent)
Kinetic Energy (mv2/2α)
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• Work done by surroundings to push the fluid into the system
P S
d
PVSVPSFdWPV
PV Work (PV)
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• Net heat passing through the boundary of the system– Positive if heat is transferred to the system from
the surroundings – Negative if system to the surroundings
• Excludes heat generated by friction
Heat (Q)
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• Net work done on the system by the surroundings
• Convention (IUPAC)– Positive if work done on the system– Negative if work done by the system
Shaft Work (Ws)
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Energy balance from point 1 to point 2:
Datum/reference plane
U1, v1, P1, V1, S1
U2, v2, P2, V2, S2
z1
z2
Q
Ws
Total Energy Balance
21
1 1 1 1
22
2 2 2 2
2
2( )
s
vm U gz PV
Q Wvm U gz PV
dmUdt
a
a
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Energy balance from point 1 to point 2:
Total Energy Balance
2 21 2
1 1 1 1 2 2 2 2( )
2 2sv v dmUm U gz PV Q W m U gz PV
dta a
2( )2 s
dmU vm U gz PV Q Wdt a
2( )2 s
dmU vm H gz Q Wdt a
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Water at 93.3°C is being pumped from a large storage tank at 1 atm abs at a rate of 0.189 m3/min by a pump. The motor that drives the pump supplies energy at the rate of 1.49 kW. The water is pumped through a heat exchanger, where it gives up 704 kW of heat and is then delivered to a large open storage tank 15.24 m above the first tank. What is the final temperature of the water to the second tank?
Total Energy Balance
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A modification of the total energy balance- shaft work- kinetic energy- potential energy- flow work (PV)
Does not include heat and internal energy. - Why?
Energy converted to heat is lost work- loss of mechanical energy by friction
Mechanical Energy Balance
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No shear stress; zero viscosity
For isothermal flow and Q=WS=0,
Bernoulli Equation
2
22
221
21
11 22gzvVPgzvVP
aa
2
22
2
21
21
1
1
22z
gv
gPz
gv
gP
aa
Ideal Fluids
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Restrictions:
1. Valid only for incompressible fluids 2. No devices that add/remove energy should
be between points 1 and 23. No heat transfer occurring in the system4. No loss of energy due to friction
2
22
2
21
21
1
1
22gz
gv
gPz
gv
gP
aa
Bernoulli Equation
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• Friction losses: SF (energy dissipation)• Total heat absorbed by the fluid• Total work done by fluid,
-W = -WS + SF– Additional work must be done by the fluid to
overcome fluid friction
Real Fluids
Q Q F
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• Note: energy per mass units• kJ/kg or ft-lbf/lbm
• For incompressible flow:
2
2 Sv Pg z F Wa
S
2
( ) 2SWv Q FU PV gz
m ma S
Real Fluids
2
2sWv QU gz PV
m ma
Q Q FQ Q F