Post on 13-Feb-2018
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UNIVERSITY OF PRETORIADepartment of Mechanical and Aeronautical Engineering
Exam 2012
THERMOFLUIDS (MTV 310)
Lecturer: Mr. Johnathan VadaszExternal Examiner: Prof. G. Ziskind (Ben-Gurion Universityof the Negev)Time: 180 min.Total: 100 marks
Check that there are 18 question pages including this one. Answer all questions andshow all working and assumptions.Question paper must be handed in with answer bookletincluding moody diagram!!!
Name:________________________________Surname: _____________________________Student Number:________________________
Question 1: please turn the pageplease turn the page
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Question 1 (15)
In the figure below gate AB is 3 m wide into the paper and is connected by a rod andpulley to a concrete sphere (SG = 2.4). What sphere diameter is just right to close thegate? Note B is the pivot point.SG is relative to water, or air for a gas.
Evaluation Matrix:Has the student used a correct and logical method to solve the problem. √ XHas the student shown all working in his/her derivation to achieve the solution. √ XHas the student made all the necessary assumptions to solve the problem. √ XHas the student explained his/her reasoning behind using the relevant equations/method to solve theproblem.
√ X
Has the student considered if the solution is viable in practice under physics laws in the physical world. √ XHas the student made correct numerical approximations. √ X
Total √ X
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Question 2 (10)
A spar buoy is a buoyant rod weighted to float and protrude vertically, as in figure below.It can be used for measurements or markers. Suppose that the buoy is maple wood (SG =0.6), 2 cm by 2 cm by 120 cm, floating in seawater (SG = 1.025). How many kilogramsof steel (SG = 7.85) should be added to the bottom end so that h = 18 cm?SG is relative to water, or air for a gas.
Evaluation Matrix:Has the student used a correct and logical method to solve the problem. √ XHas the student shown all working in his/her derivation to achieve the solution. √ XHas the student made all the necessary assumptions to solve the problem. √ XHas the student explained his/her reasoning behind using the relevant equations/method to solve theproblem.
√ X
Has the student considered if the solution is viable in practice under physics laws in the physical world. √ XHas the student made correct numerical approximations. √ X
Total √ X
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Question 3 (15)
An 8 m internal diameter spherical tank made of 1.5 cm thick stainless steel (k = 15W/m⋅°C) is used to store iced water at 0°C. The tank is located in a room whosetemperature is 25°C. The walls of the room are also at 25°C. The outer surface of the tankis black (emissivity ε = 1), and heat transfer between the outer surface of the tank and thesurroundings is by natural convection and radiation. The convection heat transfercoefficients at the inner and outer surfaces of the tank are 80 W/m2⋅°C and 10 W/m2⋅°C,respectively. Determine (a) the rate of heat transfer to the iced water in the tank and (b)the amount of ice at 0°C that melts during a 24 hour period. The heat of fusion of water atatmospheric pressure is hif = 333.7 kJ/kg.hrad = εσ T2
2 + Tsurr2( ) T2 + Tsurr( )
Evaluation Matrix:Has the student used a correct and logical method to solve the problem. √ XHas the student shown all working in his/her derivation to achieve the solution. √ XHas the student made all the necessary assumptions to solve the problem. √ XHas the student explained his/her reasoning behind using the relevant equations/method to solve theproblem.
√ X
Has the student considered if the solution is viable in practice under physics laws in the physical world. √ XHas the student made correct numerical approximations. √ X
Total √ X
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Question 4 (15)
Given is a three-pipe series system as in the figure below. The total pressure drop ispA-pB = 150 kPa, and the elevation drop is zA-zB = 7 m. The pipe data are:
Pipe L, m d, cm ε, mm ε/d1 100 2 0.2 0.012 150 5 0.3 0.0063 80 7 0.07 0.001
The fluid is water, ρ = 998 kg/m3 and υ = 0.000001 m2/s. Calculate the flow rate Q inm3/h through the system.You MUST use the Moody diagram attached to solve this problem!!!
Evaluation Matrix:Has the student used a correct and logical method to solve the problem. √ XHas the student shown all working in his/her derivation to achieve the solution. √ XHas the student made all the necessary assumptions to solve the problem. √ XHas the student explained his/her reasoning behind using the relevant equations/method to solve theproblem.
√ X
Has the student considered if the solution is viable in practice under physics laws in the physical world. √ XHas the student made correct numerical approximations. √ X
Total √ X
6
Question 5 (15)
The Asphalted cast-iron pipes are laid in parallel with these dimensions:
Pipe 1 L1 = 600 m d1 = 12 cmPipe 2 L2 = 800 m d2 = 6 cmPipe 3 L3 = 900 m d3 = 20 cm
The total flow rate is 200 m3/h of water at 20°C. Determine (a) the flow in each pipe; and(b) the pressure drop across the system.You MUST use the Colebrook’s formula to solve this problem!!!
Evaluation Matrix:Has the student used a correct and logical method to solve the problem. √ XHas the student shown all working in his/her derivation to achieve the solution. √ XHas the student made all the necessary assumptions to solve the problem. √ XHas the student explained his/her reasoning behind using the relevant equations/method to solve theproblem.
√ X
Has the student considered if the solution is viable in practice under physics laws in the physical world. √ XHas the student made correct numerical approximations. √ X
Total √ X
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Question 6 (15)
Consider a 5 m high, 8 m long, and 0.22 m thick wall whose representative cross sectionis given in the figure below. The thermal conductivities of various materials used, inW/m⋅°C, are kA = kF = 2, kB = 8, k`C = 20, kD = 15, and kE = 35. The left and right surfacesof the wall are maintained at uniform temperatures of 300°C and 100°C, respectively.Assuming heat transfer through the wall to be one-dimensional, determine (a) the rate ofheat transfer through the wall; (b) the temperature at the point where the sections B, D,and E meet; and (c) the temperature drop across the section F. Disregard any contactresistances at the interfaces.
Evaluation Matrix:Has the student used a correct and logical method to solve the problem. √ XHas the student shown all working in his/her derivation to achieve the solution. √ XHas the student made all the necessary assumptions to solve the problem. √ XHas the student explained his/her reasoning behind using the relevant equations/method to solve theproblem.
√ X
Has the student considered if the solution is viable in practice under physics laws in the physical world. √ XHas the student made correct numerical approximations. √ X
Total √ X
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Question 7 (15)
Consider a 2 m high electric hot-water heater that has a diameter of 40 cm and maintainsthe hot water at 55°C. The tank is located in a small room whose average temperature is27°C, and the heat transfer coefficients on the inner and outer surfaces of the heater are50 and 12 W/m2⋅°C, respectively. The tank is placed in another 46 cm diameter sheetmetal tank of negligible thickness, and the space between the two tanks is filled withfoam insulation (k = 0.03 W/m⋅°C). The thermal resistances of the water tank and theouter thin sheet metal shell are very small and can be neglected. The price of electricity is$0.08/kWh, and the home owner pays $280 a year for water heating. Determine thefractions of the hot-water energy cost of this household that is due to the heat loss fromthe tank.Hot-water tank insulation kits consisting of 3 cm thick fibre-glass insulation (k = 0.035W/m⋅°C) large enough to wrap the entire tank are available in the market for about $30. Ifsuch an insulation is installed on this water tank by the home owner himself, how longwill it take for this additional insulation to pay for itself?
Evaluation Matrix:Has the student used a correct and logical method to solve the problem. √ XHas the student shown all working in his/her derivation to achieve the solution. √ XHas the student made all the necessary assumptions to solve the problem. √ XHas the student explained his/her reasoning behind using the relevant equations/method to solve theproblem.
√ X
Has the student considered if the solution is viable in practice under physics laws in the physical world. √ XHas the student made correct numerical approximations. √ X
Total √ X
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Information:
Fluid Specific weight γat 68ºF = 20ºC
(N/m3)Air (at 1 atm) 11.8Ethyl alcohol 7733SAE 30 oil 8720Water 9790Seawater 10050Glycerin 12360Carbon tetrachloride 15570Mercury 133100
Thermodynamic Properties of a Fluid:V x, y, z,t( ) = iu x, y, z,t( ) + jv x, y, z,t( ) + kw x, y, z,t( )
p = ρRTR = cp − cv = gas const.
τ = µ dθdt
= µ dudy
Δp = ϒ R1−1 + R2
−1( )Pressure Distribution in a Fluid:
Liquids p2 − p1 = −γ z2 − z1( )or z1 − z2 =
p2γ
− p1γ
p = pa 1−
BzT0
⎛⎝⎜
⎞⎠⎟
g/ RB( )
where gRB
= 5.26(air)
Hydrostatic Forces on Plane Surfaces:F = paA + γ hCGA = pa + γ hCG( )A = pCGA
yCP = −γ sinθ IxxpCGA
= − sinθ IxxhCGA
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xCP = −γ sinθIxypCGA
= − sinθIxyhCGA
Integral Relations for a Control Volume:
ddt(Bsyst ) =
ddt
βρdVCV∫
⎛
⎝⎜⎞
⎠⎟+ βρV cosθdAout
CS∫ − βρV cosθdAin
CS∫
ddt(Bsyst ) =
ddt
βρdVCV∫
⎛
⎝⎜⎞
⎠⎟+ βρ V ⋅n( )dA
CS∫
ddt(Bsyst ) =
ddt
βρdVCV∫
⎛
⎝⎜⎞
⎠⎟+ βρ Vr ⋅n( )dA
CS∫
∂ρ∂tdV
CV∫ + ρiAiVi( )out
i∑ − ρiAiVi( )in
i∑ = 0
F∑ − areldm
CV∫ = d
dtVρdV
CV∫
⎛
⎝⎜⎞
⎠⎟+ Vρ Vr ⋅n( )dA
CS∫
where arel =
d 2Rdt 2
+ dΩdt
× r + 2Ω×V +Ω× Ω× r( )
M0∑ = ∂
∂tr ×V( )ρdV
CV∫
⎡
⎣⎢
⎤
⎦⎥ + r ×V( )ρ V ⋅n( )dA
CS∫
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Q − Ws − Wv =∂∂t
u + 12V 2 + gz⎛
⎝⎜⎞⎠⎟ ρdV
CV∫
⎡
⎣⎢
⎤
⎦⎥
+ h + 12V 2 + gz⎛
⎝⎜⎞⎠⎟ ρ V ⋅n( )dA
CS∫
Friction and Shaft Work in Low-Speed Flow:
pγ+ V
2
2g+ z
⎛⎝⎜
⎞⎠⎟ in
= pγ+ V
2
2g+ z
⎛⎝⎜
⎞⎠⎟ out
+ hfriction − hpump + hturbine
Kinetic Energy Correction Factor:
pγ+ α2gV 2 + z
⎛⎝⎜
⎞⎠⎟ in
= pγ+ α2gV 2 + z
⎛⎝⎜
⎞⎠⎟ out
+ hturbine − hpump + hfriction
∂V∂tds
1
2
∫ + dpρ1
2
∫ + 12V22 −V1
2( ) + g z2 − z1( ) = 0
Dimensional Analysis and Similarity:
FρV 2L2
= g ρVLµ
⎛⎝⎜
⎞⎠⎟
CF = g Re( )Viscous Flow in Ducts:
hf = f L
dV 2
2g where f = fcn Red ,
εd,duct shape⎛
⎝⎜⎞⎠⎟
flam =8τw,lamρV 2 =
8 µV / d( )ρV 2 = 64
ρVd / µ= 64Red
1f 1/2
= −2 log ε / d3.7
+ 2.51Red f
1/2
⎛⎝⎜
⎞⎠⎟
Δhtot = hf + hm∑ = V2
2gfLd
+ K∑⎛⎝⎜
⎞⎠⎟
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Heat Conduction:
Qn = −kA ∂T
∂n
∂∂x
k ∂T∂x
⎛⎝⎜
⎞⎠⎟+ egen = ρc ∂T
∂t∂2T∂x2
+egenk
=1α∂T∂t
1r∂∂r
rk ∂T∂r
⎛⎝⎜
⎞⎠⎟+ egen = ρc ∂T
∂t1r∂∂r
r ∂T∂r
⎛⎝⎜
⎞⎠⎟+egenk
=1α∂T∂t
1r2
∂∂r
r2k ∂T∂r
⎛⎝⎜
⎞⎠⎟+ egen = ρc ∂T
∂t1r2
∂∂r
r2 ∂T∂r
⎛⎝⎜
⎞⎠⎟+egenk
=1α∂T∂t
∂∂x
k ∂T∂x
⎛⎝⎜
⎞⎠⎟+
∂∂y
k ∂T∂y
⎛⎝⎜
⎞⎠⎟+
∂∂z
k ∂T∂z
⎛⎝⎜
⎞⎠⎟+ egen = ρc ∂T
∂t
1r∂∂r
kr ∂T∂r
⎛⎝⎜
⎞⎠⎟+1r2
∂∂φ
k ∂T∂φ
⎛⎝⎜
⎞⎠⎟+
∂∂z
k ∂T∂z
⎛⎝⎜
⎞⎠⎟+ egen = ρc ∂T
∂t
1r2
∂∂r
kr2 ∂T∂r
⎛⎝⎜
⎞⎠⎟+
1r2 sin2θ
∂∂φ
k ∂T∂φ
⎛⎝⎜
⎞⎠⎟+
1r2 sinθ
∂∂θ
k sinθ ∂T∂θ
⎛⎝⎜
⎞⎠⎟+ egen = ρc ∂T
∂tT 0( ) = T1 T L( ) = T2
−kdT 0( )dx
= q0 =
0 → Insulationh1 T∞1 − T 0( )⎡⎣ ⎤⎦ → Convection
ε1σ Tsurr ,14 − T 0( )4⎡⎣ ⎤⎦ → Radiation
⎧
⎨⎪⎪
⎩⎪⎪
⎫
⎬⎪⎪
⎭⎪⎪
−kdT L( )dx
= qL =0 → Insulationh2 T L( ) − T∞2⎡⎣ ⎤⎦ → Convection
ε2σ T L( )4 − Tsurr ,24⎡⎣ ⎤⎦ → Radiation
⎧
⎨⎪⎪
⎩⎪⎪
⎫
⎬⎪⎪
⎭⎪⎪
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egen =Egen,electric
Vwire=I 2Reπr0
2L
Ts = T∞ +egenVhAs
Ts, plane wall = T∞ +egenLh
Ts, cylinder = T∞ +egenr02h
Ts, sphere = T∞ +egenr03h
ΔTmax, cylinder = T0 − Ts =egenr0
2
4kTcenter = T0 = Ts + ΔTmax
ΔTmax, plane wall =egenL
2
2k
Tmax, sphere =egenr0
2
6k
kavg =k(T )dT
T1
T2∫T2 − T1
Qplane wall = kavgAT1 − T2
L=AL
k(T )dTT2
T1∫Qcylinder = 2πkavgL
T1 − T2
ln(r2 / r1)=
2πLln(r2 / r1)
k(T )dTT2
T1∫Qsphere = 2 = 4πkavgr1r2
T1 − T2
r2 − r1=
4πr2r1r2 − r1
k(T )dTT2
T1∫
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Steady Heat Conduction:
Qcond, wall = kAT1 − T2
L=T1 − T2
RwallQconv =
Ts − T∞
Rconv
Rconv =1hAs
Qrad = εσAs Ts4 − Tsurr
4( ) = hradAs Ts − Tsurr( ) = Ts − TsurrRrad
Rrad =1
hradAs
Q =T∞1 − T∞2
Rtotal
Rtotal = Rconv,1 + Rwall + Rconv,2 =1h1A
+LkA
+1h2A
ΔT = QRQ =UAΔT
UA =1Rtotal
hc =Q / A
ΔTinterface
Rc =1hc
=ΔTinterfaceQ / A
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Q =T1 − T2Rtotal
1Rtotal
=1R1
+1R2
Qcond ,cyl =T1 − T2Rcyl
Rcyl =ln r2 / r1( )2πLk
Qcond ,sphere =T1 − T2Rsphere
Rsphere =r2 − r14πr1r2k
Q =T∞1 − T∞2
RtotalRtotal = Rconv,1 + Rcyl ,1 + Rcyl ,2 + Rcyl ,3 + Rconv,2
Rtotal =1h1A1
+ln(r2 / r1)2πLk1
+ln(r3 / r2 )2πLk2
+ln(r4 / r3)2πLk3
+1
h2A4
rcr, cylinder =kh
rcr, sphere =2kh
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d 2θdx2 − m2θ = 0
m2 =hpkAc
θ = T − T∞
Long fin: T (x) − T∞
Tb − T∞
= e−mx = e− x hp /kAc
Long fin: Qlong fin = −kAcdTdx x=0
= hpkAc Tb − T∞( )
Adiabatic fin tip: T (x) − T∞
Tb − T∞
=coshm(L − x)
coshmL
Adiabatic fin tip: Qadiabatic tip = −kAcdTdx x=0
= hpkAc Tb − T∞( ) tanhmL
Corrected fin length: Lc = L +Acp
Lc, rectangular fin = L +t2
Lc, cylinderical fin = L +D4
Qfin,max = hAfin (Tb − T∞ )
η fin =Qfin
Qfin,max
ε fin =Qfin
Qno fin
=Qfin
hAb Tb − T∞( ) =Afin
Abη fin
εlong fin =Qfin
Qno fin
=hpkAc Tb − T∞( )hAb Tb − T∞( ) =
kphAc
ε fin,overall =Qtotal , fin
Qtotal ,no fin
=h Aunfin +η finAfin( ) Tb − T∞( )
hAno fin Tb − T∞( )
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Heat Transfer Ratio: Qfin
Qlong fin
=hpkAc Tb − T∞( ) tanhmL
hpkAc Tb − T∞( ) = tanhmL
Qfin =Tb − T∞
R= hAfinη fin Tb − T∞( )
Q = Sk T1 − T2( )S = conduction shape factor
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Name:________________________________Surname: _____________________________Student Number:________________________