THERMOFLUIDS (MTV 310) - Varsity · PDF file1 UNIVERSITY OF PRETORIA Department of Mechanical...

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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.


√ 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( )


√ 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!!!


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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!!!


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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.


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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?


√ 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

⎛⎝⎜


∂t1r∂∂r

r ∂T∂r

⎛⎝⎜

⎞⎠⎟+egenk

=1α∂T∂t

1r2

∂∂r

r2k ∂T∂r

⎛⎝⎜


∂t1r2

∂∂r

r2 ∂T∂r

⎛⎝⎜

⎞⎠⎟+egenk

=1α∂T∂t

∂∂x

k ∂T∂x

⎛⎝⎜

⎞⎠⎟+

∂∂y

k ∂T∂y

⎛⎝⎜

⎞⎠⎟+

∂∂z

k ∂T∂z

⎛⎝⎜


∂t

1r∂∂r

kr ∂T∂r

⎛⎝⎜

⎞⎠⎟+1r2

∂∂φ

k ∂T∂φ

⎛⎝⎜

⎞⎠⎟+

∂∂z

k ∂T∂z

⎛⎝⎜


∂t

1r2

∂∂r

kr2 ∂T∂r

⎛⎝⎜

⎞⎠⎟+

1r2 sin2θ

∂∂φ

k ∂T∂φ

⎛⎝⎜

⎞⎠⎟+

1r2 sinθ

∂∂θ

k sinθ ∂T∂θ

⎛⎝⎜


∂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:________________________

THERMOFLUIDS (MTV 310) - Varsity · PDF file1 UNIVERSITY OF PRETORIA Department of Mechanical...

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