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    (4 marks)

    Solution Q1 (b)

    AnalysisNoting that the cross-sectional area of the spoon is constant and measuring x from

    the free surface of water, the variation of temperature along the spoon can be expressed as

    T x T

    T T

    a L x

    aLb

    ( ) cosh ( )

    cosh

    where

    2

    m000026.0)m100/2.0m)(100/3.1(

    m03.0)m100/2.0m100/3.1(2

    cA

    p

    1-

    2

    2

    m162.36)m000026.0)(CW/m.15(

    )m03.0)(C.W/m17(

    ckA

    hpa

    Noting that x =L = 18/100 = 0.18m at the tip and substituting, the tip

    temperature of the spoon is determined to be

    C25.2=5.335

    1)25(95+C25=)18.0162.36cosh(

    0cosh)25(95+C25=

    cosh

    )(cosh)()(

    aL

    LLaTTTLT b

    Therefore, the temperature difference across the exposed section of the spoon handle is

    C69.8 C)2.2595(tipTTT b

    Tb

    L = 18 cm

    (2 marks)

    (2 marks)

    (1 marks)

    (1 marks)

    (4 marks)

    (2 marks)

    (4 marks)

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    Q1(B)

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    Solution Q2 (a)

    The combination of criteria for a fin gives rise to a dimensionless group ht/k, termed thefin

    Biot number, Bi.

    For a good fin, Bi no. has to be small.

    Heat has to flow through half the fin thickness in the transverse direction. Thus t in the fin

    Biot number is half the fin thickness. For cylindrical fins of diameter 2r, t = r.

    (4 marks)

    Solution Q2 (b)

    PropertiesThe thermal conductivity, density, and specific heat

    of the milk at 20C are k = 0.607 W/m.C, = 998 kg/m3, and

    Cp= 4.182 kJ/kg.C (Table A-9).

    AnalysisThe characteristic length and Biot number for the glass of milk are

    1.0>076.2)CW/m.607.0(

    )m0105.0)(C.W/m120(

    m01050.0m)03.0(2+m)m)(0.0703.0(2

    m)07.0(m)03.0(

    22

    2

    2

    2

    2

    2

    k

    hLBi

    rLr

    Lr

    A

    V

    L

    c

    oo

    o

    sc

    For the reason explained above we can use the lumped system analysis to determine how long

    it will take for the milk to warm up to 38C:

    min5.8s348

    teeTT

    TtT

    LC

    h

    VC

    hAb

    tbt

    i

    cpp

    s

    )s002738.0(

    1-

    3

    2

    -1

    603

    6038)(

    s002738.0m)C)(0.0105J/kg.4182)(kg/m(998

    C.W/m120

    (3 marks)

    (3 marks)

    (2 marks)

    (2 marks)

    (2 marks)

    (2 marks)

    (2 marks)

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    Therefore, it will take about 6 minutes to warm the milk from 3 to 38C.

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    Q2(a)

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    Q2(b)

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    Solution Q3 (a)

    (i)

    Explain how does the convection heat transfer coefficient,hdiffer from the thermal

    conductivity, kof a fluid.

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    The value of the convection heat transfer coefficient depends on the fluid motion as well as the

    fluid properties. Thermal conductivity is a fluid property, and its value does not depend on the

    flow. (2 m)

    (ii)

    For the parallel flow over the flat plate, sketch the temperature profile for the surface

    temperature larger than fluid temperature. (3 m)

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    Q3(b)

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    Solution Q4 (a)

    Define natural convection and explain the situation of natural convection with an appropriate

    illustration. (6 marks)

    Natural convectionis a convection heat ransfer occurs due to temperature differences which affect

    the density, and thus relative buoyancy, of the fluid. (2 m)

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    or any else which suitable (1m)

    - Heat from hot object (hot boiled egg) will heat up the colder air adjacent to the hot object

    surface. (1 m)

    - These air becomes lighter (less dense) and will rise up this leading to bulk fluid movement.

    (2m)

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    Solution Q5 (a)

    (i) The heat exchangers usually operate for long periodsof time with no change in their operating conditions,

    and then they can be modeled as steady-flow devices.

    (ii) The mass flow rate of each fluid remains constant and the fluid properties such as temperature and

    velocity at any inlet and outlet remain constant.

    (iii) The kinetic and potential energy changes are negligible. The specific heat of a fluid can be treated as

    constant in a specified temperature range.

    (iv) Axial heat conduction along the tube is negligible.

    (v) The outer surfaceof the heat exchanger is assumed to be perfectly insulatedso that there is no heat lossto

    the surroundingmedium and any heat transfer thus occurs is between the two fluids only.

    Solution Q5 (b)

    Glycerin is heated by hot water in a 1-shell pass and 13-tube passes heat exchanger. The mass flow

    rate of glycerin and the overall heat transfer coefficient of the heat exchanger are to be determined.

    Properties The specific heats of water and glycerin are given to be 4.18 and 2.48 kJ/kg.C,

    respectively. [ ( )] (5Q mC T T p in out water kg / s)(4.18 kJ / kg. C)(100 C C) = 940.5 kW55

    The mass flow rate of the glycerin is determined from:

    kg/s9.5

    C]15CC)[(55kJ/kg.(2.48

    kJ/s5.940

    )(

    )]([

    glycerin

    glycerin

    inoutp

    inoutp

    TTC

    Qm

    TTCmQ

    (2 marks)

    The logarithmic mean temperature difference for counter-flow

    arrangement and the correction factor F are

    T T TT T T

    h in c out

    h out c in

    1

    2

    100 5555 15

    , ,

    , ,

    C C = 45 CC C = 40 C

    Glyceri

    n

    55C

    (2 marks)

    (2 marks)

    (1 marks)

    (1 marks)

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    T

    T T

    T Tlm CF,

    ln( / ) ln( / ).

    1 2

    1 2

    45 40

    45 4042 5 C

    77.0

    89.0100555515

    53.010015

    10055

    12

    21

    11

    12

    F

    ttTTR

    tT

    ttP

    The heat transfer surface area is

    2m0.94=m)m)(2015.0(10 DLnAs

    Then the overall heat transfer coefficient of the heat exchanger is determined to be

    C.kW/m30.6 2

    C)5.42)(77.0)(m94.0(

    kW5.9402

    ,,

    CFlmsCFlms

    TFA

    QUTFUAQ

    (2 marks)

    (2 marks)

    (1 marks)

    (1 marks)

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    Q5(B)

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    Solution Q6 (a)

    The value of effectiveness increases slowly with a large values of NTU (usually larger than 3). Therefore,

    doubling the size of the heat exchanger will not save much energy in this case since the increase in the

    effectiveness will be very small.

    Solution Q6 (b)

    Cold water is heated by hot water in a heat exchanger. The net rate of heat transfer and the heat transfer

    surface area of the heat exchanger are to be determined.

    Properties The specific heats of the cold and hot water are given to be 4.18 and 4.19 kJ/kg.C, respectively.

    Analysis The heat capacity rates of the hot and cold fluids are

    C m C

    C m C

    h h ph

    c c pc

    ,

    (0.25 kg / s)(4180 J / kg. C) W/ C

    (3 kg / s)(4190 J / kg. C) W/ C

    1045

    12 570

    Therefore, CW/1045min cCC

    and C C

    C min

    max ,.

    1045

    12 5700083

    Then the maximum heat transfer rate becomes

    ( ) ,max min , ,Q C T T h in c in (1045 W/ C)(100 C -15 C) W88 825

    The actual rate of heat transfer is

    W31,350 )C15C45)(CW/1045()( ,, outhinhh TTCQ

    Then the effectiveness of this heat exchanger becomes

    Q

    Qmax

    ,

    ,.

    31350

    88825035

    The NTU of this heat exchanger is determined using the relation in Table 13-5 to be

    438.01083.035.0

    135.0ln

    1083.0

    1

    1

    1ln

    1

    1

    CCNTU

    Then the surface area of the heat exchanger is determined from

    2m0.482

    C.W/m950

    )CW/1045)(438.0(2

    min

    min U

    CNTUA

    C

    UANTU

    Hot Water

    Cold Water

    15C

    45C

    (5 marks)

    (1 mark)

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    Q6(a)

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    Q6(b)