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Heat Transfer Test 2

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Heat Transfer Test 2
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  • Question 1
    2 / -0.33
    Two finned surfaces with long fins are identical, except that the convention heat transfer coefficient for the first finned surface is twice that of second one. Then the correct statement for the efficiency and effectiveness of the first finned surface relative to second one.
    Solution

    Concept:

    The efficiency of a fin is defined as the ratio of the actual heat transferred by the fin to the maximum heat transferable by fin, if entire fin area were at base temperature, for infinitely long fin:

    η=QfinQmax=PhkAc(tota)hPL(tota)=kAchP.1L

    η=1mLη1h

    As, h1 = 2h2

    ⇒ η1 < η2

    Effectiveness of the fin is the ratio of the fin heat transfer rate to the heat transfer rate that would exist without fin.

    ϵfin=QwithfinQwithoutfin=PhkAc(tota)hAc(tota)=PkhAc

    ϵfin1h,h1=2h2ϵ1<ϵ2

    Calculation:

    η1handϵ1h

    h1=2h2ϵ1<ϵ2andη1<η2

  • Question 2
    2 / -0.33
    A metal sphere of surface area 0.0225 m2 is an evacuated enclosure whose walls are held at very low temperature. Electric current is passed through resistors embedded in the sphere causing electrical energy to be dissipated at rate of 75 watts. If sphere surface temperature is measured to be 560 K while is steady state. The reflectivity of sphere is
    Solution

    Explanation:

    Heat energy lost by radiation = Electrical energy to be dissipated

    σϵAT4 = 75

    ϵ × 5.67 × 10-8 × 0.0225 × (560)4 = 75

    ϵ = 0.598

    Now,

    For gray surface, ϵ = α = 0.598

    In this case,

    α + ρ = 1

    ∴ ρ = 1 - α = 0.402
  • Question 3
    2 / -0.33
    A fluid (Prandtl number, Pr = 0.7) at 500 K flows over a flat plate 500 mm length, maintained at 300 K. The velocity of the fluid is 10 m/s. Assuming kinematic viscosity = 30×106m2/s, then which of the following statements are true regarding the boundary layer thickness at 5 mm from the leading edge?
    Solution

    Concept:

    δhδt=Pr1/3

    Here,

    δh = Hydrodynamic boundary layer thickness, δt = Thermal boundary layer thickness

    Calculation:

    Given:

    Pr = 0.7, x = 5 mm

    Kinematicviscosity(γ)=30×106m2s

    Velocity u() = 10 m/s

    δh=5xRex

    Rex=uxγ=10×5×10330×106=1666.66

    δh=5xRex=5(5)1666.66

    δh = 0.6123 mm

    Now,

    δhδt=Pr1/3

    0.6123δt=(0.7)1/3

    ∴ δt = 0.6896 mm
  • Question 4
    2 / -0.33
    It is proposed to coat a 1 mm diameter wire with enamel paint (k = 0.1 W/mK) to increase the heat transfer with air. If the air side heat transfer coefficient is 100 W/m2K, the optimum thickness of enamel paint should be
    Solution

    Concept:

    For maximum heat dissipation, the thickness of insulation should be critical thickness.

    The critical thickness for wire is given as, rc=kh

    Calculation:

    Given:

    k = 0.1 W/mK, h = 100 W/m2K

    rc=kh=0.1100=0.001 m=1 mm

    Thickness of insulation = rc - r

    ⇒ 1 – 0.5 = 0.5 mm

  • Question 5
    2 / -0.33
    The effectiveness of a counter-flow heat exchanger has been estimated as 0.25. Hot gases enter at 200°C and leave at 75°C. Cooling air enters at 40°C. The temperature of the air leaving the unit will be:
    Solution

    Concept:

    Heat exchanger effectiveness:

    ϵ=QactualQmax

    Qactul=m˙hcph(Th1Th2)=m˙ccpc(Tc2Tc1)

    Qmax=Cmin(Th1Th2)

    ϵ=QactualQmax=Ch(Th1Th2)Cmin(Th1Th2)=Cc(Tc2Tc1)Cmin(Th1Tc1)

    m˙nCpn=m˙cCpcCh=Cc=Cmin=Cmax

    η=Ch(Th1Th2)Cmin(Th1TC1)=Th1Th2Th1Tc1

    η=Ch(Th1Th2)Cmin(Th1TC1)=Th1Th2Th1Tc1

    Calculation:

    ϵ=(TceTci)(ThiTci)

    0.25=Tce4020040

    Tce - 40 = 40

    Tce = 80°C
  • Question 6
    2 / -0.33
    Air at 1 atmospheric pressure and 27 °C blows across a 12 mm diameter sphere at a a small heater inside the sphere maintains the surface temperature at 77 °C. With k = 0.026 W/m (kelvin) and with (Nu) = 31.4, the heat loss by the sphere would be
    Solution

    Concept:

    Heat loss through the convection is given by

    Qloss = h × A × (∆T)

    Nusselt number Nu=hlck

    Where l­c is the characteristic length, in case of heater it is equal to the diameter of the sphere

    Calculation:

    Given, Nu = 31.4, k = 0.026 W/mK, D = 12 mm ⇒ r = 6 × 10-3 m , Surface temperature Ts = 27°C and atmospheric temperature T = 77°C

    Nu=hlck

    ⇒ h=Nu×klc

    h=31.4×0.0260.012=68.03W/m2K

    Surface area of sphere A = 4 × π × r2 = 4 × 3.14 × (6 × 10-3)2

    Qloss = 68.03 × 4 × 3.14 × (6 × 10-3)2 × (77 – 27) = 1.54 J/s

  • Question 7
    2 / -0.33
    A sphere with radius 20 cm has density 2500 kg/m3 and specific heat of 0.52 kJ/kg K has a uniform heat generation rate of 100 kW/m3. If heat received over it’s surface in 240 W, the rate of change of temperature of solid is ______ (°C/sec)
    Solution

    Explanation:

    Given:

    ρ = 2500 kg/m3, Cp = 0.52 kJ/kg k, Q = 1000 kW/m3, a = 240 W

    Volume=43πr3=43π×(0.2)3=0.03351m3

    Now,

    Heat received + total heat generated = Rate of change of internal energy

    Q+q×Volume=ρCpVTt

    Tt=Q+q×VolumeρCpV

    Tt=240+100×103×0.033512500×520×0.03351

    Tt=0.082
  • Question 8
    2 / -0.33

    An air stream flows with a velocity of 3 m/s. The temperature of which is to be measured by a thermocouple which can be approximated as a sphere of diameter 0.25 cm. Initially the junction and air are at a temperature of 25°C. The air temperature suddenly changes and is maintained at 215°C. The time required for the thermocouple to indicate a temperature of 165°C is:

    Take ρ = 8750 kg/m3, c = 380 J/kg°C, k (thermocouple) = 28 W/m°C and h = 145 W/m2°C
    Solution

    Concept:

    For a lumped parameter problem, the time-temperature dependence is given by:

    ln[TTTiT]=hAmct=hAρVct

    TTTiT=ehAρVct=e1τt

    TTT0T=etτ

    Where, T = final temperature, T = air temperature, Ti = initial temperature, t = time, τ = time constant = ρVc/hA

    Calculation:

    Ti = 25°C, T = 215°C, T = 165°C

    ρ = 8750 kg/m3, c = 380 J/kg°C, k (thermocouple) = 28 W/m°C and h = 145 W/m2°C

    For a sphere:

    VA=43πR34πR2=R3=D6

    Time constant:

    τ=ρVchA=ρcD6h=8750×380×0.00256×145=9.554

    Putting the values in the equation:

    TTT0T=etτ

    16521525215=et9.554

    t = 12.754 sec
  • Question 9
    2 / -0.33
    A fin has 5 mm diameter and 100 mm length with insulated tip. The thermal conductivity of fin material is 400 Wm-1K-1. One end of the fin is maintained at high temperature and its remaining surface is exposed to ambient air. If the convective heat transfer coefficient is 40 Wm-2K-1, then
    Solution

    Concept:

    Efficiency of fin is calculated as

    η=QfinQmax

    Qmax=hAs(TbTa)

    As = Surface area of fin = πDL, Tb = Base temperature, Ta = Ambient temperature

    Effectiveness of fin is calculated as

    ε=QfinQnofin

    Qnofin=hAc(TbTa)

    Ac=Crosssectionalareaoffin=πD24

    Calculation:

    Given:

    D = 5 mm, L = 100 mm, K= 400 Wm-1K-1, h = 40 Wm-2K-1,

    Now,

    Perimeter (P) = πD = 0.0157 m

    Ac=πD24=1.9634×105m2

    m=hPKAC=8.944m1

    mL = 0.894

    Now,

    η=QfinQmax=hPkAc(TbTa)tanh(mL)hAs(TbTa)=tanh(mL)mL=79.93%

    ε=QfinQnofin=hPkAc(TbTa)tanh(mL)hAc(TbTa)

    ∴ ϵ = 63.66
  • Question 10
    2 / -0.33

    A long 15 cm diameter shaft of stainless steel (K = 14.9 W/mK), ρ = 7900 kg/m3 C = 477 J/kgK and α = 3.95 × 10-6 m2/s is at a uniform temperature of 450 °C. The shaft is allowed to cool slowly in a chamber at 150°C with heat transfer coefficient of h = 85 W/m2K. Calculate the temperature (℃) at the end of 25 min after the cooling starts.

    Use (θ0 = 0.49 at Bi-1 = 2.337 and Fo = 1.053) and (θ0 = 0.88 at Bi-1 = 4.672 and Fo = 1.053)
    Solution

    Concept:

    Biot number = hL/K

    If biot > 0.1

    Then lumped system analysis is not applicable and we use Heisler charts.

    For Heisler charts Bi = hR/K

    Fourier number = αt/L2

    θ0=TTTiT

    Ti, T= initial temperature and surrounding temperature respectively.

    T= temperature at time‘t’

    After finding out θ0 we can calculate the temperature ‘T’

    Calculation:

    L = 1m, R = 0.075 m, α = 3.95 × 10-6 m2/s, k = 14.9 W/mK

    C = 477 J/kgK, ρ = 7900 kg/m3

    Ti = 450°C, T = 150°C

    h = 85 W/m2k, t = 1500 s

    Bi=hR/2k=0.214>0.1

    (Lumped analysis not applicable)  

    For Heisler chart,

    Bi = hR/k = 0.428

    1Bi=2.337

    F0=αtL2=αtR2=1.053

    ∴ θ0 = 0.49

    0.49=TTTiT

    T = T + 0.49 (Ti – T)

    T = 297°C

  • Question 11
    2 / -0.33
    A cylindrical insulated pipe of diameter 0.3m has a surface temperature of 330K and is exposed to surrounding s at 298K. The emissivity and absorptivity of the pipe material are 0.89 and 0.92 respectively. If the surrounding are assumed to be transparent, find the % contribution of radiation to total heat transfer rate to surroundings. (Take convective heat transfer coefficient = 4.5 W/m2K)
    Solution

    Concept:

    Heat transfer due to radiation, Q=σA(ϵT14αT24)

    Heat transfer due to convection, = hAΔT 

    Calculation:

    d = 0.3 m, Ts = 3.30 K = T1, T = 298K = T2, E = 0.89, α = 0.92

    σ = 5.67 × 10-8 W/m2 k4

    Q1 = σA ( T14E - α T24 )

    Q1 = 187.078 A

    Q2 = hA ΔT

    Q2 = 4.5 A (330 - 298)

    Q2 = 144 A

    %ofradiation=θ1θ1+θ2×100

    %ofradiation=187.078331.078×100

    % of radiation = 56.5057     

  • Question 12
    2 / -0.33

    Liquid metal flows through a circular tube in which the velocity and temperature profiles are governed by:

    U(r)=A;T(r)TW=B[1(rR)2]

    Where, TW and R are the wall radius and temperature respectively, and A, B are constants. The bulk mean temperature of water TM is given by:
    Solution

    Concept:

    Just as the absence of a free stream velocity requires use of a mean velocity to describe an internal flow, the absence of a fixed free stream temperature necessitates using a mean temperature.

    The mean (or bulk) temperature of a fluid at a given cross section is defined in terms of the thermal energy transported by the fluid it moves past the cross section.

    The bulk mean temperature of internal flow is given by the expression:

    TM=2UMR20RT(r)U(r)rdr

    Calculation:

    Given:

    As, U(r) = A, so the mean velocity UM = A

    T(r)TW=B[1(rR)2]

    TM=2AAR20R{TW+B[1(rR)2]}rdr

    TM=2R20R{TWr+B[r1R2r3]}dr

    TM=2R2[{TWr22+B[r221R2r44]}]0R

    TM=2R2{TWR22+B[R22R24]}=2R2{TWR22+B(R24)}=TW+B2

    TM=TW+B2
  • Question 13
    2 / -0.33
    A fluid (k = 0.6 W/m°C and ν = 10-5 m2/s) enters a 10mm diameter and 15 m long tube at 298 K with a velocity of 0.5 m/s and leave the tube at 330 K. The tube is subjected to a uniform heat flux of 3 kW/m2 on its surface. Calculate the temperature of the surface at the exit in (K)
    Solution

    Concept:

    ReynoldNumber(Re)=ρvDμ

    For constant heat flux, Nusselt number (Nu) = 4.36 and Nu = hd/k

    Heat flux (Q/A) = h(∆T)

    Calculation:

    Given:

    ν = 10-5 m2/s, D = 0.01 m, L = 15 m, V = 0.5 m/s

    Twi = 298 K = 25°C, Two = 330 K = 57°C

    Now,

    Re=VDν=0.5×0.01105=500<2300(laminar)

    For constant heat flux in laminar flow

    Nu = 4.36

    hdk=4.36

    ∴ h = 4.36 × 0.6/0.01

    ∴ h = 261.6 W/m2°C

    Heat flux = h (ΔT)

    ΔT=3000261.6=11.467

    Ts=11.467+57=68.46C

    ∴ Ts = 68.46 °C = 341.46 K
  • Question 14
    2 / -0.33

    A plate of dimensions 35 cm × 15cm (width) is kept vertically. The plate is initially at a temperature of 413 K and the ambient temperature may be taken as 293K. Calculate the maximum possible rate of heat dissipation (W) from both sides neglecting any radiation effects and use the following data for reference.

    Nu = 0.60(Ra) 0.25, ϑ = 41.09 × 10-6 m2/s, Pr = 0.723, k = 0.05W/mK
    Solution

    Concept:

    Rayleigh number (Ra) is given by,

    Ra=Gr.Pr=gβΔTL3Prϑ2

    Where, β=1Tf and Tf = Mean temperature

    Nusselt number, Nu = 0.60(Ra)0.25 = hL/K

    We will calculate ‘h’ from this

    And then calculate the heat dissipation as, Q = 2hA∆T

    Calculation:

    Ra = Gr – Pr =gβΔTL3V2Pr

    β = 1/Tf, and

    ∵  Tf=293+4132=353K=80C

    Now,

    Ra=9.81×1353×120×(0.35)3(41.09×106)2×0.723

    Ra= 61227382.83

    Nu = 0.60 (Ra) 0.25 = 53.07 = hL/k

    h=53.07×0.050.35

    h = 7.58 W/m2K

    Now,

    Q = 2hA ΔT

    ∴ Q = 95.53 watts.
  • Question 15
    2 / -0.33
    A long rod 20 mm in diameter has one end maintained at 110°C. The surface of the rod is exposed to ambient air at 20°C with convection coefficient of 5 W/m2k. Calculate the heat loss (in W) from the rod having thermal conductivity of 15 W/mK.
    Solution

    m=hpkAc

    Concept:

    Heat dissipation from an infinitely long fin:

    Qlongfin=kAcdTdx|x=0=hpkAc(TbT)

    TTToT=emx

    Heat dissipation form a fin insulated at the flip negligible heat loss fin tip.

    θθo=TTToT=cosh(m(lx))cosh(ml)

    Qinsulatedfin=kAcdTdx|x=0=hPkAc(ToT)tanh(mL)

    Calculation:

    A long rod applies that it is an infinity long fin

    Given: D = 20 mm = 0.02 m, Tb = 110°C, T = 20°C

    h = 5 W/m2K , k = 15 W/mK

    P = πd = π (0.02) = 0.063 m

    A=π4d2=π4(0.02)2=3.14×104m2

    Q=hPkA(TbT)

    Q=5×0.063×3.14×104×15(11020)

    Q = 0.03852 × 90 = 3.47 W
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