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MEC203 > Chapter 3 Steady heat conduction > Flashcards

Chapter 3 Steady heat conduction Flashcards

1
Q

How can the wall of a house be modeled

A

steady and one dimensional, and thus the temperature expressed as T(x)

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2
Q

Heat transfer through a wall equation (words)

A

rate of heat transfer into the wall - rate of heat transfer out of the wall = rate of change of the energy of the wall

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3
Q

Heat transfer through a wall equation and how does this simplify in steady conditions

A

Q.in - Q.out = dEwall/dt

dEwall/dt = 0

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4
Q

How can fouriers law of heat conduction be applied to a wall

A

steady 1D Q.conwall = -kA dT/dx

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5
Q

under steady conditions the temperature distribution through a wall is and what does this mean

A

a straight line dT/dx = const, therefore T1-T2/L = dT/dx

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6
Q

For the thermal resistance concept what does current translate as

A

rate of heat transfer ie Q.

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7
Q

For the thermal resistance concept what does voltage translate as

A

temperature difference ie T1-T2

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8
Q

Thermal resistance for conduction

A

L/kA = Rcond

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9
Q

Thermal resistance for convection

A

Rconv = 1/hAs

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10
Q

what happens to Rconv as h increases

A

the thermal resistance tends towards zero, surface offers no rsistance to convection and thus it does not slow down the heat transfer process

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11
Q

Thermal resistance for radiation

A

Rrad = 1/hrad*As

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12
Q

What does hrad =

A

eff * simga *(Ts^2 +Tsurr^2)(Ts + Tsurr)

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13
Q

when does hcombined = hconv + h rad

A

as Tsurr approx Tinf

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14
Q

V=Ir for thermal resistance

A

T1-T2 = Q. * Rtotal

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15
Q

How does thermal resistance compare to electrical resistance

A

exactly the same, you can sum in series and do 1/R for parallel

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16
Q

what are the two assumptions in solving complex multi dimensional heat transfer problems as one dimensional using thermal resistance network

A

any plane normal to the x axis is isothermal (temp variation only in x)
any plane parallel to the x axis is adiabatic

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17
Q

Conduction resistance of a cylinder layer

A

Rcyl = ln(R2/R1)/2Pi()L*k

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18
Q

How to find the conduction resistance of a cylinder

A

use fouriers law -> integrate with respect to dr and dT, convert area to 2PiR*L, to find equation in terms of heat transfer, temperature difference, and geometrey of the cylinder

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19
Q

How does insulation work

A

decreases heat transfer as it increases the thermal resistance

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20
Q

what happens when adding insulation to a cylindrical pipe or spherical shell

A

additional insulation increases the conduction resistance of the insulation layer but decreases the convection resistance of the surface because of the increase in the outer surface area for convecction

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21
Q

Will heat transfer decrease or increase when you add insulation

A

it may do either depending on whether the convection resistance decreases dominates or the conduction resistance increases

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22
Q

How do you calculate the critical radius for a cylinder

A

Rcr = k/h

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23
Q

How do you calculate the critical radius for a sphere

A

Rcr = 2k/h

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24
Q

Draw the critcial radius heat transfer diagram

A

Heat transfer on y axis, radius on x axis, Q.bare and Q.max, increases from Q.bare to Q.max before decreasing again

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25
Q

Why do we use fins to increase heat transfer rates

A

by Newtons law of cooling if all variables are fixed only way to increase heat transfer is via increasing area

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26
Q

What is the fin equation (words)

A

Rate of heat conduction into the element at x = rate of heat conduction from the element at x+delta x + rate of heat convection from the element

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27
Q

what is the fin equation

A

Q.cond = Q.cond x+deltax + Q.conv

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28
Q

How is the area of convection determined for a fin

A

it is the perimeter of the fin times by delta x as looking for surface area in contact with the fluid

29
Q

General equation of a fin

A

d^2*theta/dx^2 - m^2 * theta = 0

30
Q

In the general equation for a fin what does theta stand or

A

T - Tinf

31
Q

In the general equation for a fin what does m stand or

A

m^2 = hp/kAc (Ac = the cross sectional area of a fin)

32
Q

In the general equation for a fin which term refers to conduction

A

d^2 theta / d x^2 (t

33
Q

In the general equation for a fin which term refers to convection

A

m^2 * theta

34
Q

Derive the general solution for the fin equation

A

see exam prep notebook

35
Q

What is the general solution of the differential equation

A

Theta(x) = C1e^mx + C2e^-mx

36
Q

What is the boundary condition at the fin base

A

Theta(0) = Thetab = Tb - Tinf

no heat loss has occured yet so simply temperature of the rest of the base

37
Q

What is the boundary condition given by infinitely long fin

A

Tfin tip = Tinf

Theta(L) = T(L) - Tinf = 0 as L -> inf (relative just long enough such that temperature has reached surrounding)

38
Q

Prove the Q. of a long fin

A

see book, should be equal to Q.longfin = sqrt(h p k Ac) *(Tb - Tinf)

39
Q

How do you calculate the heat transfer from an entire fin

A

Use fouriers law Q. = -k Ac dT/dx, so once you have an equation in theta(x), differentiate it and take it when x =0 to get heat transfer for the entire fin

40
Q

Under steady conditions what is heat transfer equal to in a fin

A

heat transfer from the exposed surfaces of the fin is equal to heat conduction to the fin at the base

41
Q

What does negligible heat loss from the fin tip mean

A

Qfin tip = 0 therefore dtheta/dx at x=L will =0

42
Q

What is the equation for heat transfer for an adiabatic fin tip

A

Q. = Q.longfin = sqrt(h p k Ac) *(Tb - Tinf) * tanh(mL)

43
Q

What does the specified temperature boundary condition mean for fin

A

Tfintip = TL therefore theta(L) = thetaL = TL - Tinf

44
Q

What is the equation for heat transfer for specified fin tip temperature

A

Q.longfin = sqrt(h p k Ac) *(Tb - Tinf) * (cosh(mL) - (TL-Tinf)/(Tb - Tinf)) / sinh(mL)

45
Q

What is the most realistic boundary condition for fin tip

A

convection from the fin tip maybe including radiation, this can be used to determine heat transfer using energy balance at the fin tip

46
Q

Energy balance for fin tip with convection

A

Fouriers at x=L = Newtons

-kAc dT/dx = hAc(T(L) - T inf)

47
Q

What is the heat transfer from a fin with convection fin tip

A

Q.longfin = sqrt(h p k Ac) (Tb - Tinf) * (sinh(mL) + (h/mk)cosh(mL)) / (cosh(mL) + (h/mk) * sinh(mL))

48
Q

What is a more practical way for accounting for heat loss through convection at a fin tip

A

Replace the fin length L with a corrected version where the fin tip can be counted as insulated
such that the eat transfer from a fin of length Lc with insulated tip is equal to heat transfer from the actual fin of length L with convection at the fin tip

49
Q

What is the corrected fin length defined as (equation)

A

Lc = L + Ac/p
Lc rectangular = L + t/2 (t = thickness of in)
Lc cylindrical = L + D/4

50
Q

What is the maximum heat transfer from a thin defined as and what doe sthis assume

A

Q.fin max = h Afin (Tb - Tinf)

zero thermal resistance or infinite thermal conductivity (Tfin = Tb) entire fin at base temp

51
Q

How do you calculate the efficiency of a fin

A 
eff = Q.fin / Q.finmax therefore 
Q.fin = eff *hAfin(Tb-Tinf)
52
Q

What is the efficiency of a long fin =

A

1/mL

53
Q

What is the efficiency of a adiabatic tip

A

tanh(mL)/mL

54
Q

Triangle fins are better than rectangular fins as

A

they contain less material and are more efficient than rectangular profiles

55
Q

Why is fin length limited

A

as efficiency is proportional to 1/L therefore as you increase L the less efficient the fin is

56
Q

What is fin effectiveness

A

It compares how well a fin conducts away heat compared to just the plain surface of the heat source

57
Q

What does fin effectiveness =

A

Q.fin / Q.no fin = Q.fin / hAb(Tb - Tinf) = heat transfer rate from the fin of base area Ab, heat transfer rate from the surface of area Ab

58
Q

Relationship between fin effectiveness and efficiency

A

effective = (Afin/Ab) * efficiency

59
Q

What is the effectiveness of a long fin =

A

sqrt (kp/hAc)

60
Q

What should the effectiveness of a fin always be

A

greater than 1

61
Q

What does the equation for effectiveness of a fin tell us about fin design

A

thermal conductivity should be high as possible (copper, aluminium)
the ratio of the perimeter of the fin to cross sectional area p/Ac should be high as possible -> use slender pin fins
Low convection heat transfer coefficient -> place fins on gas (air) side, (should be external not internal)

62
Q

what is the total rate of heat transfer from a finned surface equal to

A

Q.total fin = Q.unfin + Q.fin

=h(Aunfin + efficiency(fin) * Afin) (Tb - Tinf)

63
Q

What is the overall effectiveness for a finned surface =

A

Q.total fin/ Q. total no fin

64
Q

What is the overall effectiveness equal to

A

depends on fin density, number of fins per unit length as well as the effectiveness of the individual fins

65
Q

How best to judge effectiveness of a finned surface

A

using overal effectiveness not effectiveness of individual fins

66
Q

Why does the region near the fin tip make little or no contribution to heat transfer

A

the gradual temperature drop along the fin means the end there is only a small temperature difference

67
Q

At what value can a fin be considered infinitely long and why

A

when mL = 5, as makes Q.fin / Q.fin long = 1

68
Q

What is the most effective value of mL

A

=1 as offers a good compromise between heat transfer performance and fin size

69
Q

What does Q.fin/Q.finlong =

A

tanh(mL)

MEC203 (12 decks)

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