1

NAME: 1.AMAN KUMAR BHAKAT(10)

2.ARKA PROVA MRIDHA(11)

2

HEAT TRANSFER:Heat transfer is thermal energy in transit due to a spatial temperature difference. Whenever there exists a tempearture difference in a medium or between medium heat transfer must occur.There are three different modes of heat transfer-1.conduction 2.convection 3.radiation.

3

CONDUCTION-When a temperature gradient exists in a stationary medium, which may be solid or fluid, heat transfer by conduction will occur across the medium.

CONVECTION-It occurs between a surface and a moving fluid at different temperature.

RADIATION-In the absence of any intervening medium ,the net heat transfer between two surfaces at different temperature in the form of em waves is called radiation.

4

Forced convection-In forced convection the fluid movement results from external surface forces such as fan or pump. It is typically used to increase the rate of heat exchange.Natural convection-Natural convection occurs due to temperature difference which affect the density and hence the relative buyoancy of the fluid.  Heavier (more dense) components will fall while lighter (less dense) components rise, leading to bulk fluid movement. Natural convection can only occur, therefore, in a gravitational field.

5

The cooling of a boiled egg in a cooler enviornment is by natural convection. The temperature of the air adjacent to the egg is higher and thus the density is lower. Thus we have a situation in which some low density gas is surrounded by a high density gas and the light air rises. The space vacated by the warmer air in the vicinity of the egg is replaced by the cooler air and the presence of cooler air in the vicinity of egg speeds up the cooling process. The motion that results from the continual replacement of the heated air by cooler air is called natural convection current and the heat transfer due to this is called natural convection heat transfer.

6

The average nusselt number for free convection on a vertical cylinder is the same as that for avertical plate if the thickness of the thermal boundary layer is much smaller than the cylinder radius. Therefore for an for an vertical cylinder the average nusselt number can be found from vertical plate relations.Nu =0.59* RaL

1/4 when range of Ra is 104-109

Nu=0.1*RaL1/3 when range of Ra is 1010-1013

7

For fluids having Prandtl number 0.7 and higher avertical cylinder may be treated as avertical plate when (L/D)/GrL

1/4<0.025 where D is the diameter of the cylinder.If the above criterion is not satisfied from the plot of ᶓ vs (Num)cyl/(Num)fp for several different values of Pr, (Num)cyl is determined, where ᶓ =(2√2/Grx

1/4)*(x/R)

8

The apparatus consists of a vertical stainless steel tube enclosed in a rectangular duct. Front side of the duct is made of transparent section to facillate visual observation. An electrical heating element embedded in a copper tube acts as a heat source. The surface temperatures is measured at different heights using thermocouples. The surface of the tube is polished to minimize radiation losses . Avoltmeter and an ammeter used for determination of wattage dissipated by water.

9

10

To determine the convective heat transfer co-efficient for heated vertical cylinder losing heat to the ambient by free convection.To find the theoritical convective heat transfer coefficient and to compare with the experimental value.

11

Nu= nusselt number=hLc/kK= thermal conductivity of fluidβ=volume expansion coefficient=(1/v).(δv/δt)p =1/T(for ideal gases)GrL= Grashof number=gβ(Ts-T∞)Lc

3/υ2

RaL=Gr

L*Pr=Rayleigh number

g= gravitational acceleration

Ts=temperature of the surface, T

∞= ambient temperature.

Lc=characteristic length of the geometry, υ= kinematic viscosity

Pr=prandtl number=μcp

/k= υ/α

Tf= (T

s+T

∞)/2=film temperature

SL.NO.

HEAT INPUT TEMP. ON THE SURFACE OF THE PIPE

T(avg)

Amb.Temp.

V A WATTS

T1 T2 T3 T4 T5 T6 T7

1 50

0.25

12.5 47.7

57.6

65.5

87.4

99.4

105 102.6

81.17

32.3

2 60

0.29

17.4 48.8

59.7

72.2

93.2

106 111.7

108.4

85.7 32.6

SERIAL NO. 1: q=hA(Ts-T∞)----(1)

TS=81.17◦C

T∞=32.3◦C

∆T=(TS-T∞) =(81.17-32.3) =48.87

From (1) we get h=q⁄A(TS-T∞)

A=∏dl m2

where d=diameter of pipe=0.045m

l=length of the pipe=0.45m

h=12.5⁄3.14*0.045*0.45*48.87=4.02watt⁄m

 

hexp=4.02 watt⁄ m2kagain,Tf=(T∞+TS)⁄2=(81.17+32.3)⁄3=56.7◦C=330Kᵝ=1⁄Tf=3.03*10-3

Properties of air at 56.7◦cυ=18.9*10-6 m2⁄secPr=0.705 , k=28.52*10-3watt⁄mkGrL Pr = (g ß l3 ∆T Pr)/ᵧ2 =9.81*3.03*10-3*(.45)3*48.87*0.705 =2.61*108 Nu=0.59*(Pr*GrL)0.25 =75htheo = (NuK)/L=(75*28.52*10-3*)/0.45 = 4.75Since, (l/d)/ Gr L

0.25=0.072 > 0.025 =(2√2 / Gr ᶓ L

1/4)(L/R)=0.4from the graph (Num)cyll ⁄ (Num)fp = 1.1(Num)cyll = 1.1*75 =82.5 h = Nu*k⁄l =82.5*30*10-3⁄0.4 =5.5 watt⁄m2*K

15

Weather events such as a thunderstormGlider planesRadiator heatersHot air balloonHeat flow through and on outside of a double pane windowOceanic and atmospheric motions

16

17

18