#2. In a few weeks, we will be covering convection in more detail, including how
ID: 2074170 • Letter: #
Question
#2. In a few weeks, we will be covering convection in more detail, including how to Metal foil -0.02 estimate convection heat transfer 200 coefficients. In many cases, these Fluid. T:-20°C coefficients are calculated using correlations that were developed by Slab collecting experimental data. In one experimental set-up, a very thin sheet of metal foil with low emissivity (0.02) is attached onto the top surface of a 25-mm thick slab with low thermal conductivity (0.023 Wm-K). The bottom of the slab has a temperature of 20°C. The upper surface of the metal foil is exposed to convection heat transfer with a fluid flowing over the foil at 20°C flow and to radiation heat transfer with the surroundings at 20°C. The metal foil is heated electrically with a uniform heat flux of 5000 W/m2. The surface area of the slab (and the metal foil) is 0.09 m2 25 mm a) With air flowing over the metal foil, the surface temperature of the foil (T) is measured at 150°C. What is the convection heat transfer coefficient? (37 W/m2-K) b) With water flowing over the metal foil, the surface temperature of the foil (Ti) is measured at 30°C. What is the convection heat transfer coefficient? (499 W/m2-K) For the results in part b), compare the convection, conduction, and radiation thermal resistances. Based on these resistance values, explain why the experimental set-up uses a foil with low emissivity and a slab with low thermal conductivity c)Explanation / Answer
If Q represent heat flux in metallic foil then
Q=heat loss due to conduction+ heat loss due to convection+heat loss due to radiation
a) with air flowing over the foil
Conduction heat loss per unit area= -K(dT/dx)
-0.023(20-150)/0.025
=119.6w/m^2 (a1)
Convection heat loss per unit area=h(150-20) (b1)
Radiation heat loss per unit area= e×k×((150+273)^4-(273+20)^4)
Where e is emissivity
and k is boltzmann constant
Which is 5.67×10^(-8)
substituting values of e and K
We get radiation loss-27.948w/m^2 (c1)
Now Q= a1+b1+c1
5000=119.6+ h(130)+ 27.948
4852.452=130h
h=37.326w/m^2/k
b) with water flowing over foil
Conduction heat loss per unit area = -K(dT/dx)
-0.023×(20-30)/.025
=9.2w/m^2
Convection heat loss per unit area= h(30-20)
=10h
Radiation heat loss per unit area= .02×5.67×10^(-8)×((303)^4-(293)^4)
=1.200w/m^2
Again
Q= conduction loss+ convection loss+ radiation loss
Substituting values
5000=9.2+ 10h +1.2
10h=4989.6
h=498.96w/m^2/k
For part b
Conduction resistance=dx/KA
= 0.025/(0.023×0.09)
12.077 k/w
Convection resistance= 1/hA
= 1/(498.96×0.09)
=0.02222k/w
Radiation resistance= 1/e×A
=1/((0.02×0.09)
555.555k^4/w
The above results show that resistance in convection is much lesser than the heat transfer through either conduction or radiation mode. So better result can be obtained for heat tetransf coefficient.
The reason behind low emissivity and low conductivity is to keep radiation and conduction resistances on higher side respectively and then major heat loss can approximated to convection resistance
Q= Q(conduction) + Q(convection) +Q(radiation)
When Q(conduction) and Q(radiation) are small than Q can be approximated to Q(convection) and hence more accurately heat transfer coefficient can be obtained.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.