An Omega Engineering cartridge heater is embedded within a circular, stainless-s

Question

An Omega Engineering cartridge heater is embedded within a circular, stainless-steel (SS304) rod. The cartridge heater has a diameter of ¼ inch, a length of 10 inches and provides a total power of 500W. The steel rod has a length of 10 inches, and an outer diameter of ½ inch. The purpose of the steel rod is to heat water flowing across it at a temperature of 22C, with a heat transfer coefficient of 250 W/(m2-K). The steel rod is attached to a wall, which is maintained at a temperature of 22C. You may include the following assumptions: a) steady state exists, b) heat conduction within the cartridge heater can be neglected, c) constant properties, d) no power from the cartridge heater is dissipated from its ends, only on its peripheral surface, e) you may neglect the presence of the cartridge heater itself, besides being the source of the heat, and f) radial variation of the temperature can be neglected. Determine the maximum temperature of the steel rod, and provide a sketch of its spatial variation. You must include a derivation of an O.D.E. for the temperature variation along the steel rod based on a differential control volume and provide the details of its solution

Explanation / Answer

solution:

1)one dimensional steady state heat cnduction equation with heat generation in cylidrical coordinate is given by

(1/r)(d/dr)(r*(dT/dr))+g/K=0

g=heat generation per unit volume

K=thermal conductivity of ss304 steel=14 w/mk from reference

here hat generation takes place inside heater and no heat generated inside steel rod so above equation becomes

(1/r)(d/dr)(r*(dT/dr))=0

on differentiating we get that

dT/dr=C1/r

and T=C1ln(r)+C2

3)here boundary conditions are such that at outer surface to heat get convected so we can write that

Q=h*Ao*(To-Ts)

500=250*(pi*.0128*.256)(To-295)

To=489.28 K

outer surface of cylinder is To=489.28 K

4)where at inner surface total heat get conducted so we can write that

Q=-K*Ai*(dT/dr)

500=-14*(2*pi*ri*.256)*(C1/ri)

C1=-22.2035

5)for getting second constant C2 we can write that for outer surfac as

and T=C1ln(r)+C2

To=-22.2035ln(ro)+C2

ro=.0128/2 m

we get C2=377.1199

6)so temperature distribution inside steel rod is given by

T=-22.2035ln(r)+377.1199

where maximum temperature occure at inner surface so we get maximum temperature as

T=-22.2035ln(ri)+377.1199

at r=ri=6.4*10^-3/2

Tmax=504.67 K

7)so maximum temperature occure at inner surface and that is

Tmax=504.67 k

and temperature distribution is given by

T=-22.2035ln(r)+377.1199

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.