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Flow of Heat in a Body Ralph Marwin Corpuz BSME 1310928 College of Engineering and Technology New Era University, Quezon City [email protected]

Heat is very essential in every day’s living and in fact, is also plays a great impact in studying thermodynamics. It is defined, scientifically, as an energy in process of transfer between a system and its surroundings, other than as work or with the transfer of matter. Using the methods of solving differential equations, we get an expression for the heat flowing through the body having an area A normal to the direction of the heat “the rate of flow of heat through a cylindrical shell is directly proportional to the difference of the inner and outer temperatures and inversely proportional to the natural logarithm of the ratio of the outer to the inner radius.” we have observed the performances how heat flows in a body using the derive equation. The constant quantity of heat flowing through the body is equal to the negative of the product of constant proportionality, which is a property of the making up the body and is known as the “thermal conductivity”, and the cross-sectional area and the temperature gradient with respect to direction x normal to the direction of the flow.

Introduction Heat is very essential in every day’s living and in fact, is also plays a great impact in studying thermodynamics. It is defined, scientifically, as an energy in process of transfer between a system and its surroundings, other than as work or with the transfer of matter. Heat is transferred by Conduction, Convection, Radiation, or more normally, by a combination of two or more of them. In this study, we must consider the Laws on Heat Flow in a Body (Derive from experimental Studies): (a) “The quantity of heat (calorie/seconds) flowing in body is proportional to its mass (in grams) and to its temperature (ᵒC).” (b) “Heat flows from higher to a lower temperature.” (c) “The rate of flow (calorie/second) across an area (square centimeter) is proportional to the area and to the temperature gradient (i.e. the rate of

change of the temperature relative to the direction perpendicular to the area.)” Then, if Laws (a) and (b) are applied to a body under “steady-state” heat flow conditions (flow rate of heat is independent of time) the following relation is obtained,

q=−KA

du dx

where, q is the constant quantity of heat flowing through the body having an area A (square centimeter) normal to the direction of the heat (calorie/second). The temperature at any point on the body (ᵒC) is denoted by u.

du dx

is the temperature gradient relative to

the direction x (cm.) normal to the direction of flow. K is the constant of proportionality which is a property of the material making up the body and is known as the “thermal conductivity” (calorie/cm.deg.sec.)

Methodology If the above relation is applied to a “steady-state” heat flow of steam, in particular, through a cylindrical pipe with an insulating material of thermal conductivity K, then for length L (cm) of the pipe and the inner and outer radii of the insulating material x1 and x2 (cm), we have,

q=−KA

du dx

with A is equal to 2

π

(1)

multiply by the

length L and x (x1≤x≤x2).

A=2 πLx

Substituting this to Equation (1),

q=−2 πKLx

du dx

And, divide both side by

dx x

MAT 311 - Applications of Differential Equations

q

dx =−2 πKLdu . x

Then by integrating both side, x2

q∫ x1

u2

dx =−2 πKL ∫ du x u1

we obtain,

q=

2 πKL(u 1−u 2) ln ⁡( x 2/x 1)

which states that “the rate of flow of heat through a cylindrical shell is directly proportional to the difference of the inner and outer temperatures and inversely proportional to the natural logarithm of the ratio of the outer to the inner radius.”

Results and Discussions The Heat Flow concept is very huge when in terms of applications because we can simply observe it from our surroundings and environment. Best way to show its application is by solving this real-scenario problems: (1) A pipe 20 cm in diameter contains steam at 100ᵒC and is covered with a certain insulation 5 cm in thickness. The outside temperature is kept at 40ᵒC. By how much should the thickness of insulation be increased in order that the rate of heat loss shall be decreased by 20%? (2) A steam pipe having a radius of 3cm and with a steam temperature of 100ᵒC is wrapped with a 1 cm layer of insulation of thermal conductivity 0.0003 calorie/cm.deg.sec. and then that layer is wrapped with a 2cm layer of insulation of thermal conductivity 0.0002 calorie/cm/deg.sec. At what temperature must the outside surface be maintained in order that 0.008 calorie/sec will flow from each square cm of pipe surface? (3) A pipe 10 cm in diameter contains steam at 100ᵒC. is to be covered with two coats of insulation, each 2.50 cm thick, the inner with k= 0.00017 calorie/cm.deg.sec. and the outside surface temperature is 30ᵒC, find the heat loss per hour for a meter length of pipe.

1. For the solution in problem (1), we use the derived basic equation:

q=

2 πKL(u 1−u 2) ln ⁡( x 2/x 1)

where, from the given values,: x1 = 10 cm x2 = 15 cm u1 = 100ᵒC (constant) u2 = 40ᵒC (fixed) Under the given condition, the original amount of heat flow rate is,

q=

2 πKL(100−40) ln ⁡( 15/10) q=

or

120 πKL ln ⁡1.5

For the rate of heat loss to decrease by 20% the new rate is, qn = 0.80 qo

2 πKL(100−40) 120 πKL =0.80 ln ⁡( x /10) ln 15 Or

0.80 ( lnx−ln10 )=ln 1.5 lnx=2.81

x=16.6 cm So, the thickness of the insulation must be increased by

16.60−15=1.60 cm

1.60 cm. 2. Here, two (2) insulations are involved. Basic Equation:

q=

2 πKL(u 1−u 2) x2 ln ⁡( ) x1

For the first insulation, X1 = 3 cm ; u1 = 100ᵒC X2 = 4cm ; u2 =? K1 = 0.0003 cal/cm.deg.sec. L=

1 2 π (3)

or 2πL = 1/3 cm

q = 0.0080 cal/sec substitute the known values,

MAT 311 - Applications of Differential Equations

4 ¿ 3 ¿¿ ¿ ln ¿

q=

q=0. 120 π (u 2−30)

1 ( )(0.0003)(100−u 2) 3 0.0080= ¿ From which, u2 = 77ᵒC For the second insulation, X2 = 4cm; u2 = 77ᵒC X3= 6cm; u3 = ? K2= 0.0002 cal/cm.deg.sec. 2πL= 1/3 cm (same as in the first) q= 0.0080 cal/sec substitute known values,

6 ln ⁡( ¿¿ 4) ( 1/3)(0.00020)(77−u 3) 0.0080= ¿

Simplify to get

u 3=28.3 ᵒC .

3. Problem number three (3) is similar to problem two (2). Use the basic equation:

q=

2 πKL ( u 1−u 2 ) ln ( x 2/ x 1 )

For the first insulation, X1 = 5 cm ; u1 = 100ᵒC X2 = 7.5 cm ; u2 =? K1 = 0.00060 cal/cm.deg.sec. L=

1 m(100 cm)

Substitute to the basic relation,

2 π (0.00017)(100)(100−u2) q= ln ⁡( 10 /7.5) q=0.30 π (100−u 2) . For the second insulation, X1 = 7.5 cm ; u2 = ? X2 = 10 cm ; u3 = 30ᵒC K1 = 0.00017 cal/cm.deg.sec. L=

2 π (0.00017)(100)(u 2−30) ln ⁡( 10 /7.5)

1 m(100 cm)

Substitute known values,

Equate the q’s of the previous solutions,

0.30 π ( 100 u 2 )=0.120 π (u 2−30) From which,

u2 = 80ᵒC.

So, using any of the two values of q,

q=0.30 π ( 100−80 )=18.85 cal /sec = (67,860 cal/hr) Conclusions and Recommendations In summary, we have observed the performances how heat flows in a body using the derive equation. The constant quantity of heat flowing through the body is equal to the negative of the product of constant proportionality, which is a property of the making up the body and is known as the “thermal conductivity”, and the crosssectional area and the temperature gradient with respect to direction x normal to the direction of the flow. The first problem is implies in a circular cylindrical steam pipe concerning about the increased of the insulation in order that the rate of heat loss be decreased. So, first, they have to substitute the initial conditions equating it to the expression made for the rate of heat loss to decrease. In the second problem, two (2) insulations are involved finding the temperature outside surface in order that certain heat will flow from each square cm of pipe surface. The third problem is solved the same way as the second (2nd) because they’re both having pipes with two insulating material. References [1] https://www.teachengineering.org/view_lesso n.php? url=collection/cub_/lessons/cub_housing/cub_ housing_lesson01.xml (Retrieved Oct. 12, 2015).

MAT 311 - Applications of Differential Equations

[2] R.A.D., H.T.T., A.D.T., Engineering Mathematics Made Easy:Theorethical Applications of Ordinary Differential Equations. [3] http://www.engr.sjsu.edu/trhsu/Chapter %203%20First%20order%20DEs.pdf (Retrieved Oct. 12, 2015).

MAT 311 - Applications of Differential Equations