Heat Equations of Change II

7. Heat Equations of Change

7.1. Derivation of Basic Equations

7.1.1. Differential Equation for Heat Conduction

7.1.2. Energy Equation

7.1.3. Buckingham Pi Method

7.2. Unsteady-state Conduction

7.2.1. Gurney-Lurie Charts

7.2.2. Lumped Systems Analysis

Outline

Quiz – 2014.02.14

An electrically heated resistance wire has a diameter of 2 mm and a resistance of 0.10 ohm per foot of wire. The thermal conductivity of the wire is 20 W/m·K. At a current of 100 A, calculate the steady state temperature difference between the center and surface of the wire.

TIME IS UP!!!

From the previous lecture…

A 10-cm diameter nickel-steel sphere has a thermal conductivity, k = 10 W/m-K. Within the sphere, 800 W/m3 of heat is being generated. The surrounding air is at 20°C and the heat transfer coefficient from the surroundings to the surface of the sphere is 10 W/m2-K. What is the temperature at the center of the sphere?

Exercise!

A Comparison Between Methods

Energy Equation:Solution!

We are left with: 0=𝑘[ 1𝑟2 𝜕𝜕𝑟 (𝑟2 𝜕𝑇𝜕𝑟 )]+𝑔Because T depends on r only: 0=

𝑑𝑑𝑟 (𝑟2 𝑑𝑇𝑑𝑟 )+𝑔𝑟

2

𝑘

𝜌𝑐𝑝𝐷𝑇𝐷𝑡 =𝑘𝛻2𝑇 +𝑔

𝜌𝐶𝑝𝐷𝑇𝐷𝑡 =− (𝛻 ∙𝑞 )− (𝜏 ∙𝛻 𝒗 )

In spherical coordinates:

From the Energy Equation

For solids with generation

+𝑔

1𝑟2

𝜕𝜕𝑟 (𝑟 2 𝜕𝑇𝜕𝑟 )+ 1

𝑟2 sin𝜃𝜕𝜕𝜃 (sin𝜃 𝜕𝑇𝜕𝜃 )+ 1

𝑟 2sin 2𝜃𝜕2𝑇𝜕𝜙2

+𝑔𝑘=

1𝛼𝜕𝑇𝜕𝑡

A Comparison Between Methods

Differential Equation of Heat Conduction:

Solution! Another way!!

We are left with: 0= 1𝑟2

𝜕𝜕𝑟 (𝑟2 𝜕𝑇𝜕𝑟 )+𝑔𝑘

Because T depends on r only:

𝛻2𝑇+𝑔𝑘=

1𝛼𝜕𝑇𝜕𝑡

In spherical coordinates:

From the Diff. Eqn of Heat Conduction0= 𝑑

𝑑𝑟 (𝑟2 𝑑𝑇𝑑𝑟 )+𝑔𝑟2

𝑘

A Comparison Between Methods

Overall Shell Heat Balance:

Solution! Another way again!! The Shell:

Rate of Heat IN:

Rate of Heat OUT:

Generation:

(4𝜋 𝑟2𝑞𝑟❑)|𝑟(4𝜋 𝑟2𝑞𝑟❑)|𝑟+∆𝑟(4𝜋𝑟 2∆𝑟 )𝑔

(𝑟 2𝑞𝑟❑)|𝑟+∆𝑟− (𝑟 2𝑞𝑟❑ )|𝑟∆𝑟 =𝑔𝑟2

Adding the terms and dividing :

𝑑𝑑𝑟 (𝑟 2𝑞𝑟 )=𝑔𝑟2

A Comparison Between Methods

Overall Shell Heat Balance:

Solution! Another way again!! The Shell:

Rate of Heat IN:

Rate of Heat OUT:

Generation:

(4𝜋 𝑟2𝑞𝑟❑)|𝑟(4𝜋 𝑟2𝑞𝑟❑)|𝑟+∆𝑟(4𝜋𝑟 2∆𝑟 )𝑔

− 𝑑𝑑𝑟 (𝑟2𝑘 𝑑𝑇𝑑𝑟 )=𝑔𝑟2

Inputting Fourier’s Law:

𝑑𝑑𝑟 (𝑟 2𝑞𝑟 )=𝑔𝑟2

A Comparison Between Methods

Overall Shell Heat Balance:

Solution! Another way again!! The Shell:

Rate of Heat IN:

Rate of Heat OUT:

Generation:

(4𝜋 𝑟2𝑞𝑟❑)|𝑟(4𝜋 𝑟2𝑞𝑟❑)|𝑟+∆𝑟(4𝜋𝑟 2∆𝑟 )𝑔

Inputting Fourier’s Law:

From the Overall Shell Heat Balance0= 𝑑

𝑑𝑟 (𝑟2 𝑑𝑇𝑑𝑟 )+𝑔𝑟2

𝑘− 𝑑𝑑𝑟 (𝑟2𝑘 𝑑𝑇𝑑𝑟 )=𝑔𝑟2

A Comparison Between Methods

Solution! Apparently…

0= 𝑑𝑑𝑟 (𝑟2 𝑑𝑇𝑑𝑟 )+𝑔𝑟

2

𝑘From the Overall

Shell Heat Balance

From the Diff. Eqn of Heat Conduction0= 𝑑

𝑑𝑟 (𝑟2 𝑑𝑇𝑑𝑟 )+𝑔𝑟2

𝑘

0= 𝑑𝑑𝑟 (𝑟2 𝑑𝑇𝑑𝑟 )+𝑔𝑟

2

𝑘From the

Energy Equation

Using any method below, we can obtain the same ODE to solve!!

Increasing range of

applicability of the method

A Comparison Between Methods

Solution!

0= 𝑑𝑑𝑟 (𝑟2 𝑑𝑇𝑑𝑟 )+𝑔𝑟

2

𝑘From the Overall

Shell Heat Balance

From the Diff. Eqn of Heat Conduction0= 𝑑

𝑑𝑟 (𝑟2 𝑑𝑇𝑑𝑟 )+𝑔𝑟2

𝑘

0= 𝑑𝑑𝑟 (𝑟2 𝑑𝑇𝑑𝑟 )+𝑔𝑟

2

𝑘From the

Energy Equation

The next step is to solve the ODE and define boundary conditions.

Integrating twice:

𝑇 (𝑟 )=− 𝑔𝑟2

6𝑘 −𝐶1

𝑟 +𝐶2

Boundary conditions:

𝑎𝑡 𝑟=0 , 𝑞𝑟 𝑖𝑠 𝑓𝑖𝑛𝑖𝑡𝑒

¿𝑎𝑡 𝑟=𝑅 , 𝑘 𝑑𝑇𝑑𝑟 =h (𝑇 𝑎𝑖𝑟−𝑇 (𝑅 ) )

A Comparison Between Methods

Solution!

Applying the boundary conditions:

𝑇 (𝑟 )=𝑇 𝑎𝑖𝑟+𝑔𝑅2

6 𝑘 [1−( 𝑟𝑅 )2]+𝑔𝑅3h

𝑘=10 𝑊𝑚𝐾h=10 𝑊

𝑚2𝐾

𝑅=0.05𝑚𝑔=800𝑊

𝑚3

𝑇 𝑎𝑖𝑟=20 °𝐶

Final Answer:

21.37°C

7. Heat Equations of Change

7.1. Derivation of Basic Equations

7.1.1. Differential Equation for Heat Conduction

7.1.2. Energy Equation

7.1.3. Buckingham Pi Method

7.2. Unsteady-state Conduction

7.2.1. Gurney-Lurie Charts

7.2.2. Lumped Systems Analysis

Outline

Example!

1-D Transient Heat Conduction

Consider a flat slab of thickness L with an initial temperature of T0. It is later submerged in a large fluid with temperature T1 (> T0). Neglecting surface resistance (infinite h), determine the temperature at the center of the slab at any time t.

Example!

1-D Transient Heat Conduction

What do we expect to happen?

• The surface temp. is held at T1. (because of infinite h)

• Temperature profile at any time is symmetrical

• Unidirectional (x only) heat flow only

• The temp. at the center, T0, slowly approaches T1.

T0

T1

Example! Solution:

1-D Transient Heat Conduction

T0

T1

Differential Equation of Heat Conduction:

𝛻2𝑇+𝑔𝑘=

1𝛼𝜕𝑇𝜕𝑡

With the assumptions:𝜕𝑇𝜕𝑡 =𝛼 𝜕

2𝑇𝜕𝑥2

Initial and Boundary conditions:

Initial:

Boundary:

Example! Solution:

1-D Transient Heat Conduction

𝜕𝑇𝜕𝑡 =𝛼 𝜕

2𝑇𝜕𝑥2

To facilitate solving, we will make this dimensionless!

Let: Y = dimensionless temp.

𝑌=𝑇1−𝑇𝑇 1−𝑇 0

*The choice of definition of Y is arbitrary.

𝜕𝑌𝜕𝑡 =

−1𝑇 1−𝑇0

𝜕𝑇𝜕𝑡

Implications:

𝜕𝑌𝜕𝑥 =

−1𝑇 1−𝑇0

𝜕𝑇𝜕 𝑥

𝜕2𝑌𝜕 𝑥2

=−1

𝑇 1−𝑇0𝜕2𝑇𝜕𝑥2

− (𝑇1−𝑇 0 ) 𝜕𝑌𝜕𝑡 =−𝛼 (𝑇 1−𝑇 0 ) 𝜕2𝑌𝜕 𝑥2

1-D Transient Heat Conduction

𝜕𝑌𝜕𝑡 =𝛼 𝜕

2𝑌𝜕 𝑥2

To facilitate solving, we will make this dimensionless!

Let: X = dimensionless length

𝑋=𝑥𝐿

*The choice of definition of X is arbitrary.

𝜕 𝑋=1𝐿 𝜕𝑥

Implications:

𝜕𝑌𝜕𝑡 =

𝛼𝐿2𝜕2𝑌𝜕 𝑋 2

𝜕 𝑋 2=( 1𝐿𝜕𝑥)2

=1𝐿2𝜕𝑥2

Example! Solution:

1-D Transient Heat Conduction

𝜕𝑌𝜕𝑡 =

𝛼𝐿2𝜕2𝑌𝜕 𝑋 2

To facilitate solving, we will make this dimensionless!

Let: τ = dimensionless time

𝜏=𝛼𝑡𝐿2

*The choice of definition of τ is arbitrary.

𝜕𝜏= 𝛼𝐿2𝜕𝑡

Implications:

𝜕𝑌𝜕𝜏 =

𝜕2𝑌𝜕 𝑋 2

Example! Solution:

1-D Transient Heat Conduction

𝜕𝑌𝜕𝜏 =

𝜕2𝑌𝜕 𝑋 2

DIMENSIONLESS FORM:

Previous IC & BC: New IC & BC:

Easier to solve and no worries with units!

𝑌=1 , 𝜏=0 , 𝑋=𝑋𝑌=0 , 𝜏=𝜏 , 𝑋=0𝑌=0 , 𝜏=𝜏 , 𝑋=1

𝜏=𝛼𝑡𝐿2

𝑋=𝑥𝐿𝑌=

𝑇1−𝑇𝑇 1−𝑇 0

Example! Solution:

A new dimensionless number…

1-D Transient Heat Conduction

𝜏=𝛼𝑡𝐿2

𝑋=𝑥𝐿𝑌=

𝑇1−𝑇𝑇 1−𝑇 0

Dim. Group Ratio EquationFourier, Fo Rate of heat conduction/

Rate of heat storage

Rate of heat conduction:

𝑘𝐿3

Rate of heat storage: 𝜌𝑐𝑝𝐿3/𝑡

Going back…

1-D Transient Heat Conduction

𝜕𝑌𝜕𝜏 =

𝜕2𝑌𝜕 𝑋 2

DIMENSIONLESS FORM:

New IC & BC:

𝑌=1 , 𝜏=0 , 𝑋=𝑋𝑌=0 , 𝜏=𝜏 , 𝑋=0𝑌=0 , 𝜏=𝜏 , 𝑋=1

𝜏=𝛼𝑡𝐿2

𝑋=𝑥𝐿𝑌=

𝑇1−𝑇𝑇 1−𝑇 0

T0

T1

Going back…

1-D Transient Heat Conduction

𝜕𝑌𝜕𝜏 =

𝜕2𝑌𝜕 𝑋 2

DIMENSIONLESS FORM:

New IC & BC:𝑌=1 , 𝜏=0 , 𝑋=𝑋𝑌=0 , 𝜏=𝜏 , 𝑋=0𝑌=0 , 𝜏=𝜏 , 𝑋=1

𝜏=𝛼𝑡𝐿2

𝑋=𝑥𝐿𝑌=

𝑇1−𝑇𝑇 1−𝑇 0

Solution: (A Fourier series)

𝑌 (𝜏 ,𝑋 )= 4𝜋∑𝑛=0

∞ 𝑒− (2𝑛+1)2 𝜋2𝜏

2𝑛+1sin [ (2𝑛+1 ) 𝜋 𝑋 ]

Gurney-Lurie Charts

Gurney-Lurie Charts

𝜏=𝛼𝑡𝐿2

𝑌=𝑇1−𝑇𝑇 1−𝑇 0

- plots of the dimensionless temperature Y against Fo with varying Bi and X for different geometries.

- Each point in the curves are solutions to the PDE involving heat conduction + convection.

𝑋=𝑥𝐿

For flat plates with convection

Gurney-Lurie Charts

𝜏=𝛼𝑡𝐿2

𝑌=𝑇1−𝑇

𝑇1−𝑇0

Bi

Geankoplis, Figure 5.3-6

For long cylinders with convection

Gurney-Lurie Charts

𝜏=𝛼𝑡𝐿2

𝑌=𝑇1−𝑇

𝑇1−𝑇0

Bi

Geankoplis, Figure 5.3-8

For solid spheres with convection

Gurney-Lurie Charts

𝜏=𝛼𝑡𝐿2

𝑌=𝑇1−𝑇

𝑇1−𝑇0

Bi

Geankoplis, Figure 5.3-10

Gurney-Lurie Charts

How do we compute the Biot Numbers for different geometries?

Recall: Biot Number 𝐵𝑖=h 𝑥1𝑘

Characteristic length: Volume/Surface Area

From a while ago…

A 10-cm diameter nickel-steel sphere has a thermal conductivity, k = 10 W/m-K. Within the sphere, 800 W/m3 of heat is being generated. The surrounding air is at 20°C and the heat transfer coefficient from the surroundings to the surface of the sphere is 10 W/m2-K. What is the temperature at the center of the sphere?

Recall this exercise:

From a while ago…

A 10-cm diameter nickel-steel sphere has a thermal conductivity, k = 10 W/m-K. Within the sphere, 800 W/m3of heat is being generated.The sphere, initially at T0 = 30°C, is suddenly submerged into a fluid…The surrounding air is at and the heat transfer coefficient from the surroundings to the surface of the sphere is 10 W/m2-K. What is the temperature at the center of the sphere across time?

Recall this exercise:

From a while ago…

To solve the new problem:Differential Equation of Heat Conduction:

𝛻2𝑇+𝑔𝑘=

1𝛼𝜕𝑇𝜕𝑡

1𝑟2

𝜕𝜕𝑟 (𝑟2 𝜕𝑇𝜕𝑟 )+ 1

𝑟2 sin𝜃𝜕𝜕𝜃 (sin𝜃 𝜕𝑇𝜕𝜃 )+ 1

𝑟 2sin2𝜃𝜕2𝑇𝜕𝜙2

+𝑔𝑘=

1𝛼𝜕𝑇𝜕𝑡

In spherical coordinates:

Steps:1. Turn the remaining PDE into

dimensionless form.2. Solve analytically for T(r, t).

But there is an easier way!

Lumped Systems Analysis

Let’s assume that the sphere is too small for conduction to matter. The temperature distribution inside the sphere can, therefore, be assumed uniform!

Heat Balance: ( 𝑅𝑎𝑡𝑒𝑜𝑓 h𝑒𝑎𝑡 𝑓𝑙𝑜𝑤𝑖𝑛𝑡𝑜𝑠𝑜𝑙𝑖𝑑𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒𝑉 h h𝑡 𝑟𝑜𝑢𝑔𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑠𝑢𝑟𝑓𝑎𝑐𝑒𝑠 𝐴 )=(𝑅𝑎𝑡𝑒𝑜𝑓 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑜𝑓𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓

𝑠𝑜𝑙𝑖𝑑𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒𝑉 )h𝐴 (𝑇 ∞−𝑇 (𝑡 ) )=𝜌𝑐𝑝𝑉

𝑑𝑇 (𝑡)𝑑𝑡

Initial Condition:

𝑇 (𝑡 )=𝑇 0 , 𝑡=0𝑑𝑇 (𝑡)𝑑𝑡 = h𝐴

𝜌𝑐𝑝𝑉(𝑇 ∞−𝑇 (𝑡 ) )

Lumped Systems Analysis

𝑑𝑇 (𝑡)𝑑𝑡 = h𝐴

𝜌𝑐𝑝𝑉(𝑇 ∞−𝑇 (𝑡 ) )Initial Condition:

𝑇 (𝑡 )=𝑇 0 , 𝑡=0𝑑𝑇 (𝑡 )

𝑇 ∞−𝑇 (𝑡)= h𝐴𝜌𝑐𝑝𝑉

𝑑𝑡

− ln(𝑇 ∞−𝑇 (𝑡)𝑇∞−𝑇0 )= h𝐴 𝑡

𝜌𝑐𝑝𝑉Integrating and plugging the IC:

Rearranging: 𝑇 (𝑡 )−𝑇∞𝑇 0−𝑇 ∞

=exp ( − h𝐴 𝑡𝜌𝑐𝑝𝑉 )

Lumped Systems Analysis

Rearranging: 𝑇 (𝑡 )−𝑇∞𝑇 0−𝑇 ∞

=exp ( − h𝐴 𝑡𝜌𝑐𝑝𝑉 )

(−h (𝑉 / 𝐴 )𝑘 )( 𝑘

𝜌𝑐𝑝𝑡

(𝑉 / 𝐴 )2 )

(−𝐵𝑖 ) (𝐹𝑜 )Temperature Profile at the center of the sphere across time:

𝑇 (𝑡 )−𝑇∞𝑇 0−𝑇 ∞

=𝑒−𝐵𝑖𝐹𝑜

Lumped Systems Analysis

Temperature Profile at the center of the sphere across time:

𝑇 (𝑡 )−𝑇∞𝑇 0−𝑇 ∞

=𝑒−𝐵𝑖𝐹𝑜

This Lumped System Analysis is sufficiently accurate only when

Bi < 0.1.

𝑘=10 𝑊𝑚𝐾h=10 𝑊

𝑚2𝐾

𝑅=0.05𝑚𝑇 0=30 °𝐶𝑇 ∞=20 °𝐶

𝐵𝑖=10 𝑊𝑚2𝑘

10 𝑊𝑚𝐾( 0.05𝑚3 )=0.0167<0.1

Given:

Valid!!Checking:

Lumped Systems Analysis

For review:A person is found dead at 5 PM in a room where T = 20°C. The temperature of the body was measured at 25°C when found. The heat transfer coefficient is estimated to be 0.8 W/m2K. Estimate the time of death assuming:

(1) The body can be modeled as a 30-cm diameter, 1.7-m long cylinder.(2) The thermal properties of the body and the heat transfer are

constant.(3) The body temperature was 37°C at the time of death.(4) Since the human body has almost the same properties of water at

the average temperature of 31°C: k = 0.617 W/mK, density = 996 kg/m3, and cp = 4178 J/kgK.

(5) Radiation effects are negligible.