Heat Transfer –

Heat Transfer –

Practical Lecture 6 (Solved Problems)

40. A solid with volume V and surface area A is at the temperature T_∞ and is immersed in a fluid at the same temperature. At a given instant t= 0, heat starts to be released in the solid at the rate per unit mass ˙q₀exp (−βt), where ˙q₀ and β are constants. Assuming constant properties and neglecting the internal temperature gradients, deduce an expression for the temperature in the solid as a function of time for t >0.

Solution:

Since the (spatial) temperature gradients in the solid are negligible during the transient heat transfer process, the temperature of the solid at any time instant (t >0) can be determined by formulating an overall energy balance on the solid.

˙

E_in−E˙_out+ ˙E_g = ˙E_st ⇔ −E˙_out+ ˙E_g = ˙E_st ⇔

⇔ −hA(T −T∞) +ρVq˙₀exp (−βt) = ρV cdT dt ⇔

⇔ dT

dt + hA

ρV c(T −T∞)− q˙₀

cexp (−βt) = 0

(1)

Considering the temperature difference θ = (T −T∞) and realizing that (hA/ρV c) = τ_t⁻¹, whereτ_t corresponds to the thermal time constant, Equation (1) can be written as follows – see Equation (2).

dθ dt + θ

τt

−q˙₀

cexp (−βt) = 0 (2)

Equation (2) is a linear, first-order, nonhomogeneous differential equation whose solution can be found as the sum of a homogeneous and a particular solution considering the method of undetermined coefficients – see Equation (3).

θ(t) = θh(t) +θp(t) (3)

The homogeneous solution – solution obtained for Equation (2) assuming a negligible nonho- mogeneous term – is presented in Equation (4).

dθ_h dt + θ_h

τ_t = 0⇒θ_h(t) = c₁exp

−t τ_t

(4)

Since the nonhomogeneous term (( ˙q₀/c) exp (−βt)) is comprised by an exponential function, the particular solution is assumed to be of the form:

θ_p(t) = c₂exp (−βt) (5)

The constant c₂ is calculated substituting Equation (5) in Equation (2) – see Equation (6).

dθp

dt + θp

τ_t − q˙0

cexp (−βt) = 0⇔ −c₂βexp (−βt) + 1

τ_tc₂exp (−βt)− q˙0

cexp (−βt) = 0⇔

⇔c₂ = q˙0

c

1 [(1/τ_t)−β]

(6)

Replacing the expression obtained forc₂ (Equation (6)) in Equation (5), the particular solution is written in accordance to Equation (7).

θ_p(t) = q˙0

c

1

[(1/τ_t)−β]exp (−βt) (7)

Replacing the expressions for θ_h (Equation (4)) and θ_p (Equation (7)) in Equation (3), the following equation is obtained.

θ(t) =c₁exp

−t τ_t

+ q˙₀

c

1

[(1/τ_t)−β]exp (−βt) (8) The constant c1 in the previous equation can be evaluated considering the initial condition (condition at t= 0) for which T(t = 0) =T∞, and consequently, θ(t= 0) =T(t = 0)−T∞= 0 – see Equation (9).

0 = c₁exp

−0 τ_t

+q˙0

c

1

[(1/τ_t)−β]exp (−β×0)⇔

⇔0 =c₁+q˙₀ c

1

[(1/τ_t)−β] ⇔c₁ =−q˙₀ c

1 [(1/τ_t)−β]

(9)

Finally, replacing c₁ evaluated in the previous equation in Equation (8) the temperature differ- ence – and consequently, the temperature of the solid as a function of time – is obtained – see Equation (10).

θ(t) = q˙₀ c

1 [(1/τ_t)−β]

exp (−βt)−exp

−t τ_t

(10)

If the constant β were equal to zero, the thermal energy generation in the solid, ˙Eg would be constant along the transient process – ˙E_g = ρVq˙₀. In such situation, Equation (2) would be written as shown by Equation (11).

dθ dt + θ

τ_t − q˙₀

c = 0 (11)

The solution of Equation (11) can be found with a different procedure than that considered before – method of undetermined coefficients – since the nonhomogeneous term is no longer dependent on t. Therefore, according to such procedure, the equation nonhomogeneity can be eliminated considering the following variable change – see Equation (12).

θ⁰ =θ−τ_tq˙₀

c (12)

Taking into account Equation (12), Equation (11) can be written in accordance to Equation (13).

dθ⁰ dt + θ⁰

τ_t = 0 (13)

Separating variables and integrating from the initial condition (t = 0) for which θ⁰ = θ_i⁰ the solution of Equation (13) is obtained – see Equation (14).

θ⁰

θ⁰_i = exp

− t τ_t

(14) Considering Equation (12) in Equation (14) and considering that (T_i−T∞) = 0 – in accordance with the problem statement – the temperature difference as a function of the elapsed from the beginning of the transient process is given by Equation (15).

θ⁰

θ_i⁰ = exp

−t τ_t

⇔ T −T∞−τ_tq˙₀/c

T_i−T∞−τ_tq˙₀/c = exp

−t τ_t

⇔

⇔T −T_∞= (T_i−T_∞) exp

−t τ_t

+τ_tq˙₀/c

1−exp

−t τ_t

⇒

⇒θ(t) = τ_tq˙₀ c

1−exp

−t τ_t

(15)

Note that Equation (15) is equal to Equation (10) considering β equal to zero.

43. The diffuser wall in the exhaust of a rocket motor has a thickness L = 25 mm and consists of a steel alloy whose properties are ρ = 8000 kg m⁻³, c= 500 J kg⁻¹K⁻¹, and k = 25 W m⁻¹K⁻¹. During a fire-resistance test, the wall is at a uniform initial temperature of T_i = 25^◦C and is exposed to the hot gases resulting from the combustion, whose temperature is T∞ = 1750^◦C.

The outer surface of the wall is insulated. The wall should be maintained at a temperature of at least 100^◦C below the material melting temperature, which is equal to 1600^◦C. Assume that the diffuser diameter is much larger than the wall thickness and that the convection coefficient on the hot gases side is equal to 500 W m⁻²K⁻¹.

(a) Determine the temperature on the surface of the wall in contact with gases after 30 s.

Solution:

The nozzle (diffuser) wall can be approximated as a plane wall because the nozzle diameter is much larger than the nozzle wall thickness. To evaluate the relevance of spatial temper- ature gradients along the transient heat conduction process, the Biot number is calculated in Equation (16). Since the Biot number is not lower than 0.1 the temperature gradients in the wall are relevant and must be considered, and consequently, the lumped capacitance method is not recommended to solve the problem. Instead, an approximate solution to the appropriate formulation of heat diffusion equation should be considered to solve the problem.

Bi= hL

k ⇔Bi= 500×0.025

25 ⇔Bi = 0.5 (16)

The analytical solution of the (dimensionless) temperature distribution is given in the form of an infinite series. However, if the Fourier number (F o) is higher that 0.2, the infinite series solution can be approximated by the first term of the series. Equation (17) shows that for the time instant in consideration (t = 30 s) – and the thermophysical and geometrical properties of the wall – the Fourier number is equal to 0.3 (>0.2), and consequently, the one-term approximation solution that is given by Equation (18) can be applied. In Equation (18), θ^∗ (= (T(x^∗, F o)−T∞)/(T_i−T∞)) is the dimensionless temperature difference, x^∗ (= x/L) is the dimensionless spatial coordinate and ζ₁ and C₁ are constants that for a plane wall only depend on the Biot number.

F o= αt

L² ⇔F o = kt

ρcL² ⇔F o = 25×30

8000×500×0.025² ⇔

⇔F o= 0.3

(17)

θ^∗(x^∗, F o) =C₁exp −ζ₁²F o

cos (ζ₁x^∗) (18)

For a Biot number equal to 0.5 (Equation (16)), the tabulated values of ζ₁ andC₁ are given in Equations (19) and (20), respectively – values obtained from Table 5.1 of the textbook

“Fundamentals of Heat and Mass Transfer”, Sixth/Seventh Edition.

ζ₁(Bi= 0.5) = 0.6533 rad (19)

C₁[ζ₁(Bi= 0.5)] = 1.0701 (20) Finally, considering in Equation (18) ζ₁, C₁, andF o equal to 0.6533 rad, 1.0701, and 0.3, respectively, the temperature on the surface of the nozzle wall – i.e., atx =L, or equiva- lently, at x^∗ = 1 – after 30 s of the beginning of the transient process,T (x^∗ = 1, F o= 0.3), is calculated as follows – see Equation (21).

θ^∗(x^∗ = 1, F o= 0.3) = 1.0701×exp −0.6533²×0.3

×cos (0.6533×1)⇔

⇔ T (x^∗ = 1, F o= 0.3)−T∞

T_i−T∞

= 1.0701×exp −0.6533²×0.3

×cos (0.6533×1)⇔

⇔T (x^∗ = 1, F o= 0.3) = 1.0701×exp −0.6533²×0.3

×cos (0.6533×1)× (25−1750) + 1750⇔ T (x^∗ = 1, F o= 0.3) = 460.350^◦C

(21)

The figure below presents the dimensionless temperature difference (θ^∗) and temperature (T) profiles along the thickness of the nozzle wall after 30 s of the beginning of the tran- sient heat conduction process. The temperature profile shows significant (non-negligible) temperature gradients along the wall thickness – in full agreement with Bi > 0.1 (see Equation (16)). Since the wall is receiving thermal energy from the exhaust gases during the transient heating process, the wall external surface temperature (x=L) will be higher than the temperatures within the wall thickness, i.e., T (0< x < L)< T (x=L).

Dim. Temp. Diff., θ^* Temperature, T

Position, x [mm]

0 5 10 15 20 25

Temperature,T[ºC]

100 200 300 400 500

DimensionlessTemp.Difference,θ* [-]

0.75 0.80 0.85 0.90 0.95

Dimensionless Position, x^* [-]

0.0 0.2 0.4 0.6 0.8 1.0

Note that the dimensionless temperature difference (θ^∗) varies from 1 (at t = 0 – initial condition for which T = T_i) and 0 (at t = ∞ – thermal equilibrium for which T = T∞), independently of the direction of the energy transfer – to or from the solid experiencing a temporal change in its stored thermal energy, i.e., either for a transient heating or cooling problem. Since heat transfer to/from a solid occurs at the external surface of the solid (interface between the solid and the adjoining fluid), the solid surface will be closer to

the thermal equilibrium with the surrounding fluid than the inner regions of the solid for any time instant (0 < t < ∞). Consequently, θ^∗(x^∗) > θ^∗(x^∗+dx^∗) i.e., θ^∗ decreases as the distance to the external solid surface decreases (or, equivalently, as x^∗ increases) – in full accordance with the decreasing behavior of cos (ζ₁x^∗) in Equation (18) as x^∗ increases. For transient heating problems (such as the current problem), the temperature as a contrary trend to that exhibited by the dimensionless temperature difference: the temperature increases as x^∗ increases. This is due to the definition of the dimensionless temperature difference (θ^∗ = (T −T∞)/(T_i−T∞)) because considering a transient heating problem T_i < T_∞, and consequently, (T_i−T_∞) <0 which leads to an increasing trend of T (x^∗) as θ^∗(x^∗) decreases.

(b) Determine the time at which the maximum permissible temperature is reached.

Solution:

At any time instant during the transient heating process the maximum temperature in the wall is observed at the wall inner surface (x = L). Equation (22) calculates the Fourier number corresponding to the time instant when the wall inner surface equals the maximum permissible temperature, T_max (= 1500^◦C).

θ^∗(x^∗ = 1, F o) =C₁exp −ζ₁²F o

cos (ζ₁ ×1)⇔

⇔ T (x^∗ = 1, F o)−T∞

T_i−T_∞ =C₁exp −ζ₁²F o

cos (ζ₁ ×1)⇔

⇔ 1500−1750

25−1750 = 1.0701×exp −0.6533²F o

×cos (0.6533×1)⇔

⇔F o≈4.144

(22)

Finally, having evaluated the Fourier number the corresponding time instant is computed in Equation (23).

F o= kt

ρcL² ⇔t= F oρcL²

k ⇔t= 4.144×8000×500×0.025²

25 ⇔

⇔ t ≈414.4 s (≈6.9 min)

(23)

46. (Homework) A steel ball (k = 36.4 W m⁻¹K⁻¹, ρ = 7750 kg m⁻³ and c = 486 J kg⁻¹K⁻¹) with diameter of 8 cm is heated in a furnace until it reaches a uniform temperature of 800^◦C. It is then cooled by immersion in a bath maintained at 300^◦C until the temperature in the center of the sphere reaches 500^◦C. Determine the time required for this cooling, assuming a very high convection coefficient.

Solution:

Since a very high convection heat transfer coefficient is considered, the Biot number can be approximated as equal to infinity. Therefore, the spatial temperature gradients within the ball are relevant and the solution of the proper form of the heat diffusion equation has to be considered – instead of formulating an overall energy balance on the ball, i.e., instead of applying the lumped capacitance method. (A very high convection coefficient (or a very high Biot number) implies that during the transient process the surface temperature of the sphere is approximately equal to the fluid temperature since the thermal resistance between the solid and the fluid (convection resistance) is small or negligible.)

The one-term approximation solution for the dimensionless temperature difference at the sphere centerpoint (r = 0) is given by Equation (24).

θ^∗₀ =C₁exp −ζ₁²F o

(24) For a sphere with Bi = ∞ (very high convection heat transfer coefficient), the values of the constants ζ₁ and C₁ – extracted from Table 5.1 of the textbook “Fundamentals of Heat and Mass Transfer”, Sixth/Seventh Edition – are presented in Equations (25) and (26), respectively.

ζ₁(Bi=∞) = 3.1415 rad (25)

C₁[ζ₁(Bi=∞)] = 2 (26) When the temperature at the sphere centerpoint (r^∗ =r/r_o = 0) is equal to 500^◦C, the corre- sponding dimensionless temperature difference is calculated in Equation (27) taking into account the initial (uniform) sphere temperature, T_i, and the fluid temperature, T∞.

θ^∗₀(F o) = T(r^∗ = 0, F o)−T∞

T_i−T∞

⇔θ^∗₀(F o) = 500−300

800−300 ⇔θ₀^∗(F o) = 0.4 (27) Considering Equations (25) to (27) in Equation (24), the Fourier number corresponding to the time instant at which the centerpoint temperature reaches 500^◦C is obtained – see Equation (28).

F o=−1 ζ₁²ln

θ₀^∗ C₁

⇔F o≈0.163 (28)

The time required for cooling the sphere from the initial condition – at which the sphere tem- perature was uniform and equal to 800^◦C – to the instant when the temperature at the sphere centerpoint reaches a value equal to 500^◦C is calculated in Equation (29) once the corresponding Fourier number (dimensionless time) is known (see Equation (28)).

F o= αt

r²_o ⇔t= 1

αr²_oF o ⇔t =ρc k

r²_oF o⇔

⇔t = 7750×486

36.4 ×(0.08/2)²×0.163 ⇔ t≈26.986 s

(29)

Note that the calculated Fourier number (Equation (28)) is not higher than 0.2, and conse- quently, the one-term approximation solution considered to solve the problem (Equation (24)) may only provide a rough estimation of the required time – the one-term approximation solu- tion of the infinite series (exact) solution is not recommended for F o ≤ 0.2. For an accurate prediction of the required time, the exact (analytical) solution or numerical solutions should be obtained.

55. (Homework) A glass of water at 300 K with 8 cm in diameter and 12 cm in height is placed in a refrigerator, which maintains the air temperature at 277 K. The convection coefficient is 5 W m⁻²K⁻¹. After 6 hours the glass is removed from the refrigerator. Estimate the average water temperature at this time, assuming that there is only heat conduction in the water.

Solution:

An estimate of the thermophysical properties of water – ρ, c, and k (required to solve the problem) – can be obtained considering saturated liquid data (Table A.6 of the textbook “Fun- damentals of Heat and Mass Transfer”, Sixth/Seventh Edition) at an average temperature T = (T_i+T∞)/2 = (300 + 277)/2 = 288.5 K. By linear interpolation, the stated thermophys- ical properties have the following values: ρ(T) = 999.3 kg m⁻³, c(T) = 4185.5 J kg⁻¹K⁻¹, and k(T) = 0.5956 W m⁻¹K⁻¹.

The relevance of the spatial temperature gradients in the water during the transient cooling process – which decides the solution procedure for the problem (lumped capacitance method (for 0D problems) or one-term approximation solutions (for 1D to 3D problems)) – is assessed through the evaluation of the Biot number (Biot Criterion) – see Equation (30).

Bi = hL_c

k ⇔Bi= hV

kA_s ⇔Bi= h(πr_o²H)

k(2πr_oH+ 2πr_o²) ⇔

⇔Bi= h(r_oH)

2k(H+r_o) ⇔Bi= 5×(0.04×0.12)

2×0.5956×(0.12 + 0.04) ⇔Bi≈0.126

(30)

The lumped capacitance method cannot be applied to obtain the (average) water temperature after 6 hours of the beginning of the transient process because the Biot number is not lower than 0.1.

Therefore, the spatial temperature gradients are relevant and need to be taken into account.

Since the glass height (H) is comparable to the glass diameter (D) – short cylinder –, multidi- mensional (two-dimensional) heat conduction in the water has to be considered. (If the glass height were much greater than the glass diameter (H/D ≥ 10), the heat conduction in the water could be approximated as one-dimensional heat diffusion – exclusively along the radial direction – and the solution procedure for the infinite cylinder could be applied.)

From the definition of the dimensionless (fractional) energy transfer from the water to adjoining fluid it is possible to obtain an expression to compute the average water temperature – see Equation (31).

Q Q₀ = 1

V Z

(1−θ^∗)dV ⇔ Q

Q₀ = 1− 1 V

Z

θ^∗dV ⇔ Q

Q₀ = 1−θ^∗ ⇔

⇔ Q Q0

= 1− T −T∞

Ti−T∞

⇔T =T_i− Q Q0

(T_i−T_∞)

(31)

Since the cooling problem solution procedure has to be formulated considering two-dimensional transient heat conduction, the corresponding dimensionless energy transfer is calculated taking into account contributions of a plane wall and an infinite cylinder in accordance to Equation

(32).

Q Q₀ =

Q Q₀

glass

= Q

Q₀

Plane Wall

+ Q

Q₀

Inf.Cylind.

− Q

Q₀

Plane Wall

× Q

Q₀

Inf.Cylind.

(32)

In the following, the dimensionless energy transfer for the plane wall and infinite cylinder are calculated assuming the one-term approximation solution – Equations (38) and (44). The procedure required the computation of the Biot number in order to determine the value of the constants ζ₁ and C₁, the Fourier number and finally the dimensionless energy transfer.

• Plane wall

Q Q0

Plane Wall

= 1−sin (ζ₁) ζ1

θ₀^∗ ⇔

⇔ Q

Q₀

Plane Wall

= 1− sin (ζ₁)

ζ₁ C₁exp −ζ₁²F o

(33)

Bi= hL

k ⇔Bi= h(H/2)

k ⇔Bi= 5×(0.12/2)

0.5956 ⇔Bi≈0.504 (34)

ζ₁(Bi = 0.504)≈0.655 rad (35)

C₁[ζ₁(Bi= 0.504)] = 1.071 (36)

F o = αt

L² ⇔F o= kt

ρc(H/2)² ⇔F o= 0.5956×6×3600

999.3×4185.5×(0.12/2)² ⇔

⇔F o ≈0.854

(37)

Q Q₀

Plane Wall

= 1−sin (ζ₁)

ζ₁ C₁exp −ζ₁²F o

⇔

⇔ Q

Q₀

Plane Wall

= 1−sin (0.655)

0.655 ×1.071×exp −0.655²×0.854

⇔ Q

Q₀

Plane Wall

≈0.310

(38)

• Infinite Cylinder

Q Q₀

Inf.Cylind.

= 1− 2θ^∗₀

ζ₁ J₁(ζ₁)⇔

⇔ Q

Q₀

Inf.Cylind.

= 1− 2C₁exp (−ζ₁²F o) ζ₁ J₁(ζ₁)

(39)

Bi= hro

k ⇔Bi = 5×(0.08/2)

0.5956 ⇔Bi≈0.336 (40)

ζ₁(Bi = 0.336)≈0.784 rad (41)

C₁[ζ₁(Bi= 0.336)] = 1.079 (42)

F o= αt

r²_o ⇔F o= kt

ρc(D/2)² ⇔F o = 0.5956×6×3600

999.3×4185.5×(0.08/2)² ⇔

⇔F o≈1.922

(43)

Q Q₀

Inf.Cylind.

= 1− 2C₁exp (−ζ₁²F o)

ζ₁ J₁(ζ₁)⇔

⇔ Q

Q₀

Inf.Cylind.

= 1− 2×1.079×exp (−0.784²×1.922)

0.784 J₁(0.784)⇔

⇔ Q

Q₀

Inf.Cylind.

≈0.694

(44)

Substituting the values obtained in Equations (38) and (34) in Equation (32), the dimensionless energy transfer from the water to the air after 6 h of the beginning of the cooling process is obtained – see Equation (45).

Q Q₀ =

Q Q₀

Plane Wall

+ Q

Q₀

Inf.Cylind.

− Q

Q₀

Plane Wall

× Q

Q₀

Inf.Cylind.

⇔

⇔ Q

Q₀ = 0.310 + 0.694−0.310×0.694 ⇔ Q

Q₀ ≈0.789

(45)

Finally, substituting the value calculated forQ/Q₀ (Equation (45)) in Equation (31), the average

temperature of the water after 6 h of cooling is obtained – see Equation (46).

T =T_i− Q Q0

(T_i−T_∞)⇔T = 300−0.789×(300−277)⇔

⇔ T ≈281.853 K

(46)

An alternative procedure to determine the dimensionless energy transfer for the plane wall ((Q/Q₀)_{Plane Wall}) and infinite cylinder ((Q/Q₀)Inf.Cylind.) consists in using the graphical repre- sentation of one-dimensional, transient conduction (one-term approximation solutions) for the plane wall and infinite cylinder – Heisler Charts.

For the plane wall with the geometrical parameters (L = H/2), thermophysical properties (ρ, c,k, andh), and elapsed time (t) under consideration, Bi≈0.504 (Equation (34)), F o≈0.854 (Equation (37)), and consequently, Bi²F o = 0.504² ×0.854 ≈ 0.217. Considering the proper Heisler Chart (Q/Q₀ chart) for the plane wall along withBi²F o= 0.22 andBi= 0.50, it follows that (Q/Q₀)_{Plane Wall} ≈0.31 – see figure below. (In this figure Q/Q_max is equal to Q/Q₀ and τ corresponds to the Fourier number (F o).)

For the infinite cylinder with the geometrical parameters (r_o =D/2), thermophysical properties (ρ, c, k, and h), and elapsed time (t) under consideration, Bi ≈ 0.336 (Equation (40)), F o ≈ 1.922 (Equation (43)), and consequently, Bi²F o = 0.336² ×1.922 ≈ 0.217. Considering the proper Heisler Chart (Q/Q₀ chart) for the infinite cylinder along with Bi²F o = 0.22 and Bi = 0.34, it follows that (Q/Q₀)_Inf.Cylind.≈0.69 – see figure below.

The values of Q/Q₀ obtained using the Heisler Charts for the plane wall (≈0.31) and infinite cylinder (≈ 0.69) are in full agreement with the corresponding values calculated previously – see Equations (38) and (44), respectively.