Complete Disc Temperature Analysis in a Single Stop

In the previous section, brake temperature was computed for one location on the rotor only, namely at the friction surface. Convective cooling was ignored. In this section brake temperatures will be computed for any location beneath the fric-tion surface, and as a funcfric-tion of time. Convective cooling is included (Ref. 3.3).

t s h

The physical system to be analyzed is illustrated in Fig. 3-4. The temperature formulation results in the following set of equations in terms of the

transformation θ = Τ − Τ_¥:

Differential equation: δθ/δt = a (δ²θ/δz²)

Boundary conditions: δθ/δz (o,t) = 0; δθ/δz (L,t) + (h/k)θ(L,t) = q′′₀ /k Initial condition: θ(z,0) = θ_i = T_i − T_¥

Because the temperature reaches a finite value as time approaches infinity, the solution may be obtained by a superposition of the form: θ(z,t) = ψ(z,t) + Φ(z).

The function Φ(z) satisfies the following conditions:

δ²Φ(z)/δz² = 0 δΦ(L)/δz + (h/k) Φ = q′′₀ /k

δΦ(0)/δz = 0 The function ψ(z,t) satisfies the following conditions:

δψ/δt = a (δ²ψ/δz²) δψ(L,t)/δz + (h/k) ψ(L,t) = 0 ψ(z,0) = θ_i − Φ(z) at time t = 0.

Figure 3-4. Physical system representing solid rotor.

Solving the differential equations is accomplished by first deriving the

temperature response due to a constant heat flux as observed during constant-speed downhill braking. The final temperature expression is obtained from the constant heat flux temperature and the application of Duhamel’s theorem using a time-varying heat flux.

The solid non-ventilated rotor is illustrated in Fig. 3-4. Both sides of the rotor are heated by the heat flux ′′q₍₀₎, and are cooled by convection h_R. For the solid rotor, the conditions permit an analytical solution for a constant heat flux (Ref. 3.3).

(3-26)

== convective heat transfer coefficient, Nm/h K m (BTU/h°² FF ft ) L = one-half rotor thickness, m (ft)

n = numerals 1,

2

2, 3, ...

q = average heat flux into rotor, Nm/h m (BTU₀′′ ² //h ft ) t = time, h

T (z, t) = transient temperature distri

2

0 bbution in rotor due to a constant heat flux, K (°F) T = ini iitial temperature, K (°F)

T = ambient temperature, K (°F) z

∞

= horizontal distance measured from midplane of rotor, m (ft) (z, t) = (z, t) , relative temperature of brake r

θ0 T0 −T_∞ eesulting from constant heat flux, K (°F)

=T = initia

i i

θ −T_a ll temperature difference between brake and ambient, K (°F)) λn=n Lπ/ , / ( / )1 m 1 ft

The value of λ_nL is determined from the transcendental equation

Inherent in the derivation of Eq. (3-26) are the following assumptions

1. The temperature is only a function of the coordinate normal to the fric-tion surface and time t.

2. The heat transfer coefficient h_R is constant.

3. The heat flux is in the direction normal to the friction surface.

4. The thermal properties of both friction partners are constant and evaluated at some mean temperature.

5. The ambient temperature T_∞ is constant.

6. Radiation heat transfer is included in terms of an equivalent radiation heat transfer coefficient (see Section 3.1.9).

A few solutions to the transcendental equation of practical importance for typical brakes are presented in Table 3-2. For different hL/k-values, the roots of the transcendental equation can easily be solved by trial and error. The use of Eq. (3-26) requires the calculation of a summation term (Σ). For braking

θ θ λ

times greater than approximately 0.5 to 1 second, only the first two terms (n = 1 and 2) in the summation are required for an acceptable accuracy of the brake temperatures calculated.

Solid rotors are frequently used on rear axles of cars and pickup trucks as well as in drive shaft-mounted parking brakes of heavy construction vehicles.

Example 3-3: Pickup Truck Brake Temperature - Solid Rotor – Constant Heat Flux

Calculate the brake temperatures at the mid plane and swept surface at the end of the braking process for the following data: Vehicle weight 40,034 N (9000 lb), deceleration 0.9 g, speed 128 km/h (80 mph), tire slip 10%, rotating mass factor k = 1, brake force distribution Φ = 0.40, solid rotor rear disc 25.4 mm (1 in.) thick, outer diameter 355.6 mm (14 in.), convective heat transfer coefficient h_R = 278,038 Nm/hKm² (13.4 BTU/h°Fft²), ambient temperature 278 K (40°F), initial brake temperature 311 K (100°F), heat distribution into rotor γ

= 0.95, swept rotor surface area of one rear rotor side 0.0645 m² (100 in²), rotor material thermal diffusivity a_t = 0.058 m²/h [0.62 ft²/h], thermal conductivity k

= 174,465 Nm/(hKm) [28 BTU/(h^oFft)].

Equation (3-26) applies, however limited to only one term of the summation, that is, λ₁ only. As is shown in Ref. 3.3, one term only is sufficient for

acceptable accuracy for the solid rotor case. We will carry out the calculation to demonstrate the somewhat complicated mathematics involved. We first must calculate the temperature response achieved by a constant heat flux braking power equal to the braking power of the time-varying heat flux, as illustrated in Fig. 3-2.

With the data given, Eq. (3-8) [or Eq. (3-13) with imperial units] yields an average braking power for the entire vehicle q₀ = 2.07´10⁹ Nm/h (1,978,082 BTU/h). The average heat flux into one rear rotor side equals ′′q₀ = 3.21´10⁹ Nm/(hm²) [284,848 BTU/(hft²)].

Next, we calculate h_RL/k to obtain the value for the λ₁L term given in Table 3-2.

For the data specified, we have (278,038)(0.0127)/(174,465) = 0.0202 [(13.4) (0.5/12)/28 = 0.0199] or 0.02. Consequently, Table 3-2 yields λ₁L = 0.1410.

However, if, for example, the heat transfer coefficient had been 373,484 Nm/(hKm²) [18 BTU/(h^oFft²)], then h_RL/k = 0.02678. The roots of the transcendental equation are solved by trial-and-error. Inspection of Table 3-2 reveals the value to be 0.141 < λ₁L < 0.1987. After several tries, using 0.1629 yields: (0.1629)tan(0.1629)(57.29) = 0.026771, which is close enough to λ₁L = 0.02678.

In Equation (3-26) the arguments for sine and cosine must be entered as angles in degrees, resulting in 0.141´57.3 = 8.08 degrees.

TABLE 3-2

Roots of Transcendental Equation (λ_nL)tan(λ_nL) = h_RL/k

h_RL/k λ₁L λ₂L λ₃L

0.008 0.0893 3.1441 6.2845

0.01 0.0998 3.1448 6.2848

0.02 0.1410 3.1479 6.2864

0.04 0.1987 3.1543 6.2895

0.06 0.2425 3.1606 6.2927

0.08 0.2791 3.1668 6.2959

0.10 0.3111 3.1731 6.2991

In Eq. (3-26) the exponent of e, λ₁ must be used. It is calculated from one-half rotor thickness L as

λ₁ = λ₁L/L = (0.141)/(0.0127) = 11.1 m⁻¹ [(0.141) (12)/(0.5) = 3.384 ft⁻¹] The stopping time is 35.7/(0.9´9.81) = 4.05 s [117.28/(0.9 ´ 32.2) = 4.05 s] or 0.00114 h. The temperatures to be calculated are located at z = 0 ft (mid plane), and z = 0.00127 m [0.0417 ft] (swept surface).

Substituting into Equation (3-26) for z = 0.00127 m [0.0417 ft] yields for n = 1 using the imperial system:

Θ₀(L, 4 s) = T₀ – T_¥

= (284,848/13.4){2[(100 − 40)(13.4)/(142,422) − 1]

´ [sin8.08/(0.141 + sin8.08cos8.08)]

´ (e-0.62(3.384)^2(0.0011)) cos[(3.384)(0.0417)(57.3)] + 1} = 368°F The swept surface temperature increase over the ambient temperature is 460K [368°F] when only one λ₁ value is used. The summation term for n = 1 equals 0.49274. For n = 2, the summation term equals 0.49274 + 0.0000339 = 0.49277.

Consequently, in this particular example n = 1 is sufficient.

To calculate the temperature at the mid plane, only cos(λ₁z) changes to cos(0)

= 1. The term 0.49274 changes to 0.49768, and the temperature increase at the mid plane is 343 K [158 °F] at the moment the vehicle comes to a complete stop. The results indicate that after four seconds of braking, the temperature difference between swept surface and mid plane is 372 K [210°F].

For very large braking times, the exponential term in Equation (3-26) approaches zero, and the theoretical maximum swept surface temperature is

determined by the ratio of ′′q₀/h . No automotive brake system would be able to withstand the extremely large brake temperature obtained for very long braking times with input data of Example 3.3.

Eq. (3-26) computes the temperature response resulting from a constant heat flux at the rotor surface. When the vehicle decelerates, the heat flux varies with time. In most cases a linearly decreasing heat flux is assumed. The temperature response of the brake rotor for the time-dependent heat flux may be obtained directly from the temperature solution shown by Eq. (3-26) associated with the constant heat flux ′′q₀ by application of Duhamel’s theorem or superposition integral. The temperature response from a time-varying heat flux is (Ref. 3.4):

(3-27)

where d = differential operator

q = time-varying heat fl₀′′ uux into rotor at time t = 0, Nm/hm (BTU/h ft ) q ( ) = t

2 2

′′ τ iime-varying heat flux, Nm/hm (BTU/h ft ) t = time, h θ₀ −− T = relative temperature of brake resulting from constannt heat flux, K ( F)_∞ T (z, t) = transient temperature dist₀

° rribution in rotor due to a constant heat flux, K ( F)

= t

° τ iime, h

If a time-varying heat flux [Eq. (3-5)]

(3-28) is assumed, where t = time, h, and t_s = braking time to a stop, h, then integration of Eq. (3-27) with Eq. (3-26) and θ_i = 0 yields the temperature response in a solid disc brake resulting from a time-varying heat flux:

(3-29)

(z, t) = relative temperature of brake resulti

θ0 nng from constant heat flux, K (°F),

Example 3-4: Linearly decreasing heat flux for Example 3-3.

For the data of Example 3-3, compute the temperature response at the swept surface. Inspection of Eq. (3-29) indicates that the brake temperature for the braking process with decreasing velocity, that is linearly decreasing heat flux, uses the constant heat flux case modified by the negative term of Eq. (3-29).

The instantaneous heat flux at the beginning of braking is ′′q₀= 6.42x10⁹ Nm/h [569,696 BTU/(hft²)][Eq. (3-14)]. The temperature at the end of the braking process after 4 s of braking existing at the swept surface (z = L) with imperial units becomes [Eq. (3-29)]:

Θ (L, 4 s) = T – T_in

= (569,696)(368)/(284,848) – 569,696/[(0.01124)(13.4)]{(0.00112 – 2[sin8.08/(0.141 + sin8.08cos8.08)] x

(1 – e-0.62(3.384^2)(0.0011))/[0.62(3.384²)] cos8.08}

= 443^oF.

The temperature increase at the swept surface (z = L) is 501K (443 ^oF) at the end of the braking process when the vehicle is stationary.

Eq. (3-29) was evaluated for a particular test vehicle with a solid disc brake having an outer diameter of 317.5 mm (12.5 in.) and a rotor thickness of 12.7 mm (0.5 in.). The heat flux into one rotor side at the onset of braking is ′′q₀ = 5.56 × 10⁹ Nm/hm² (489,500 BTU/hft²). The convective heat transfer coefficient is 255,553 Nm/hKm² (12.5 BTU/h°Fft²).

Figure 3-5. Surface temperature computed from Eq. (3-29).

The surface-temperature response computed from Eq. (3-29) is shown in Fig.

3-5, using a vehicle deceleration of 0.46 g and speeds of 80 and 97 km/h (50 and 60 mph). The braking times are approximately 5 and 6 s, respectively.

With the same input parameters, the temperature distribution in the rotor computed for a stop from 97 km/h (60 mph) is illustrated in Fig. 3-6. Inspection of Fig. 3-6 reveals that the temperature is nearly uniformly distributed across the width of the rotor after 5 seconds of braking. The temperature gradient existing at the surface after 1.0 s is approximately 45 K/mm (1580°F/in.).

Figure 3-6. Temperature distribution in the rotor – 97 km/h (60 mph) stop.

Equations (3-26) and (3-29) apply to a solid rotor, that is, extremely simple geometry. Later the finite difference temperature formulation will be used to analyze the temperature response of complex brake geometries as well as randomly varying heat flux.