Brake energy and drum temperature rise

Data

$m = 1400$ kg, $v = 90$ km/h $= 25$ m/s, $t = 6$ s.

(a) Energy and power

Kinetic energy to be absorbed:

$$E = \tfrac12 m v^2 = \tfrac12\times1400\times25^2 = 437\,500\ \text{J}$$

Per brake: $E_1 = 437\,500/4 = 109\,375$ J $= 109.4$ kJ.

Average retardation: $a = v/t = 25/6 = 4.17\ \text{m/s}^2$.

Stopping distance: $s = \dfrac{v t}{2} = \dfrac{25\times6}{2} = 75$ m.

Average power per brake: $P_1 = \dfrac{E_1}{t} = \dfrac{109\,375}{6} = 18.2$ kW (with uniform retardation the power at the start is twice this value).

(b) Temperature rise (no heat loss)

$$E_1 = m_d\,c\,\Delta T \;\Rightarrow\; \Delta T = \frac{109\,375}{7\times500} = 31.3\ ^\circ\text{C}$$

For repeated stops, the cooling between applications must also be checked; a heavier or finned drum keeps the temperature lower.

(c) Descending a slope

A slope of 1 in 20 gives $\tan\theta = 1/20$, so $\sin\theta = 0.04994 \approx 0.05$; the speed is $v = 36$ km/h $= 10$ m/s. At steady speed the brakes absorb the loss of potential energy:

$$P = m g\,v\sin\theta = 1400\times9.81\times10\times0.04994 = 6859\ \text{W}$$

Per brake: $6859/4 = 1715$ W.

At steady state, heat generated equals heat dissipated, $Q = hA\,\Delta T$:

$$\Delta T = \frac{1715}{18} = 95.3\ ^\circ\text{C}$$

This is acceptable for cast-iron drums, but on long descents the temperature may rise further and cause brake fade (the coefficient of friction falls as the temperature rises), so engine braking is used.

Answer: (a) 109.4 kJ per brake, retardation 4.17 m/s², distance 75 m, 18.2 kW; (b) 31.3 °C; (c) 1.72 kW per brake, steady rise 95 °C.