Open flat belt power and cross belt length

Data: $d = 0.4$ m, $D = 0.8$ m, $C = 2.5$ m, $n = 600$ rpm, $b = 0.15$ m, $t = 0.008$ m, $\rho = 1000$ kg/m$^3$, $\sigma = 2$ MPa, $\mu = 0.3$.

(a) Open belt

Length

$$L = 2C + \frac{\pi}{2}(D + d) + \frac{(D - d)^2}{4C} = 5 + 1.885 + 0.016 = 6.901\ \text{m}$$

Angle of contact on the small pulley

$$\sin\alpha = \frac{D - d}{2C} = \frac{0.4}{5} = 0.08, \qquad \alpha = 4.589^\circ$$ $$\theta = 180^\circ - 2\alpha = 170.82^\circ = 2.981\ \text{rad}$$

Speed and masses

$$v = \frac{\pi d n}{60} = \frac{\pi \times 0.4 \times 600}{60} = 12.57\ \text{m/s}$$ $$m = \rho b t = 1000 \times 0.15 \times 0.008 = 1.2\ \text{kg/m}$$

Tensions

$$\begin{aligned} T_{max} &= \sigma b t = 2 \times 150 \times 8 = 2400\ \text{N} \\ T_c &= m v^2 = 1.2 \times 12.57^2 = 189.5\ \text{N} \\ T_1 - T_c &= 2400 - 189.5 = 2210.5\ \text{N} \end{aligned}$$

The tight side tension is the maximum permissible value, $T_1 = T_{max} = 2400$ N. Using $\dfrac{T_1 - T_c}{T_2 - T_c} = e^{\mu\theta}$:

$$e^{\mu\theta} = e^{0.3 \times 2.981} = e^{0.894} = 2.446$$ $$T_2 - T_c = \frac{2210.5}{2.446} = 903.7\ \text{N} \quad\Rightarrow\quad T_2 = 903.7 + 189.5 = 1093.2\ \text{N}$$

Power

$$P = (T_1 - T_2)\,v = (2400 - 1093.2) \times 12.57 = 16\,420\ \text{W} = 16.4\ \text{kW}$$

(b) Cross belt

$$\sin\alpha = \frac{D + d}{2C} = \frac{1.2}{5} = 0.24, \qquad \alpha = 13.89^\circ$$ $$\theta = 180^\circ + 2\alpha = 207.8^\circ$$ $$L = 2C + \frac{\pi}{2}(D + d) + \frac{(D + d)^2}{4C} = 5 + 1.885 + 0.144 = 7.029\ \text{m}$$

The angle of contact is the same on both pulleys, and larger than for the open belt, so a cross belt can transmit more power, but the belt rubs where it crosses and wears faster.

Answer: (a) L = 6.90 m; theta = 170.8 degrees; Tc = 189.5 N; T1 = 2400 N, T2 = 1093 N; P = 16.4 kW. (b) L = 7.03 m; theta = 207.8 degrees.