Sprocket and Pulley Speed Ratios: 22-Agric-B5 Power Units for Agricultural, Biosystems, and Food Industries, May 2015, Q1

22-Agric-B5 Power Units for Agricultural, Biosystems, and Food Industries · May 2015

Problem 1: Sprocket and Pulley Speed Ratios (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) Chain drive: driven (pump) sprocket $T_{\text{pump}}=18$ teeth turning at $N_{\text{pump}}=2100$ rpm; driver (PTO) sprocket turning at $N_{\text{PTO}}=540$ rpm. (b) Belt drive: motor pulley $D_{\text{motor}}=6.5$ cm turning at $N_{\text{motor}}=1725$ rpm; fan pulley must turn at $N_{\text{fan}}=500$ rpm.

Find. (a) The number of teeth on the PTO (driver) sprocket. (b) The fan pulley diameter, and the ratio of pulley diameters vs. the ratio of pulley speeds.

Figure 1a — PTO (driver) sprocket, larger and slower, chain-coupled to the pump (driven) sprocket, smaller and faster; sprocket size is inversely proportional to speed.

Approach. Both are constant-linear-speed drives (chain pitch line / belt surface speed matched at both wheels), so the driver and driven elements obey $N_1 X_1 = N_2 X_2$ where $X$ is teeth count for a chain drive and diameter for a belt drive; solve the one unknown in each sub-part directly.

(a) PTO sprocket size from equal chain pitch-line speed. The source states the rpm–teeth product is equal for both sprockets: $$N_{\text{PTO}}\,T_{\text{PTO}} = N_{\text{pump}}\,T_{\text{pump}}$$ $$T_{\text{PTO}} = \frac{N_{\text{pump}}\,T_{\text{pump}}}{N_{\text{PTO}}} = \frac{(2100)(18)}{540} = \boxed{70\ \text{teeth}}$$ Discussion: the PTO (driver) turns slower than the pump (driven), so it needs a larger sprocket (70 vs. 18 teeth) — a chain drive is a fixed-ratio speed-up here, the mechanical equivalent of gearing 540 rpm up to 2100 rpm (a 3.89:1 step-up).
(b) Fan pulley size from equal belt surface speed. A flat/V-belt has one linear speed at both pulley rims, $\pi D_1 N_1 = \pi D_2 N_2$, i.e. $D_1N_1=D_2N_2$: $$D_{\text{fan}} = \frac{D_{\text{motor}}\,N_{\text{motor}}}{N_{\text{fan}}} = \frac{(6.5\ \text{cm})(1725)}{500} = \boxed{22.4\ \text{cm}}$$
(b, cont'd) Diameter ratio vs. speed ratio. $$\frac{D_{\text{fan}}}{D_{\text{motor}}} = \frac{22.425}{6.5} = 3.45 \qquad \frac{N_{\text{motor}}}{N_{\text{fan}}} = \frac{1725}{500} = 3.45$$ $$\boxed{\dfrac{D_{\text{fan}}}{D_{\text{motor}}} = \dfrac{N_{\text{motor}}}{N_{\text{fan}}} = 3.45}$$ Discussion: the two ratios are identical, confirming $D_1N_1=D_2N_2$ — the fan pulley is 3.45× the motor pulley's diameter because the fan must turn 3.45× slower; belt speed reduction always trades pulley size for speed in exact inverse proportion, with no gearing losses assumed.

Quantity	Result
(a) PTO (driver) sprocket size	70 teeth
(b) Fan pulley diameter	22.4 cm
(b) Diameter ratio $D_{\text{fan}}/D_{\text{motor}}$	3.45
(b) Speed ratio $N_{\text{motor}}/N_{\text{fan}}$	3.45

Concept & Theory

Topic. Constant-linear-speed mechanical power transmission — chain (sprocket) and belt (pulley) drives.

Key relations. Chain drive: $N_1T_1=N_2T_2$ (pitch-line speed match via tooth count). Belt drive: $D_1N_1=D_2N_2$ (rim surface speed match via diameter).

Why this works. A chain's pitch-line linear speed is the same on both sprockets it engages ($v=\pi D N$, and pitch diameter is proportional to tooth count for a fixed chain pitch, so $T$ substitutes directly for $D$); a belt's surface speed is likewise identical on both pulleys it wraps. Equating the two elements' linear speeds and cancelling the shared constant ($\pi\times$pitch, or $\pi$ itself) leaves the simple product relation used above. Both are the mechanical-engineering analogue of a gear pair, where speed ratio is inversely proportional to the size (tooth count or diameter) ratio.

Common pitfall. Assuming the driver is always the smaller wheel: here the PTO (driver) is the LARGER sprocket because it turns slower — size always scales with the inverse of speed, never with "driver vs. driven" role alone.

Source. Srivastava, Buckmaster & Hull, Engineering Principles of Agricultural Machines, Ch. 2 (power transmission); Kepner, Bainer & Barger, Principles of Farm Machinery, Ch. 2.

Question 1 of 4: Sprocket and Pulley Speed Ratios

Problem 1: Sprocket and Pulley Speed Ratios (equal value)