Belt and Chain Drive Sizing: 22-Agric-B5 Power Units for Agricultural, Biosystems, and Food Industries, May 2016, Q1

22-Agric-B5 Power Units for Agricultural, Biosystems, and Food Industries · May 2016

Problem 1: Belt and Chain Drive Sizing (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) Belt drive: motor pulley $D_{\text{motor}}=6.5$ cm at $N_{\text{motor}}=1725$ rpm; fan must turn at $N_{\text{fan}}=500$ rpm. (b) Chain drive: pump (driven) sprocket $T_{\text{pump}}=18$ teeth at $N_{\text{pump}}=2100$ rpm; PTO (driver) turns at $N_{\text{PTO}}=540$ rpm.

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

Figure 1 — (a) belt/pulley drive: the slower fan pulley must be larger in exact inverse proportion to its speed. (b) chain/sprocket drive: the slower PTO (driver) sprocket must be larger than the faster pump (driven) sprocket.

Approach. Both are constant-linear-speed drives — a belt's rim surface speed and a chain's pitch-line speed are identical at both wheels — so driver and driven elements satisfy $D_1N_1=D_2N_2$ (belt) or $T_1N_1=T_2N_2$ (chain); solve the one unknown in each sub-part.

Check: this exact pulley/sprocket data set (500↔1725 rpm belt drive, 6.5 cm motor pulley.

(a) Fan pulley size from equal belt surface speed. $\pi D_1 N_1=\pi D_2 N_2 \Rightarrow D_1N_1=D_2N_2$: $$D_{\text{fan}} = \frac{D_{\text{motor}}\,N_{\text{motor}}}{N_{\text{fan}}} = \frac{(6.5\ \text{cm})(1725)}{500} = \boxed{22.4\ \text{cm}}$$
(a, cont'd) Diameter ratio vs. speed ratio. $$\frac{D_{\text{fan}}}{D_{\text{motor}}} = \frac{22.425}{6.5} = 3.45 \qquad \frac{N_{\text{motor}}}{N_{\text{fan}}} = \frac{1725}{500} = 3.45$$ $$\boxed{\dfrac{D_{\text{fan}}}{D_{\text{motor}}} = \dfrac{N_{\text{motor}}}{N_{\text{fan}}} = 3.45}$$ The two ratios are numerically identical (both equal 3.45) because pulley size and pulley speed are exact inverse proportions of one another for a fixed belt speed.
(b) PTO sprocket size from equal chain pitch-line speed. $N_{\text{PTO}}\,T_{\text{PTO}} = N_{\text{pump}}\,T_{\text{pump}}$: $$T_{\text{PTO}} = \frac{N_{\text{pump}}\,T_{\text{pump}}}{N_{\text{PTO}}} = \frac{(2100)(18)}{540} = \boxed{70\ \text{teeth}}$$ The PTO (driver) turns slower than the pump (driven), so it needs the larger sprocket — a fixed 3.89:1 step-up.

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

Concept & Theory

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

Key relations. Belt drive: $D_1N_1=D_2N_2$ (rim surface speed match). Chain drive: $T_1N_1=T_2N_2$ (pitch-line speed match via tooth count, which is proportional to pitch diameter for a fixed chain).

Why this works. A belt's surface speed $v=\pi DN$ is identical on both pulleys it wraps; a chain's pitch-line speed is identical on both sprockets it engages, and pitch diameter scales with tooth count for a fixed chain pitch, so $T$ substitutes directly for $D$. Cancelling the shared constant leaves the simple product relation used above — the mechanical-engineering analogue of a gear pair, where size and speed are always in inverse proportion.

Common pitfall. Assuming the driver is always the smaller wheel: here the PTO (driver) is the larger sprocket because it turns slower — size 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: Belt and Chain Drive Sizing

Problem 1: Belt and Chain Drive Sizing (equal value)