Section IV : Steam Engine

Coal Washery Plan

With some basic dimensions taken from the drawing .. and note that these are VERY basic as the drawing was blurry, small and not necessarily even close to the right size. Still .. it is what I have to work with and should allow me to find a steam engine of similar size.

I assumed an engine speed of 250 RPM and modified the Crusher pulley Dia to get that RPM. From the text a single Jig requires about 10 HP and the entire unit with belts etc. about 20 HP per jig. The Crusher requires 45 HP at 100 RPM – so I assume that is the size engine I need.

The Pulley Info below will be moved to their own page at some point .. this is a work in progress after all .. I am having fun massaging the numbers at the moment.

Pump Drive Pulley
$D=74"$
$C=232.48"$ – $C=19.37'$
$v_{m}=C_f*RPM$ – $v_{m}=19.37*250$ – $v_{m}=4,842.5$

45 HP : $P=\frac{33000*H}{v_m}$ – $P=\frac{33000*45}{4843}$ – $P=307$
20 HP : $P=\frac{33000*H}{v_m}$ – $P=\frac{33000*20}{4843}$ – $P=136$

From the ‘Belt Width Table‘ we can see that at 45 Hp the belt — b — needs to be 3-1/2″ wide and at 20 HP only 1-1/2″ wide. Pulley width — B — is taken as $B=\frac{5}{4}b$ .. pulley widths are $4\frac{3}{8}$ and $1\frac{7}{8}$ respectively.

Running the calculations for where in the text it refers to the Stewart Washery having five jigs with 100 HP .. my ‘assumption’ is that the draftsman shows a single jig for simplicity while showing the engine and pulleys for the full five-jig Washery.

100 HP : $P=\frac{33000*H}{v_m}$ – $P=\frac{33000*100}{4843}$ – $P=681$

This requires a 8″ wide belt or a 10″ wide pulley. This is well within the error of me trying to trace a small, blurry image. Since I am modeling a single-jig I am happy to go with the $4\frac{3}{8}$ wide pulley calculated for 45 HP.

Jackshaft Drive Pulley

$D=30"$
$C=94.25"$ – $C=7.85'$
$v_{m}=C_f*RPM$ – $v_{m}=7.85*250$ – $v_{m}=1962.5$

45 HP : $P=\frac{33000*H}{v_m}$ – $P=\frac{33000*45}{1963}$ – $P=757$
20 HP : $P=\frac{33000*H}{v_m}$ – $P=\frac{33000*20}{1963}$ – $P=336$

From the ‘Belt Width Table‘ we can see that at 45 Hp the belt — b — needs to be 9″ wide and at 20 HP only 4″ wide. Pulley width — B — is taken as $B=\frac{5}{4}b$ .. pulley widths are $11\frac{1}{8}$ ” and $5$ ” respectively.

Once again, running calculations for a five-jig Stewart Washery at 100 HP

100 HP : $P=\frac{33000*H}{v_m}$ – $P=\frac{33000*100}{1963}$ – $P=1681$

This requires a 18″ wide belt or a 22.5″ wide pulley. This is (again) well within the error of me trying to trace a small, blurry image. Since I am modeling a single-jig I am happy to go with the $11\frac{1}{8}$ ” wide pulley calculated for 45 HP.

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