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	<updated>2026-05-13T08:16:09Z</updated>
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		<id>https://wiki.opensourceecology.org/index.php?title=Talk:50_hp_Power_Cube_Calculations&amp;diff=124904</id>
		<title>Talk:50 hp Power Cube Calculations</title>
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		<updated>2015-01-05T03:58:35Z</updated>

		<summary type="html">&lt;p&gt;James Sullivan: Hydraulic system power calculation explanation&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Power in a hydraulic system is flow time pressure.  &lt;br /&gt;
Starting with a 50 HP engine and a 3000 psi system pressure, to find volume flow rate first convert horse power into ft lbs / sec and then in lb / sec: 50 HP * 550 = 2750 ft lb / sec = 330,000 in lb /sec.  Now dividing power by pressure gives flow: 330,000 in lb / sec / 3000 lb / in² = 110 in³ / sec = 6600 in³ / min.  If this is direct drive from the engine at 2700 RPM, we can compute the require displacement of the pump: 6600 in³/min / 2700 rev/min = 2.44 in³/rev.  (This is based on engine revolutions.)&lt;br /&gt;
&lt;br /&gt;
Alternatively, given the 1.84 CIPR, we can compute maximum system pressure: Pressure = Work / Disp. / RPM = 1,980,000 in.lb/min / 1.84 in³/rev / 2700 rev/min = 3985 lb/in².&lt;br /&gt;
&lt;br /&gt;
If you already have decided on a system pressure of 3000 psi and acquired a pump with a 1.84 CIPR displacement, then you have a few options.  First you can just run what you have which will give a large margin for error and also reduce fuel consumption.  Alternatively, you can insert a gear or chain drive between the engine and pump to give a speed increase to the pump.  To find the transmission ratio needed we divide the computed displacement based on engine speed by the actual pump displacement 2.44 / 1.84 = 1.32.  I&#039;d put a 10 tooth sprocket on the pump and a 13 on the engine to achieve this ratio, or maybe even a 12 to give some room for error.&lt;/div&gt;</summary>
		<author><name>James Sullivan</name></author>
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