18.4: Calculate the Retract Speed of a Cylinder Given its Size and a Flow Rate
- Page ID
- 117079
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)When a double-acting hydraulic cylinder retracts, fluid enters the rod end rather than the full bore area. Because the rod takes up space inside the cylinder, only the volume resulting from the annular area of the piston is filled. This is why retract speeds are typically faster than extend speeds for the same flow rate.
1. Calculate the annular area of each cylinder.
The annular area is calculated using the formula:
A = 0.7854 × (D² − d²)
Where:
- D is the piston (bore) diameter
- d is the rod diameter
Given:
- Large Bore Cylinder: D = 1.5 in, d = 0.44 in
- Small Bore Cylinder: D = 1.125 in, d = 0.31 in
Large Bore Cylinder Annular Area:
A = 0.7854 × (1.5² − 0.44²)
= 0.7854 × (2.25 − 0.1936)
= 0.7854 × 2.0564
=1.615 in²
Small Bore Cylinder Annular Area:
A = 0.7854 × (1.125² − 0.31²)
= 0.7854 × (1.266 − 0.0961)
= 0.7854 × 1.1699
= 0.918 in²
2. Using the areas above, calculate the retract speeds for the given flow rates.
Use the same speed formula:
Rod Speed (in/min) = Flow Rate (in³/min) / Annular Area (in²)
Remember:
- 1 GPM = 231 in³/min
Where:
- Flow Rate = 3 GPM = 924 in³/min
Rod Speed for large bore (in/min) = 693 in³/min / 1.615 in²
≈ 429 in/min
Rod Speed for small bore (in/min) = 693 in³/min / 0.918 in²
≈755 in/min
3. Compare these speeds with the extend speeds.
In both cylinders, the retract speeds are significantly higher than the extend speeds at the same flow rates. This is because the rod reduces the volume that needs to be filled during retraction, leaving a smaller area for oil to act on, resulting in higher speeds. This difference is more pronounced in cylinders with larger rod diameters.
4. Determine the cylinder size needed for a specific retract speed.
Given:
Target Retract Speed = 110 in/min
Rod Diameter = 1 in
Flow Rate = 3 GPM = 693 in³/min
First, find the annular area:
A = Flow Rate / Rod Speed
= 693 in³/min / 110 in/min =6.3 in²
Now solve for bore diameter:
6.3 in²= 0.7854 × (D²−1²)
⇒D²−1 = 6.3 / 0.7854 = 8.02 in²
⇒D²= 8.02 in²
⇒D≈ 2.83 in
Answer:
Cylinder Bore Diameter = 2.83 inches
5. Determine the flow rate needed to retract a cylinder at a speed of 30 in/min.
Given:
Speed = 30 in/min
Bore Diameter (D) = 6 in
Rod Diameter (d) = 1.15 in
Annular Area = 0.7854 × (D² − d²)
A = 0.7854 × (6² − 1.15²)
= 0.7854 × (36 − 1.3225)
= 0.7854 × 34.6775
= 27.24 in²
Volume = Rod Speed × Area = 30 in/min × 27.24 in² = 817.2 in³/min
Convert to GPM:
Flow Rate = 817.2 (in³/min) / 231 (in³/gal) ≈ 3.54 GPM
Answer:
Flow Rate Needed = 3.54 GPM