A gear pump has a 75-mm outside diameter, a 55-mm inside diameter, and a 30-mm width. If the actual pump flow rate at 2700 rpm and rated pressure is 0.0025 m³/s. What is the volumetric efficiency?
A gear pump has a 75-mm outside diameter, a 55-mm inside diameter, and a 30-mm width. If the actual pump flow rate at 2700 rpm and rated pressure is 0.0025 m³/s. What is the volumetric efficiency?
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![### Gear Pump Volumetric Efficiency Calculation
**Problem Statement:**
A gear pump has the following specifications:
- Outside diameter: 75 mm
- Inside diameter: 55 mm
- Width: 30 mm
The actual pump flow rate at 2700 rpm and rated pressure is 0.0025 m³/s. What is the volumetric efficiency?
**Solution Steps:**
1. **Determine the Theoretical Flow Rate:**
To find the theoretical flow rate (\(Q_{th}\)), we use the formula for the volumetric displacement of the gear pump:
\[
V = \frac{\pi}{4} \cdot (D_o^2 - D_i^2) \cdot W
\]
Where:
- \( V \) is the volumetric displacement per revolution (in \( m^3 \))
- \( D_o \) is the outside diameter
- \( D_i \) is the inside diameter
- \( W \) is the width
Given values:
- \( D_o \) = 75 mm = 0.075 m
- \( D_i \) = 55 mm = 0.055 m
- \( W \) = 30 mm = 0.03 m
Substituting these values into the formula:
\[
V = \frac{\pi}{4} \cdot (0.075^2 - 0.055^2) \cdot 0.03
\]
\[
V \approx \frac{\pi}{4} \cdot (0.005625 - 0.003025) \cdot 0.03
\]
\[
V \approx \frac{\pi}{4} \cdot 0.0026 \cdot 0.03
\]
\[
V \approx 6.136 \times 10^{-5} \, \text{m}^3
\]
2. **Convert Volumetric Displacement to Flow Rate:**
The theoretical flow rate \(Q_{th}\) is the product of the volumetric displacement and the rotational speed (in revolutions per minute or rpm):
\[
Q_{th} = V \cdot \text{rpm}
\]
Converting rpm to revolutions per](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f8a6c8f-9c9e-4a3f-8cfc-fb848b18acc8%2F99ae8ae2-a7ec-4881-9414-55879c1f26eb%2Fb1pviaf_processed.gif&w=3840&q=75)
Transcribed Image Text:### Gear Pump Volumetric Efficiency Calculation
**Problem Statement:**
A gear pump has the following specifications:
- Outside diameter: 75 mm
- Inside diameter: 55 mm
- Width: 30 mm
The actual pump flow rate at 2700 rpm and rated pressure is 0.0025 m³/s. What is the volumetric efficiency?
**Solution Steps:**
1. **Determine the Theoretical Flow Rate:**
To find the theoretical flow rate (\(Q_{th}\)), we use the formula for the volumetric displacement of the gear pump:
\[
V = \frac{\pi}{4} \cdot (D_o^2 - D_i^2) \cdot W
\]
Where:
- \( V \) is the volumetric displacement per revolution (in \( m^3 \))
- \( D_o \) is the outside diameter
- \( D_i \) is the inside diameter
- \( W \) is the width
Given values:
- \( D_o \) = 75 mm = 0.075 m
- \( D_i \) = 55 mm = 0.055 m
- \( W \) = 30 mm = 0.03 m
Substituting these values into the formula:
\[
V = \frac{\pi}{4} \cdot (0.075^2 - 0.055^2) \cdot 0.03
\]
\[
V \approx \frac{\pi}{4} \cdot (0.005625 - 0.003025) \cdot 0.03
\]
\[
V \approx \frac{\pi}{4} \cdot 0.0026 \cdot 0.03
\]
\[
V \approx 6.136 \times 10^{-5} \, \text{m}^3
\]
2. **Convert Volumetric Displacement to Flow Rate:**
The theoretical flow rate \(Q_{th}\) is the product of the volumetric displacement and the rotational speed (in revolutions per minute or rpm):
\[
Q_{th} = V \cdot \text{rpm}
\]
Converting rpm to revolutions per
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