Oil of relative density 0.85 issues from a 50 mm diameter orifice under a pressure of 100 kPa (gauge). The diameter of the vena contracta is 39.5 mm and the discharge is 18 L ·s'. What is the coefficient of velocity?
Oil of relative density 0.85 issues from a 50 mm diameter orifice under a pressure of 100 kPa (gauge). The diameter of the vena contracta is 39.5 mm and the discharge is 18 L ·s'. What is the coefficient of velocity?
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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![**Problem Statement:**
Oil of relative density 0.85 issues from a 50 mm diameter orifice under a pressure of 100 kPa (gauge). The diameter of the vena contracta is 39.5 mm and the discharge is 18 L·s⁻¹. What is the coefficient of velocity?
**Solution:**
To determine the coefficient of velocity, let's explore the given data and the relevant concepts. The coefficient of velocity (\(C_v\)) is used to describe how actual velocity of a fluid jet compares to its theoretical velocity.
**Step-by-step Calculation:**
1. **Theoretical Discharge Velocity (\(v_t\)):**
The theoretical discharge velocity can be calculated using Torricelli’s theorem:
\[
v_t = \sqrt{\frac{2 \times P}{\rho}}
\]
where \(P = 100 \text{ kPa} = 100,000 \text{ Pa}\) and \(\rho\) is the density of oil. The oil's density can be found from the relative density:
\[
\rho = \text{relative density} \times \text{density of water} = 0.85 \times 1000 \text{ kg/m}^3 = 850 \text{ kg/m}^3
\]
Substituting in the values:
\[
v_t = \sqrt{\frac{2 \times 100,000}{850}} \text{ m/s}
\]
2. **Actual Discharge Velocity (\(v_a\)):**
To find the actual velocity, we use the discharge formula:
\[
Q = A \times v_a
\]
where \(Q = 18 \, \text{L/s} = 0.018 \text{ m}^3/\text{s}\), and the area \(A\) is based on the vena contracta diameter:
\[
A = \frac{\pi \times (39.5 \times 10^{-3})^2}{4} \, \text{m}^2
\]
Solving for \(v_a\):
\[
v_a = \frac{Q}{A}
\]
3. **Coefficient of Velocity (\(C_v\)):**
Using \(v_a\) and \(v_t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b5ca759-da4a-4759-9a39-f30a7bb2e4bf%2F73888307-48ad-4702-9b59-c1f47767c331%2Fbx5wi5d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Oil of relative density 0.85 issues from a 50 mm diameter orifice under a pressure of 100 kPa (gauge). The diameter of the vena contracta is 39.5 mm and the discharge is 18 L·s⁻¹. What is the coefficient of velocity?
**Solution:**
To determine the coefficient of velocity, let's explore the given data and the relevant concepts. The coefficient of velocity (\(C_v\)) is used to describe how actual velocity of a fluid jet compares to its theoretical velocity.
**Step-by-step Calculation:**
1. **Theoretical Discharge Velocity (\(v_t\)):**
The theoretical discharge velocity can be calculated using Torricelli’s theorem:
\[
v_t = \sqrt{\frac{2 \times P}{\rho}}
\]
where \(P = 100 \text{ kPa} = 100,000 \text{ Pa}\) and \(\rho\) is the density of oil. The oil's density can be found from the relative density:
\[
\rho = \text{relative density} \times \text{density of water} = 0.85 \times 1000 \text{ kg/m}^3 = 850 \text{ kg/m}^3
\]
Substituting in the values:
\[
v_t = \sqrt{\frac{2 \times 100,000}{850}} \text{ m/s}
\]
2. **Actual Discharge Velocity (\(v_a\)):**
To find the actual velocity, we use the discharge formula:
\[
Q = A \times v_a
\]
where \(Q = 18 \, \text{L/s} = 0.018 \text{ m}^3/\text{s}\), and the area \(A\) is based on the vena contracta diameter:
\[
A = \frac{\pi \times (39.5 \times 10^{-3})^2}{4} \, \text{m}^2
\]
Solving for \(v_a\):
\[
v_a = \frac{Q}{A}
\]
3. **Coefficient of Velocity (\(C_v\)):**
Using \(v_a\) and \(v_t
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