VIPSolutionsGive step-by-step solution with explanation and final answer: Fluid Mechanics in Mechanical Engineering – Calculation Based Questions A liquid of density 9 0 0 kg / m ^ 3 flows through a pipe with a cross - sectional area of 0 . 0 5 m ^ 2 at a velocity of 1 . 6 m / s . Calculate the mass flow rate.
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Give step-by-step solution with explanation and final answer: Fluid Mechanics in Mechanical Engineering – Calculation Based Questions A liquid of density 9 0 0 kg / m ^ 3 flows through a pipe with a cross - sectional area of 0 . 0 5 m ^ 2 at a velocity of 1 . 6 m / s . Calculate the mass flow rate.
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Asked by: Alok Kumar
Created at: 2026-01-24 03:11:53
Asked by: Alok Kumar
Created at: 2026-01-24 03:11:53
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# Fluid Mechanics: Mass Flow Rate Calculation
## Given Data
- **Density of liquid (\(\rho\))**: \(900 \ \mathrm{kg/m^3}\)
- **Cross-sectional area of pipe (\(A\))**: \(.05 \ \mathrm{m^2}\)
- **Velocity of (\(v))**: \(1.6 \ \mathrm{m/s}\)
## Formula for Mass Flow Rate
The mass flow rate (\(\dot{m}\)) is given by:
\[
\dot{m} = \rho \cdot A \cdot v
\]
where:
- \(\dot{m}\) = mass flow rate (\(\mathrm{kg/s}\))
- \(\rho\) = density (\(\mathrm{kg/m^3}\))
- \(A\) = cross-sectional area (\(\mathrm{m^2}\))
- \(v\) = velocity (\(\mathrm{m/s}\))
## Step-by-Step Calculation
1. **Substitute the values:**
\[
\dot{m} = 900 \times .05 \times 1.6
\]
2. **Calculate \(.05 \times 1.6\):**
\[
.05 \times 1.6 = .08
\]
3. **Multiply by density:**
\[
900 \times .08 = 72
\]
## Final Answer
\[
\boxed{72\ \mathrm{kg/s}}
\]
**The mass flow rate is \(72\ \mathrm{kg/s}\).**
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