Consider steady, one-dimensional compressible flow of a gas in a pipe of diameter 1 m. At one location in the pipe, the density and velocity are 1 kg/m3 and 100 m/s, respectively. At a downstream location in the pipe, the velocity is 170 m/s. If the pressure drop between these two locations is 10 kPa, the force exerted by the gas on the pipe between these two locations is _______ N.
750π
Problem Overview: Gas Flow in a Pipe
This problem involves analyzing the force exerted by a gas on a pipe during steady, one-dimensional compressible flow. We are provided with initial conditions (density and velocity) at one location and the final velocity at a downstream location. Additionally, the pressure drop between these two points is given. The core task is to determine the net force the gas exerts on the pipe segment.
Key Concepts: Momentum Equation for Fluid Flow
To solve this problem, we need to apply the integral form of the momentum equation for a control volume. For steady, one-dimensional flow in a pipe with constant cross-sectional area, the momentum equation can be simplified.
The momentum equation states that the sum of external forces acting on a control volume equals the net rate of momentum outflow from the control volume. For our pipe section, the forces involved are:
- Pressure Forces: Acting on the inlet and outlet areas.
- Force exerted by the pipe on the fluid (or vice-versa): This is the force we need to find, often due to friction and pressure difference over the pipe wall.
The general momentum equation for steady flow in the x-direction is: $$ \sum F_x = \dot{m}(V_{2x} - V_{1x}) $$ Where:
- $ \sum F_x $ is the sum of forces in the x-direction acting on the fluid within the control volume.
- $ \dot{m} $ is the mass flow rate.
- $ V_{1x} $ is the x-component of velocity at the inlet.
- $ V_{2x} $ is the x-component of velocity at the outlet.
Considering the forces: $$(P_1 A_1) - (P_2 A_2) - F_{\text{pipe on fluid}} = \dot{m}(V_2 - V_1)$$ Here, $P_1$ and $P_2$ are the pressures at the inlet and outlet, $A_1$ and $A_2$ are the cross-sectional areas. $F_{\text{pipe on fluid}}$ is the force exerted by the pipe wall on the fluid. The force exerted by the gas on the pipe is the reaction force, $F_{\text{gas on pipe}} = - F_{\text{pipe on fluid}}$.
Therefore, substituting this into the equation: $$ (P_1 A_1) - (P_2 A_2) + F_{\text{gas on pipe}} = \dot{m}(V_2 - V_1) $$ Rearranging to solve for $F_{\text{gas on pipe}}$: $$ F_{\text{gas on pipe}} = \dot{m}(V_2 - V_1) - (P_1 A_1) + (P_2 A_2) $$ Since the pipe has a constant diameter, $A_1 = A_2 = A$. $$ F_{\text{gas on pipe}} = \dot{m}(V_2 - V_1) - (P_1 - P_2)A $$ Let $ \Delta P = P_1 - P_2 $ be the pressure drop. $$ F_{\text{gas on pipe}} = \dot{m}(V_2 - V_1) - \Delta P \cdot A $$
Step-by-Step Force Calculation
Let's break down the calculation into clear steps using the provided data for the compressible flow of gas.
1. Identifying Given Parameters
- Pipe diameter, $D = 1 \text{ m}$
- Density at location 1, $ \rho_1 = 1 \text{ kg/m}^3 $
- Velocity at location 1, $ V_1 = 100 \text{ m/s} $
- Velocity at location 2, $ V_2 = 170 \text{ m/s} $
- Pressure drop between locations, $ \Delta P = P_1 - P_2 = 10 \text{ kPa} = 10,000 \text{ Pa} $
2. Calculating Cross-sectional Area of the Pipe
The cross-sectional area $A$ of the pipe is given by the formula for the area of a circle: $$ A = \frac{\pi D^2}{4} $$ Substituting the diameter $D = 1 \text{ m}$: $$ A = \frac{\pi (1)^2}{4} = \frac{\pi}{4} \text{ m}^2 $$
3. Determining Mass Flow Rate ($ \dot{m} $)
For steady flow, the mass flow rate remains constant throughout the pipe. It can be calculated at location 1 using the density, area, and velocity: $$ \dot{m} = \rho_1 A V_1 $$ Substituting the values: $$ \dot{m} = (1 \text{ kg/m}^3) \cdot \left(\frac{\pi}{4} \text{ m}^2\right) \cdot (100 \text{ m/s}) $$ $$ \dot{m} = 25\pi \text{ kg/s} $$
4. Applying the Momentum Equation to Find Force
Now, we can use the derived momentum equation to find the force exerted by the gas on the pipe: $$ F_{\text{gas on pipe}} = \dot{m}(V_2 - V_1) - \Delta P \cdot A $$ Substitute the calculated and given values: $$ F_{\text{gas on pipe}} = (25\pi \text{ kg/s})(170 \text{ m/s} - 100 \text{ m/s}) - (10,000 \text{ Pa}) \cdot \left(\frac{\pi}{4} \text{ m}^2\right) $$ $$ F_{\text{gas on pipe}} = (25\pi \text{ kg/s})(70 \text{ m/s}) - (2500\pi \text{ N}) $$ $$ F_{\text{gas on pipe}} = 1750\pi \text{ N} - 2500\pi \text{ N} $$ $$ F_{\text{gas on pipe}} = -750\pi \text{ N} $$
The negative sign indicates that the force exerted by the gas on the pipe is in the opposite direction to the assumed positive flow direction (which is usually the direction of $V_2 - V_1$). However, the question asks for the magnitude of the force.
The magnitude of the force exerted by the gas on the pipe is $750\pi \text{ N}$.
| Parameter | Symbol | Value | Units |
|---|---|---|---|
| Pipe Diameter | $D$ | 1 | m |
| Inlet Density | $ \rho_1 $ | 1 | $ \text{kg/m}^3 $ |
| Inlet Velocity | $ V_1 $ | 100 | m/s |
| Outlet Velocity | $ V_2 $ | 170 | m/s |
| Pressure Drop | $ \Delta P $ | 10,000 | Pa |
| Cross-sectional Area | $ A $ | $ \pi/4 $ | $ \text{m}^2 $ |
| Mass Flow Rate | $ \dot{m} $ | $ 25\pi $ | kg/s |
| Force by Gas on Pipe | $ F_{\text{gas on pipe}} $ | $ -750\pi $ | N |
Final Answer Determination
Based on the calculations, the magnitude of the force exerted by the gas on the pipe between the two locations is $750\pi \text{ N}$.
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