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.

The correct option is 750π.

Step-by-step Explanation:

To find the force exerted by the gas on the pipe between the two locations, we apply the principles of conservation of mass and the linear momentum equation to a control volume of the gas in the pipe.

1. Cross-sectional Area of the Pipe:
The pipe has a uniform circular cross-section with a diameter $D=1\mathrm{m}$.
The cross-sectional area $A$ is given by:
$$A=\frac{\pi }{4}{D}^2=\frac{\pi }{4}{\left(1\right)}^2=\frac{\pi }{4}{\mathrm{m}}^2$$

2. Mass Flow Rate:
For a steady flow, the mass flow rate ($\dot{m}$) remains constant at all sections of the pipe.
At the upstream location (Section 1):
Density, ${\rho }_1=1{\mathrm{kg}/\mathrm{m}}^3$
Velocity, $V_1=100\mathrm{m}/\mathrm{s}$
Thus, the mass flow rate is:
$$\dot{m}={\rho }_{1}A{V}_1=1\times \frac{\pi }{4}\times 100=25\pi \mathrm{kg}/\mathrm{s}$$

3. Linear Momentum Equation:
For steady, one-dimensional flow along the horizontal x-axis, the control volume momentum equation is:
$$\sum F_x=\dot{m}(V_2-V_1)$$
Here, $V_2=170\mathrm{m}/\mathrm{s}$ is the velocity at the downstream location (Section 2).

The external forces acting on the gas control volume in the direction of flow are:
- The pressure force at Section 1 acting in the direction of flow: $P_{1}A$
- The pressure force at Section 2 acting opposing the flow: $-P_{2}A$
- The force exerted by the pipe wall on the gas: $F_{\text{wall on gas}}$

Substituting these forces into the momentum equation:
$$P_{1}A-P_{2}A+F_{\text{wall on gas}}=\dot{m}(V_2-V_1)$$
$$(P_1-P_2)A+F_{\text{wall on gas}}=\dot{m}(V_2-V_1)$$

4. Calculating the Force:
The pressure drop is given as:
$P_1-P_2=10\mathrm{kPa}=10,000\mathrm{Pa}$

Substituting the values into the momentum balance equation:
$$(10,000)\times \left(\frac{\pi }{4}\right)+F_{\text{wall on gas}}=\left(25\pi \right)\times (170-100)$$
$$2500\pi +F_{\text{wall on gas}}=25\pi \times 70$$
$$2500\pi +F_{\text{wall on gas}}=1750\pi$$
$$F_{\text{wall on gas}}=1750\pi -2500\pi =-750\pi \mathrm{N}$$

By Newton's third law, the force exerted by the gas on the pipe ($F_{\text{gas on pipe}}$) is equal in magnitude and opposite in direction to the force exerted by the pipe wall on the gas:
$$F_{\text{gas on pipe}}=-F_{\text{wall on gas}}=-(-750\pi )=750\pi \mathrm{N}$$

Consequently, the magnitude of the force exerted by the gas on the pipe between the two locations is 750π N.