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In a frictionless, constant-area flow of a perfect gas, the inlet conditions are \(\rho_{1}=100 \mathrm{kPa}, \rho_{1}=1 \mathrm{~kg} / \mathrm{m}^{3}\), and \(u_{1}=100 \mathrm{~m} / \mathrm{s}\). At a downstream station the gas is at \(200 \mathrm{~m} / \mathrm{s}\). Assuming the medium is air, calculate the corresponding static pressure and density, \(p_{2}\) and \(p_{2}\), respectively. $$ \left[R_{\text {air }}=287 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \gamma_{\text {air }}=1.4\right] $$

Short Answer

Expert verified
The corresponding static pressure and density at the downstream station are \(p_2=125 \, kPa\) and \(\rho_2 = 0.5 \, kg/m^3\), respectively.

Step by step solution

01

Formulate Isentropic Flow Relations

The continuity equation for one-dimensional isentropic flow gives the relationship between density and velocity: \(\rho_1 u_1 = \rho_2 u_2\). Similarly, the energy equation leads to the relation: \(\frac{u_1^2}{2} + \frac{p_1}{\rho_1} = \frac{u_2^2}{2} + \frac{p_2}{\rho_2}\). These equations allow us to relate the conditions at the two points in the flow.
02

Calculate Intermediate Velocity Ratio

First determine the velocity ratio \( \frac{u_2}{u_1} = 2 \). Substituting this value into the continuity equation will allow us to find the ratio \(\frac{\rho_2}{\rho_1}\).
03

Calculate Density Ratio

From the continuity equation we find \(\frac{\rho_2}{\rho_1} = \frac{1}{2}\). This gives us the density at the downstream point, \( \rho_2 = 0.5 \, kg/m^3 \).
04

Calculate Pressure Ratio

Substitute the values of upstream and downstream velocities and densities into the energy equation. Simplifying will give the ratio \(\frac{p_2}{p_1} = \frac{5}{4}\). Hence we find the static pressure at point 2 to be \(p_2=125 \, kPa\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity Equation
In fluid dynamics, the continuity equation is a fundamental principle that describes the conservation of mass in a flow system. For one-dimensional isentropic flow, the continuity equation states that the product of density (\( \rho \)) and velocity (\( u \)) remains constant along the flow. This is expressed as \( \rho_1 u_1 = \rho_2 u_2 \). This relationship ensures that the mass flow rate is the same at all points in the system.
  • Initial conditions are used to calculate downstream properties. For example, if velocity increases downstream, density must decrease to maintain mass conservation.
  • In our problem, we have an initial velocity and density, allowing us to find the downstream density using the continuity equation.
Understanding the continuity equation helps us grasp how fluid behaves when moving through various environments. It's especially useful when determining changes in speed and density, which are key in engineering applications like jet engines or nozzles.
Energy Equation
The energy equation for fluid flow incorporates principles of energy conservation. In isentropic (no energy loss due to heat transfer or friction) and adiabatic (no heat exchange) processes, the energy in the flow is conserved across the system. This is represented by the Bernoulli equation, adapted for compressible flows, as: \[ \frac{u_1^2}{2} + \frac{p_1}{\rho_1} = \frac{u_2^2}{2} + \frac{p_2}{\rho_2} \] Here, the kinetic energy per unit mass and the flow work term contribute to the total energy at any point in the system. Energy conservation helps us relate pressure and velocity changes across the flow.
  • This equation allows us to relate the input conditions to the output conditions, particularly finding how pressure changes when velocity increases without external work being done.
  • Using this equation, we understand that when velocity doubles, downstream pressure can be calculated, keeping the system isolated from external influences.
In isentropic flows, knowing upstream conditions allows calculation of downstream conditions, essential for designing systems that utilize gas dynamics efficiently.
Density Ratio
The density ratio in isentropic flow reveals how the density of a fluid changes as it moves through a system with varying velocities. Using the continuity equation, for our scenario, we found that the density changes in response to velocity changes. Since \( \rho_1 u_1 = \rho_2 u_2 \), we can rearrange to find \( \rho_2 = \rho_1 \frac{u_1}{u_2} \).
  • Given the velocity ratio \( \frac{u_2}{u_1} = 2 \), the density ratio is \( \frac{\rho_2}{\rho_1} = \frac{1}{2} \).
  • Thus, the downstream density becomes half of the initial density, highlighting how density is inversely related to velocity in a conservation setup.

  • The density ratio indicates that as fluid speeds up, its density decreases, a critical understanding in applications like air intakes in aircraft.
    Pressure Ratio
    The pressure ratio is a crucial quantity in analyzing flow changes in systems where pressure and velocity vary. Isentropic relations help us in connecting the changes in the fluid properties as it progresses downstream.
    We use the energy equation to derive the pressure ratio in our example. With velocity doubling, we found:
    • The calculated pressure ratio from the energy equation is \( \frac{p_2}{p_1} = \frac{5}{4} \).
    • This means the downstream pressure is 1.25 times the initial pressure, resulting in \( p_2 = 125 \, \text{kPa} \).
    Understanding pressure ratios in isentropic flow is vital for predicting how gases behave under compression or expansion, such as in turbines or compressors, and ensures we have the design data needed for these complex systems.

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    Most popular questions from this chapter

    A scramjet combustor has a supersonic inlet condition and a choked exit. The combustor flow area increases linearly in the flow direction, as shown. The inlet and exit flow conditions are $$ \begin{aligned} M_{1} &=3.0 \\ p_{1} &=1 \mathrm{bar} \\ T_{1} &=1000 \mathrm{~K} \\ A_{1} &=1 \mathrm{~m}^{2} \\ M_{2} &=1.0 \\ A_{2} &=1.4 \mathrm{~m}^{2} \\ \gamma &=1.4, R=287 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \end{aligned} $$ The total heat release due to combustion, per unit flow rate in the duct, is initially assumed to be \(15 \mathrm{MJ} / \mathrm{kg}\). If we divide the combustor into three constant-area sections, with stepwise jumps in the duct area, we may apply Rayleigh flow principles to each segment, as shown. The heat release per segment is then \(1 / 3\) of the total heat release in the duct, i.e., \(5,000 \mathrm{~kJ} / \mathrm{kg}\). As the exit condition of a segment needs to be matched to the inlet condition of the following segment, we propose to satisfy continuity equation at the boundary through an isentropic step area expansion, i.e., \(p_{1}, T_{\mathrm{t}}\) remain the same and only the Mach number jumps isentropically through area expansion. If we march from the inlet condition toward the exit with the assumed heat release rates, we calculate the exit Mach number \(M_{2}\). Since the exit flow is specified to be choked, then we need to adjust the total heat release in order to get a choked exit. Calculate the critical heat release in the above duct that leads to thermal choking of the flow.

    In a Fanno flow problem with a supersonic inlet Mach number of \(M_{1}=2.0\), a normal shock appears at \(M_{x}=\) 1.2. Assuming the average friction coefficient in the pipe is \(C_{t}=0.005\) and \(\gamma=1.4\), calculate (a) \(L_{\mathrm{x}} / D\) (b) \(L / D\) (c) percentage total pressure loss, \(\Delta p_{\mathrm{t}} / p_{\mathrm{u} 1}\)

    A scramjet combustor is modeled as a frictionless flow of a perfect gas in a constant-area duct with heating (Rayleigh flow). The inlet Mach number to the burner is \(M_{1}=2.8\) with \(\gamma=1.4\). Critical heating of the gas (i.e., by burning fuel) in the combustor achieves a sonic exit state. Assuming the gas is calorically perfect, calculate (a) the nondimensional critical heat flux, \(q_{1}^{*} / c_{p} T_{\mathrm{t}}\), to achieve thermal choking of the flow (b) the percentage static pressure rise, \(\Delta p / p_{1}\)

    A supersonic combustion is modeled as Rayleigh flow. The inlet Mach number is \(M_{1}=3.0\), the total temperature and pressure at the inlet are \(T_{\mathrm{tl}}=1,500 \mathrm{~K}\) and \(p_{\mathrm{t} 1}=\) \(100 \mathrm{kPa}\), respectively. Calculate (a) minimum \(q\) to choke the flow at the exit (b) exit temperature, \(T_{2}\) (c) if the fuel is hydrogen with a heating value of \(120,000 \mathrm{~kJ} / \mathrm{kg}\), calculate the fuel-to-air ratio, \(f\). Assume \(\gamma=1.4\) and \(c_{p}=1,004 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\)

    A perfect gas flows in a well-insulated, constant-area pipe with friction at \(M_{1}=2.0\). For an average wall friction coefficient of \(C_{f}=0.004\), calculate (a) the maximum length of the pipe that can transmit the flow (b) total pressure loss at this length Assume the pipe cross-section is rectangular and has the dimensions of \(10 \times 20 \mathrm{~cm}\).

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