Question

Air as an ideal gas with $k=1.4$ undergoes a normal shock. The upstream conditions are $p_{\mathrm{x}}=0.5$ bar, $T_{\mathrm{x}}=280\mathrm{K}$, and $M_{\mathrm{x}}=1.8.$ Determine (a) the pressure $p_y$, in bar. (b) the stagnation pressure $p_{\mathrm{ox}}$, in bar. (c) the stagnation temperature $T_{\mathrm{ax}}$, in $\mathrm{K}$. (d) the change in specific entropy across the shock, in $\mathrm{kJ}/\mathrm{kg}\cdot \mathrm{K}$. (e) Plot the quantities of parts (a)-(d) versus $M_x$ ranging from $1.0$ to $2.0$. All other upstream conditions remain the same.

Short answer

a) Pressure downstream $p_y=1.805\text{ bar}$. b) Stagnation pressure upstream $p_{ox}\approx 2.3775\text{ bar}$. c) Stagnation temperature $T_{ax}\approx 461.44\text{ K}$. d) Entropy change $\mathrm{\Delta }s\approx 1.358\text{ kJ/kg.K}$. e) Plot each term versus $M_x$.

Step-by-step solution

Define upstream conditions

Given upstream conditions are: $p_x=0.5\text{ bar}$, $T_x=280\text{ K}$, $M_x=1.8$, $k=1.4$ (ratio of specific heats).

Calculate downstream pressure ($p_y$)

Using the normal shock relations for pressure, $$\frac{p_y}{p_x}=1+\frac{2k}{k+1}(M_x^2-1)$$ Substitute the values: $$\frac{p_y}{0.5}=1+\frac{2(1.4)}{1.4+1}({1.8}^2-1)$$ Calculate: $$\frac{p_y}{0.5}=1+\frac{2.8}{2.4}(2.24)$$ $$\frac{p_y}{0.5}=1+2.61$$ $$\frac{p_y}{0.5}=3.61$$ Thus, $$p_y=1.805\text{ bar}$$

Calculate upstream stagnation pressure ($p_{ox}$)

Using the isentropic relations: $$p_{ox}=p_x{(1+\frac{k-1}{2}{M}_x^2)}^{\frac{k}{k-1}}$$ Substitute the values: $$p_{ox}=0.5{(1+\frac{1.4-1}{2}(1.8{)}^2)}^{\frac{1.4}{1.4-1}}$$ Calculate: $$p_{ox}=0.5{(1+0.2(3.24))}^{3.5}$$ $$p_{ox}=0.5{(1+0.648)}^{3.5}$$ $$p_{ox}=0.5{\left(1.648\right)}^{3.5}$$ $$p_{ox}\approx 0.5\cdot 4.755$$ $$p_{ox}\approx 2.3775\text{ bar}$$

Calculate stagnation temperature ($T_{ax}$)

For an ideal gas, the stagnation temperature is related to the Mach number by: $$T_{ax}=T_x(1+\frac{k-1}{2}{M}_x^2)$$ Substitute the values: $$T_{ax}=280(1+\frac{1.4-1}{2}(1.8{)}^2)$$ Calculate: $$T_{ax}=280(1+0.2(3.24))$$ $$T_{ax}=280(1+0.648)$$ $$T_{ax}=280\cdot 1.648$$ $$T_{ax}\approx 461.44\text{ K}$$

Calculate the change in specific entropy across the shock

The change in specific entropy across the shock can be calculated using: $$\mathrm{\Delta }s=c_p\ln \left(\frac{p_y}{p_x}\right)-R\ln \left(\frac{T_y}{T_x}\right)$$ First, find downstream temperature ($T_y$): $$\frac{T_y}{T_x}=\frac{(2k{M}_x^2-(k-1))((k-1)M_x^2+2)}{(k+1{)}^2{M}_x^2}$$ Substitute the values: $$\frac{T_y}{280}=\frac{(2(1.4)(1.8{)}^2-(1.4-1))((1.4-1)(1.8{)}^2+2)}{(1.4+1{)}^2(1.8{)}^2}$$ Calculate: $$\frac{T_y}{280}=\frac{(5.04-0.4)(1.744+2)}{6.76(1.8{)}^2}$$ $$\frac{T_y}{280}=\frac{4.64(3.744)}{21.96}$$ $$\frac{T_y}{280}=0.79$$ $$T_y\approx 221.2\text{ K}$$ Finally, calculate the change in entropy: $$\mathrm{\Delta }s=c_p\ln (3.61)-R\ln (0.79)$$ Let's take $c_p=1.005\text{ kJ/kg K}$ and $R=0.287\text{ kJ/kg K}$: $$\mathrm{\Delta }s=1.005\ln (3.61)-0.287\ln (0.79)$$ Calculate: $$\mathrm{\Delta }s\approx 1.005(1.284)-0.287(-0.236)$$ $$\mathrm{\Delta }s\approx 1.29+0.068$$ $$\mathrm{\Delta }s\approx 1.358\text{kJ/kgK}$$

Plot quantities (a)-(d) versus Mach number (Mx)

Using a graphing tool, plot the data: y-axis represents each quantity (pressure, stagnation pressure and temperature, and entropy change), and x-axis the Mach number ($M_x$) from 1.0 to 2.0. Compute these values iteratively by substituting different $M_x$ values and plotting curves for each quantity.

Key concepts

Ideal Gas Behavior

Understanding the behavior of ideal gases is crucial for solving problems involving compressible flow, such as normal shocks. Ideal gases follow the ideal gas law, which is given by:
$$pV=nRT$$
Here, $p$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the universal gas constant, and $T$ is the temperature.
These gases also adhere to specific heat capacities at constant pressure ($c_p$) and constant volume ($c_v$). The ratio of these, denoted as $k$ or the adiabatic index, is important in isentropic and shock calculations, especially in thermodynamic relations. For air, $k$ is typically 1.4, meaning air doesn't deviate significantly under normal conditions, making it a good candidate for ideal gas approximation.

Stagnation Pressure

Stagnation pressure represents the pressure a fluid attains when brought to rest isentropically. Essentially, it assumes all kinetic energy of the moving fluid is converted to pressure energy without any losses.
For an ideal gas undergoing isentropic flows, the stagnation pressure $p_{ox}$ can be calculated using the following relation:
$$p_{ox}=p_x{(1+\frac{k-1}{2}{M}_x^2)}^{\frac{k}{k-1}}$$
Here, $p_x$ is the upstream static pressure, $k$ is the ratio of specific heats, and $M_x$ is the Mach number. This relation helps to determine how the pressure changes in the flow field and is particularly useful in high-speed aerodynamics and gas dynamics problems.

Stagnation Temperature

Similar to stagnation pressure, stagnation temperature represents the temperature a fluid reaches if brought to rest isentropically. This concept is particularly useful in the analysis of high-speed flows where kinetic energies are significant.
The formula to determine stagnation temperature $T_{ax}$ for an ideal gas is:
$$T_{ax}=T_x(1+\frac{k-1}{2}{M}_x^2)$$
In this equation, $T_x$ denotes the upstream static temperature, $k$ is the ratio of specific heats, and $M_x$ is the Mach number. This equation shows how the temperature in the flow changes as the flow speed (Mach number) changes, providing insight into thermal effects in compressible flow regimes.

Specific Entropy Change

Entropy is a measure of a system's disorder, and its change, $\mathrm{\Delta }s$, across a shock can provide insights into the irreversibility and energy dissipation of the process. For an ideal gas experiencing a normal shock, the specific entropy change can be calculated with:
$$\mathrm{\Delta }s=c_p\ln \left(\frac{p_y}{p_x}\right)-R\ln \left(\frac{T_y}{T_x}\right)$$
Here, $c_p$ is the specific heat at constant pressure, $R$ is the specific gas constant, $p_y$ and $p_x$ are the downstream and upstream pressures, respectively, and $T_y$ and $T_x$ are the downstream and upstream temperatures.
The change in entropy indicates the degree of disorder introduced due to the shock and helps in understanding energy losses. A positive entropy change signifies an irreversible process with energy dispersed as heat.