Find the pressure that a liquid jet of prescribed speed exerts on a rigid wall. Follow three links in order. Theory gives a closed-form reference. A one-dimensional finite-volume code calculates the planar history. An axisymmetric COMSOL model verifies the finite jet.

求给定速度的液体射流作用于刚性壁面的压力。依次沿三个环节推进。理论给出封闭形式的参照。一维有限体积程序计算平面历程。轴对称 COMSOL 模型核验有限射流。

Fix the coordinates first. The rigid wall lies at $z=0$. The liquid occupies $z>0$ and moves toward $-z$. The radial coordinate $r$ measures distance from the jet axis. Time $t=0$ is the first contact of the liquid front with the wall. The jet radius $a$ and the incoming speed $U$ are supplied.

先固定坐标。刚性壁面位于 $z=0$。液体占据 $z>0$ 并朝 $-z$ 运动。径向坐标 $r$ 为到射流轴线的距离。时间 $t=0$ 为液体前沿首次接触壁面的时刻。射流半径 $a$ 与入射速度 $U$ 为给定值。

State the assumptions. The liquid is the only phase. There is no gas cushion, splash, wall elasticity or cavitation. The wall is flat, rigid and frictionless. The jet side is free, so its pressure is released to the reference pressure $p_j$. Only the first arrival is modelled, up to about one sound-crossing time of the jet radius.

说明假设。液体是唯一的相。不含气垫、飞溅、壁面弹性或空化。壁面平坦、刚性且无摩擦。射流侧面为自由面，其压力释放到参考压力 $p_j$。仅模拟首次到达，时长约为声波穿过射流半径所需时间的一倍。

Chapter 3 bounds the speed a source can give a finite jet. This chapter maps a given speed to a wall pressure. Whether a real source reaches that speed is not decided here. Physical validation against an experiment: Not established. Missing: a measured wall pressure for a real jet.

第三章界定了能源能给有限射流的速度上限。本章把给定速度映射为壁面压力。真实能源能否达到该速度，本章不作判断。对照实验的物理验证：尚未确立。缺少：真实射流的壁面压力实测。

Step 1 — Adopt a pressure law that depends on density alone. Eq. (C5-E01) is the stiffened-gas form. It is an adopted constitutive law, not a derivation from first principles. [R17] Both solvers below integrate this same law.

步骤 1——采用仅依赖于密度的压力律。式（C5-E01）为刚化气体形式。它是采用的本构律，并非由第一性原理推导。 [R17] 下文两个求解器积分的都是这同一条定律。

Step 2 — Differentiate the law. The factor $(\rho/\rho_j)^{\gamma}$ gives $dp/d\rho=\gamma(p_j+\Pi)\rho^{\gamma-1}/\rho_j^{\gamma}$. Replace $(p_j+\Pi)(\rho/\rho_j)^{\gamma}$ by $p+\Pi$. This gives $c^2=\gamma(p+\Pi)/\rho$. At the reference state, $c_0^2=4.4\times6.00101325\times10^{8}/998.1539919$. Hence $c_0=1626.447$ m s⁻¹. The check is dimensional: pressure over density is a squared speed.

步骤 2——对该律求导。因子 $(\rho/\rho_j)^{\gamma}$ 给出 $dp/d\rho=\gamma(p_j+\Pi)\rho^{\gamma-1}/\rho_j^{\gamma}$。将 $(p_j+\Pi)(\rho/\rho_j)^{\gamma}$ 替换为 $p+\Pi$，得到 $c^2=\gamma(p+\Pi)/\rho$。在参考状态下，$c_0^2=4.4\times6.00101325\times10^{8}/998.1539919$，故 $c_0=1626.447$ m s⁻¹。量纲检验：压力除以密度是速度的平方。

Step 3 — Take the rigid-wall limit of the Chapter 3 contact pressure. Chapter 3 matched liquid and receiver impedances and found the early pressure of Eq. (C3-E24). A rigid wall has $Z_r\to\infty$. The common contact velocity then tends to zero. The pressure tends to $Z_0U$ with $Z_0=\rho_jc_0=1.6234\times10^{6}$ Pa s m⁻¹. The opposite limit $Z_r\to0$ gives zero pressure, as a free surface should.

步骤 3——取第三章接触压力的刚性壁面极限。第三章匹配了液体与接收体的阻抗，得到式（C3-E24）的早期压力。刚性壁面有 $Z_r\to\infty$。共同接触速度趋于零，压力趋于 $Z_0U$，其中 $Z_0=\rho_jc_0=1.6234\times10^{6}$ Pa s m⁻¹。相反的极限 $Z_r\to0$ 给出零压力，符合自由面。

At $U=99.027$ m s⁻¹ the excess pressure $Z_0U$ is 160.765 MPa. The absolute value $p_A$ is 160.867 MPa. This is a linear estimate. It holds for $U/c_0\ll1$. The next section removes that restriction.

在 $U=99.027$ m s⁻¹ 时，超压 $Z_0U$ 为 160.765 MPa，绝对值 $p_A$ 为 160.867 MPa。这是线性估计，适用于 $U/c_0\ll1$。下一节去掉这一限制。

Step 4 — Describe the stopping shock. At first contact a shock leaves the wall and travels toward $+z$ at lab speed $S$. Liquid behind it is at rest against the wall. Liquid ahead of it moves toward the wall at speed $U$. Its speed relative to the incoming liquid is $U_s=U+S$. Work in the frame of the shock. Liquid enters at speed $U_s$ and leaves at speed $S=U_s-U$.

步骤 4——描述停止激波。首次接触时，激波离开壁面，以实验室速度 $S$ 朝 $+z$ 传播。激波后的液体相对壁面静止，激波前的液体以速度 $U$ 朝壁面运动。它相对入射液体的速度为 $U_s=U+S$。取激波参考系：液体以速度 $U_s$ 流入，以速度 $S=U_s-U$ 流出。

Mass balance across the front gives $\rho_jU_s=\rho_1S$. Momentum balance gives $p_j+\rho_jU_s^2=p_1+\rho_1S^2$. Substitute $\rho_1S=\rho_jU_s$ into the second relation. Then $p_1-p_j=\rho_jU_s(U_s-S)=\rho_jU_sU$. These are the two relations of Eq. (C5-E03). They are exact for a planar steady shock and need no equation of state. [R16]

跨波前的质量平衡给出 $\rho_jU_s=\rho_1S$。动量平衡给出 $p_j+\rho_jU_s^2=p_1+\rho_1S^2$。把 $\rho_1S=\rho_jU_s$ 代入第二式，得 $p_1-p_j=\rho_jU_s(U_s-S)=\rho_jU_sU$。这就是式（C5-E03）的两个关系。它们对平面定常激波精确成立，且不需要状态方程。 [R16]

Step 5 — Close the jump with energy conservation. The specific internal energy of the stiffened-gas liquid is $e=(p+\gamma\Pi)/[(\gamma-1)\rho]$. Energy conservation across the shock requires $e_1-e_j=\tfrac12(p_1+p_j)(1/\rho_j-1/\rho_1)$. Write $\rho_1=\rho_j/(1-x)$ with $x=U/U_s$. Put $\Delta=p_1-p_j$ and $B=p_j+\gamma\Pi$. Multiply the energy relation by $(\gamma-1)\rho_j$. It becomes $\Delta(1-x)-Bx=\tfrac12(\gamma-1)x(2p_j+\Delta)$.

步骤 5——用能量守恒封闭跳跃。刚化气体液体的比内能为 $e=(p+\gamma\Pi)/[(\gamma-1)\rho]$。跨激波的能量守恒要求 $e_1-e_j=\tfrac12(p_1+p_j)(1/\rho_j-1/\rho_1)$。令 $\rho_1=\rho_j/(1-x)$，其中 $x=U/U_s$。记 $\Delta=p_1-p_j$、$B=p_j+\gamma\Pi$。将能量关系乘以 $(\gamma-1)\rho_j$，得 $\Delta(1-x)-Bx=\tfrac12(\gamma-1)x(2p_j+\Delta)$。

Collect the terms in $\Delta$. Then $\Delta\,[1-(\gamma+1)x/2]=\gamma(p_j+\Pi)\,x$. Momentum gives $\Delta=\rho_jU^2/x$. Insert it and use $c_0^2=\gamma(p_j+\Pi)/\rho_j$. The result is $x^2+\tfrac12(\gamma+1)M^2x-M^2=0$ with $M=U/c_0$. Its positive root is $x=-k+\sqrt{k^2+M^2}$ with $k=(\gamma+1)M^2/4$. The Hugoniot pressure follows as $p_H=p_j+\rho_jU^2/x$. [R16]

合并 $\Delta$ 的项，得 $\Delta\,[1-(\gamma+1)x/2]=\gamma(p_j+\Pi)\,x$。动量关系给出 $\Delta=\rho_jU^2/x$。代入并使用 $c_0^2=\gamma(p_j+\Pi)/\rho_j$，得到 $x^2+\tfrac12(\gamma+1)M^2x-M^2=0$，其中 $M=U/c_0$。其正根为 $x=-k+\sqrt{k^2+M^2}$，$k=(\gamma+1)M^2/4$。Hugoniot 压力为 $p_H=p_j+\rho_jU^2/x$。 [R16]

Check the limits and the number. For $M\to0$, $x\to M$ and $p_H\to p_j+\rho_jc_0U$, which is Eq. (C5-E02). At $U=99.02724$ m s⁻¹, $M=0.060886$, $k=0.0050046$ and $x=0.056086$. Hence $U_s=1765.62$ m s⁻¹, $\rho_1/\rho_j=1.0594$ and $p_H=174.6229$ MPa. A separate bisection on the energy residual in the C++ code gives the same 174.622895 MPa.

检验极限与数值。当 $M\to0$ 时，$x\to M$，$p_H\to p_j+\rho_jc_0U$，即式（C5-E02）。在 $U=99.02724$ m s⁻¹ 时，$M=0.060886$，$k=0.0050046$，$x=0.056086$。于是 $U_s=1765.62$ m s⁻¹，$\rho_1/\rho_j=1.0594$，$p_H=174.6229$ MPa。C++ 程序对能量残差另作二分求根，得到相同的 174.622895 MPa。

Step 6 — Keep the density-only law and drop the energy equation. This is the model that both solvers integrate. Mass and momentum still give $\rho_1=\rho_j(U+S)/S$ and $p_1-p_j=\rho_j(U+S)U$. The law of Eq. (C5-E01) must also give $p_1$ at the density $\rho_1$. Solve this condition for the root $S_B$.

步骤 6——保留仅含密度的定律并舍去能量方程。这是两个求解器积分的模型。质量与动量仍给出 $\rho_1=\rho_j(U+S)/S$ 与 $p_1-p_j=\rho_j(U+S)U$。式（C5-E01）的定律还须在密度 $\rho_1$ 处给出 $p_1$。对该条件求根 $S_B$。

At the nominal speed $S_B=1661.86$ m s⁻¹ and $p_B=174.155$ MPa. This lies 0.268% below $p_H$. The gap is the effect of dropping the energy equation. The acoustic estimate of Eq. (C5-E02) lies 7.9% below $p_H$ in absolute pressure. The nonlinear factor is $(p_H-p_j)/(Z_0U)=U_s/c_0=1.0856$.

在名义速度下 $S_B=1661.86$ m s⁻¹，$p_B=174.155$ MPa，比 $p_H$ 低 0.268%。该差值是舍去能量方程的影响。式（C5-E02）的声学估计在绝对压力上比 $p_H$ 低 7.9%。非线性因子为 $(p_H-p_j)/(Z_0U)=U_s/c_0=1.0856$。

Step 7 — Define what is measured. The planar result of Section 3 holds under a column that fills the wall. A real jet has a free side from the first instant. Report the wall pressure as a mean over a fixed disk. Eq. (C5-E06) uses the axisymmetric area weight $2r\,dr/R^2$. The disk radius $R$ is $a$ for the full sensor and $a/2$ for the centre. This follows the fixed-observer rule of Eq. (C3-E26).

步骤 7——定义被测量。第 3 节的平面结果适用于充满壁面的液柱。真实射流自首次接触起就有自由侧边。将壁面压力报告为固定圆盘上的平均值。式（C5-E06）采用轴对称面积权重 $2r\,dr/R^2$。圆盘半径 $R$ 对全传感区取 $a$，对中心区取 $a/2$。这遵循式（C3-E26）的固定观测量规则。

Step 8 — Estimate the effect of the free side. Pressure release starts at the edge $r=a$ at first contact. Let it advance inward at $c_0$. After the dimensionless time $\tau=c_0t/a$ it has reached the annulus $a(1-\tau)<r\le a$. The fraction of the disk area in that annulus is $f_{\mathrm{rel}}=1-(1-\tau)^2$. If the annulus keeps no excess pressure and the core keeps $p_H$, the mean falls to $p_H(1-\tau)^2$. This is a scale estimate. It fixes the sign and size of the gap, not its value.

步骤 8——估计自由侧边的影响。压力释放自首次接触起从边缘 $r=a$ 开始，设其以 $c_0$ 向内推进。经无量纲时间 $\tau=c_0t/a$ 后，它到达环带 $a(1-\tau)<r\le a$。该环带占圆盘面积的比例为 $f_{\mathrm{rel}}=1-(1-\tau)^2$。若环带不保留超压而核心保持 $p_H$，平均值降为 $p_H(1-\tau)^2$。这是尺度估计，它确定差值的符号与大小，而不确定其数值。

Step 9 — Count the energy that crosses a plane upstream of the wall. The plane $z=2a$ is hit by the incident liquid before the reflected shock arrives. The reflected shock needs about $2a/(U_s-U)=1.20$ ns to climb back for $a=1$ μm. Use the window $t_w=1$ ns. The kinetic-energy flux $\dot K=\tfrac12\rho_jU^3\pi a^2$ is 1.5226 mW. Energy through the plane in 1 ns is 1.5226 pJ. This is liquid energy crossing a plane. A rigid wall does no work on the liquid, so it receives none.

步骤 9——计量穿过壁面上游平面的能量。入射液体在反射激波到达之前先经过平面 $z=2a$。对 $a=1$ μm，反射激波返回该平面约需 $2a/(U_s-U)=1.20$ ns。取时间窗 $t_w=1$ ns。动能通量 $\dot K=\tfrac12\rho_jU^3\pi a^2$ 为 1.5226 mW，1 ns 内穿过平面的能量为 1.5226 pJ。这是穿过平面的液体能量。刚性壁面不对液体做功，因此也不接收功。

Step 10 — State the validity range. The nominal Mach number is $M=0.0609$. The tested range is $0.031\le M\le0.092$. The sound-crossing time of the jet radius is $a/c_0=0.615$ ns for $a=1$ μm. The compression behind the shock is only 5.9%. That is why the barotropic law stays within 0.27% of the energy law. No conclusion is claimed outside this range.

步骤 10——说明适用范围。名义马赫数为 $M=0.0609$，检验范围为 $0.031\le M\le0.092$。$a=1$ μm 时，声波穿过射流半径的时间 $a/c_0$ 为 0.615 ns。激波后的压缩仅 5.9%，因此正压律与能量律相差不超过 0.27%。超出此范围不作任何结论。

Step 11 — Write the planar problem as conservation laws. Eq. (C5-E10) holds mass and momentum in conservative form. Conservative form lets a shock form and move at the correct speed. The wall is a mirrored state with $u=0$. The inflow boundary holds the state $(\rho_j,-U)$. The initial state is the uniform inflow state.

步骤 11——把平面问题写成守恒律。式（C5-E10）以守恒形式表示质量与动量。守恒形式使激波得以形成并以正确速度运动。壁面为 $u=0$ 的镜像状态。入流边界保持状态 $(\rho_j,-U)$，初态为均匀入流状态。