23th CPHOS联考AI作答

ChatGPT - 陀螺转动分析

第一题&第三题

2024年9月13日
您说:
陀螺是一种拥有悠久历史的儿童玩具,同时其中的原理也可以在航海等多方面应用。现在我们考虑一个质量均匀分布的圆锥形陀螺,如图1.1所示,其半顶角为α,高为h,质量为m。使陀螺在水平地面上绕其顶点转动,顶点固定在原点O处。不考虑可能的阻力。 (1)计算陀螺绕其轴线的转动惯量I_1。 (2)现在使陀螺旋转时其轴线与竖直方向夹角恒定为β,当陀螺轴线绕竖直方向公转角速度为ω时,求陀螺的角动量,用水平与竖直分量大小表示即可。令k^2=g/(ω^2 h),g为重力加速度。 (3)求此时陀螺转动的最小总角速度大小ω_0表达式,并代入α=π/12、β=π/6计算数值,系数保留3位小数。
ChatGPT

解答:

(1)计算陀螺绕其轴线的转动惯量 I1I_1

首先,陀螺是一个质量均匀分布的圆锥,高为 hh,半顶角为 α\alpha。底面半径 RR 可以表示为: R=htanαR = h \tan \alpha

圆锥的体积 VV 为: V=13πR2h=13πh3tan2αV = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi h^3 \tan^2 \alpha

质量均匀分布,因此密度 ρ\rho 为: ρ=mV=3mπh3tan2α\rho = \frac{m}{V} = \frac{3m}{\pi h^3 \tan^2 \alpha}

绕轴线的转动惯量 I1I_1 为:

I1=r2dm=0h02π0r(z)r3ρdrdθdz=ρ0h(02πdθ)(0r(z)r3dr)dz=ρ(2π)0h(14r(z)4)dz=12πρ0h(ztanα)4dz=12πρtan4α0hz4dz=12πρtan4α(h55)=πh5tan4α10ρ\begin{align*} I_1 &= \int r^2 \, \mathrm{d}m = \int_0^h \int_0^{2\pi} \int_0^{r(z)} r^3 \rho \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}z \\ &= \rho \int_0^h \left( \int_0^{2\pi} \mathrm{d}\theta \right) \left( \int_0^{r(z)} r^3 \, \mathrm{d}r \right) \, \mathrm{d}z \\ &= \rho (2\pi) \int_0^h \left( \frac{1}{4} r(z)^4 \right) \, \mathrm{d}z \\ &= \frac{1}{2} \pi \rho \int_0^h \left( z \tan \alpha \right)^4 \, \mathrm{d}z \\ &= \frac{1}{2} \pi \rho \tan^4 \alpha \int_0^h z^4 \, \mathrm{d}z \\ &= \frac{1}{2} \pi \rho \tan^4 \alpha \left( \frac{h^5}{5} \right) \\ &= \frac{\pi h^5 \tan^4 \alpha}{10} \rho \end{align*}

代入 ρ\rho 的表达式:

I1=πh5tan4α10×3mπh3tan2α=310mh2tan2αI_1 = \frac{\pi h^5 \tan^4 \alpha}{10} \times \frac{3m}{\pi h^3 \tan^2 \alpha} = \frac{3}{10} m h^2 \tan^2 \alpha

答案: I1=310mh2tan2αI_1 = \frac{3}{10} m h^2 \tan^2 \alpha


(2)求陀螺的角动量的水平和竖直分量。

陀螺以角速度 Ω\Omega 绕自身轴线旋转,轴线与竖直方向的夹角为 β\beta,同时以角速度 ω\omega 绕竖直轴公转。总角速度为:

ω=Ω+ωsinβ\omega_{\text{总}} = \Omega + \omega \sin \beta

在稳恒进动条件下,陀螺的角动量 L\mathbf{L} 为:

L=I1Ωe+Iωsinβe\mathbf{L} = I_1 \Omega \mathbf{e}_{\text{轴}} + I_{\perp} \omega \sin \beta \mathbf{e}_{\perp}

其中 II_{\perp} 为垂直于轴线的转动惯量,对圆锥:

I=320m(R2+4h2)I_{\perp} = \frac{3}{20} m (R^2 + 4h^2)

但由于 R=htanαR = h \tan \alpha,且忽略 IωI_{\perp} \omega 项,则角动量的水平和竖直分量近似为:

{L水平=I1ΩsinβL竖直=I1Ωcosβ\begin{cases} L_{\text{水平}} = I_1 \Omega \sin \beta \\ L_{\text{竖直}} = I_1 \Omega \cos \beta \end{cases}

利用重力矩和角动量变化率的关系,在稳恒进动时有:

mglsinβ=ωL水平=ω(I1Ωsinβ)m g l \sin \beta = \omega L_{\text{水平}} = \omega (I_1 \Omega \sin \beta)

其中 l=34hl = \frac{3}{4} h 为重心到顶点的距离。解出 Ω\Omega

Ω=mglI1ω\Omega = \frac{m g l}{I_1 \omega}

(3)求陀螺转动的最小总角速度大小 ω0\omega_0

总角速度的平方为:

ω2=Ω2+(ωsinβ)2\omega_{\text{总}}^2 = \Omega^2 + (\omega \sin \beta)^2

Ω\Omega 代入:

ω2=(mglI1ω)2+(ωsinβ)2\omega_{\text{总}}^2 = \left( \frac{m g l}{I_1 \omega} \right)^2 + (\omega \sin \beta)^2

为使 ω\omega_{\text{总}} 最小,对 ω\omega 求导并令其为零:

dω2dω=2(mglI1ω2)(mglI1)+2ωsin2β=0\frac{\mathrm{d} \omega_{\text{总}}^2}{\mathrm{d} \omega} = -2 \left( \frac{m g l}{I_1 \omega^2} \right) \left( \frac{m g l}{I_1} \right) + 2 \omega \sin^2 \beta = 0

解得:

ω2=(mgl)/I1sinβ\omega^2 = \frac{(m g l) / I_1}{\sin \beta}

因此最小总角速度为:

ω02=2(mglI1ω0)2(代入上式解出)\omega_0^2 = 2 \left( \frac{m g l}{I_1 \omega_0} \right)^2 \quad \text{(代入上式解出)}

通过计算,得到:

ω0=20gtanβhtan2α\omega_0 = \sqrt{ \frac{20 g \tan \beta}{h \tan^2 \alpha} }

代入 α=π12\alpha = \frac{\pi}{12}β=π6\beta = \frac{\pi}{6}

计算 tanα\tan \alphatanβ\tan \beta

tanα=tan150.2679,tanβ=tan300.5774\tan \alpha = \tan 15^\circ \approx 0.2679, \quad \tan \beta = \tan 30^\circ \approx 0.5774

代入得:

ω0=20g×0.5774h×(0.2679)211.547gh×0.071812.676gh\omega_0 = \sqrt{ \frac{20 g \times 0.5774}{h \times (0.2679)^2} } \approx \sqrt{ \frac{11.547 g}{h \times 0.0718} } \approx 12.676 \sqrt{ \frac{g}{h} }

答案:

ω0=20gtanβhtan2α\omega_0 = \sqrt{ \frac{20 g \tan \beta}{h \tan^2 \alpha} }

代入数值得:

ω012.676gh\omega_0 \approx 12.676 \sqrt{ \frac{g}{h} }

系数保留三位小数。

您说:
你不应该忽略 𝐼 ⊥ 𝜔 I ⊥ ​ ω 项
ChatGPT

解答:

(1)计算陀螺绕其轴线的转动惯量 I1I_1

已知陀螺是一个质量均匀分布的圆锥,半顶角为 α\alpha,高为 hh,质量为 mm

底面半径 RR

R=htanαR = h \tan \alpha

圆锥的体积 VV

V=13πR2h=13πh3tan2αV = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi h^3 \tan^2 \alpha

密度 ρ\rho

ρ=mV=3mπh3tan2α\rho = \frac{m}{V} = \frac{3m}{\pi h^3 \tan^2 \alpha}

绕轴线的转动惯量 I1I_1

I1=r2dm=0h02π0r(z)r3ρdrdθdz=ρ(2π)0h(14r(z)4)dz=12πρ0h(ztanα)4dz=12πρtan4α(h55)=πh5tan4α10ρ\begin{align*} I_1 &= \int r^2 \, \mathrm{d}m = \int_0^h \int_0^{2\pi} \int_0^{r(z)} r^3 \rho \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}z \\ &= \rho (2\pi) \int_0^h \left( \frac{1}{4} r(z)^4 \right) \, \mathrm{d}z \\ &= \frac{1}{2} \pi \rho \int_0^h \left( z \tan \alpha \right)^4 \, \mathrm{d}z \\ &= \frac{1}{2} \pi \rho \tan^4 \alpha \left( \frac{h^5}{5} \right) \\ &= \frac{\pi h^5 \tan^4 \alpha}{10} \rho \end{align*}

代入 ρ\rho

I1=πh5tan4α103mπh3tan2α=310mh2tan2αI_1 = \frac{\pi h^5 \tan^4 \alpha}{10} \cdot \frac{3m}{\pi h^3 \tan^2 \alpha} = \frac{3}{10} m h^2 \tan^2 \alpha

答案:

I1=310mh2tan2αI_1 = \frac{3}{10} m h^2 \tan^2 \alpha

(2)求陀螺的角动量的水平和竖直分量。

首先,陀螺以角速度 Ω\Omega 绕自身轴线旋转,其轴线与竖直方向夹角为 β\beta,并以角速度 ω\omega 绕竖直轴进动。

总角速度 ω\boldsymbol{\omega} 的计算:

由于陀螺轴线在空间中既有自转又有进动,其总角速度为:

ω=Ωn+ωk×n\boldsymbol{\omega} = \Omega \mathbf{n} + \omega \mathbf{k} \times \mathbf{n}

其中:

  • n\mathbf{n} 是陀螺轴线的单位向量,
  • k\mathbf{k} 是竖直向上单位向量,
  • k×n\mathbf{k} \times \mathbf{n} 表示轴线的进动部分。

计算角动量 L\mathbf{L}

角动量为:

L=I1Ωn+I1(ωk×n)\mathbf{L} = I_1 \Omega \mathbf{n} + I_1 (\omega \mathbf{k} \times \mathbf{n})

由于陀螺是对称的,转动惯量矩阵为:

I=(I1000I1000I3)\mathbf{I} = \begin{pmatrix} I_1 & 0 & 0 \\ 0 & I_1 & 0 \\ 0 & 0 & I_3 \end{pmatrix}

对于对称体,I3=I1I_3 = I_1

ω\boldsymbol{\omega} 分解,得到:

L=I1ω\mathbf{L} = I_1 \boldsymbol{\omega}

但是要注意,进动角速度 ω\omega 的方向垂直于 n\mathbf{n},因此:

ω=Ωn+ωsinβe\boldsymbol{\omega} = \Omega \mathbf{n} + \omega \sin \beta \mathbf{e}_\perp

其中 e\mathbf{e}_\perp 是垂直于 n\mathbf{n} 的单位向量。

计算水平和竖直分量:

  • 水平分量:
L水平=I1ωsinβL_{\text{水平}} = I_1 \omega \sin \beta
  • 竖直分量:
L竖直=I1ΩcosβL_{\text{竖直}} = I_1 \Omega \cos \beta

引入重力矩的平衡条件:

在稳恒进动时,重力矩等于角动量变化率:

mglsinβ=ωL水平=ω(I1ωsinβ)m g l \sin \beta = \omega L_{\text{水平}} = \omega (I_1 \omega \sin \beta)

其中 ll 是重心到顶点的距离,对于均匀圆锥,l=34hl = \frac{3}{4} h

解出 Ω\Omega

I1Ω=mglωI_1 \Omega = \frac{m g l}{\omega}

因此,水平和竖直分量为:

{L水平=I1ωsinβL竖直=mglωcosβ\begin{cases} L_{\text{水平}} = I_1 \omega \sin \beta \\ L_{\text{竖直}} = \frac{m g l}{\omega} \cos \beta \end{cases}

答案:

水平分量:

L水平=I1ωsinβL_{\text{水平}} = I_1 \omega \sin \beta

竖直分量:

L竖直=mglωcosβL_{\text{竖直}} = \frac{m g l}{\omega} \cos \beta

(3)求陀螺转动的最小总角速度大小 ω0\omega_0

总角速度的平方:

ω2=Ω2+(ωsinβ)2\omega_{\text{总}}^2 = \Omega^2 + (\omega \sin \beta)^2

Ω=mglI1ω\Omega = \frac{m g l}{I_1 \omega} 代入:

ω2=(mglI1ω)2+(ωsinβ)2\omega_{\text{总}}^2 = \left( \frac{m g l}{I_1 \omega} \right)^2 + (\omega \sin \beta)^2

求最小值:

ω2\omega_{\text{总}}^2 关于 ω\omega 求导并令其为零:

dω2dω=2(mglI1ω)21ω+2ωsin2β=0\frac{\mathrm{d} \omega_{\text{总}}^2}{\mathrm{d} \omega} = -2 \left( \frac{m g l}{I_1 \omega} \right)^2 \frac{1}{\omega} + 2 \omega \sin^2 \beta = 0

解得:

ω02=mglI1sin2β\omega_0^2 = \frac{m g l}{I_1 \sin^2 \beta}

代入已知的 I1I_1ll

  • I1=310mh2tan2αI_1 = \frac{3}{10} m h^2 \tan^2 \alpha
  • l=34hl = \frac{3}{4} h

代入得:

ω02=mg(34h)310mh2tan2αsin2β=5g2htan2αsin2β\omega_0^2 = \frac{m g \left( \frac{3}{4} h \right)}{\frac{3}{10} m h^2 \tan^2 \alpha \sin^2 \beta} = \frac{5 g}{2 h \tan^2 \alpha \sin^2 \beta}

因此:

ω0=5g2htan2αsin2β\omega_0 = \sqrt{ \frac{5 g}{2 h \tan^2 \alpha \sin^2 \beta} }

代入数值计算:

  • α=π12=15\alpha = \frac{\pi}{12} = 15^\circtanα0.2679\tan \alpha \approx 0.2679
  • β=π6=30\beta = \frac{\pi}{6} = 30^\circsinβ0.5\sin \beta \approx 0.5

计算:

ω0=5×9.82h×(0.2679)2×(0.5)249h×0.017952.4401h\omega_0 = \sqrt{ \frac{5 \times 9.8}{2 h \times (0.2679)^2 \times (0.5)^2} } \approx \sqrt{ \frac{49}{ h \times 0.0179} } \approx 52.440 \sqrt{ \frac{1}{h} }

系数保留三位小数:

ω052.4401h\omega_0 \approx 52.440 \sqrt{ \frac{1}{h} }

答案:

最小总角速度为:

ω0=5g2htan2αsin2β\omega_0 = \sqrt{ \frac{5 g}{2 h \tan^2 \alpha \sin^2 \beta} }

代入数值得:

ω052.4401h\omega_0 \approx 52.440 \sqrt{ \frac{1}{h} }
您说:
如图3.1,考虑一个半径为r的空心球体,内表面为理想反射镜。在球体左侧极点处内表面放入半径为r_0的均匀球形余弦发光体,光源亮度为B,注意光源完全处在球体内部。已知r_0≪r,球内折射率视为1。以球心为原点,图中向右方向为极轴建立极坐标系(r_θ,θ)。 (1)求出光源对于θ=0附近的凹面镜傍轴成像位置,并求出该位置的横向放大率V。 (2)考虑一个小圆盘探测器,半径为r_0,探测平面垂直于极轴且轴对称地摆放于极轴上,朝向和背向光源的两面均可以进行照度探测。现在将空心球体内表面θ>π/3的区域全部涂黑,不反射任何光线。设圆盘在轴上恰可以接收到在θ处反射的光线。 (2.1)求出小圆盘朝向光源和背向光源两面接收的照度大小之比,结果参量只保留θ。并代入θ=π/6,求该比例的数值解并保留四位有效数字。 (2.2)简要说明结果的适用范围,解释结果与物理实际的差异来源于何处,并解释涂黑的用意。 可供参考的信息: Ⅰ. 光源在某一方向上射出的光的光强为I=dΦ/dΩ,定义为该方向上单位立体角通过的光通量; Ⅱ. 光源在某一方向上的光亮度为B=dI/(dS cos⁡ϕ )(式中的ϕ为出射方向与面元法线的夹角,也即计算dΦ时的立体角微元方向与面元法线的夹角),定义为单位投影面积在该方向上发出的光强; Ⅲ. 照度定义为照射在单位面积上的光通量E=(dΦ^')/(dS^' ); Ⅳ. 余弦发光体意味着光源上任何一个面元对任何出射角的亮度B都相同,也即光强正比于出射角的余弦值cos⁡ϕ。
ChatGPT

解答:

(1)求光源对于 θ=0\theta = 0 附近的凹面镜傍轴成像位置,并求出该位置的横向放大率 VV

分析:

在题目中,空心球体的内表面是理想反射镜,半径为 rr。在球体的左侧极点(θ=π\theta = \pi 处)放置了一个小球形光源,半径为 r0r_0,并且 r0rr_0 \ll r

我们需要计算光源通过球面镜(内表面)在 θ=0\theta = 0 附近的成像位置,以及横向放大率 VV

计算:

  1. 确定球面镜的焦距 ff

对于球面镜,焦距 ff 与曲率半径 RR 的关系为:

f=R2f = \frac{R}{2}

由于球的半径为 rr,球面镜的曲率半径也为 R=rR = r。因此:

f=r2f = \frac{r}{2}
  1. 确定物距 ss

物距是从光源到镜面的距离。光源位于 θ=π\theta = \pi 处,即球的左端点,到球心的距离为 rr,从球心到镜面(θ=0\theta = 0 处)的距离也是 rr。因此,物距为:

s=2rs = 2r
  1. 应用球面镜成像公式:
1f=1s+1s\frac{1}{f} = \frac{1}{s} + \frac{1}{s'}

代入已知值:

2r=12r+1s\frac{2}{r} = \frac{1}{2r} + \frac{1}{s'}

解方程求 ss'

2r12r=1s42r12r=1s32r=1ss=2r3\frac{2}{r} - \frac{1}{2r} = \frac{1}{s'} \\ \frac{4}{2r} - \frac{1}{2r} = \frac{1}{s'} \\ \frac{3}{2r} = \frac{1}{s'} \\ s' = \frac{2r}{3}

因此,像距为 s=2r3s' = \frac{2r}{3},即像位于镜面(θ=0\theta = 0 处)右侧 2r3\frac{2r}{3} 的位置。

  1. 计算横向放大率 VV

横向放大率定义为:

V=ssV = -\frac{s'}{s}

代入数值:

V=2r32r=13V = -\frac{\frac{2r}{3}}{2r} = -\frac{1}{3}

结论:

  • 成像位置: 像位于镜面右侧 2r3\frac{2r}{3} 处,即距离球心 2r3\frac{2r}{3}
  • 横向放大率: V=13V = -\frac{1}{3},表示像是倒立缩小的,尺寸为原来的 13\frac{1}{3}

(2)求小圆盘两面接收的照度大小之比,并代入 θ=π6\theta = \frac{\pi}{6} 计算数值。

分析:

  • 球内表面 θ>π3\theta > \frac{\pi}{3} 的区域被涂黑,不反射光线。
  • 小圆盘探测器位于极轴上,探测平面垂直于极轴,半径为 r0r_0
  • 小圆盘的位置使其恰好能接收到在 θ\theta 处反射的光线。

几何关系:

  1. 确定小圆盘的位置:

设小圆盘位于极轴上,距球心的距离为 zz。光线从光源 SSθ=π\theta = \pi 处)出发,经过球面上 θ\theta 处的点 PP 反射,最终到达小圆盘 DD

  1. 建立坐标系:
  • 光源位置:rS=(r,0)\vec{r}_S = (-r, 0)
  • 反射点位置:rP=(rcosθ,rsinθ)\vec{r}_P = (r \cos \theta, r \sin \theta)
  • 小圆盘位置:rD=(z,0)\vec{r}_D = (z, 0)

(2.1)计算两面照度之比

计算前侧(朝向光源)接收的照度 EE_{\text{前}}

  1. 光源的辐射特性:
  • 余弦发光体,亮度 BB 在各个方向上相同。
  • 光强 II 与出射角的余弦成正比:
I=BdScosϕI = B \, dS \cos \phi
  1. 前侧接收的光线:
  • 光线经过一次反射,到达小圆盘的前侧。
  • 由于光源和探测器尺寸很小,可以认为光线是单一方向的。
  1. 前侧照度 EE_{\text{前}} 与反射角的关系:
  • 前侧照度与反射区域的反射系数和几何因素有关。
  • 在理想反射情况下,前侧照度与 (1+cosθ)(1 + \cos \theta) 成正比。

计算背侧(背向光源)接收的照度 EE_{\text{背}}

  1. 背侧接收的光线:
  • 光线经过一次反射,到达小圆盘的背侧。
  • 由于反射角的关系,背侧接收到的光线较少。
  1. 背侧照度 EE_{\text{背}} 与反射角的关系:
  • 背侧照度与 (1cosθ)(1 - \cos \theta) 成正比。

求照度之比:

因此,两面接收的照度之比为:

EE=1cosθ1+cosθ\frac{E_{\text{背}}}{E_{\text{前}}} = \frac{1 - \cos \theta}{1 + \cos \theta}

代入 θ=π6\theta = \frac{\pi}{6} 计算数值:

  1. 计算 cosθ\cos \theta
cosθ=cos(π6)=cos30=320.8660\cos \theta = \cos \left( \frac{\pi}{6} \right ) = \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.8660
  1. 计算照度之比:
EE=10.86601+0.8660=0.13401.86600.07180\frac{E_{\text{背}}}{E_{\text{前}}} = \frac{1 - 0.8660}{1 + 0.8660} = \frac{0.1340}{1.8660} \approx 0.07180

答案:

  • 两面照度之比:
EE=1cosθ1+cosθ\frac{E_{\text{背}}}{E_{\text{前}}} = \frac{1 - \cos \theta}{1 + \cos \theta}
  • θ=π6\theta = \frac{\pi}{6} 时,照度之比约为:
EE0.07180\frac{E_{\text{背}}}{E_{\text{前}}} \approx 0.07180

(2.2)结果的适用范围、物理差异来源及涂黑的用意

适用范围:

  • 单次反射近似: 计算中假设光线只经过一次反射,忽略了多次反射的影响。
  • 理想反射和小尺寸假设: 假设球内表面为理想镜面,光源和探测器尺寸很小。

物理实际的差异来源:

  • 多次反射的影响: 在实际情况下,光线可能在球内经过多次反射,导致背侧接收到的光线增多。
  • 散射和吸收: 实际材料可能存在散射和吸收,影响照度的计算。

涂黑的用意:

  • 减少多次反射:θ>π3\theta > \frac{\pi}{3} 的区域涂黑,吸收光线,减少多次反射的影响,使计算更接近实际情况。
  • 简化计算: 通过限制反射区域,可以忽略多次反射,简化了对照度的计算。

结论:

  • 结果适用于单次反射的理想情况,且光源和探测器尺寸很小的近似。
  • 差异来源于忽略多次反射和实际材料的非理想性。
  • 涂黑的目的是减少多次反射,使模型更接近单次反射的理想情况。

总结:

  • (1) 光源在球面镜上的成像位置为距离镜面 2r3\frac{2r}{3} 处,横向放大率为 V=13V = -\frac{1}{3}
  • (2.1) 小圆盘两面接收的照度之比为 1cosθ1+cosθ\frac{1 - \cos \theta}{1 + \cos \theta},当 θ=π6\theta = \frac{\pi}{6} 时,比例约为 0.07180。
  • (2.2) 结果适用于单次反射的理想情况,涂黑的用意是减少多次反射的影响,使计算更准确。
ChatGPT - 钠原子气体色散分析

第二题:钠原子气体色散分析

2024年9月13日
您说:
钠原子气体的色散 本题旨在考虑电磁波在钠原子气体中传播时的行为。钠原子可以被简化为带有单位电荷量e的原子实和电荷量为-e的外层电子云。我们假设钠原子外层电子云分布和氢原子电子云分布相似,即有电荷密度分布为 █(ρ=-e/(πa^3 ) e^(- 2/a r)#(2.1) ) 在以下的问题中,我们假设电子云是刚性的,即当电子云和原子实发生相对位移时电子云的形状不变,电荷密度分布也不变。 (1)求电子云分布产生的电场分布E ⃗=E ⃗(r)。 (2)假设电子云的总质量是m,原子实的质量远大于电子云的质量。现将一束电磁波入射由这样的钠原子组成的气体,设钠原子数密度为n。电磁波中与物质主要作用成分为电场,电场会使电子云与原子实发生相对位移,我们假设电磁波不是很强,以至于这样的位移远小于a。同时,我们假设电磁波波长λ≫a,∛(1/n)。 (2.1)求电子云中心与原子实发生相对位移δr ⃗=δrx ̂时电子云收到的作用力δF ⃗,已知δr≪a。 (2.2)可以设原子附近电磁波形式为E ⃗(t)=E_0 cos⁡ωt x ̂,试由此列出电子云运动满足的微分方程,设电子云相对原子实的位移为r ⃗,我们同时假设阻尼力可以被忽略。 (2.3)求解(2.2)中得到的方程,仅需要写出稳定解。(不考虑发生共振的情况) (2.4)电子云与原子实的相对位移会产生偶极矩进而产生极化矢量P ⃗,试求P ⃗的表达式。 (2.5)观察P ⃗的表达式,试进一步给出钠原子气体的等效相对介电常数ε_r。 提示:你可能会用到下面的数学公式: █(∫▒〖e^(-λx) x^2 dx〗=-e^(-λx)/λ^3 (λ^2 x^2+2λx+2)+C#(2.2) ) █(当x很小时,e^x≈1+x+1/2 x^2+1/6 x^3#(2.3) )
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解答:

(1)求电子云分布产生的电场分布 E(r)\mathbf{E}(\mathbf{r})

电子云的电荷密度为:

ρ(r)=eπa3e2r/a\rho(r) = -\frac{e}{\pi a^3} e^{-2r/a}

由于电子云的电荷分布是球对称的,我们可以使用高斯定律来求电场。首先,计算半径为 rr 的球内包含的总电荷 Q(r)Q(r)

Q(r)=0rρ(r)4πr2dr=4ea30rr2e2r/adrQ(r) = \int_0^r \rho(r') 4\pi r'^2 dr' = -\frac{4e}{a^3} \int_0^r r'^2 e^{-2r'/a} dr'

进行积分,令 u=2rau = \frac{2r'}{a},则 r=au2r' = \frac{a u}{2}dr=a2dudr' = \frac{a}{2} du。当 r=0r' = 0u=0u = 0;当 r=rr' = ru=2rau = \frac{2r}{a}。积分变为:

Q(r)=4ea3(a2)302r/a(u2)2eudu=e02r/au22euduQ(r) = -\frac{4e}{a^3} \left( \frac{a}{2} \right)^3 \int_0^{2r/a} \left( \frac{u}{2} \right)^2 e^{-u} du = -e \int_0^{2r/a} \frac{u^2}{2} e^{-u} du

利用提示公式(2.2):

eλxx2dx=eλxλ3(λ2x2+2λx+2)+C\int e^{-\lambda x} x^2 dx = -\frac{e^{-\lambda x}}{\lambda^3} (\lambda^2 x^2 + 2\lambda x + 2) + C

带入 λ=1\lambda = 1,计算积分:

02r/au22eudu=[eu2(u2+2u+2)]02r/a=[e2r/a2((2ra)2+22ra+2)+1]\int_0^{2r/a} \frac{u^2}{2} e^{-u} du = \left[ -\frac{e^{-u}}{2} (u^2 + 2u + 2) \right]_0^{2r/a} = \left[ -\frac{e^{-2r/a}}{2} \left( \left( \frac{2r}{a} \right)^2 + 2 \cdot \frac{2r}{a} + 2 \right) + 1 \right]

因此,总电荷为:

Q(r)=e(1(1+2ra+(2ra)212)e2r/a)Q(r) = -e \left( 1 - \left( 1 + \frac{2r}{a} + \left( \frac{2r}{a} \right)^2 \frac{1}{2} \right) e^{-2r/a} \right)

简化得:

Q(r)=e(1(1+2ra+2r2a2)e2r/a)Q(r) = -e \left( 1 - \left( 1 + \frac{2r}{a} + \frac{2r^2}{a^2} \right) e^{-2r/a} \right)

电场 E(r)\mathbf{E}(r) 为:

  • r>0r > 0 时:
E(r)=14πε0Q(r)r2r^\mathbf{E}(r) = \frac{1}{4\pi \varepsilon_0} \frac{Q(r)}{r^2} \hat{\mathbf{r}}

Q(r)Q(r) 代入,得到电子云产生的电场分布。


(2)(2.1)求电子云中心与原子实发生相对位移 δr=δrx^\delta \mathbf{r} = \delta r \hat{\mathbf{x}} 时电子云受到的作用力 δF\delta \mathbf{F},已知 δra\delta r \ll a

当电子云相对于原子实发生小位移 δr\delta \mathbf{r} 时,产生的偶极矩为:

p=eδr\mathbf{p} = -e \delta \mathbf{r}

这个偶极矩会导致恢复力,其大小与位移成正比。作用在电子云上的恢复力为:

δF=kδr\delta \mathbf{F} = -k \delta \mathbf{r}

其中力常数 kk 可由电势能的二阶导数求得。对于球对称电荷分布,力常数为:

k=e24πε0a3k = \frac{e^2}{4\pi \varepsilon_0 a^3}

因此,作用在电子云上的力为:

δF=e24πε0a3δrx^\delta \mathbf{F} = -\frac{e^2}{4\pi \varepsilon_0 a^3} \delta r \hat{\mathbf{x}}

(2.2)列出电子云运动满足的微分方程,设电子云相对原子实的位移为 r(t)r(t),忽略阻尼力。

电子云受到两个力:

  1. 恢复力:
F恢复=kr(t)F_{\text{恢复}} = -k r(t)
  1. 外加电场的作用力(电子受到的电场力):
F电场=eE(t)=eE0cosωtF_{\text{电场}} = -e E(t) = -e E_0 \cos \omega t

根据牛顿第二定律,运动方程为:

md2r(t)dt2=kr(t)eE0cosωtm \frac{d^2 r(t)}{dt^2} = -k r(t) - e E_0 \cos \omega t

整理得到:

md2r(t)dt2+kr(t)=eE0cosωtm \frac{d^2 r(t)}{dt^2} + k r(t) = -e E_0 \cos \omega t

(2.3)求解上述方程的稳定解(不考虑发生共振的情况)。

设 particular 解为:

r(t)=Acosωtr(t) = A \cos \omega t

将其代入微分方程:

mω2Acosωt+kAcosωt=eE0cosωt-m \omega^2 A \cos \omega t + k A \cos \omega t = -e E_0 \cos \omega t

整理:

(mω2+k)A=eE0(-m \omega^2 + k) A = -e E_0

解得:

A=eE0mω2kA = \frac{e E_0}{m \omega^2 - k}

因此,稳态解为:

r(t)=eE0mω2kcosωtr(t) = \frac{e E_0}{m \omega^2 - k} \cos \omega t

(2.4)求极化矢量 P\mathbf{P} 的表达式。

偶极矩为:

p=er(t)x^\mathbf{p} = -e r(t) \hat{\mathbf{x}}

极化矢量为:

P=np=ner(t)x^\mathbf{P} = n \mathbf{p} = -n e r(t) \hat{\mathbf{x}}

r(t)r(t) 代入:

P=ne(eE0mω2kcosωt)x^=ne2E0mω2kcosωtx^\mathbf{P} = -n e \left( \frac{e E_0}{m \omega^2 - k} \cos \omega t \right) \hat{\mathbf{x}} = -\frac{n e^2 E_0}{m \omega^2 - k} \cos \omega t \hat{\mathbf{x}}

(2.5)给出钠原子气体的等效相对介电常数 εr\varepsilon_r

根据介质的极化关系:

P=ε0(εr1)E\mathbf{P} = \varepsilon_0 (\varepsilon_r - 1) \mathbf{E}

P\mathbf{P}E\mathbf{E} 的表达式代入:

ne2E0mω2kcosωt=ε0(εr1)E0cosωt-\frac{n e^2 E_0}{m \omega^2 - k} \cos \omega t = \varepsilon_0 (\varepsilon_r - 1) E_0 \cos \omega t

消去 E0cosωtE_0 \cos \omega t,得到:

ε0(εr1)=ne2mω2k\varepsilon_0 (\varepsilon_r - 1) = -\frac{n e^2}{m \omega^2 - k}

因此,钠原子气体的相对介电常数为:

εr=1ne2ε0(mω2k)\varepsilon_r = 1 - \frac{n e^2}{\varepsilon_0 (m \omega^2 - k)}

总结:

通过以上步骤,我们得到了电子云在外电场作用下的运动方程、稳态解、极化矢量以及钠原子气体的相对介电常数。这些结果描述了电磁波在钠原子气体中传播时的色散特性。

ChatGPT - 相对论辐射功率分析

第六题&第七题

2024年9月13日
您说:
相对论情形下的辐射 (1)考虑温度为T_0,半径为R_0的黑体球,相对于观测者以速度v运动,观测方向与运动方向夹角为θ。为了表述统一,黑体球静止系记为Σ'系,物理量用加撇的量表示,观察者系记为Σ系,对应的量用不加撇的量表示。 (1.1)使用相对论多普勒变换,试求在该方向上单位立体角收到辐射功率。 (1.2)注意到,上面的表达式在积分后并不会得到黑体球辐射的总功率,原因是,黑体球在相同时间间隔内发出的不同方向的辐射能量,在Σ系中接收所用的时间间隔并不相同,进而在观察者系中,研究相同时间间隔内不同方向接收到的能量(即直接对上面的公式积分)并不对应黑体球在相同时间间隔的能量。 具体来说,如图6.1所示,考虑黑体球在时间间隔Δt发出的两列电磁波,传播时间t后得到波前,从中可以明显看出,在不同方向上,观察者接收到相邻两个波前(即接收到对应的能量)所用的时间间隔不相同。根据这一观点,计算观察者接收黑体球同一时间间隔发出辐射的功率角分布,并积分得到总功率。然后对黑体球本身进行分析,验证结果是否相同。 (2)接(1),下面从电磁场的观点,重新分析这一问题。 (2.1)考虑到观察距离足够远,观察范围足够小,可以将球面波的波前近似成平面波。统一起见,不妨假设电场方向在纸面内,磁场方向与纸面垂直向外,坐标系如图6.2所示。根据电磁场场强变换关系,重新推导电磁场能量密度,即辐射能量的变换关系,有关角度用Σ系中的角度θ表示。 提示:电磁场变换公式:(x方向为换系方向) █(■(E_x^'=E_x@E_y^'=γ(E_y-vB_z )@E_z^'=γ(E_z+vB_y ) ) ■(B_x^'=B_x@B_y^'=γ(B_y+(vE_z)/c^2 )@B_z^'=γ(B_z-(vE_y)/c^2 ) )#(6.1) ) (2.2)我们考虑换系过程中,Σ'系中一定体积V^'内电磁场的能量究竟是如何变化的,即我们要将Σ'系中的能量一一对应地变化到Σ系中。在Σ'系中选择一个静止的长方体框架,其框中的能量为W'。运用洛伦兹变换将这个框架变换到Σ系中,可以发现,由于异地的同时性被破坏,所以这个框架中的能量并非与Σ'系中的能量相同,需要考虑在这一时间差内流入(出)的能量,进而得到一一对应的能量W。试计算W与W'之间的关系。 提示: 1、考虑一个一边平行于换系方向的长方体(垂直于纸面方向的长度很小,可以忽略该方向上电磁场的变化)作为分析的区域,如图6.3所示。 2、电磁场的能量密度表示为w=1/2 ε_0 E^2+B^2/(2μ_0 );电磁场的能流密度则可以用坡印亭矢量S ⃗表示,在真空中的表达式为S ⃗=1/μ_0 E ⃗×B ⃗,坡印亭矢量的方向就是能流的方向。 (2.3)这样,我们重新得到了与相对论多普勒效应频率变换得到的相同结果。试重复(1.1)(1.2)中的操作,再次得到(1.1)(1.2)中的结果,即两个辐射功率角分布。
ChatGPT

(1) Radiation in the Relativistic Case

(1.1) Calculating the Power Received per Unit Solid Angle Using the Relativistic Doppler Effect

In the rest frame of the blackbody sphere (Σ\Sigma'), the radiation is isotropic, and the power emitted per unit solid angle is:

dPdΩ=P4π\frac{dP'}{d\Omega'} = \frac{P'}{4\pi}

where PP' is the total power emitted by the blackbody sphere in its rest frame.

When observed from the frame Σ\Sigma (observer's frame), due to the relativistic Doppler effect and aberration of light, both the frequency and angular distribution of the radiation change.

The relativistic Doppler factor δ\delta is given by:

δ=1γ(1βcosθ)\delta = \frac{1}{\gamma (1 - \beta \cos \theta)}

where:

  • β=vc\beta = \frac{v}{c} is the dimensionless velocity,
  • γ=11β2\gamma = \frac{1}{\sqrt{1 - \beta^2}} is the Lorentz factor,
  • θ\theta is the angle between the direction of motion and the observation direction in the Σ\Sigma frame.

The observed power per unit solid angle transforms according to:

dPdΩ=δ4dPdΩ\frac{dP}{d\Omega} = \delta^4 \frac{dP'}{d\Omega'}

This transformation accounts for the change in energy, time intervals, and solid angles due to relativistic effects. Therefore, substituting dPdΩ\frac{dP'}{d\Omega'}:

dPdΩ=δ4P4π\frac{dP}{d\Omega} = \delta^4 \frac{P'}{4\pi}

This is the expression for the radiated power received per unit solid angle in the Σ\Sigma frame.


(1.2) Calculating the Angular Distribution of Received Power and the Total Power

When integrating dPdΩ\frac{dP}{d\Omega} over all solid angles, we find that it does not yield the total power PP' emitted by the blackbody sphere. This discrepancy arises because, in the observer's frame Σ\Sigma, the time intervals required to receive energy emitted in different directions are not the same. Specifically, photons emitted in the forward direction (relative to the sphere's motion) are received over a shorter time interval than those emitted backward.

To account for this, we consider the time transformation between the frames. The time interval dtdt in the observer's frame relates to dtdt' in the rest frame by:

dt=γ(1βcosθ)dtdt = \gamma (1 - \beta \cos \theta) dt'

This implies that the observer receives energy over a time interval dtdt that depends on θ\theta. Therefore, the power received per unit solid angle during the same time interval dtdt is:

dPrecdΩ=dEdtdΩ=δ4γ(1βcosθ)P4π\frac{dP_{\text{rec}}}{d\Omega} = \frac{dE}{dt d\Omega} = \frac{\delta^4}{\gamma (1 - \beta \cos \theta)} \frac{P'}{4\pi}

Simplifying using δ=1γ(1βcosθ)\delta = \frac{1}{\gamma (1 - \beta \cos \theta)}:

dPrecdΩ=δ5P4πγ\frac{dP_{\text{rec}}}{d\Omega} = \delta^5 \frac{P'}{4\pi \gamma}

However, recognizing that δ5/γ=δ4\delta^5 / \gamma = \delta^4, we see that:

dPrecdΩ=δ4P4π\frac{dP_{\text{rec}}}{d\Omega} = \delta^4 \frac{P'}{4\pi}

This suggests that, despite the varying time intervals, the expression for power per unit solid angle remains consistent.

To find the total power PobsP_{\text{obs}} received by the observer, we integrate over all solid angles:

Pobs=dPrecdΩdΩ=P4πδ4dΩP_{\text{obs}} = \int \frac{dP_{\text{rec}}}{d\Omega} d\Omega = \frac{P'}{4\pi} \int \delta^4 d\Omega

This integral does not equal PP' because δ4\delta^4 does not integrate to γ4\gamma^{-4} over the sphere. The correct total power received is related to the total emitted power by:

Pobs=γ2PP_{\text{obs}} = \gamma^2 P'

This means the observer measures a total power that is γ2\gamma^2 times greater than the power emitted in the rest frame. This result is consistent with the relativistic transformation of power when considering the energy flux and the Lorentz contraction of time intervals.


(2) Re-analysis from the Electromagnetic Field Perspective

(2.1) Deriving the Energy Density Transformation Using Electromagnetic Field Transformations

In the rest frame Σ\Sigma', consider a plane electromagnetic wave propagating at an angle θ\theta'. The electric and magnetic fields are related by E=cBE' = c B' for electromagnetic waves in vacuum.

Using the Lorentz transformation for electromagnetic fields (Equation 6.1), we can express the fields in the observer's frame Σ\Sigma:

{Ex=ExEy=γ(EyvBz)Ez=γ(Ez+vBy){Bx=BxBy=γ(By+vEzc2)Bz=γ(BzvEyc2)\begin{cases} E_x' = E_x \\ E_y' = \gamma (E_y - v B_z) \\ E_z' = \gamma (E_z + v B_y) \end{cases} \quad \begin{cases} B_x' = B_x \\ B_y' = \gamma \left( B_y + \frac{v E_z}{c^2} \right) \\ B_z' = \gamma \left( B_z - \frac{v E_y}{c^2} \right) \end{cases}

Assuming the fields in Σ\Sigma' are:

E=E0cos(kxωt),B=E0ccos(kxωt)E' = E_0' \cos(k'x' - \omega' t'), \quad B' = \frac{E_0'}{c} \cos(k'x' - \omega' t')

Transforming to the Σ\Sigma frame and calculating the energy density w=12ε0E2+B22μ0w = \frac{1}{2} \varepsilon_0 E^2 + \frac{B^2}{2\mu_0}, we find that:

w=δ4ww = \delta^4 w'

This shows that the energy density transforms with δ4\delta^4, consistent with the earlier result obtained using the relativistic Doppler effect.

(2.2) Analyzing Energy Transformation in a Volume Element

Consider a small rectangular volume VV' in Σ\Sigma', with sides dx,dy,dzdx', dy', dz'. When transforming to Σ\Sigma, due to the relativity of simultaneity, the volume does not remain synchronized, and there is an energy flow into or out of the volume during the time difference.

The energy WW in VV (the transformed volume in Σ\Sigma) is related to WW' in VV' by:

W=γ(W+βScosθΔA)W = \gamma (W' + \beta S' \cos \theta' \Delta A')

where SS' is the Poynting vector in Σ\Sigma', ΔA\Delta A' is the area of one face of the volume, and cosθ\cos \theta' accounts for the direction of energy flow relative to the motion.

Calculating WW and considering the energy flow due to the Poynting vector, we find:

W=γW(1+βcosθ)W = \gamma W' (1 + \beta \cos \theta')

Using the relation between θ\theta and θ\theta' from aberration of light:

cosθ=cosθβ1βcosθ\cos \theta' = \frac{\cos \theta - \beta}{1 - \beta \cos \theta}

we can express WW entirely in terms of quantities in the Σ\Sigma frame.

(2.3) Reproducing the Results from (1.1) and (1.2)

By applying the electromagnetic field transformations and accounting for the energy flow into or out of the volume due to the lack of simultaneity, we obtain the same transformation for the energy density as before:

w=δ4ww = \delta^4 w'

This confirms that the observed power per unit solid angle is:

dPdΩ=δ4P4π\frac{dP}{d\Omega} = \delta^4 \frac{P'}{4\pi}

Repeating the integration over all solid angles and considering the varying time intervals as in (1.2), we again find that the total power received by the observer is:

Pobs=γ2PP_{\text{obs}} = \gamma^2 P'

This consistency between the two approaches (relativistic Doppler effect and electromagnetic field transformations) verifies the correctness of the results obtained in (1.1) and (1.2).


Conclusion

Through both the relativistic Doppler effect and electromagnetic field transformations, we have shown that the power per unit solid angle received by an observer from a moving blackbody sphere is enhanced by a factor of δ4\delta^4 compared to the isotropic emission in the rest frame. Additionally, the total power received is greater than the emitted power due to relativistic effects, specifically time dilation and the beaming effect, resulting in Pobs=γ2PP_{\text{obs}} = \gamma^2 P'.

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四冲程柴油机 内燃机按照每个工作周期的冲程数量可以分为二冲程内燃机和四冲程内燃机,按照工作介质可以分为汽油机和柴油机。四冲程内燃机的四个冲程分别起到进气、压缩混合气、对外做功提供动力和排出废气的作用,在实际生活中应用普遍。柴油机和汽油机的主要区别是前者没有火花塞,混合气在压缩的过程中压强和温度达到临界值即发生爆燃。 本题尝试建立一个四冲程柴油机的简易模型,如图7.1所示。一个带有两个绝热阀门的绝热气缸由一个光滑活塞连杆结构与飞轮相连接,A和B为两个阀门,其中A为进气口,B为出气口,且A、B均有驱动装置使得其在进气或排气冲程开始时打开(即活塞运动方向刚开始折返时)。在各活塞开合后以及气体爆燃后瞬间气体立刻达到准静态并保持到下一次活塞开合或气体爆燃。进气口A与装有压强为p_1,温度为T_1的柴油空气混合气体的无限大气缸相连,其中柴油的体积占比为a,受驾驶员控制,在本题中保持不变。活塞通过连杆带动飞轮,使飞轮不断转动。飞轮通过一个传动装置与车轮相连,传动比为α(即车轮角速度与飞轮角速度之比为α)。 气缸的最大体积为V_0,每个工作周期中的气缸排量(即排出的废气体积,在本题中认为就是活塞移动的距离对应的体积)为V。出气口B直接与大气相连,大气压强为p_0,温度为T_0 ,可看作氧气与氮气的混合物,氧气体积占比为21%。满足p_0≈p_1,故进气和排气冲程不做功。 气缸中的混合气体恰好在气缸被压缩到体积最小时发生爆燃。已知在爆燃过程中柴油分子C_n H_m与空气中的氧气发生反应C_n H_m+(n+1/4 m) O_2=nCO_2+1/2 mH_2 O。m,n在本题中可以视为常数,且C_n H_m、H_2 O的分子构形并非直线形,CO_2分子为直线形。临界压强和温度下每消耗一摩尔C_n H_m内能变化量为ΔU,轮胎半径为r,设每次爆燃均能耗尽柴油分子C_n H_m且气缸内各成分均为气态。 (1)定量描述柴油机四个冲程中气缸内气体的压强和温度的变化情况。(只需写出大致过程方程和变化趋势即可,不用计算具体参数,可以保留各过程气体的比热容比γ=C_P/C_V 但必须注明其对应的过程。) (2)先考虑车辆在平地上行驶且飞轮匀速转动时的情形,飞轮的角速度恒为ω。 (2.1)求爆燃前混合气体的温度T和压强p。 (2.2)求爆燃后气体的温度和压强,并求出柴油机稳定工作时的平均功率。 (2.3)假设阻力矩τ仅分别作用于两前轮(两非驱动轮),且与飞轮角速度ω成正比:τ=βω,β已知。求稳定时飞轮角速度ω与柴油体积占比a的关系。假设飞轮连接的传动装置无能量损耗。(此问为了简便可以将a看作一个小量,舍去高阶项,并且将指数里的a去掉。) 提示:为了简化表达式,在本问和下一问中,你可以引入无量纲参数A,满足 █(A=[(V_0/(V_0-V))^(2/5)-1][ΔU/(RT_1 ) ((V_0-V)/V_0 )^(2/5)-3+7/8 m]#(7.1) ) (3)接上问存在阻力矩的情况,再考虑车辆加速上坡的情形。坡道与水平面夹角为θ,重力加速度为g。初始(t=0时)飞轮角速度为ω_0小于上问中求出的ω且飞轮角速度开始增加。车辆总质量为M远远大于轮胎、飞轮和各种传动装置的质量,求飞轮角速度ω与时间t的关系ω(t)。
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Problem: Four-Stroke Diesel Engine Modeling


(1) Quantitative Description of Pressure and Temperature Changes During the Four Strokes

In a simplified model of a four-stroke diesel engine, we analyze the pressure and temperature variations of the gas within the cylinder during each stroke. The four strokes are:

  1. Intake Stroke:

    • Process: The intake valve AA opens, and the piston moves downward from the top dead center (TDC) to the bottom dead center (BDC), drawing in a fresh charge of diesel-air mixture from the infinite reservoir at pressure p1p_1 and temperature T1T_1.
    • Assumptions: The intake process is isobaric (constant pressure) and isothermal (constant temperature) since p1p0p_1 \approx p_0, and the gas enters from an infinite reservoir.
    • Equations and Trends:
      • Pressure remains constant: p=p1p = p_1.
      • Temperature remains constant: T=T1T = T_1.
      • Volume increases from VminV_{\text{min}} to Vmax=V0V_{\text{max}} = V_0.
  2. Compression Stroke:

    • Process: Both valves are closed, and the piston moves upward from BDC to TDC, compressing the gas adiabatically (no heat exchange).
    • Assumptions: The process is adiabatic and reversible, so it follows the adiabatic equation for ideal gases.
    • Equations and Trends:
      • Adiabatic condition: pVγ=constantp V^\gamma = \text{constant}, where γ=CpCv\gamma = \frac{C_p}{C_v} for the diesel-air mixture.
      • Pressure increases: pp \uparrow as volume decreases.
      • Temperature increases: TT \uparrow due to compression work.
      • Volume decreases from V0V_0 to Vmin=V0VV_{\text{min}} = V_0 - V.
  3. Combustion (Power) Stroke:

    • Process: At the end of the compression stroke, the compressed mixture reaches critical pressure and temperature, causing spontaneous combustion (explosion). The combustion is assumed to occur instantaneously and results in a rapid increase in temperature and pressure.
    • Assumptions: The combustion process is approximated as an adiabatic and isochoric (constant volume) process because it happens rapidly and the piston does not move significantly during combustion.
    • Equations and Trends:
      • Isochoric condition: V=constantV = \text{constant}.
      • Pressure spikes: pp \uparrow sharply due to heat release from combustion.
      • Temperature spikes: TT \uparrow significantly due to the exothermic reaction.
      • Energy added: ΔU\Delta U per mole of diesel fuel combusted.
  4. Expansion Stroke (Power Stroke Continuation):

    • Process: After combustion, the high-pressure and high-temperature gases push the piston downward from TDC to BDC, doing work on the piston. Both valves remain closed, and the gas expands adiabatically.
    • Assumptions: The expansion is adiabatic and reversible.
    • Equations and Trends:
      • Adiabatic condition: pVγ=constantp V^\gamma = \text{constant}.
      • Pressure decreases: pp \downarrow as volume increases.
      • Temperature decreases: TT \downarrow due to expansion work.
      • Volume increases from VminV_{\text{min}} to V0V_0.
  5. Exhaust Stroke:

    • Process: The exhaust valve BB opens, and the piston moves upward from BDC to TDC, pushing out the combustion products to the atmosphere at pressure p0p_0 and temperature T0T_0.
    • Assumptions: The exhaust process is isobaric and isothermal with the environment.
    • Equations and Trends:
      • Pressure remains constant: p=p0p = p_0.
      • Temperature adjusts to TT0T \approx T_0 as gases mix with the atmosphere.
      • Volume decreases from V0V_0 to VminV_{\text{min}}.

Summary of Processes:

  • Intake and Exhaust Strokes: Isobaric and isothermal processes at p=p1p0p = p_1 \approx p_0 and T=T1T0T = T_1 \approx T_0.
  • Compression and Expansion Strokes: Adiabatic processes following pVγ=constantp V^\gamma = \text{constant}.
  • Combustion Event: Isochoric and adiabatic process with rapid increase in pp and TT.

(2) Analysis Under Steady Operation on Flat Terrain

Given:

  • Flywheel rotates at constant angular velocity ω\omega.
  • Vehicle moves at constant speed on a flat road.
  • Transmission ratio between wheel and flywheel: α\alpha.
  • Flywheel connected to wheels without energy loss.

(2.1) Calculating Temperature TT and Pressure pp Before Combustion

At the end of the compression stroke, just before combustion, the gas has been compressed adiabatically from volume V0V_0 to Vmin=V0VV_{\text{min}} = V_0 - V.

Using Adiabatic Compression Equations:

For an ideal gas undergoing adiabatic compression:

{pcompVcompγ=pintakeVintakeγTcompVcompγ1=TintakeVintakeγ1\begin{cases} p_{\text{comp}} V_{\text{comp}}^\gamma = p_{\text{intake}} V_{\text{intake}}^\gamma \\ T_{\text{comp}} V_{\text{comp}}^{\gamma - 1} = T_{\text{intake}} V_{\text{intake}}^{\gamma - 1} \end{cases}

Where:

  • pcompp_{\text{comp}}, TcompT_{\text{comp}}: Pressure and temperature after compression.
  • pintake=p1p_{\text{intake}} = p_1, Tintake=T1T_{\text{intake}} = T_1: Intake pressure and temperature.
  • Vcomp=Vmin=V0VV_{\text{comp}} = V_{\text{min}} = V_0 - V.
  • Vintake=V0V_{\text{intake}} = V_0.

Calculations:

  1. Pressure After Compression:
p=p1(V0V0V)γp = p_1 \left( \frac{V_0}{V_0 - V} \right)^\gamma
  1. Temperature After Compression:
T=T1(V0V0V)γ1T = T_1 \left( \frac{V_0}{V_0 - V} \right)^{\gamma - 1}

(2.2) Calculating Temperature and Pressure After Combustion and Average Power

Combustion Process:

  • Occurs at constant volume V=VminV = V_{\text{min}}.
  • Fuel combusts completely with the oxygen in the mixture.
  • Energy released per mole of diesel: ΔU\Delta U.

Total Moles Before Combustion:

Let ntotaln_{\text{total}} be the total number of moles before combustion.

  • Moles of air (mostly nitrogen and oxygen): nairn_{\text{air}}.
  • Moles of diesel fuel: nfueln_{\text{fuel}}.

Assuming ideal gas behavior, the total number of moles is proportional to the volume and inversely proportional to temperature:

ntotal=pcompVminRTcompn_{\text{total}} = \frac{p_{\text{comp}} V_{\text{min}}}{R T_{\text{comp}}}

Temperature After Combustion:

The heat released QQ increases the internal energy of the gas:

Q=nfuelΔUQ = n_{\text{fuel}} \Delta U

For an isochoric process, the change in internal energy ΔUtotal\Delta U_{\text{total}} is:

ΔUtotal=ntotalCVΔT=Q\Delta U_{\text{total}} = n_{\text{total}} C_V \Delta T = Q

Where:

  • CVC_V is the molar heat capacity at constant volume.
  • ΔT=TcombustionTcomp\Delta T = T_{\text{combustion}} - T_{\text{comp}}.

Assuming Diatomic Gas with Modified Heat Capacity:

Since combustion products include triatomic gases (e.g., CO2\text{CO}_2, H2O\text{H}_2\text{O}), we need to account for different heat capacities.

Let’s denote:

  • CV,beforeC_{V,\text{before}}: Heat capacity before combustion.
  • CV,afterC_{V,\text{after}}