Hydraulic jump type energy dissipator
import numpy as np def main(): g=9.8 q=42.0 h0=554.1-321.0 v1=26.5 b=4.0 d1=np.round(q/b/v1,decimals=3) fr=np.round(v1/np.sqrt(g*d1),decimals=3) d1d2=np.round(1/2*(np.sqrt(1+8*fr**2)-1),decimals=3) d2=np.round(d1*d1d2,decimals=3) print('v1=',v1) print('d1=',d1) print('Fr=',fr) print('d1d2=',d1d2) print('d2=',d2) c=1.704 c1=(1+2*fr**2)*np.sqrt(1+8*fr**2)-1-5*fr**2 c2=1+4*fr**2-np.sqrt(1+8*fr**2) cc=np.round(c1/c2-(np.sqrt(g)/c*fr)**(2/3),decimals=3) ww=np.round(d1*cc,decimals=3) print('cc=',cc) print('W=',ww) print('L=',4.5*d2) #============== # Execution #============== if __name__ == '__main__': main()
Calculation results are shown below.
v1= 26.5 d1= 0.396 Fr= 13.452 d1d2= 18.531 d2= 7.338 cc= 10.31 W= 4.083 L= 33.021
Thank you.
Normal depth of steep channel
(known) discharge | |
(known) width of rectangular open channe | |
(known) Manning's roughness coefficient | |
(known) gradient of channel invert | |
(unknown) normal depth |
The critical depth of steep channel is calculated as following equation.
# normal depth and critical depth of rectangular cross section import numpy as np from scipy import optimize def cal_hc(q,b,cs): # critical depth g=9.8 hc=(q**2/g/b**2/cs)**(1/3) return hc def func(h,q,b,n,sn): f=q-b*h/n*(b*h/(b+2*h))**(2/3)*sn**(1/2) return f def main(): q=42.0 # discharge b=4.0 # channel width n=0.014 # Manning's roughness coefficient theta=np.radians(37.0) sn=np.sin(theta) cs=np.cos(theta) h1=0.0 h2=10.0 hh=optimize.brentq(func,h1,h2,args=(q,b,n,sn)) v=q/b/hh print('hn=',hh) # normal depth print('v=',v) hc=cal_hc(q,b,cs) print('hc=',hc) # critical depth #============== # Execution #============== if __name__ == '__main__': main()
Calculation results are shown below.
hn= 0.3962334595851377 v= 26.499528866122652 hc= 2.415097316241126
Thank you.
Soil pressure coefficient considering earthquake by Mononobe-Okabe
Active soil pressure coefficient considering earthquake by Mononobe-Okabe can be obtained by following equation. (USACE EM 1110-2-2100 Appendix G)
import numpy as np kh=0.077 # horizontal seismic coefficient #kh=0 # horizontal seismic coefficient kv=0 # vertical seismic coefficient phi=np.radians(30.0) # internal friction angle of soil alpha=0 # gradient of wall surface beta=0 # gradient of ground surface delta=0 # friction angle of wall surface theta=np.arctan(kh/(1-kv)) # seismic inertia angle c1=np.cos(phi-alpha-theta) c2=np.cos(theta) * (np.cos(alpha))**2 * np.cos(alpha+delta+theta) c3=1 + np.sqrt(np.sin(phi+delta) * np.sin(phi-beta-theta) / np.cos(alpha+beta+theta) / np.cos(alpha-beta)) Ka=c1**2/c2/c3**2 print(Ka)
Calculation results are shown below.
Ka=0.381 # earthquake (kh=0.077, kv=0) Ka=0.333 # normal (kh=0, kv=0)
That's all. Thak you.
Python: calculation of required plate thickness of exposed type penstock against buckling pressure by Brent method
A critical buckling pressure of exposed type penstocks can be obtained by following equation.
critical buckling pressure of penstock | |
plate thickness excluding corrosion allowance | |
design external diameter | |
elastic modulus of penstock (=206,000MPa) | |
Poisson's modulus of penstock (=0.3) |
usually, a corrosion allowance is set to , and the design plate thickness including corrosion allowance and the design external diameter can be calculated as follows.
Where, means the design internal diameter.
In case of exposed type penstocks, it is required to withstand the buckling pressure of 0.02 MPa which includes a safety factor of 1.5, and the plate thickness with this capacity shall be defined as a lower limit of plate thickness.
The equation to be solved by Brent method is shown below. The equation has a shape of .
Required two initial values for Brent method are given as t1=1.0, t2=50.0
in the program.
Program code is shown below.
import numpy as np from scipy import optimize def func(t,d0,eps,pk): Es=206000 # elastic modulus of steel po=0.3 # Poisson's ratio of steel f=pk-2*Es/(1-po**2)*(t/(d0+t+eps))**3 return f def main(): d0=4000.0 # internal diameter of penstock eps=1.5 # corrosion alloowance pk=0.02 # critical buckling pressure t1=1.0 # initial value for Brent method t2=50.0 # initial value for Brent method tt=optimize.brentq(func,t1,t2,args=(d0,eps,pk)) t0=np.ceil(tt+eps) # required plate thickness print('t + eps=',tt+eps) print('t0=',t0) #============== # Execution #============== if __name__ == '__main__': main()
The calculation results are shown below.
t + eps= 15.695548237490522 t0= 16.0
That's all. Thank you.
Non-uniform flow analysis by Python: calculation of water surface frofile of settling basin
A program to calculate water surface profile of settling basin which has an open channel with varied width and varied invert level. Since the condition is subcritical flow, the calculation will be carried out from downstream section to upstream section.
Basic equation of non-uniform flow analysis is shown below.
Discharge (constant value) | |
Properties of upstream section (distance, invert level, flow section area, hydraulic radius, Manning's roughness coefficient, water depth) | |
Properties of downstream section (distance, invert level, flow section area, hydraulic radius, Manning's roughness coefficient, water depth) |
Where, subscripu '1' is for downstream section and subscript '2' is for upstream section.
The equations for calculations of section area and hydraulic radius for rectangular cross section are shown below.
where, means channel width, means water depth.
Since this case is a condition under subcritical flow, the conditions of downstream with subscript of '1' are all known values, and calculation is carried out to obtain the water depth of using the know conditions.
The section properties are defined as a function of x
in def sec(x,h)
.
Since the calculation model has many varied section, the part of definition of section properties (def sec(x,h)
) is long.
However, the core part for bi-section method is simple.
Two required initial values for bi-section method are defined as ha=3.0
and hb=7.0
in the function of def bisection(h1,x1,x2,q)
. These initial values should be changed appropriately, depending on the problem to be solved.
A function def sec(x,h)
has a fuction to output section properties (z, A, R, n
) from input data of distance (x
) and water depth (h
). When this function def sec(x,h)
is changed, this program can be applied to a non-uniform flow analysis of open channel with various cross section.
A program is shown below.
# Non-Uniform Flow Analysis (Subcritical flow) import numpy as np def sec(x, h): # definition of section property # x : distance # h : water depth # zz : invert level # aa : secion area # rr : hydraulic radius # nn : Manning's roughness coefficient n0=0.014 # roughness coefficient (normal value) zz,aa,rr,nn=0,0,0,0 if 0.0 <= x < 11.0: ds=11 nn=n0 z1=562.2; b1=4.0 z2=560.5; b2=26.0 zz=z1-(z1-z2)/ds*x bb=b1+(b2-b1)/ds*x aa=bb*h rr=bb*h/(bb+2*h) if 11.0 <= x < 19.0: ds=8.0 nn=n0 z1=560.5; b1=12 z2=560.5; b2=12 zz=z1-(z1-z2)/ds*(x-11) bb=b1+(b2-b1)/ds*(x-11) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 if 19.0 <= x < 47.0: ds=29.0 nn=n0 z1=560.5 ; b1=12.0 z2=z1+ds*0.02; b2=12.0 zz=z1-(z1-z2)/ds*(x-19) bb=b1+(b2-b1)/ds*(x-19) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 if 47.0 <= x < 62.0: ds=15.0 nn=n0 z1=561.06 ; b1=12.0 z2=z1+ds*0.02; b2=6.0 zz=z1-(z1-z2)/ds*(x-47) bb=b1+(b2-b1)/ds*(x-47) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 if 62.0 <= x < 69.0: ds=7.0 nn=n0 z1=561.36 ; b1=6.0 z2=z1+ds*0.02; b2=6.0 zz=z1-(z1-z2)/ds*(x-62) bb=b1+(b2-b1)/ds*(x-62) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 if 69.0 <= x < 69.5: ds=0.5 nn=n0 z1=561.5; b1=6.0 z2=562.0; b2=6.0 zz=z1-(z1-z2)/ds*(x-69) bb=b1+(b2-b1)/ds*(x-69) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 if 69.5 <= x < 73.5: ds=4.0 nn=n0 z1=562.0; b1=6.0 z2=562.0; b2=6.0 zz=z1-(z1-z2)/ds*(x-69.5) bb=b1+(b2-b1)/ds*(x-69.5) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 if 73.5 <= x < 74.0: ds=0.5 nn=n0 z1=562.0; b1=6.0 z2=561.5; b2=6.5 zz=z1-(z1-z2)/ds*(x-73.5) bb=b1+(b2-b1)/ds*(x-73.5) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 if 74.0 <= x < 77.0: ds=3.0 nn=n0 z1=561.5; b1=6.5 z2=561.5; b2=6.5 zz=z1-(z1-z2)/ds*(x-74.0) bb=b1+(b2-b1)/ds*(x-74.0) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 if 77.0 <= x < 96.625: ds=19.625 nn=n0 z1=561.5; b1=6.5 z2=563.0; b2=6.5 zz=z1-(z1-z2)/ds*(x-77.0) bb=b1+(b2-b1)/ds*(x-77.0) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 if 96.625 <= x <= 102.125: ds=5.5 nn=n0 z1=563.0; b1=6.5 z2=563.0; b2=6.5 zz=z1-(z1-z2)/ds*(x-96.625) bb=b1+(b2-b1)/ds*(x-96.625) ah=bb*h rh=bb*h/(bb+2*h) aa=ah*2 rr=rh*2 return zz,aa,rr,nn def func(h2,h1,x1,x2,q): g=9.8 z1,a1,r1,n1=sec(x1,h1) z2,a2,r2,n2=sec(x2,h2) e2=q**2/2/g/a2**2+h2+z2 e1=q**2/2/g/a1**2+h1+z1 e3=0.5*(n1**2*q**2/r1**(4/3)/a1**2 + n2**2*q**2/r2**(4/3)/a2**2)*(x2-x1) return e2-e1-e3 def bisection(h1,x1,x2,q): ha=3.0 # lower initial value for bisection method hb=7.0 # upper initial value for bisection method for k in range(100): hi=0.5*(ha+hb) fa=func(ha,h1,x1,x2,q) fb=func(hb,h1,x1,x2,q) fi=func(hi,h1,x1,x2,q) if fa*fi<0: hb=hi if fb*fi<0: ha=hi #print(fa,fi,fb) if np.abs(hb-ha)<1e-10: break return hi def main(): g=9.8 # gravity acceleration q=42.0 # discharge # starting point (sub-critical flow) h1=3.9 # water depth at starting point x1=0.0 # origin of distance coordinate z1,a1,r1,n1=sec(x1,h1) # section property v1=q/a1 # flow velocity print('{0:>8s}{1:>8s}{2:>8s}{3:>8s}{4:>10s}{5:>8s}'.format('x','z','h','z+h','Energy','v')) print('{0:8.3f}{1:8.3f}{2:8.3f}{3:8.3f}{4:10.5f}{5:8.3f}'.format(x1,z1,h1,z1+h1,z1+h1+v1**2/2/g,v1)) # calculation point #xp=np.arange(1,103,1) xp=np.array([11,19,47,62,69,69.5,73.5,74,77,96.625,102.125]) # water level calculation to upstream direction for x2 in xp: h2=bisection(h1,x1,x2,q) z2,a2,r2,n2=sec(x2,h2) v2=q/a2 print('{0:8.3f}{1:8.3f}{2:8.3f}{3:8.3f}{4:10.5f}{5:8.3f}'.format(x2,z2,h2,z2+h2,z2+h2+v2**2/2/g,v2)) x1=x2 # distance h1=h2 # water depth #============== # Execution #============== if __name__ == '__main__': main()
Calculated results are shown below. A water level drop is larger than expected.
x z h z+h Energy v 0.000 562.200 3.900 566.100 566.46982 2.692 11.000 560.500 5.971 566.471 566.47522 0.293 19.000 560.500 5.971 566.471 566.47523 0.293 47.000 561.060 5.410 566.470 566.47528 0.323 62.000 561.360 5.091 566.451 566.47541 0.687 69.000 561.500 4.950 566.450 566.47553 0.707 69.500 562.000 4.444 566.444 566.47554 0.788 73.500 562.000 4.444 566.444 566.47562 0.788 74.000 561.500 4.954 566.454 566.47563 0.652 77.000 561.500 4.954 566.454 566.47567 0.652 96.625 563.000 3.431 566.431 566.47615 0.942 102.125 563.000 3.431 566.431 566.47634 0.942
That's all. Thank you.
Python : Calculation of normal depth of open channel with rectangular cross section by 'scipy.optimize.brentq'
Calculation program of normal depth of open channel with rectangular cross sectionby Brent's method is introduced in this post. Basic equations for normal depth calculation of open channel with rectangular cross section are shown below.
(known) discharge | |
(known) width of rectangular open channe | |
(known) Manning's roughness coefficient | |
(known) gradient of channel invert | |
(unknown) normal depth |
Following equation is solved for .
The critical depth calculation of is extra.
# normal depth and critical depth of rectangular cross section import numpy as np from scipy import optimize def cal_hc(q,b): # critical depth g=9.8 hc=(q**2/g/b**2)**(1/3) return hc def func(h,q,b,n,i): f=q-b*h/n*(b*h/(b+2*h))**(2/3)*i**(1/2) return f def main(): q=42.0 # discharge b=4.0 # channel width n=0.014 # Manning's roughness coefficient i=0.001 # invert gradient h1=0.0 h2=10.0 hh=optimize.brentq(func,h1,h2,args=(q,b,n,i)) print('hn=',hh) # normal depth hc=cal_hc(q,b) print('hc=',hc) # critical depth #============== # Execution #============== if __name__ == '__main__': main()
Calculation results are shown below.
hn= 3.866645305835682 hc= 2.2407023732785825
That's all. Thank you.
Jupyter の ipynb ファイルの中身を見たい
2020.09.20投稿
必要性
Jupyter notebook 上でコードを書いていて、「あのとき作った Jupyter のコードをコピーしたい!」ということがよくある。その ipynb ファイルが、たまたま今開いている Jupyter のディレクトリの下にあればいいのだが、そうでない場合は、一度 Jupyter notebook を終了・再起動し所要の ipynb ファイルを開いてエディタにコピー、もとの作業に戻るため Jupyter notebook を再起動するという手間をかけていた。 なんとかならないかと思いながら調べていたら、効率よく他の ipynb ファイルの中身を確認する以下の3つの方法があることがわかったので、紹介しておく。 もしここに記載の方法を使ったことにより不都合なことが起きても、筆者は知りません。 自分の責任においてトライしてください。
VS code を使う
こちらのサイトに方法が記載されている。
私は VS Cod を使う方法を選択し使ってみた。
Mac の Get info の Open with: で、VS code をデフォルトにしておけば、ipynb ファイルをダブルクリックすることで開くことができる。
Jupyter lab を使う
Jupyter Lab なら複数のタブを開ける! そこでさっそくインストールして調子を見ているところです。 私のマシンではなんとなく動きがもっさりしている感じ。
複数のブラウザで Jupyter を開く
Jupyter notebook でも これ(http://localhost:8888/?token=xxxxxx......
)を url に入れてやると違うブラウザで Jupyter notebook を開ける。
例えば、私の場合、default は Safari であるが、terminal に出てくるこれ(http://localhost:8888/?token=xxxxxx......
)を Firefox や Chrome の url にいれることにより、それぞれのブラウザで notebook を開ける。
以 上