# Dimension of flow field ; velocity components at (1,2) ; magnitude and direction of velocity from math import * # Given # V = 4*Xi-4Yj x=1 # x co-ordinate y=2 # y co-ordinate # Solution print "(a) u = 4*X; v = -4*Y " u = 4*x v=- 4*y print "(b) u=",round(u,0),"m/s and v=",round(v,0),"m/s" R =sqrt(u**2+v**2) ang = atan(v/u)*180/pi print "(c) Magnitude of velocity =",round(R,2),"m/s and angle of resultant velocity = ",round(ang,1),"deg"
(a) u = 4*X; v = -4*Y (b) u= 4.0 m/s and v= -8.0 m/s (c) Magnitude of velocity = 8.94 m/s and angle of resultant velocity = -63.4 deg
# Discharge and mass flow rate from math import * from __future__ import division # Given d = 0.3 # diameter of pipe in m v = 15 # velocity in m/s rho = 997.1 # density in kg/m**3 A = pi*d**2/4 # Solution Q=A*v print "(a) Discharge =",round(Q,2),"m**3/s" mdot = rho*Q print "(b) Mass flow rate = ",round(mdot,2),"kg/s"
(a) Discharge = 1.06 m**3/s (b) Mass flow rate = 1057.21 kg/s
# Mean Velocity from math import * from __future__ import division from scipy import integrate # Given Vo = 10 # velocity in m/s r1 = 0 ro = 0.1 # radius in m N = 1 # Solution R = lambda r: (10*r-1000*r**3) R1,err=integrate.quad(R,r1,ro) Q = R1*2*pi A = pi*(0.1)**2 V = Q/A print "Mean velocity of the flow =",round(V,0),"m/s"
Mean velocity of the flow = 5.0 m/s
# Sketch the stream lines in the first quadrant import matplotlib.pyplot as plt from math import * from scipy import integrate import numpy as np from sympy import * #init_printing(use_unicode=False, warp_line=False, no_global=True) # Given # V = 4*y(m)i+2(m)j x = Symbol('x') U = integrate(2,x) #print u y = Symbol('y') V = integrate(-4*y,y) #print V Zhi = U + V print Zhi # for x and y =0 we get C = 0 X = [5,6,7,8,9,10,11,12,13,14,15,16,17] Y = [0,1.414,2,2.449,2.828,3.16,3.46,3.741,4,4.242,4.472,4.69,4.898] b1=plt.plot(X,Y) X1 = [2.5,3,4,5,6,7,8,9,10,11,12,13,14,15] Y1 = [0,1,1.732,2.23,2.645,3,3.31,3.60,3.87,4.123,4.35889,4.5825,4.795,5] b2=plt.plot(X1,Y1) X2 = [0.5,1.5,2.5,3.5,4.5,5.5,6.5,7.5,8.5,9.5,10.5,11.5,12.5,13.5,14.5,15.5] Y2 = [0,1.414,2,2.449,2.828,3.162,3.462,3.741,4,4.242,4.472,4.69,4.898,5.099,5.29,5.4772] b3=plt.plot(X2,Y2) plt.xlabel("x") plt.ylabel("y") plt.title("Streamline plot") plt.legend(["zhi=10","zhi=5","zhi=1"]) plt.show()
2*x - 2*y**2
# magnitude and direction of flow field from math import * from sympy import * import numpy as np # Given x = 2 # X co-ordinate Y = 4 # Y co-ordiante # Solution y = Symbol('y') zhi = 4*x*y zhiprime = zhi.diff(y) u = zhiprime x = Symbol('x') zhi = 4*x*Y zhiprime = zhi.diff(x) v = zhiprime R=sqrt(u**2+v**2) theta = atan(v/u)*180/pi print "Resutant velocity magnitude = ",round(R,2),"m/s" print "Angle =",round(theta,1),"deg with the X-axis in the 4th quadrant"
Resutant velocity magnitude = 17.89 m/s Angle = 63.4 deg with the X-axis in the 4th quadrant
# Determine velocity ; convective accleration from math import * from __future__ import division from sympy import * import numpy as np from scipy import integrate # Given d1 = 0.09 # diameter in cm d2 = 0.025 # diameter in cm rho = 1000 # density in kg/m**3 mdot = 25 # mass flow rate in kg/s # Solution x = Symbol('x') A1 = pi*d1**2/4 A2 = pi*d2**2/4 AA = A1 - ((A1-A2)/40)*10 # from figure V = mdot/(rho*AA) print "(a) Velocity =",round(V,1),"m/s" AX = (A1 - ((A1-AA)/40)*x) v = 25*10**4/(rho*AX) vprime = v.diff(x) V1 = vprime # at x = 0.1 m we get dv/dx = 0..09 VPrime = 0.09 Acx = V*VPrime print "(b) Convective acceleration =",round(Acx,3),"m**2/s"
(a) Velocity = 5.1 m/s (b) Convective acceleration = 0.46 m**2/s
# Is the flow irrotational from math import * # Given # w = (16y-12x)i +(12y-9x)j # Solution y = Symbol('y') U = 16*y-12*x zhiprime = U.diff(y) u = zhiprime x = Symbol('x') V = 12*y-9*x zhiprime1 = V.diff(x) v = zhiprime1 #Vx = -9 # differentiate V wrt x #Vx = -9 # differentiate V wrt x #Uy = 16 # differentiate U wrt y z = v-u print "z = ",round(z,0) print "Hence the flow is rotational"
z = -25.0 Hence the flow is rotational
# Velocity in larger section from math import * # Given d1 = 0.1 # diameter in m d2 = 0.3 # diameter in m V1 = 30 # velocity in m/s # Solution V2 = (d1**2/d2**2)*V1 print "Velocity at the larger cross section = ",round(V2,2),"m/s"
Velocity at the larger cross section = 3.33 m/s