Script to estimate the effectiveness of using boat engines as water propeller (using energy budget).

flow.py

t_flow = 3000.#m^3/s total chaopraya flow
t_area = 4000.#m^2 total crossection of chaopraya
rho = 1000.#kg/m^3 water density

#non interference
#P is power per boat  (watts)
#A is effective area (m^3)
#n is number of boats 
def flowdiff(P,A,n):
    a_area = A #affected Area
    a_flow = t_flow/t_area*A #flow in affected area
    f_new = pow(a_flow**3 + 2*P*A**2/rho,1./3.) #new flow in affected area
    print 'f_new-a_flow',(f_new-a_flow) #flow generated per boat
    un_flow = t_flow/t_area*(t_area-n*A)#unaffected flow
    new_total = f_new*n+un_flow #new total flow
    flow_diff = new_total - t_flow
    return flow_diff

#interfered flow
#P is power per boat (watts)
#A is effective area (m^3)
#n is number of boats
#assuming total affected area = n*A
#actually if you write downt he expression there is a magic factor of n^3
#which makes it correct to add flow of 1000 boats
def interfered_flowdiff(P,A,n):
    a_area = A #affected Area
    a_flow = t_flow/t_area*(n*A) #flow in affected area
    f_new = pow(a_flow**3 + 2*n*P*(n*A)**2/rho,1./3.) #new flow in affected area
    print 'f_new-a_flow',(f_new-a_flow) #flow generated per boat
    un_flow = t_flow/t_area*(t_area-n*A)#unaffected flow
    new_total = f_new+un_flow #new total flow
    flow_diff = new_total - t_flow
    return flow_diff

#using estimated power and effective from K.Matipon
print '1000 small boats non interference'
print flowdiff(P=1000.,A=0.1,n=1000.) #198m^3/s
print '10 big ones non interference'
print flowdiff(P=2000000.,A=1.,n=10.) #151m^3/s


print '1000 small boats interfere'
print interfered_flowdiff(P=1000.,A=0.1,n=1000.) #198m^3/s
print '10 big ones interfere'
print interfered_flowdiff(P=2000000.,A=1.,n=10.) #151m^3/s