# 7.36 Hydraulic Resistance of a Diverging Wye

Given that 60 deg F water flows through a 6” sched 80 45 degree wye with equal leg diameters, 250 gpm flows through the branch leg and 400 gpm flows through the straight leg.

Find the loss coefficients for the straight leg, and the branch leg, as well as the loss coefficients for each flow path.

    [1]:

    from fluids.units import *
    from math import pi
    NPS, Di, Do, t = nearest_pipe(NPS=6, schedule='80')

    A = 0.25*pi*Di**2
    beta = 1 # same diameters

    rho = 998*u.kg/u.m**3

    Q_tot = (250+400)*u.gal/u.min
    Q_straight = 400*u.gal/u.min
    Q_branch = 250*u.gal/u.min

    v_combined = Q_tot/A
    print('The combined velocity is %s' %(v_combined.to(u.ft/u.s)))
    branch_flow_ratio = Q_branch/Q_tot

    v_main = Q_straight/A
    v_leg = Q_branch/A

    K_branch = K_branch_diverging_Crane(D_run=Di, D_branch=Di, Q_run=Q_straight, Q_branch=Q_branch, angle=45.0*u.degrees)
    print('The branch loss coefficient is %s' %(K_branch))
    K_run = K_run_diverging_Crane(D_run=Di, D_branch=Di, Q_run=Q_straight, Q_branch=Q_branch, angle=45.0*u.degrees)
    print('The run loss coefficient is %s' %(K_run))

    head_loss_branch = 0.5*rho*v_combined**2*K_branch/(1*u.gravity*rho)
    print('The branch head loss is %s' %(head_loss_branch.to(u.ft)))

    head_loss_run = 0.5*rho*v_combined**2*K_run/(1*u.gravity*rho)
    print('The run head loss is %s' %(head_loss_run.to(u.ft)))

    The combined velocity is 7.996972275001512 foot / second
    The branch loss coefficient is 0.46461592374032235 dimensionless
    The run loss coefficient is -0.06827492034592626 dimensionless
    The branch head loss is 0.4617528263268458 foot
    The run head loss is -0.06785419058213756 foot

The values presented in crane match very nearly exactly; this type of a problem does not require any iteration, unless the density of the fluid is variable.