I am trying to program code in Python that will generate the blue line in the graph. I have already sucessfully generated the red line but my code for the blue one is not working. I will attach my code down below once I include the problem statement:

Problem: A centrifugal pump is used to pump water at $25\backslash$deg C $(\backslash$rho $=997\mathrm{.}0$ kg$/$m$^3,\backslash$mu $=8\mathrm{.}91$x$10^(-4)$ kg$/($m$$s$),$P$\_$v$=3\mathrm{.}169$ kPa$)$ from a reservoir whose surface is $2\mathrm{.}2$ m below the centerline of the pump inlet. The pipe is PVC plastic $($smooth$)$ with an inner diameter of $24\mathrm{.}0$ mm$.$ The pipe length from the submerged pipe inlet to the pump inlet is $2\mathrm{.}8$ m$.$ There are only two minor losses in the piping system from the pipe inlet to the pump inlet: a sharp$-$edged reentrant inlet $($K$\_$L$=0\mathrm{.}85)$ and a flanged smooth $90\backslash$deg regular elbow $($K$\_$L$=0\mathrm{.}3)($we don$$t include a loss associated with the pump inlet itself$$we$$re doing the balance at the point right before the pump entrance$).$ The pump manufacturer provides the required net positive suction head as a curve fit: NPSH$\_$required$=2\mathrm{.}2$m$+(0\mathrm{.}0013$ m$/($Lpm$^2))$ Q$^2$ where Q is the volumetric flow rate in LPM$.$ Plot NPSH and NPSH$\_$required for this pump and estimate the maximum volume flow rate $($in units of Lpm$)$ that can be pumped without cavitation. You can approximate the kinetic energy correction factory as $1.$

Code thus far:

import numpy as np

import scipy.optimize as opt

from scipy.optimize import fsolve

import matplotlib.pyplot as plt

from math import pi$,$ sqrt$,$ log$10$

Problem $1$

# Parameters

rho $=997\mathrm{.}0$ #kg$/$m$^3$

mu $=8\mathrm{.}91$e$-4$ #kg$/$m$*$s

Pv $=3169$ #Pa

z $=2\mathrm{.}2$ #m

KL$\_$tot $=1\mathrm{.}15$

L $=2\mathrm{.}8$ #m

D $=0\mathrm{.}024$ #m

epsilon $=0$

P$\_$atm $=101325$ #Pa

g $=9\mathrm{.}81$ #m$/$s$^2$

## Plot Required NPSH

# Generate flow rates

Q $=$ np$.$linspace$(0,100,100)$ # L$/$min

# Calculate required NPSH for each flow rate

NPSH$\_$required$\_$values $=2\mathrm{.}2+0\mathrm{.}0013*$ Q $**2$

# Plot

plt$.$plot$($Q$,$ NPSH$\_$required$\_$values$)$

plt$.$xlabel$(\text{'}$Flow Rate $($L$/$min$)\text{'})$

plt$.$ylabel$(\text{'}$NPSH $($m$)\text{'})$

plt$.$title$(\text{'}$NPSH vs Flow Rate'$)$

plt$.$grid$($True$)$

plt$.$xlim$(0,100)$

plt$.$ylim$(0,12)$

plt$.$show$()$

# Colebrook equation

def colebrook$\_$eq$($f$,$ Re$,$ epsilon, D$)$:

try:

return $1/$ sqrt$($f$)+2*$ log$10(2\mathrm{.}51/($Re $*$ sqrt$($f$)))$

except $($ZeroDivisionError$,$ ValueError$)$:

return float$(\text{'}$inf$\text{'})$

def calculate$\_$NPSH$\_$avail$($f$,$ v$)$:

return $(($P$\_$atm $-$ Pv$)/($rho$*$g$))-$ z $-((($f $*$ L$)/$D $+$ KL$\_$tot$)*(($v$**2)/2*$g$))$

# Generate flow rates

Q$\_$m$3$s $=$ np$.$linspace$(0\mathrm{.}1,200,100)$

# Calculate NPSH available for each flow rate

NPSH$\_$avail$\_$values $=[]$

for Q in Q$\_$m$3$s:

v $=$ Q $/($pi $*($D $/2)**2)$

Re $=($rho $*$ v $*$ D$)/$ mu

f $=$ opt.fsolve$($colebrook$\_$eq$,0\mathrm{.}02,$ args$=($Re$,$ epsilon, D$))[0]$

NPSH$\_$avail $=$ calculate$\_$NPSH$\_$avail$($f$,$ v$)$

NPSH$\_$avail$\_$values.append$($NPSH$\_$avail$)$

# Convert to Lpm

Q$\_$Lpm $=$ Q$\_$m$3$s #$*60*1000$

# Print the first few values of Q$\_$Lpm and NPSH$\_$avail$\_$values

print$("$Flow Rates $($L$/$min$)$:$",$ Q$\_$Lpm$[$:$5])$

print$("$NPSH Available $($m$)$:$",$ NPSH$\_$avail$\_$values$[$:$5])$

# Plot

plt$.$plot$($Q$\_$Lpm$,$ NPSH$\_$avail$\_$values$)$

plt$.$xlabel$(\text{'}$Flow Rate $($L$/$min$)\text{'})$

plt$.$ylabel$(\text{'}$NPSH Available $($m$)\text{'})$

plt$.$title$(\text{'}$NPSH Available vs Flow Rate'$)$

plt$.$grid$($True$)$

plt$.$xlim$(0,200)$

plt$.$ylim$(0,20)$

plt$.$show$()$

$**$Please verify that your answer generates the required graph$**$