<jupyter_start><jupyter_text>Prelim 2   Name this file: Lastname_Firstname_Prelim2  Multiple Choice (4 points each - 12 points)Make your answer **bold** by typing `**` before and after the correct answer.1. In class we talked about the floc volcano phenomenon of the San Nicolas plant. The floc volcano phenomenon occurs when water entering the sed tank is warmer than the water in the sed tank, and stops when the water entering the sed tank becomes colder than the water in the sed tank. Why does colder water stop the floc volcano phenomenon?    1. Colder influent water has a higher viscosity and thus reduces the velocity of the floc volcanoes    1. **Colder influent water is more dense than the water in the sedimentation tank and thus the density stratification in the sedimentation tank is stable**     1. Flow rate through the plant decreases when the water is colder, and thus floc volcano formation is averted1. Which of the following changes could AguaClara do to further reduce the **depth** of the sedimentation tanks? For each change assume that the other design parameters are held constant while that one change is made.    1. Increase the vertical velocity    1. **Increase the capture velocity**    1. Increase the spacing between plate settlers (hold capture velocity constant)    1. Increase the water velocity exiting the diffusers 1. What does the volume of water in a vertical flow hydraulic flocculator do when the flow rate inceases?    1. Decreases    1. Remains constant    1. **Increases** ***  Short Answer (3 points each - 18 points) Explain your answers as thoroughly as possible. Double click on this cell to write your answers 1.  Give at least 3 reasons why water treatment plants should be enclosed in a building     1. Protect the plant operator from the weather    1. Shield plastic components from UV damage    1. Reduce unequal heating of water in the sedimentation tanks    1. Provide security for the plant operator at night1. List at least 3 problems that can cause sedimentation tanks to perform poorly    1. Accumulated sediment degrades anaerobically, produces methane, and causes flocs to float to the surface    1. Temperature fluctuations in the raw water cause the sedimentation tank to be unstable    1. Failure to resuspend settled flocs caused by incorrect bottom geometry    1. inlet channels can have velocities that are too low such that flocs settle out before arriving in the sedimentation tank    1. Sunshine on sedimentation tank can cause heating and thermally driven water currents    1. generic something about poor flow distribution...1. Sort the following list according to the amount of solids each unit process removes from highly turbid (for example, 500 NTU) raw water. A is the most solids removal and C is the least solids removal:     1. Flocculation    1. Sedimentation    1. Filtration1. List three reasons why the Sedimentation tank inlet manifold has diffuser pipes.    1. Head loss to improve flow distribution    1. Straighted the flow to eliminate horizontal velocity component that would cause large scale floc circulation in the sedimentation tank    1. High velocity (momentum) jet to resuspend settled flocs1. How does accumulated sludge in a sedimentation tank cause poor performance?    1. Within accumulated sludge there is an anaerobic zone in which bacteria can grow. Anaerobic bacteria can produce methane, which bubbles out of the sedimentation tank. These bubbles take sediment/flocs with them up through the sed tank. The sediment/flocs then escape through the exit launder, which causes poor performance.1. Why is it important to be able to send poorly flocculated water to waste BEFORE it enters the sedimentation tank?        1.  Poorly flocculated water contains very small diameter flocs which are the ones that tend to escape a sedimentation tank and continue on to the filters. By wasting poorly flocculated water, we are removing the small flocs that are more likely to make it through the entire treatment plant. Purging poorly flocculated water thereby decreases the plant's effluent turbidity.         Additionally, a sedimentation tank with poorly flocculated water will continue to release small flocs for up to multiple residence times (since the sed tank has properties of a CSTR). This means that you will have poor quality-sedimented effluent for a significant amount of time after the flocculator performance has stabilized.   Design Challenges (70 points)Assign the given inputs to variables. Write the equation in python using variables. ** Make sure to use print statements which put your answer in context, and to display your answer in appropriate units **  1) Flocculator (27 points)A 60 L/s water treatment plant has a vertical flow hydraulic flocculator. The channels are 50 cm wide, 6 m long, and 2 m deep at the end of the flocculator. The head loss through the flocculator is 40 cm. The minimum water temperature is 15°C.There are 4 channels. The minor loss coefficient for each 180° bend is 2.5.Answer A through F. State any assumptions that you make.<jupyter_code># define your parameters and variables given in the problem statement here
# head loss through the flocculator
headloss_floc = 40 * u.cm

# water depth at the end of flocculator where it flows into the inlet channel of the sedimentation tank
height_floc_end = 2 * u.m

width_floc = 50 * u.cm

length_floc = 6 * u.m

N_channels = 4

flow_floc = 60 * u.L/u.s

# expansion minor loss coefficient for 180 degree bend
K_e = 2.5

# this is the minimum temperature of the raw water
T_water = 15* u.degC<jupyter_output>1.137e-06 meter ** 2 / second<jupyter_text>A. (5 points)What is the hydraulic residence time in the flocculator if you neglect the volume change due to head loss?<jupyter_code>volume_floc = width_floc * height_floc_end * length_floc * N_channels
t_floc = (volume_floc / flow_floc).to(u.s)
print('The hydraulic residence time is' , t_floc)<jupyter_output>The hydraulic residence time is 400 second<jupyter_text>B. (5 points)What is the average velocity gradient in the flocculator?<jupyter_code>Gave_floc = (np.sqrt(pc.gravity * headloss_floc/(t_floc * pc.viscosity_kinematic(T_water) ))).to(1/u.s)
print('The average velocity gradient in the flocculator is' , Gave_floc)<jupyter_output>The average velocity gradient in the flocculator is 92.87 / second<jupyter_text>C. (5 points)What is the collision potential $\big(G\theta\big)$  for this flocculator?<jupyter_code>Gtheta_floc = Gave_floc * t_floc
print('The flocculator collision potential is' , Gtheta_floc)<jupyter_output>The flocculator collision potential is 3.715e+04 dimensionless<jupyter_text>D. (5 points)What is the maximum distance between expansions for this flocculator? The maximum ratio of H/S ($\Pi _{H{S_{Max}}}$) is 6 ${H_{{e_{Max}}}} = {\bigg[ {\frac{{{K_e}}}{{2\nu {{\bar G}^2}}}{{\left( {\frac{{Q{\Pi _{H{S_{Max}}}}}}{W}} \right)}^3}} \bigg]^{\frac{1}{4}}}$<jupyter_code>Pi_HS_max = 6
Lmax_expansions = ((K_e / (2 * pc.viscosity_kinematic(T_water) * Gave_floc**2) * 
                    (flow_floc * Pi_HS_max / width_floc)**3)**(1/4)).to(u.m)
print('The maximum distance between expansions for an efficient design is' , Lmax_expansions)<jupyter_output>The maximum distance between expansions for an efficient design is 2.626 meter<jupyter_text>E. (2 points)Is it necessary to include additional expansions in this design to maintain high efficiency? **Your Response Here:**Additional expansions are not needed in this flocculator since the distance between expansions is greater than the depth of the flocculator.  F. (5 points)What is the spacing of the baffles in this flocculator? $ S = {\left( {\frac{{{K_e}}}{{2{H_e}{G^2}\nu }}} \right)^{\frac{1}{3}}}\frac{Q}{W}$<jupyter_code>spacing_baffles = ((K_e / (2 * pc.viscosity_kinematic(T_water) * height_floc_end * Gave_floc**2))**(1/3) * flow_floc / width_floc).to(u.m)
print('The baffle spacing is' , spacing_baffles)<jupyter_output>The baffle spacing is 0.4793 meter<jupyter_text>2) Sedimentation tank area (5 points)The average water use per person in the US is 3 mL/s. How much floc blanket plan view area does one person require? The floc blanket upflow velocity is 1 mm/s.<jupyter_code>flow_per_person = 3 * u.mL/u.s
velocity_floc_blanket = 1 * u.mm/u.s
area_floc_blanket = (flow_per_person / velocity_floc_blanket).to(u.cm**2)
print('The floc blanket plan view area for one person is' , area_floc_blanket)<jupyter_output>The floc blanket plan view area for one person is 30 centimeter ** 2<jupyter_text>3) Plate settler design (15 points)Assume the upflow velocity in AguaClara sedimentation tanks is increased to 3 mm/s,the capture velocity is increased to 0.2 mm/s, the spacing between plates is maintained at 2.5 cm, and the plates are inclined at 60$^{\circ}$ from the horizontal. You may assume that the vertical velocity entering the plate settlers is 3 mm/s.Answer questions A and B.   A. (10 points)   How long would the plate settlers be? You may neglect the effect of the thickness of theplate settlers. **Create a function** for the equation that you use to solve this problem. Include all the necessary inputs in your function definition. $$L = \frac{{S\left( {\frac{{{V_{Plate \uparrow }}}}{{{V_c}}} - 1} \right)}}{{\sin \alpha \cos \alpha }}$$<jupyter_code>V_sed_up = 3 * u.mm/u.s
V_sed_capture = 0.2 * u.mm/u.s
angle_sed_plate = 60 * u.deg
s_sed_plate = 2.5 * u.cm

def L_sed_plate(s_sed_plate, V_sed_up, V_sed_capture, angle_sed_plate):
    return (s_sed_plate * ((V_sed_up/V_sed_capture)-1) / (np.sin(angle_sed_plate) * np.cos(angle_sed_plate))).to(u.m)

sed_plate_length = L_sed_plate(s_sed_plate,V_sed_up,V_sed_capture,angle_sed_plate)

print('The minimum length of the plate settlers is',sed_plate_length)<jupyter_output>The minimum length of the plate settlers is 0.8083 meter<jupyter_text>B. (5 points)   What is the ** *vertical* ** height of the plate settlers? This will tell us how much of the sedimentation tank depth these plates occupy.<jupyter_code>height_sed_plates = L_sed_plate * np.sin(angle_sed_plate) 
print('The vertical height of the plate settlers is',height_sed_plates)<jupyter_output><empty_output><jupyter_text>4) Head loss through a floc blanket (8 points) A. (5 points)What is the head loss through a 1 m deep floc blanket with a clay concentration of 3 g/L and an upflow velocity of  1 mm/s? You may assume the water temperature is 20°C. The density of clay is 2650 $\frac{kg}{m^3}$. This head loss helps ensure a uniform velocity of water leaving the water - floc blanket interface at the top of the floc blanket. $${h_L}_{_{FB}} = {H_{FB}}\left( {\frac{{{\rho _{Clay}}}}{{{\rho _{{H_2}O}}}} - 1} \right)\frac{{{C_{Clay}}}}{{{\rho _{Clay}}}}$$<jupyter_code>T_fb = 20 * u.degC
height_fb = 1 * u.m
density_clay = 2650 * u.kg/u.m**3
Concentration_clay = 3 * u.g/u.L
headloss_fb = (height_fb * (density_clay / pc.density_water(T_fb) - 1) * Concentration_clay / density_clay).to(u.mm)
print('The head loss through the floc blanket is',headloss_fb)<jupyter_output><empty_output><jupyter_text>B. (3 points)Which head loss is greater, the head loss through the floc blanket or the head loss through the plate settlers? **Your Response Here:**The head loss through the floc blanket is much greater than the head loss through the plate settlers.  5) Floc Hopper (10 points)The AguaClara 1 L/s sedimentation tank has a floc hopper that is a 3" PVC pipe that is angled at 60$^{\circ}$C from the horizontal. The raw water is 1000 NTU and the settled water turbidity is 3 NTU.A sample from the consolidated sludge in the bottom of the floc hopper was diluted by a factor of 100 and the turbidity of the diluted sample was 150 NTU. (Remember that NTU is an approximate measure of concentration and the unit NTU has been defined for you).What flow rate must the floc hopper drain valve be set at to maintain a constant level of the consolidated sludge in the floc hopper?<jupyter_code>flow_sed = 1 * u.L/u.s
turbidity_influent = 1000 * u.NTU
turbidity_effluent = 3 * u.NTU
turbidity_sludge = 100 * 150 * u.NTU
sludge_mass_rate = (turbidity_influent - turbidity_effluent) * flow_sed
sludge_flow = (sludge_mass_rate/turbidity_sludge).to(u.mL/u.s)
print('The flow rate of sludge is',sludge_flow)<jupyter_output><empty_output><jupyter_text>6) Multiple stage filtration plan view area (5 points)How much area would the unit processes of a multiple stage filtration plant require to provide the water for one person from the US?<jupyter_code># define your parameters and variables given in the problem statement here
V_net_MSF = 0.03 * u.mm/u.s
area_MSF = (flow_per_person / V_net_MSF).to(u.cm**2)
print('The multiple stage filtration plant plan view area for one person is' , area_MSF)<jupyter_output><empty_output>
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