Final Exam 2008
Open book, Walton Chapters, and one page of notes (100/106 points) b

  1. (6) Label: confined and unconfined aquifer. Is the stream gaining or losing?Show (or describe) where transpiration occurs on the figure

Answer:

2 pts each question, trees provide the transpiration, gaining stream, confined aquifer is below confining layer, unconfined above

  1. (5) Mark the location of springs:

Answer: technically the landslide area also has springs so allow that but only require the ones on the figure below

  1. (15) Assuming a hydraulic conductivity of 0.003 ft/s and a porosity of 0.23, find the discharge in cfs, the specific discharge, and the pore velocity at the 5200 foot contour. The width of the channel at the water table is 800 feet.

Answer: V=-K(h2-h1)/(x2-x1), don’t take off for sign errors, allow some slop for distance estimate, they can pick different points 1 and 2 as long as they are close to 5200; don’t see any need for partial credit but they could get a few points if the area is wrong or they just make a calculator error

V=3 e-5 ft/s, Q = V A = 0.48 cfs pore velocity = V/phi = 1.3e-4 ft/s

  1. (15) Find the discharge and velocity, draw the EGL and HGLshow the energy equation and how you simplify it. Calculate the distance between the EGL and HGL.

Points: EGL and HGL (2 pts each); energy equation 5 pts; velocity head as distance between EGL and HGL (2 pts), correct discharge and velocity 6 pts

  1. (15) Water is flowing in a street gutter at a depth of 0.4 feet. The gutter has a Manning n value of 0.017, a side slope of 20:1, and a longitudinal slope of 0.01. Determine the velocity in ft/s and the discharge rate in the gutter in cfs. Estimate the Froude number.

Answer: 10 points for V and Q, 5 pts for Fr

partial credit, give full credit for Froude number if the start with wrong velocity and/or area from first part of problem; -2 pts off for using metric value for g, two points off for wrong area, 2 points off for wrong perimeter

V = Q/A = (or direct from Manning Eq) = 2.99 ft/s

Fr = V/Sqrt[gD] D = A/T = 1.6/(20*0.4) = 0.2 ft

Fr = 2.99/Sqrt[32.2*0.2] = 1.18

  1. (10)

partial credit: take off two points for calculator error

  1. (15) A one hour rainfall beginning at time = 0 causes the following hydrograph in the 63 hectare catchment. Find the unit (1 cm) hydrograph for the one hour storm & draw it on the figure. What is the peak discharge from the unit hydrograph?

Answer: Surface runoff = ½ base * height = ½*2 hr * 3.5 m3/s = 12,600 m3

Depth of surface runoff over drainage basin is: 12,600/(630,000) = 0.02 m = 2 cm

Since depth is twice the unit amount the hydrograph is reduced by a factor of ½

3.5/2 = 1.75 m3/s

So the unit hydrograph is at the same location but starts at zero and has a peak of 1.75 m3/s

Partial credit: -5 points if they fail to take off base flow when drawing the unit hydrograph; -2 points for each simple arithmetic error

  1. (10) Flows down Kanker Creek supply the town of Kwade Ziekte with drinking water. Based upon 75 years of data the annual maximum flow is described by a normal distribution with a mean of 100 cfs and a standard deviation of 25 cfs. Estimate the probability and return period for a peak flow of 150 cfs, the capacity of the levees. Last year the peak flow was 133 cfs.

Answer: Z = (150-100)/25 =+2 giving an area on the chart of 0.977. We want the other tail so the probability is: 0.023 and the return period = 1/0.023 = 43.5 yr

partial credit: -5 points if they don’t take (1-0.977), minus two points for each minor error

  1. (15) A booster pump is installed in the pipeline between the two reservoirs shown below. If the energy added by the pump is 50 feet, determine the flow rate in the pipeline and discharge. Neglect minor losses. The friction factor is 0.02.

Answer: points: 10 points for correct energy equation, 5 points for correct answers