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Applied Hydro Test 2
| Question | Answer |
|---|---|
| The 11 items in the Water Budget. | Ocean, evaporation, atmosphere, precipitation, surface, infiltration, groundwater, evapotranspiration, depression storage, overland flow, reaches channels. |
| Hydrologic Cycle properties | Precipitation (P), Transpiration (T), Evaporation (E), Evapotranspiration (ET), Infiltration (F), Groundwater (G), direct runoff (R). |
| Precipitation | Rainfall (P) |
| Evaporation | conversion of water to water vapor (E) |
| Transpiration | loss of water through plat tissue and leaves (T) |
| Evapotranspiration | The combined loss due to evaporation and transpiration (ET) |
| Infiltration | function of soil moisture conditions and soil type. may reenter channels later as interflow or percolate to recharge shallow groundwater. |
| Groundwater | flows in porous media in the subsurface |
| Direct Runoff | the remaining portion of precipitation which becomes overland flow. |
| Three ways to measure precipitation | Through Arithmetic, Isohyedal, or Theissan models |
| Arithmetic mean | Take all gauges within the watershed boundary and average the precipitation measured. sum(Pi)/n where Pi is precipitation data and n is the # of gauge sites. |
| Thiessen Polygon Model | Allows for areal weighting of rainfall. Draw straight lines between gauges, then perpendicular cross-secting lines dividing sections into areas. (sum(PiAi))/At |
| Isohyetal Method | Draw lines at equal points of precipitation, similar to a topographic map. =sum(P*A)/sum(A). |
| Streamflow Measurement basis | Points taken at 0.2D and 0.8D, (0.6D at small depthed ends). |
| Streamflow equation | Q=sum(vi*di*deltaWi) where vi is the average velocity for that point, di is depth, and delta W is the width surrounding the line of interest. |
| Hydrographs! (definition) | A plot of flow rate vs. time. Different types of hydrographs will give different info. For example, some include the baseflow of a stream whilst others measure the effect of a storm event. |
| Unit Hydrographs | Basin outflow result from 1 inch of direct runoff during a specified period of rainfall duration. |
| Unit Hydrograph assumptions | 1 Rainfall distribution is the same for storms of equal duration-spatially and temporally 2) independent of antecedent moisture (existing conditions) 3) Excess rainfall of equal duration produce the same response 4) twice the rainfall = double H ordina |
| Important UH characteristics | Duration (D), Lag Time (tp), Time of rise (Tr), Time of Concentration (tc), Time base (Tb). |
| Duration | D. The time from start to finish of rainfall excess. the box in the upper left corner of the diagram. |
| Lag Time | tp. The time from the center of D to the peak of the hydrograph. |
| Time of Rise | Tr the time from the start of the rainfall to the peak of the hydrograph |
| Time of Concentration | tc the time for a wave of water to propogate from the most distant point in the watershed to the outlet. The time from the end of net rainfall (end of D) to the inflection point of the hydrograph. |
| Time base | Tb the total duration of the unit hydrograph. |
| How to create a unit hydrograph | 1) Remove baseflow. 2) Calculate the total volume --> V=sum(Qi*ti). Volume = flow*time or the area under the curve. 2) find the avg. excess prec. => P(ave, excess)=V/Ar. Total volume/Area. 3)Qunit = Qi/P(ave, excess) |
| Factors that affect watershed response (7) | Vegitation Soils Size Shape Land Use Cover Antecedent Moisture Conditions (AMC) |
| Total Excess Precipitation | P(excess)=V/Ar |
| Unit hydrograph peak discharge | Qunit = Qi/P(excess) |
| How to find the design hydrograph volume given UH values | On test: if a 2 hr storm produces 2.6 in of runoff. Take Vt/P(uh) * P(given, 2.6in) |
| How to find the design discharge from a UH | Qunit*P(given, 2.6in). |
| Hydrograph Synthesis Methods | S-Curve. Lagging Storm. SCS triangular method. |
| Lagging Storm Hydrograph Synthesis | Want Hydrograph for time (n) storm where n is a whole number multiple of the orginal hydrograph. 1) Draw (n) hydrographs separated by duration (D) 2) Add the ordinates - n*d storm event 3) divide those total ordinates (from step 2) by n. |
| S-curve method | Can be used to find a 2-hr storm with 1-hr data. 1)Take cumulative data 2)Lag by desired duration D 3)Subtract initial curve by lagged curve. |
| Given: 4 hr storm producing 2in of runoff. Desired: Hydrograph for an 8 hour event producing 1in of runoff in first 4 hours and 2in in last 4 hours. | This is a combination of hydrograph synthesis methods. For the first 4 hours simply half the storm data. the values for a 1in rainfall event are half for a 2in event. For the last 4 hours, lag the data by 4 hours (lagging method). Add the two graph values |
| SCS triangle Hydrograph | Given values for each hydrograph property. |
| SCS triangle Hydrograph (S) | S = 1000/CN-10 where CN is the curve number. |
| Curve Number | Numerical representation of the type of soil/cover/land development over an area. |
| SCS triangle Hydrograph (tp) | time lag --> (L^0.8(S+1)^0.7)/(1900*Y^0.5). Where L is the length to divide, Y is the average watershed slop (%), and S is based on curve number. Note: S does not equal slope. |
| SCS triangle Hydrograph (Qp) | Qp=484A/tr. And Qp=2V/(Tr+B) Qp is the peak flow rate. Note, to find B, will need to calculate volume which can be found from P*Ar where P=1in. |
| SCS triangle Hydrograph (tr) | tr=D/2+tp. tr is the time of rise.(hr) |
| SCS triangle Hydrograph (Q) | Q=(P-0.25)^2/(P+0.85). |
| Index method | By assuming a constant infiltration rate, one cna determine the volume of direct runoff. sum(delta time*(rainfall-phi))=Runoff (Volume/Area). All values of phi that would result in a negative number are ignored. Iterative process. phi~0.5 to 1.5. |
| Infiltration Solution Methods- | Index Method. Horton Method. |
| Horton Method | f(t)=fc+(fo-fc)*e^(-kt). where f is the infiltration capacity, fc is the final capacity (in/hr), fo is the initial infiltration capacity (in/hr) and k is an empirical constant (hr^-1). |
| Risk | 1-(1-1/T)^n. where T is T-year event and n is the years of interest. For example the probability of a 100 year storm occuring at least once in 5 years =(1-(1-1/100)^n) |
| Reliability | 1-risk. Or (1-1/T)^n |
| Probability | 1/T. For example, probability of a 100 year storm (T-year storm) is 1/100 or 0.01 |
| Frequency Distributions (4) | Normal, Log Normal, Gamma, and Log Pearson III. |
| Calculating probabilities via Normal Distributions. | To find the flow of a T-storm event. 1)probability = 1/T. 2)find z via table D-2 from probability. 3)Q(T)=avg(Q)+z*std. dev.(Q) |
| Calculating probabilities via Log Normal Distributions. | 1)find probability/z 2)take log of data. Find avg(log(Q)) and std. dev.(log(Q)). 3)y=avg(log)+z*stddev(log) 4)Q(T)=10^y. |
| Calculating probabilities via Gamma Distributions | 1)find skewness coefficient (Cs) of data. 2)find k via charts from Cs and recurrance interval. 3)Q(T)=avg(Q)+K*std.dev(Q) |
| Calculating probabilities via Log Pearson III | 1)find k. 2)y(T)=avg(log(Q))+K*std.dev.(log(Q)) 3)Q(T)=10^y. |
| Flood Routing Methods | Reservoir Routing (Storage Indication Method) River Routing (Muskingum Method). Firm Yield. |
| Reservoir Routing (Storage Indication Method) | St=st-1+It-Ydt-Atet-Qt |
| Muskingum Method | S2-S1=K[x(I2-I1)+(1-x)(Q2-Q1)] where K is the travel time constant for the reach, I is the inflow, x is the weighting factor which varies from 0 to 0.5. and Q is the outflow. Q2=C0*I2+C1*I1+C2*Q1. Where C0,C1, and C2 are coefficients. |
| Conservation of Mass | Continuity-->Q=v1A1=v2A2 |
| Conservation of Energy | Bernoulli equation. |
| Conservation of Momentum | sum(forces)=ro*Q*(V2-V1). fm=gamm*hc*A |
| Pumps |
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