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Problem 1: In an average year, a small area (3.0 km^2) of agricultural catchment receives 950 mm of precipitation. The catchment is drained by a stream, and a continuous record of stream discharge is available. The total amount of surface-water runoff for the year, determined from a stream discharge, is 1.1*10^6 m^3.

a) What is the volume of water (in m^3) evapotranspired for the year (assume no change in water stored in the catchment)?

b) What is the depth of water (in millimeters (mm)) evapotranspired for the year (again, assume no change in water stored in the catchment)?

c) What is the runoff ratio (rs/p) for the catchment? (p stands for average precipitation rate, and rs stands for average surface run off rate)?

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Elin HesselLv2

17 Dec 2019

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1. What you are given: (see example in notes, runoff 2, page 23-30)

3 inches of rainfall

12-acre site

Forest with good cover

Hydrological soil group B

What you need to do:

Estimate the depth (in.) and volume (ft³) of runoff using the SCS Curve number method for:

A. Antecedent moisture condition I

B. Antecedent moisture condition II

C. Antecedent moisture condition III

CN(I) = [4.2CN(II)] / [10 â 0.058CN(II)]

CN(III) = [23CN(II)] / [10 + 0.13CN(II)]

1. What you are given: (see example in notes, runoff 2, page 23-30)

3 inches of rainfall

12-acre site

Forest with good cover

Hydrological soil group B

What you need to do:

Estimate the depth (in.) and volume (ft³) of runoff using the SCS Curve number method for:

A. Antecedent moisture condition I

B. Antecedent moisture condition II

C. Antecedent moisture condition III

CN(I) = [4.2CN(II)] / [10 â 0.058CN(II)]

CN(III) = [23CN(II)] / [10 + 0.13CN(II)]

FOREST WITH GOOD COVER HYDROLOGIC GROUP B IS 55

POWER POINT PG 23-30

Â§No expression of time

ÃCan not account for rainfall duration or intensity

Â§0.2S based on data from agricultural watersheds

ÃMay not apply to urban areas with impervious areas

ÃEquations have been developed to adjust for urban areas

Â§Can not estimate runoff from:

Ã snowmelt

ÃRain on frozen ground

Â§CN procedure is less accurate for runoff < 0.5 in.

Â§CN procedure applies only to direct surface runoff and does not include:

ÃInterflow from forested areas

ÃRapid subsurface flow to streams (important in sandy soils)

ÃLateral subsurface flow in karst regions

ÃHigh groundwater levels that may come to the surface during rainfall events

Watershed with Land Use %

And HSGs Listed

HSG=D 30% bare soil

HSG=C 35% thin forest

HSG=B 35% grass

CN EXAMPLE:

Use HSG % CN (Table 3.16 Haan text)

Bare soil D 30 91 (bare soil w/o conservation)

Grass B 35 58 (grass in good condition)

Forest C 35 77 (poor cover)

Weighted CN = 0.30(91) + 0.35(58) + 0.35(77)

Weighted CN = 75

S = (1000 / CN) â 10 = (1000 / 75) â 10

S = 3.3 in.

Q = (P â 0.2S)2 / (P + 0.8S) for P > 0.2S

P > 0.2(3.3in.)

P > 0.66 in.

Q = [3 â 0.2*3.3]2 / [3+ 0.8(3.3)]

Q = 0.97in.

Q = 0.97 in. = 0.08 ft

5 acres = 217,800.9 ft2

Volume = 0.08ft x 217,800.9 ft2

Volume = 17424.4 ft3

CN(I) = 58 (dry conditions)

CN(II) = 77 (average conditions)

CN(III) = 89 (wet conditions)

1. A sludge containing 4% solid (by weight) needs to be thickened by centrifugation to â¥15% solid for disposal. The centrifuge has a high constant speed and produces a dewatered sludge containing 20% solid. To increase the processing capacity while still meeting the disposal requirement, you decide to bypass (i.e., not dewater) some of the feed sludge and blend the untreated sludge with the dewatered (20% solid) sludge, so as to produce a final sludge with a 15% solid content. Assume all sludge streams have the density of water.

(a) Give the centrate production rate (in L/min) and the percent solid (of the total coming in) discharged via the centrate?

(b) What % of the feed sludge is bypassed (i.e., not centrifuged)?

2. Eutrophication is a natural aging process all lakes undergo as nutrients and organic matter accumulate from their drainage basins. Over time the accumulated nutrients enable increased aquatic growth and stimulate biological activities. The increase in phytoplankton causes the lake to become murky, while decaying organic matter depletes the available dissolved O2. Further accumulation of silt, nutrients, and organic matter causes the eutrophic ("well-fed") lake to become shallower and warmer with more plants growing in the shallow edges. This process transforms lakes into bogs and marshes over time. Natural eutrophication is usually quite slow, often measured in thousands of years before major changes occur. What human activity has done is rapidly accelerate this process (called "cultural eutrophication") by introducing large quantities of nutrients. Sources of nutrients include municipal wastewater, industrial wastes, and runoff from fertilized agricultural land. The introduction of superfluous nutrients disturbs the natural balance of the system, causing excessive algae growth and water quality deterioration. Algal blooms are the unsightly and often malodorous clumps of rotting debris along the shoreline and thick mats of dead organic matter in the lake. A nutrient of major importance that enter fresh water bodies such as lakes from the aforementioned sources is phosphorous. C = 20% C = 15% Q = 100 L/min C = 4% bypass centrifuge C = 0.1% ("centrate") 2 Consider a phosphorus-limited lake with a surface area equal to 8 x 107 m 2 that is being fed by a stream with a flow rate of 15 m3 /s and a phosphorus concentration of 0.01 mg/L. An outfall from a wastewater treatment plant also adds 1 g/s of phosphorus to the lake, but its volumetric flow rate is small and can be ignored. The outflow rate from the lake is the same as the inflow (15 m3 /s). Because phosphorus can attach to suspended particles in lake water, the phosphorus in the lake can also be removed through particle settling (to sediment), which occurs at a rate of 10 m/yr. For the questions below, assume the lake is a cylindrical CSTR.

(a) Draw a diagram to represent the processes involved in phosphorus mass balance. List all sources and sinks. Estimate the steady-state phosphorus concentration in the lake.

(b) An acceptable level of phosphorus in the lake is 0.01 mg/L, above which eutrophication is considered likely. By what percent must the wastewater treatment plant reduce its phosphorous discharge into the lake to achieve this concentration?

3. A textile finishing process involves drying fabric that has been treated with a volatile solvent. The drying process involves evaporation of solvent and removal of solvent vapor by air. The wet fabric entering the dryer contains 50% solvent by weight. Air enters the dryer at a rate of 8 kg per kg of solvent-free fabric. The drying process is 92% efficient; i.e., 92% of entering solvent is removed from fabric and carried out by air. What is the weight % of solvent in the dried fabric? What is the mole % of solvent vapor in the exhaust? The molecular weight of the solvent is 46 g/mol.(Note: Dried fabric still contains some solvent; i.e., dried fabric is not solvent-free).

4. A process heater burns 4.8 kg of natural gas per minute, and the air supply rate is 20% higher than the stoichiometric rate. Exhaust from the heater is sent to a heat exchanger, where water vapor is condensed and removed from the exhaust. Assume natural gas is all methane (CH4), air contains 21% O2 and 79% N2, and the combustion of CH4 is complete, what is the mole % of CO2 in the cooled exhaust?