CE3030 Water Resources Engineering – Tutorial 2 August 23, 2012
(1) Estimate the evaporation for a month from a lake of 500 hectare surface area. The mean discharge from the lake is estimated to be 1.00 m3/s. The monthly rainfall is about 10 cm. A stream flows with an average discharge of 2.00 m3/s in to the lake. The water level in the lake dropped about 5 cm in the month. The seepage losses are negligible, and the variability in surface area of the pool water is considered to be not significant. Monthly Inflow= I= 2*3600*24*30 m3=51.84*105m3 Monthly out flow=O=+1*3600*24*30 m3 =25.92*105m3 Monthly rainfall=P= (10/100)*500*10000 m3 =5*105m3 Change in storage= ∆S= (5/100)*500*10000=2.5*105m3 Using equation for estimation of monthly evaporation; = (51.84+5-25.92+2.5)* 105m3
E
=33.42*105m3 =(33.42*105/500*104)100cm = 66.84cm (2) Calculate the free water surface evaporation in June using the Penman method from an area near Fortworth, Texas, whose latitude is approximately 33oN. The available data include air temperature = 30oC, wind speed at 2-m height = 10 km/h, relative humidity = 60%, mean observed sunshine hours = 12, and reflection coefficient = 0.05 Corresponding to the latitude 330N, solar radiation received at the earth’s atmosphere (I0) is 16.56mm of evaporable water and the maximum number of sunshine hours, N is 14.37h. The short wave radiation received at the earth’s surface, Rs = I0 ( a+b*n/N) Therefore, Rs = 0.18+ (0.55*12/14.37)*16.56 = 10.587mm water/day Actual vapor pressure ea = 31.82*0.60 = 110.092 mm of Hg T = 303 k 2.01*10-9 mm of water/day
Rl = 2.01*10-9*3034(0.56-0.092(110.092)0.5) ( 0.1+0.9*12/14.37) = 2.281mm water/day H = 10.587 (1-0.05)-2.281 = 7.777 mm water/day Ea = (0.175+0.0022*10)(31.82-110.092) = 2.507 mm water/day (3) Compute the daily evaporation from a class A pan if the daily rainfall and the amount of water added to bring the water level in the pan to the fixed point are as follows: Day
1
2
3
4
5
6
Rainfall (cm)
0
0.5
0.1
0
0
0.4
Water added (cm)
1.5
1.7
0.5
1.2
0.7
1.3
If a lake has a 500-hectare surface area, compute the daily lake evaporation using the above data. Assume the pan coefficient is 0.8. Compute the 6-day evaporation and loss of water in kilograms or tons. The daily pan evaporation equals the amount of water required to bring its water level to the fixed point plus the water contributed by rainfall. Day
1
2
3
4
5
6
Evaporation (cm)
1.5
2.2
0.6
1.2
0.7
1.7
If the lake has a 500 hectare surface area, Day
1
2
3
4
5
6
1.5
2.2
0.6
1.2
0.7
1.7
1.2
1.76
0.48
0.96
0.56
1.36
PanEvaporation (cm) Lake (cm)
evaporation
(4) Thirty centimeters of water evaporated from a 200-hectare vertical-walled reservoir during 24 hours. Storm water was added to the reservoir at a constant rate of 30 m3/s during this period. Determine the volume of hectare-centimeters of water released during the day
(through the bottom of the reservoir) if the water level in the reservoir was same at the beginning and end of the day. Water evaporated = 30 cm/day Area = 200 hectares Rate of water added = 30 cumecs Water to be released during the day = (30*60*60*24)– (30*10-2*200*104) = 19920 hectare cm/day (5) Given the following daily meteorological conditions, compare the open water evaporation rates given by (a) the mass-transfer, (b) the energy-balance and (c) the penman combination approach. Air temperature (0C) =22.3; Relative humidity=0.68; wind velocity in m/s=2.16; water surface temperature (0C)=23.7; atmospheric pressure(kPa)=97.3; incoming solar radiation(MJ/m2/day)=16.2; albedo(reflectivity) =0.057; incoming atmospheric long wave radiation flux (MJ/m2/day)=30.6. Given T air = 22.30c Rh = 0.68 Wind velocity, Vw = 2.16 m/sec T water = 23.70c Atm. Pressure = 97.3 KPa Incoming solar radiation = 16.2 MJ/m2/day Albedo = 0.057 Incoming long wave radiation flux = 30.6 MJ/m2/day a) Mass Transfer Method Ea= B ( es – ea) Where B = 0.622*K2*Pa*U2 / (P* Þw*(ln(23/20)2) = 3.4328*10-11 mm/dayPa es = 611 exp
= 2693.546 Pa
ea = es*Rh/100 = 1831.611 Pa Ea = 2.556 mm/day b) Energy Balance Method Here G is taken as zero
Rn = 16.2 (1 - 0.057) + 30.6 – 0.06*(4.903*10-9*2974) Er = (Rn – G)/ λ λ = 2500 – 2.36T = 2.45 KJ/Kg Er = 17.844 mm/day b) Penman Combination approach E = (Δ / Δ + ¥) * Er + (¥/ Δ + ¥) * Er
4098e s 163.789 Pa / oc (237.3 22.3) 2
CpKh p
0.622l v K w
= 64.487Pa/oc
E = 13.525 mm/day (6) Calculate by the energy balance method the evaporation rate from an open water surface, if the net radiation is 200 W/m2 and the air temperature is 250C, assuming no sensible heat or ground heat flux. Latent heat of vaporization at 250 C; lv=2500-2.36*25=2441kJ/kg Water density; Þw=997kg/m3 Er= 200/(2441x103x997) =8.22x10-8 8.22x10-8x1000x 86400mm/day =7.10mm/day (7) Calculate the evaporation rate from an open water surface by the aerodynamic method with air temperature 250C, relative humidity 40 percent, air pressure 101.3 kPa, and wind speed 3m/s, all measured at height 2m above the water surface. Assume a roughness height z0=0.03cm. Vapour transfer coefficient B= 0.622x k2xρaxu2/ (pxρw[ln(z2/z0)]2 ) =0.622x0.4x0.4x1.19x3/ (101.3x103x997{ln[2/(3x10-4) }2 ) = 4.54x 10-11 m/Pas
Ea=B(eas-ea) =4.54x10-11(3167-1267) =8.62x 10-8 m/s = 8.62x10-8x(1000mm/1m)x(86400s/day) =7.45 mm/day (8) Use the combination method to calculate the evaporation rate from an open surface subject to net radiation of 200 W/m2, air temperature 250C, relative humidity 40 percent, and wind speed 3 m/s, all recorded at height 2m, and atmosphere pressure 101.3 kPa. CpKh p
0.622l v K w
=1005*1*101.3*1000/(0.622*2441*1000) =67.1 Pa/0C
4098es = 188.7 Pa/0C 2 (237.3 25)
E = (Δ / Δ + ¥) * Er + (¥/ Δ + ¥) * Ea= 7.2 mm/day
(9) Use the Prestley-Taylor method to calculate the evaporation rate for a water body with net radiation 200 W/m2 and air temperature 250C. Priestley – Taylor method E = α *(Δ / Δ + ¥) * Er = 1.3 *0.738*7.10 = 6.8 mm/day (10)
Compute by priestly-Taylor method the evaporation rat in millimeters per day from a
lake on a winter day when the air temperature is 50C and the net radiation 50 W/m2, and on a summer day when the net radiation is 250W/m2 and the temperature is 300C. Winter λ = 2500- 2.36*5 = 248.2 KJ/Kg
4098e s * 611 = 62.97 Pa/0c (237.3 5) 2
¥ = 67.14 Pa/0c
E = 1.26 *62.97 *1.741 / 62.97+97.14) = 1.06 mm/day Summer Rn = 250 W/m2 λ = 2500- 2.36*30= 2.429MJ/Kg Er = 250 *0.0864 *1000/ (2.429*997) = 8.91 mm/day
4098es * 611 =243.44 Pa/0c 2 (237.3 30)
¥ = 67.14 Pa/0c E = 1.26 * 8.91 *243.44/( 243.44+67.14) = 8.79 mm/day
(11)
For Cairo, Egypt, in July, average net radiation is radiation is 185 W/m2, air temperature
28.50C, relative humidity 55 percent, and wind speed 2.7 m/s at height 2m. Calculate the open water evaporation rate in millimeters per day using the energy method(Er), the aerodynamic method (Ea), the combination method, and the Preistly-Taylor method. Assume standard atmospheric pressure (101.3kPa) and z0=0.03 cm. λ = (2500- 2.36*28.5)*10-3 = 2.432 MJ / Kg Energy Method Er = Rn / (λ * Þw ) = 6.592 mm/day Aerodynamic Method. B = 0.622*K2*Pa*U2 / (P* Þw*(ln(Z2/Z1)2) = 3.703*10-3 mm/day/Pa es = 611 exp
= 3892.65 Pa
ea = 0.55 *3892.62 = 2140.95 Pa Ea = B(es - ea) = 6.49 mm/day Combination Method
4098es * 611 = 225.79 Pa/0c 2 (237.3 30)
¥ = 67.14 Pa/0c E = (Δ / Δ + ¥) * Er + (¥/ Δ + ¥) * Ea = 6.804 mm/day
Priestley – Taylor Method α = 1.26 E = 1.26 *225.79 *6.592 / 225.79+67.14) = 6.402 mm/day (12)
For Cairo, in January, the average weather conditions are: net radiation 40 W/m2
temperatures 140C, relative humidity 65 percent, and wind speed 2.0 m/s measured at 2m. Calculate the open water evaporation rate by energy method(Er), the aerodynamic method (Ea), the combination method and the priestly Taylor method. Assume standard atmospheric pressure Z0. λ = 2.500- (2.36*14)*10-3 = 2.467 MJ / Kg Energy Method Er = Rn / (λ * Þw ) = 1.405 mm/day Aerodynamic Method. B = 0.622*K2*Pa*U2 / (P* Þw*(ln(Z2/Z1)2) = 2.734*10-3 mm/day/Pa es = 611 exp
= 1599.128 Pa
ea = 0.65 *1599.128 = 1039.43 Pa Ea = B(es - ea) = 1.53 mm/day Combination Method
4098es * 611 = 103.76 Pa/0c 2 (237.3 30)
¥ = 66.85 Pa/0c E = (Δ / Δ + ¥) * Er + (¥/ Δ + ¥) * Ea = 1.454 mm/day Priestley – Taylor Method α = 1.26 E = 1.26 *103.76 *1.405 / (103.76+66.85) = 1.676 mm/day
