sm_ch11

# Water Resources Engineering

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11 – 1 Chapter 11 11.3.1. T = 20 o C, 25 o C, 30 o C. From Table 11.3.1, p = 0.33, c = 1.0, K c = 0.8, C cu = 0.5. Using Blaney-Criddle formula, ET o = c[p(0.46T + 8)] i) T = 20 o C: ET o = 1.0[0.33(0.46(20) + 8)] = 5.68 mm/day E T c = K c ET o = 0.8(5.68) = 4.54 mm/day Volume of water consumptively used per acre land = 4.54 ac-mm/day = 0.015 ac-ft/day Volume of irrigation water required = (volume of water consumptively used)/C cu = 0.015/0.5 = 0.030 ac-ft/day ii) T = 25 o C: ET o = 1.0[0.33(0.46(25) + 8)] = 6.44 mm/day E T c = 0.8(6.44) = 5.15 mm/day Volume of water consumptively used per acre land = 5.15 ac-mm/day = 0.017 ac-ft/day Volume of irrigation water required = (volume of water consumptively used)/C cu = 0.017/0.5 = 0.034 ac-ft/day iii) T = 30 o C: ET o = 1.0[0.33(0.46(30) + 8)] = 7.19 mm/day E T c = 0.8(7.19) = 5.76 mm/day Volume of water consumptively used per acre land = 5.76 ac-mm/day = 0.019 ac-ft/day Volume of irrigation water required = (volume of water consumptively used)/C cu = 0.019/0.5 = 0.038 ac-ft/day Using the volume of irrigation water required when T = 20 o C as a reference value, the percent increase at the remaining temperature values may be determined for comparison purposes, as given in the following Table. T ( o C) Irrigation water required (ac-ft/day) % increase Remark 20 0.030 -- 25 0.034 13.33 30 0.038 26.67 Linear increase (note the linear equation used to determine ET o ) 11.5.1. Equation (11.5.6) was given as d = 0.00006362 P -0.267 I 1.61 ( DIF ) -0.123 W 0.0897 The derivative of d with respect to price is

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11 – 2 0897 . 0 123 . 0 61 . 1 267 . 1 ) ( ) 267 . 0 ( 00006362 . 0 d d W DIF I P P d ± ± ± 0897 . 0 123 . 0 61 . 1 267 . 0 ) ( ) 267 . 0 ( 00006362 . 0 W DIF I P P ± ± ± P d 267 . 0 ± Thus, H ± 267 . 0 d d P d d P , that is, the price elasticity of demand is –0.267. 11.5.2. Q = 64.7 + 0.00017(Inc) + 4.76(Ad) + 3.92(Ch) – 0.406(R) + 29.03(Age) – 6.42(P) P Q Q P d d ² 42 . 6 P Q ± d d (form the given equation). Thus, H = -6.42P/Q At P = 1.7241 and Q = 75.2106, H = -6.42(1.7241)/75.2106 = -0.15.
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sm_ch11 - 11 1.3.1 T = 20oC 25oC 30oC From Table 11.3.1 p =...

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