University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

80

Pages:

5
Author:

Jabin F.

At the top, the title "Transition of Water in a Channel" is written. Below it, the following assumptions are stated: -> consider a horizontal frictionless channel of width B1 and discharge per Unit width q1 at Section 1 y1 Section 2 q2, B2 Since there is No energy loss between the Two Section and bed elevation is same, therefore specific energy is same. Below this, calculations are shown: change in water surface = 1.6 - 0.1 - 1.48 = 0.02 m elevation Now Δz' = 0.5 m Δz' = E1 - Ec Δzmax = E1 - 1.256 Δzmax = 1.7146 - 1.256 Δzmax = 0.458 m Δz' > Δzmax (checked cond^n) Flow at up/st Not possible with given specific energy. ∴ specific energy at Upstream ↑ so E1' ↑ Δz' = E1' - Ec 0.5 = E1' - 1.256 E1' = 1.756 m E1' = y1' + (v1'^2 / 2g) E1' = y1' + (q1^2 / (y1'^2 * 2g)) 1.756 = y1' + (2.4)^2 / ((y1')^2 * (2 * 9.81)) 1.756 = y1' + 5.76 / (19.62 * (y1')^2) 1.756 = y1' + 0.29368 / (y1')^2 1.756 * (y1')^2 = (y1')^3 + 0.29368 (y1')^3 - 1.756 * (y1')^2 + 0.29368 = 0 y1' = 1.6479, 0.4791, -0.371 (The last two are crossed out) y1' = 1.6479 change in water level = y1' - y1 = 1.6479 - 1.6 = 0.0479 m MCQ Q.16 A Rectangular Q = 16 m^3/s y = 1.60 m n = 0.015 Q.16 find So. Q = (A/n) * R^(2/3) * So^(1/2) 16 = (6 * 1.60 / 0.015) * ((6 * 1.60) / (6 + 2 * 1.60))^(2/3) * (So)^(1/2) 16 = (9.6 / 0.015) * (9.6 / 9.2)^(2/3) * (So)^(1/2) 16 = 640 * (1.0435)^(2/3) * (So)^(1/2) 16 = 640 * 1.0288 * (So)^(1/2) 16 = 658.432 * (So)^(1/2) (So)^(1/2) = 16 / 658.432 = 0.024299 So = (0.024299)^2 = 0.00059324