A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

(a). 42.1 ft/s, (b). 3366.66 ft^3/s, (c). 0.235, (d). 18.2 ft, (e). 3.8 ft.

Explanation:

The following parameters are given in the question above and they are;

Induced hydraulic jump, j = 80 ft wide channel, and the water depths on either side of the jump are 1 ft and 10 ft. Let k1 and k2 represent each side of the jump respectively.

(a). The velocity of the faster moving flow can be calculated using the formula below;

k1/k2 = 1/2 [ √ (1 + 8g1^2) - 1 ].

Substituting the values into the equation above a s solving it, we have;

g1 = 7.416.

Hence, g1 = V1/ √(L × k1).

Therefore, making V1 the subject of the formula, we have;

V1 = 7.416× √ ( 32.2 × 1).

V1 = 42.1 ft/s.

(b). R = V1 × j × k1.

R = 42.1 × 80 × 1.

R = 3366.66 ft^3/s.

(c). Recall that R = V2 × A.

Where A = 80 × 10.

Therefore, V2 = 3366.66/ 80 × 10.

V2 = 4.21 ft/s.

Hence,

g2 = V1/ √(L × k2).

g2 = 4.21/ √ (32.2 × 10).

g2 = 0.235.

(d). (k2 - k1)^3/ 4 × k1k2.

= (10 - 1)^3/ 4 × 1 × 10.

= 18.2 ft.

(e).The critical depth;

[ (3366.66/80)^2 / 32.2]^ 1/3.

The The critical depth = 3.80 ft.

oil flow is a highly convective flow compared to water flow at normal temperature and pressure.