Question

There is one reservoir filled with water and also connected with one pipe of uniform cross-sectional diameter. Total head at section 1 is 27 m. At section 2, potential head is 3 m, gage pressure is 160 kPa, vvelocity is 4.5 m/s. Find the major head loss at section 2 in unit of m. Round to the nearest one decimal place.

Answer 1

Answer:

6.7 m

Explanation:

Total head at section 1 = 27 m

at section 2;

potential head = 3 m

gauge pressure P = 160 kPa = 160000 Pa

pressure head is gotten as

$Ph =\frac{P}{pg}$

where p = density of water = 1000 kg/m^3

g = acceleration due to gravity = 9.81 m/^2

$Ph =\frac{160000}{1000 * 9.81} = 16.309 m$

velocity = 4.5 m/s

velocity head Vh is gotten as

$Vh = \frac{v^{2} }{2g}$

$Vh = \frac{4.5^{2} }{2*9.81} = 1.03 m$

obeying Bernoulli's equation,

The total head in section 1 must be equal to the total head in section 2

The total head in section 2 = (potential head) + (velocity head) + (pressure head) + losses(L)

Equating sections 1 and 2, we have

27 = 3 + 1.03 + 16.309 + L

27 = 20.339 + L

L = 27 - 20.339

L = 6.661 ≅ 6.7 m