Answer:
(a) The gauge pressure is 4,885.3 Pa
(b) The anchoring force needed to hold the elbow in place is approximately 296.5 N
The direction of the anchoring force is approximately ≈ 134.8
Explanation:
The given parameters in the fluid dynamics question are;
Pipe elbow angle, θ = 90°
The mass flowrate of the water,
= 40 kg/s
The diameter of the elbow, D = 10 cm
The pressure at the point of discharge of the fluid = Atmospheric pressure
The elevation between the inlet and exit of the elbow, z = 50 cm
The weight of the elbow and the water = Negligible
(a) The velocity of the fluid, v₁ = v₂ = v is given as follows;
ρ·v·A =
v =
/(ρ·A)
Where;
ρ = The density of the fluid (water) = 997 kg/m³
A = The cross-sectional area of the of the elbow = π·D²/4 = π×0.1²/4 ≈ 0.00785398163
A = 0.00785398163 m²
v = 40 kg/s/(0.00785398163 m² × 997 kg/m³) ≈ 5.1083 m/s
Bernoulli's equation for the flow of fluid is presented as follows;

v₁ = v₂ for the elbow of uniform cross section
P₁ - P₂ = ρ·g·(z₂ - z₁)
P₁ - P₂ = The gauge pressure = 
z₂ - z₁ = z = 50 cm = 0.5 m
∴
= 997 kg/m³ × 9.8 m/s² × 0.5 m = 4,885.3 Pa
The gauge pressure,
= 4,885.3 Pa
(b) The forces acting on the elbow are;
+
·A = -β·
·v
= β·
·v
∴
= -β·
·v -
·A
= -1.03 × 40 kg/s × 5.1083 m/s - 4,885.3 Pa × 0.00785398163 m² ≈ -208.83 N
= 1.03 × 40 kg/s × 5.1083 m/s = 210.46196 N
The resultant force,
, is given as follows;
= √(
² +
²)
∴
= √((-208.83)² + (210.46196)²) ≈ 296.486433934
Therefore;
The anchoring force needed to hold the elbow in place,
≈ 296.5 N
The direction of the anchoring force, θ

∴ θ = arctan(210.46196/(-208.83)) = -45..2230044°
θ = 180° + -45.2230044° = 134.7769956°
∴ The direction of the anchoring force is approximately, θ ≈ 134.8°