A reducing elbow in a horizontal pipe is used to deflect water flow by an angle θ = 45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross- sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum flux correction factor to be 1.03 at both the inlet and outlet.
Answers
The anchoring force needed to hold the elbow in place is 839.5 N.
How to find anchoring force?
To determine the anchoring force needed to hold the elbow in place, use the following steps:
Apply the conservation of momentum equation to the elbow:
∑F = ˙m(V₂ - V₁)
where:
˙m = mass flow rate
V₁ and V₂ = velocities at the inlet and outlet of the elbow, respectively
F = anchoring force
The mass flow rate is given by:
˙m = ρAV
where:
ρ = density of water (1000 kg/m³)
A = cross-sectional area of the elbow
V = velocity
The velocities at the inlet and outlet of the elbow can be calculated using the following equations:
V₁ = A₁V₁
V₂ = A₂V₂
where:
A₁ and A₂ = cross-sectional areas at the inlet and outlet of the elbow, respectively
Calculate the momentum flux correction factor at the inlet and outlet of the elbow:
β₁ = 1.03
β₂ = 1.03
Substitute all of the above equations into the conservation of momentum equation:
F = ˙m(V₂ - V₁)
F = ρA₁V₁²β₁ - ρA₂V₂²β₂
Calculate the velocity at the inlet of the elbow:
V₁ = A₂V₂/A₁
V₁ = (25 cm²/150 cm²)(V₂)
V₁ = 1/6 V₂
Substitute the above equation into the conservation of momentum equation:
F = ρA₁V₁²β₁ - ρA₂V₂²β₂
F = ρA₁[(1/6 V₂)²](1.03) - ρA₂V₂²(1.03)
F = ρV₂²(1.03)(1/36 A₁ - A₂)
Calculate the anchoring force:
F = (1000 kg/m³)(V₂²)(1.03)(1/36 × 150 cm² - 25 cm²)
F = 839.5 N
Therefore, the anchoring force needed to hold the elbow in place is 839.5 N.
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The anchoring force needed to hold the elbow in place is 932 N.
The anchoring force needed to hold the elbow in place is the sum of the following forces:
The force due to the change in momentum of the water as it flows through the elbow.
The force due to the weight of the elbow and the water in it.
The force due to the buoyancy of the water in the elbow.
The force due to the change in momentum of the water can be calculated using the momentum equation:
F = mΔv
where:
F is the force
m is the mass of the fluid
Δv is the change in velocity of the fluid
In this case, the mass of the fluid is the mass of the water that flows through the elbow per second. This can be calculated using the mass flow rate equation:
m = ρAv
where:
ρ is the density of the fluid
A is the cross-sectional area of the pipe
v is the velocity of the fluid
The velocity of the fluid at the inlet can be calculated using the Bernoulli equation:
P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2
where:
P1 is the pressure at the inlet
v1 is the velocity at the inlet
P2 is the pressure at the outlet
v2 is the velocity at the outlet
In this case, the pressure at the outlet is atmospheric pressure. The velocity at the outlet can be calculated using the continuity equation:
A1v1 = A2v2
where:
A1 is the cross-sectional area at the inlet
A2 is the cross-sectional area at the outlet
The force due to the weight of the elbow and the water in it is simply the weight of the elbow and the water in it. The weight can be calculated using the following equation:
W = mg
where:
W is the weight
m is the mass
g is the acceleration due to gravity
The force due to the buoyancy of the water in the elbow is equal to the weight of the water displaced by the elbow. The weight of the water displaced by the elbow can be calculated using the following equation:
B = ρVg
where:
B is the buoyancy
ρ is the density of the fluid
V is the volume of the fluid displaced
g is the acceleration due to gravity
The volume of the fluid displaced by the elbow is equal to the volume of the elbow.
Now that we have all of the forces, we can calculate the anchoring force needed to hold the elbow in place. The anchoring force is equal to the sum of the forces in the negative x-direction. The negative x-direction is the direction in which the water is flowing.
F_anchor = F_momentum + F_weight - F_buoyancy
where:
F_anchor is the anchoring force
F_momentum is the force due to the change in momentum of the water
F_weight is the force due to the weight of the elbow and the water in it
F_buoyancy is the force due to the buoyancy of the water in the elbow
Plugging in the values for each force, we get:
F_anchor = 1030 N - 490 N + 392 N = 932 N
Therefore, the anchoring force needed to hold the elbow in place is 932 N.
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Answer:
3 min 55 sec is the solidification time if the cylinder height is doubled
7min 40 sec if the diameter is doubled
Explanation:
see the attachment
Answers
Answer:
The wavelength of the wave is 1 m
Explanation:
Given;
mass of the string, m = 20 g = 0.02 kg
length of the string, L = 3.2 m
tension on the string, T = 2.5 N
the frequency of the wave, f = 20 Hz
The velocity of the wave is given by;
where;
μ is mass per unit length = 0.02 kg / 3.2 m
μ = 6.25 x 10⁻³ kg/m
The wavelength of the wave is given by;
λ = v / f
λ = (20 m/s )/ (20 Hz)
λ = 1 m
Therefore, the wavelength of the wave is 1 m
Answers
Answers
Answer:
360 Nm
Explanation:
Torque: This is the force that tend to cause a body to rotate or twist. The S.I unit of torque is Newton- meter (Nm).
From the question,
The expression of torque is given as
τ = F×d......................... Equation 1
Where, τ = Torque, F = force, d = distance of the bar perpendicular to the force.
Given: F = 40 N, d = 9 m
Substitute into equation 1
τ = 40(9)
τ = 360 Nm.
Answer:
360Nm
Explanation:
Torque is defined as the rotational effect of a force. The magnitude of a torque τ, is given by;
τ = r F sin θ
Where;
r = distance from the pivot point to the point where the force is applied
F = magnitude of the force applied
θ = the angle between the force and the vector directed from the point of application to the pivot point.
From the question;
r = 9m
F = 40N
θ = 90° (since the force is applied perpendicular to the end of the bar)
Substitute these values into equation (i) as follows;
τ = 9 x 40 sin 90°
τ = 360Nm
Therefore the torque is 360Nm
Answers
Answer:
L > 0.08944 m or L > 8.9 cm
Explanation:
Given:
- Flux intercepted by antenna Ф = 0.04 N.m^2 / C
- The uniform electric field E = 5.0 N/C
Find:
- What is the minimum side length of the antenna L ?
Solution:
- We can apply Gauss Law on the antenna surface as follows:
Ф =
- Since electric field is constant we can pull it out of integral. The surface at hand is a square. Hence,
Ф = E.(L)^2
L = sqrt (Ф / E)
L > sqrt (0.04 / 5.0)
L > 0.08944 m
Final answer:
The area of a square antenna needed to intercept a flux of 0.040 N⋅m2/C in a uniform electric field of magnitude 5.0 N/C is 0.008 m². Consequently, each side of the antenna must be about 0.089 meters (or 8.9 cm) long.
Explanation:
The question pertains to the relationship between electric field and flux. The electric flux through an area is defined as the electric field multiplied by the area through which it passes, oriented perpendicularly to the field.
We are given that the electric field (E) is 5.0 N/C and the flux Φ must be 0.040 N⋅m2/C.
Hence, to intercept this amount of flux, the antenna must have an area (A) such that A = Φ / E.
That is, A = 0.040 N⋅m2/C / 5.0 N/C = 0.008 m².
Since the antenna is square, each side will have a length of √(0.008) ≈ 0.089 meters (or 8.9 cm).
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Answers
Answer:
14.7 psi is equal to 19051.2 pounds per square yard.
Explanation:
Dimensionally speaking, a square yard equals 1296 square inches. Therefore, we need to multiply the atmospheric pressure by 1296 to obtain its equivalent in pounds per square yard. That is:
14.7 psi is equal to 19051.2 pounds per square yard.