Answer:
(A) Because the angle of twist of a material is often used to predict its shear toughness
Explanation:
In engineering, torsion is the solicitation that occurs when a moment is applied on the longitudinal axis of a construction element or mechanical prism, such as axes or, in general, elements where one dimension predominates over the other two, although it is possible to find it in diverse situations.
The torsion is characterized geometrically because any curve parallel to the axis of the piece is no longer contained in the plane initially formed by the two curves. Instead, a curve parallel to the axis is twisted around it.
The general study of torsion is complicated because under that type of solicitation the cross section of a piece in general is characterized by two phenomena:
1- Tangential tensions appear parallel to the cross section.
2- When the previous tensions are not properly distributed, which always happens unless the section has circular symmetry, sectional warps appear that make the deformed cross sections not flat.
B. The average velocity stays the same with distance from the entrance.
C. The maximum velocity increases with distance from the entrance.
D. The maximum velocity decreases with distance from the entrance.
E. B and C
F. B and D
Answer:
D. The maximum velocity decreases with distance from the entrance.
Explanation:
This is because over time, the pressure with with the incompressible liquid enters decreases with distance from the entrance
Answer:
C. The maximum velocity increases with distance from the entrance
Explanation:
As the fluid particles moves into the pipe, the layer of fluid in contact with the surface of the pipe come to a complete stop. This layer also causes the fluid
particles in the adjacent layers to gradually slow down as a result of friction between fluid molecules, leaving the fluid at the center of the pipe with the maximum velocity.
Since the fluid is incompressible, to make up for this velocity reduction, the velocity of the fluid at the mid-
section of the pipe has to increase to keep the mass flow rate through the
pipe constant. As a result, a velocity gradient develops along the pipe and the maximum velocity which is at the center of the pipe increases with distance from entrance.
Answer:
101.42235 kPa
Explanation:
The unit inHg means "inches of mercury", Its a pressure unit commonly used by the US aviators.
The conversion value to KPa (kilopascal) is
1 inHg= 3.386389 kPa
So now we only have to multiply:
29.95 inHg * 3.386389 kPa/in Hg =101.42235 kPa
Have a nice day and Good Luck!
Answer:
Reaction bonded Silicon carbide: 2500-3500 HV
Tungsten carbide: 1800-2500 HV.
316 Stainless Steel: 152 HV
Mild steel: 130 HV
Explanation:
In order to list those seal face materials by hardness, we look up what are the values of hardness for each material in a hardness scale.
We are going to use Vickers scale, an indentation method of measuring hardness, it measures the deformation left in a sample by a constant compression load from an indenter (a diamond pyramid) with an adequate (to the material) force, as the result is independent from the test force.
1. Reaction bonded Silicon carbide: 2500-3500 HV
2. Tungsten carbide: 1800-2500 HV
3. 316 Stainless Steel: 152 HV
4. Mild steel: 130 HV
Answer: hello your question is incomplete below is the missing part
question :Determine the temperature of the cooled water exiting the cooling tower
answer : T = 43.477° C
Explanation:
Temp of water at exit = 45°C
mass flow rate of cooling tower = 15,000 kg/s
Temp of makeup water = 20°C
Assuming an atmospheric pressure of = 101.3 kPa
Determine temperature of the cooled water exiting the cooling tower
Water entering cooling tower at 45°C
Given that Latent heat of water at 45°C = 43.13 KJ/mol
Cp(wet air) = 1.005+ 1.884(y1)
where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273
Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C
First step : calculate the value of Q
Q = m*Cp*(ΔT) + W(latent heat)
Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)
Q = 95935.8547 KJ/s
Given that mass rate of water = 15000 kg/s
Hence the temperature of the cooled water can be calculated using the equation below
Q = m*Cp*∆T
Cp(water) = 4.2 KJ/Kg°C
95935.8547 = (15000)*(4.2)*(45 - T)
( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C
Answer:
Ig =7.2 +j9.599
Explanation: Check the attachment
Answer:
When a pilot pushes the top of the right pedal, it activates the brakes on the right main wheel/wheels, and when the pilot pushes the top of the left rudder pedal, it activates the brake on the left main wheel/wheels. The brakes work in a rather simple way: they convert the kinetic energy of motion into heat energy.
Explanation:
When a pilot pushes the top of the right pedal, it activates the brakes on the right main wheel/wheels, and when the pilot pushes the top of the left rudder pedal, it activates the brake on the left main wheel/wheels. The brakes work in a rather simple way: they convert the kinetic energy of motion into heat energy.