Express a pressure of 50Kpa in
a) mm of mercury
b) m of water
c)m of acetylene tetra bromide, S=2.94
We can use the equation P = hρg to calculate the pressure at each point, where P is the pressure, h is the height of the liquid column, ρ is the density of the liquid, and g is the acceleration due to gravity (9.81 m/s^2). Since all the fluids are open to the atmosphere, we can assume that the pressure at the top of each column is atmospheric pressure (i.e., 0 Pascals). The pressure difference between points A and B is equal to the difference in pressure due to the height difference between the two points in the fluid.
First, we need to convert the specific gravity of the liquids into their densities in kg/m^3:
Density of S1 = 1000 kg/m^3
Density of S2 = 950 kg/m^3
Density of S3 = 1000 kg/m^3
Next, we calculate the pressure at each point:
At point A:
PA = (h1 + h2)ρ1g
= [(280 mm) + (280 mm)](1000 kg/m^3)(9.81 m/s^2)
= 5,496.48 Pa
At point B:
PB = h3ρ3g
= (1000 mm)(1000 kg/m^3)(9.81 m/s^2)
= 9,810 Pa
The pressure difference between points A and B is then:
PA - PB = -4313.52 Pa
To convert this pressure difference to mm of water, we can use the equation:
1 Pa = 0.10197 mmH2O
Substituting, we get:
PA - PB = -4313.52 Pa * (0.10197 mmH2O/Pa) = -440.81 mmH2O
Therefore, the pressure difference between points A and B is -440.81 mm of water, which means that B is at a higher pressure than A.
The velocity of a particle which moves along a straight line in 1D is given by;
𝑣(𝑡) = 2 − 4𝑡 + 5𝑡
3
⁄2
Where t is in seconds and v is in meters per second. Evaluate the position s, velocity v,
and acceleration 𝑎 when𝑡 = 3𝑠. The particle is at the position 𝑠0 = 3𝑚 when𝑡 = 0.
a) Converting 50Kpa to mmHg:
1 atm = 101,325 Pa = 760 mmHg
50 Kpa = (50/101,325) * 760 mmHg = 0.376 mmHg (rounded to three decimal places)
Therefore, 50Kpa is equivalent to 0.376 mmHg.
b) Converting 50Kpa to meters of water:
1 atm = 101,325 Pa = 10.33 m of water
50 Kpa = (50/101,325) * 10.33 m = 0.00505 m (rounded to five decimal places)
Therefore, 50Kpa is equivalent to 0.00505 m of water.
c) Converting 50Kpa to meters of acetylene tetra bromide:
Using the formula P = h * S, where P is the pressure, h is the height of the liquid column, and S is the specific gravity of the liquid.
Rearranging the formula to solve for h:
h = P/S = 50/2.94 = 17.007 meters (rounded to three decimal places)
Therefore, 50Kpa is equivalent to 17.007 m of acetylene tetra bromide.
Determine the heights of column of
a)water;
b)kerosene, S=0.83; and
c)acetylene tetra bromide, S=2.94 equivalent to 300mmHg
We can use the formula P = hρg, where P is the pressure, h is the height of the liquid column, ρ is the density of the liquid, and g is the acceleration due to gravity (9.81 m/s^2).
a) For water:
P = hρg
h = P/(ρg) = (300 mmHg)/(1000 kg/m^3 * 9.81 m/s^2) = 0.0305 m = 30.5 mm (rounded to one decimal place)
Therefore, the height of a column of water equivalent to 300 mmHg is 30.5 mm.
b) For kerosene:
P = hρg
h = P/(ρg) = (300 mmHg)/(830 kg/m^3 * 9.81 m/s^2) = 0.0375 m = 37.5 mm (rounded to one decimal place)
Therefore, the height of a column of kerosene (with a specific gravity of 0.83) equivalent to 300 mmHg is 37.5 mm.
c) For acetylene tetra bromide:
P = hρg
h = P/(ρg) = (300 mmHg)/(2940 kg/m^3 * 9.81 m/s^2) = 0.0102 m = 10.2 mm (rounded to one decimal place)
Therefore, the height of a column of acetylene tetra bromide (with a specific gravity of 2.94) equivalent to 300 mmHg is 10.2 mm.
Question 3: discuss the other methods of
pressure measurement excluding manometer
method?
There are several other methods of pressure measurement besides the manometer method, including:
1. Bourdon tubes: These are mechanical devices that use a flattened, curved tube that is closed at one end and open at the other. When pressure is applied to the inside of the tube, it straightens out, and the movement is transmitted to a needle that indicates the pressure on a dial.
2. Diaphragm pressure sensors: These sensors use a thin metal or plastic diaphragm that flexes when pressure is applied to it. The movement of the diaphragm is measured by a strain gauge or other device, and the pressure can be determined from this measurement.
3. Capacitive pressure sensors: These sensors use the change in capacitance of a capacitor as a measure of pressure. A diaphragm or other pressure-sensitive element serves as one plate of the capacitor, while a fixed plate serves as the other. As pressure changes, the distance between the plates changes, which changes the capacitance.
4. Piezoelectric pressure sensors: These sensors use a material that generates a voltage in response to pressure. When pressure is applied to the material, it generates a charge, which can be measured and used to determine the pressure.
5. Thermal anemometers: These devices use a heated wire or other element to measure the flow of a gas. The change in temperature of the element is proportional to the flow rate, and this can be used to determine the pressure.
6. Ultrasonic flow meters: These devices use sound waves to measure flow rates and can indirectly measure pressure. By measuring the difference in time it takes for sound waves to travel upstream and downstream through a fluid, the velocity and pressure of the fluid can be determined.
These are just a few examples of the many methods available for pressure measurement beyond the manometer method. The choice of method will depend on factors such as the required accuracy, the type of pressure being measured, the environment in which the measurement will take place, and the cost and complexity of the measurement system.
4. In fig, S1=1.0, S2=0.95, S3=1.0, h1=h2=280mm and h3=1m.
Compute PA-PB
in mm of water.
Hydrostatic pressure on plane and
curved surfaces
Hydrostatic pressure is the pressure exerted by a fluid at rest on a surface that is submerged or partially submerged in that fluid. The hydrostatic pressure on a plane surface depends on the depth of the surface below the fluid level and the density of the fluid. On the other hand, the hydrostatic pressure on a curved surface depends on the shape of the surface, the depth of the surface below the fluid level, and the density of the fluid.
Hydrostatic Pressure on a Plane Surface:
Consider a plane surface of area A that is submerged in a fluid at depth h below the fluid level. The fluid exerts a force on the surface equal to the weight of the fluid above the surface. The weight of the fluid is equal to the product of its density, the acceleration due to gravity, and the volume of fluid above the surface, which is given by A * h. Hence, the force, F, exerted by the fluid on the surface is given by:
F = ρghA
where ρ is the density of the fluid, g is the acceleration due to gravity, and A is the area of the surface. The pressure, P, exerted by the fluid on the surface is given by the force per unit area, which is:
P = F/A = ρgh
This shows that the hydrostatic pressure on a plane surface submerged in a fluid increases linearly with depth and depends only on the density of the fluid and the depth of the surface below the fluid level.
Hydrostatic Pressure on a Curved Surface:
In the case of a curved surface, the hydrostatic pressure is not constant over the surface, since the shape of the surface affects the distribution of fluid pressure across the surface. The hydrostatic pressure can be calculated by dividing the surface into small horizontal or vertical sections, and calculating the pressure on each section using the equation for hydrostatic pressure on a plane surface. The total hydrostatic pressure on the curved surface is then the sum of the pressure on each section.
The distribution of hydrostatic pressure on a curved surface depends on the shape of the surface. For example, on a horizontal curved surface, the pressure is greatest at the lowest point of the surface and decreases as the surface rises. On a vertical curved surface, the pressure is greatest at the bottom of the surface and decreases as the surface rises.
In engineering, the calculation of hydrostatic pressure on a curved surface is important in the design of structures such as dams, reservoirs, and tanks. The distribution of pressure on a curved surface can affect the stability and safety of the structure, so it is important to calculate and understand this pressure distribution in order to design a safe and efficient structure.
A girl rolls a ball up an incline and allows it to return to her. For the angle 𝜃 and a ball
involved, the acceleration of the ball along the
incline is constant at 0.25g, directed down the
incline. If the ball is released with a speed of
4m/s, determine the distance it moves up the
incline before reversing its direction and the
total time t required for the ball to return to
the child’s hand.
Assuming there is no friction, we can use the conservation of energy to solve this problem. At the highest point, the kinetic energy of the ball is zero, so all the initial energy (in the form of potential energy and kinetic energy) is converted to potential energy. As the ball rolls down the incline, the potential energy is converted back to kinetic energy, but some energy is lost due to the work done by friction, which we are assuming is negligible.
Let h be the height difference between the release point and the point where the ball reverses direction, and L be the length of the incline. Then:
Initial energy = final energy
(1/2)mv^2 = mgh + (1/2)mv^2
where m is the mass of the ball, v is its speed at the release point, and g is the acceleration due to gravity. Solving for h, we get:
h = (v^2)/(2g) = (4 m/s)^2 / (2 * 9.81 m/s^2) = 0.8160 m
Therefore, the ball moves up the incline a distance of 0.8160 m before reversing its direction.
The time it takes for the ball to reach the reversal point can be calculated as:
h = (1/2)at^2
where a is the acceleration of the ball along the incline (0.25g) and t is the time taken. Solving for t, we get:
t = sqrt((2h)/(a))
t = sqrt((2 * 0.8160 m) / (0.25 * 9.81 m/s^2))
t = 1.38 s
The total time taken for the ball to return to the girl's hand is twice this time (since the ball takes the same amount of time to travel up and down the incline), so the total time is:
2t = 2 * 1.38 s = 2.76 s
Therefore, the ball moves up the incline a distance of 0.8160 m before reversing its direction, and the total time taken for the ball to return to the girl's hand is 2.76 s.
The variation in the density of water,, with temperature, T, in the range 20c<=T<=5O c is given in the following table
Density (kg/m3)
998.2
997.1
995.7
994.1
992.2
990.2
988.1
Temperature (c)
20
25
30
35
40
45
50
To estimate the density of water at temperatures between the given values, we can use interpolation. One way to do this is to use linear interpolation, assuming that the density varies linearly with temperature:
Let's interpolate the density at 22°C:
Density at 22°C = 998.2 + [(22-20)/(25-20)] * (997.1-998.2)
= 998.2 + (2/5) * (-1.1)
= 997.28 kg/m^3
Similarly, let's interpolate the density at 42°C:
Density at 42°C = 990.2 + [(42-40)/(45-40)] * (988.1-990.2)
= 990.2 + (2/5) * (-2.1)
= 989.48 kg/m^3
Therefore, the estimated density of water at 22°C is 997.28 kg/m^3, and the estimated density of water at 42°C is 989.48 kg/m^3. We can use this method to interpolate the density at any temperature between 20°C and 50°C, assuming that the density varies linearly with temperature in this range.
Three different types of liquids having volumesV1,V2 andV3 and specific gravities S1,S2 and S3 respectively are mixed. If the bulk of the liquid shrinks by 1% on mixing, express the specific gravity of the mixture in terms of volumes and relative densities.
Let's assume that the original volumes of the liquids are V1, V2, and V3, and their original specific gravities (relative densities) are S1, S2, and S3. After mixing, the total volume of the mixture is:
V = V1 + V2 + V3
Since the bulk of the liquid shrinks by 1%, the new volume of the mixture is 99% of the original volume:
V' = 0.99V
The mass of the mixture remains the same, so:
m = V1S1 + V2S2 + V3S3
Since the density of a substance is equal to its mass divided by its volume, we can write:
ρ = m/V'
Substituting for m and V', we get:
ρ = (V1S1 + V2S2 + V3S3) / (0.99V)
Simplifying, we get:
ρ = (V1S1 + V2S2 + V3S3) / (V - 0.01V)
ρ = (V1S1 + V2S2 + V3S3) / (0.99V)
Multiplying both the numerator and denominator by (1/VS1S2S3), we get:
ρ = [(V1/V)S1 + (V2/V)S2 + (V3/V)S3] / (0.99/S1S2S3)
We can express the specific gravity of the mixture in terms of volumes and relative densities by using the fact that the specific gravity of a substance is equal to its density divided by the density of water (at standard temperature and pressure):
SG = ρ / ρ_water
where SG is the specific gravity, ρ is the density of the mixture, and ρ_water is the density of water.
Therefore, the specific gravity of the mixture in terms of volumes and relative densities is:
SG = [(V1/V)S1 + (V2/V)S2 + (V3/V)S3] / [(0.99/S1S2S3)ρ_water]
Note that the specific gravity of water at standard temperature and pressure is approximately 1.
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The velocity of a particle which moves along a straight line in 1D is given by;
𝑣(𝑡) = 2 − 4𝑡 + 5𝑡3⁄2
Where t is in seconds and v is in meters per second. Evaluate the position s, velocity v,
and acceleration 𝑎 when𝑡 = 3𝑠. The particle is at the position 𝑠0 = 3𝑚 when𝑡 = 0.