Answer:
b). False
Explanation:
A refractory material is a type of material that can withstand high temperatures without loosing its strength. They are used in reactors, furnaces, kilns, etc.
Refractory materials are certain super alloys and ceramics materials.
Properties of refractory materials :
1. Refractory materials have high melting point.
2.They acts barriers between high heat zone and low heat zone.
3. The specific heat of refractory material is very low.
4. Refractories that have high bulk densities are better in quality.
Hence, Refractory materials have a very high melting temperature.
______number can be used to describe the relative growth of the hydraulic boundary layer and the thermal boundary layer. a) Reynolds b) Stanton c) Nusselt d) Prandtl e) Fourier
Answer: d) Prandtl number
Explanation: Prandtl number is basically defined as the ratio between the fluid's viscosity to the thermal conductivity.It doesn't have any sort of dimension. The fluids which are discovered with the small Prandtl numbers are considered as good fluids as they have a smooth rate of flow and as the number increases the fluid are not considered as reliable. Thus,option (d) is the correct option.
Which of the following is/are not a common crystal structure in metals (mark all that apply)? a. Face-centered cubic (FCC) b. Face-centered orthorhombic (FCOR) C. Body-centered cubic (BCC) (HCP)
Answer: b) FCOR( Face-centered orthorhombic)
Explanation: Face centered orthorhombic lattice is the lattice that has eight lattice corner points and they also have each face with center lattice point.This lattice structure is usually not common in metals.Whereas the face centered lattice structure , body structure lattice and hexagonal close packed lattice are common in metal.Thus option (b) is the correct option.
What is a Mollier diagram?
Answer: Mollier diagram is the diagram or graph representing the relation established between enthalpy ,air, moisture and temperature.
Explanation: Mollier graph represents the basic thermodynamic properties along with terms of enthalpy and entropy. The factors like air, moisture , temperature etc. are plotted on the graph which makes it easy to understand. It is used in the field building designers and engineer.They are usually plotted for the gases with purity with general pressure and temperature.
The number-average molecular weight of a poly(styrene-butadiene alternating copolymer is 1,350,000 g/mol; determine the average number of styrene and butadiene repeat units per molecule.
To find the average number of styrene and butadiene units in a poly(styrene-butadiene) copolymer with Mn of 1,350,000 g/mol, add the molecular weights of styrene and butadiene to get the molecular weight of the repeat unit (158.24 g/mol) and divide Mn by this number to get approximately 8530 repeat units, which includes about 4265 of each monomer type.
Explanation:To calculate the average number of styrene and butadiene repeat units per molecule for a poly(styrene-butadiene) copolymer with a number-average molecular weight (Mn) of 1,350,000 g/mol, we need the molecular weights of the monomer units. The molecular weight of styrene (C8H8) is approximately 104.15 g/mol, and the molecular weight of butadiene (C4H6) is approximately 54.09 g/mol. Since the copolymer is alternating, each repeat unit consists of one styrene and one butadiene unit.
To find the total molecular weight of the repeat unit, we add the molecular weights of styrene and butadiene: 104.15 g/mol + 54.09 g/mol = 158.24 g/mol. Dividing the number-average molecular weight of the copolymer by the molecular weight of the repeat unit gives us the average number of repeat units per molecule: 1,350,000 g/mol \/ 158.24 g/mol \approximately 8530 repeat units\.
The average number of styrene units per molecule will be approximately 4265, and the same for butadiene, since there is one of each in every repeat unit in an alternating copolymer structure.
What is the uncertainty in position of an electron of an atom if there is t 2.0 x 10' msec uncertainty in its velocity? Use the reduced Planck's constant and electron mass 9.19 x 103 kg.
Answer:
18931.4
Explanation:
Given : velocity of the electron = 2.0 [tex]\times[/tex]10
mass of the electron = 9.19[tex]\times[/tex] 103
we know that reduced planks constant, h = 6.5821[tex]\times[/tex] [tex]10^{-16}[/tex] eV s
We know from uncertainity principle,
[tex]\Delta \textup{x}.\Delta \textup{v} = \frac{h}{\dot{m}}[/tex]
[tex]\Delta \textup{x} = \frac{h}{\dot{m}\times \Delta \textup{v}}[/tex][tex]\Delta \textup{x} = \frac{6.5821\times 10^{-16}}{9.19\times 103\times 2.0\times 10}[/tex]
[tex]\Delta \textup{x}[/tex] = 18931.4 m
Hence, uncertainty in position of the electron is 18931.4
Which of the following components of a PID controlled accumulates the error over time and responds to system error after the error has been accumulated? a)- Proportional b)- Derivative c)- Integral d)- ON/OFF.
Answer:
D Is the answer babe....
Answer:
d)- ON/OFF.
Explanation:
ON/OFF components of a PID controlled accumulates the error over time and responds to system error after the error has been accumulated.
What is the Principle of Entropy Increase?
Answer and Explanation:
The increase in entropy principle is defined as the process in which the total change in entropy of system with its adiabatic surroundings is always positive or equal to zero. The increase in entropy mostly takes place when a solid becomes liquid because randomness is increases when solid becomes liquid so entropy is also increases.
example of increase of entropy is when solid burns and become ash, ice melting
A 12-ft circular steel rod with a diameter of 1.5-in is in tension due to a pulling force of 70-lb. Calculate the stress in the rod
Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:
[tex] A=\pi*D^2/4 [/tex]
Replacing the diameter the area results:
[tex] A= 17.76 in^2 [/tex]
Therefore the the stress results:
[tex] σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi [/tex]
For high temperature deformation, the bigger the gran sine, the higher the creep rate. a)-True b)- False
Answer:
The given statement is False.
Explanation:
This is because at high temperature the creep rate depends on grain boundary area, increasing with an increase in grain boundary area thus decreasing the grain size. Thus at higher temperatures, grain size has opposite effect , the bigger the grain size the slower the slower is the creep rate, the larger the grain boundary area.
Answer:
a
Explanation:
a
In a steady flow device, the properties of the system remains constant with time. a)True b) False
Answer:
True
Explanation:
By definition of steady flow we have
[tex]\frac{\partial f(x,y,z,t) }{\partial t}=0[/tex]
where f(x,y,z,t) is any property of the system under consideration
=> f(x,y,z,t) = constant
The purifier is a device in a)- The air separation plant b)- steam power plant c)-- fuel cell d)- refrigerator
Answer:
a). Air separation plant
Explanation:
The purifier is a device that removes the pollutants or air contaminants from air to improve the air quality in a given space.
There are basically two techniques or ways by which the air purifier makes the air contaminants free. This two techniques are active and passive.
Active air purifiers : negatively charged ions are released to the atmosphere which sticks the pollutants to the surface.
Passive air purifiers : They use air filters to eliminate pollutants and are more efficient than an active air purifiers.
Answer:
(A) Air separation plant
Explanation:
Purifier is used in air separation plant to purify the air main work of purifier is to remove the pollutants particles from the air and make the air fresh.
On the other hand it has no use team power pant because in steam power plant we don't need to purify the steam and also not used in fuel cell and refrigerator
So purifiers are used in the air separation plant.
What i s the value of a capacitor with 250 V applied and has 500 pC of charge? (a) 200 uF (b) 0.5 pF (c) 500 uF (d) 2 pF
Answer:
(d) 2 pF
Explanation: the charge on capacitor is given by the expression
Q=CV
where Q=charge
C=capacitance
V=voltage across the plate of the capacitor
here we have given Q=500 pF, V=250 volt
using this formula C=[tex]\frac{Q}{V}[/tex]
=500×[tex]10^{-12}[/tex]×[tex]\frac{1}{250}[/tex]
=2×[tex]10^{-12}[/tex]
=2 pF
Describe the steps involved in accomplishing a life-cycle cost analysis (LCCA). I have listed the steps. just describe. a)-Operational Performance. b)-Salvage externalities c)- Value vs. Risk d)- Initial expenditure e)- Maintenance implications
Explanation:
a). Operational Performance
It is defined as the parameter that describes the percentage of accuracy of performing an operation or carrying out an activity.
b). Salvage externalities
Salvage in Life cycle cost analysis is a process of estimating the value of the remaining assets in the organisation,.
c). Value Vs Risk
When we take risk in doing any activity we know the value of accomplising the activity. So value relates directly with risk. When the value of a certain task is high, the risk involve in it is also high.
d). Initial expenditure
Initial expenditure is nothing but the cost involve in starting a particular acitvity or task at the starting phase.
e). Maintenance implications
It lays emphasis in maintaining the cost of every possible parameters that are involve in the activity. It includes labour, machines, positions, energy, facilities, etc.
A disk-shaped part is to be cast out of aluminum. The diameter 500 mm and thickness = 20 mm. If the mold constant = 3.0 s/mm2 in Chvorinov's Rule, how long will it take the casting to solidify, in minutes?
Answer:
t =253.8s
Explanation:
Chvorinov's Rule can be written as:
[tex]t=B(\frac{V}{A} )^{n}[/tex]
where t is the solidification time,
V is the volume of the casting,
A is the surface area of the casting that contacts the mold,
n is a constant
B is the mold constant
The S.I. units of the mold constant B are s/m2.
According to Askeland, the constant n is usually 2.
[tex]V=\frac{\pi D^{2}h }{4} = 3.9*10^6[/tex]mm3
[tex]As=\pi D h+2\frac{\pi }{4} D^{2} =0.424*10^6[/tex] mm2
[tex]V/A=9.198[/tex]mm
[tex]t = 3.0*9.198^2[/tex] =253.8s
Answer:
Chvorinov's Rule with Askeland Method: t = 4.286694102 minutes
Chvorinov's Rule with Degarmo Method:
Minimum time required at constant n = 1.5 : t = 1.408751434 minutesMaximum time required at constant n = 2.0 : t = 4.286694102 minutesExplanation:
Data:
Aluminum disc
Diameter (D) = 500 mm
Thickness = Height (h) = 20 mm
Mold Constant (C) = 3.0 sec / [tex]mm^{2}[/tex]
Required:
Solidification time (t) in minutes = ?
Formula:
The solidification time can be found by using the Chvorinov's Rule:
[tex]t = C (\frac {V}{A})^{n}[/tex]
Where;
t = solidification time
C = mold constant
V = Volume of disc
A = Surface area of disc
n = constant
Note: According to Askeland n = 2.0 and According to Degarmo n varies 1.5 to 2.0 therefore , we will do for both method and by Degarmo method we can predict maximum and minimum solidification time.
Solution:
First, we will find the volume of the disc
disc = cylinder
therefore, Volume of cylinder is given by:
[tex]V = \frac{\pi }{4} * D^{2} * H[/tex]
Where:
V = Volume of Cylinder
H = Height of disc
D = Diameter of disc
Now, putting dimensional values in above equation
[tex]V = \frac{\pi }{4} * 500^{2} *20[/tex]
V = 3926990.817 [tex]mm^{3}[/tex]
Second, we will find the surface area of the disc
Therefore, surface area of cylinder is given by:
[tex]A = (\pi * D * H) + (2 * \frac{\pi }{4} * D^{2} )[/tex]
Where:
A = Surface area of disc
D = Diameter of disc
H = Height of disc
Now, putting dimensional values in above equation
[tex]A = (\pi * 500 * 20) + (2 * \frac{\pi }{4} * 500^{2} )[/tex]
A = 424115.0082 [tex]mm^{2}[/tex]
Finally, Moving towards the final solution
Chvorinov's Rule with Askeland Method n = 2:Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]
t = 257.2016461 seconds
Converting to minutes
t = 4.286694102 minutes
Chvorinov's Rule with Degarmo Method n = 1.5 (Minimum Solidification Time)Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{1.5}[/tex]
t = 84.52508604 seconds
Converting to minutes
t = 1.408751434 minutes
Chvorinov's Rule with Degarmo Method n = 2.0 (Maximum Solidification Time)
Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]
t = 257.2016461 seconds
Converting to minutes
t = 4.286694102 minutes
A Carnot machine operates with 25% efficiency, whose heat rejection reservoir temperature is 300K. Determine the temperature at which the machine absorbs heat.
Answer:
The temperature at which observed heat is 400 K
Explanation:
Given data:
rejection reservoir temperature at exit [tex]T_{L}[/tex] is 300 k
the efficiency of a engine is η = 25%
we know that efficiency of Carnot is given as
[tex]\eta = (1-\frac{T_{L}}{T_{H}})*100[/tex]
Putting all value to obtained temperature at which observed heat
[tex]0.25 = (1-\frac{300}{T_{H}})[/tex]
[tex]T_{H}[/tex] = 400 K
Water flows at the rate of 200 I/s upwards through a tapered vertical pipe. The diameter at Marks(3) CLO5) the bottom is 240 mm and at the top 200 mm and the length is 5m. The pressure at the bottom is 8 bar, and the pressure at the topside is 7.3 bar. Determine the head loss through the pipe. Express it as a function of exit velocity head.
Answer: 5.35m
Explanation:
By using energy equation:
[tex]\frac{P_1}{\gamma}+z_1+\frac{v_1^{2} }{2g} =\frac{P_2}{\gamma}+z_2+\frac{v_2^{2} }{2g}+h_{L}[/tex]
[tex]\gamma=specific weight[/tex]
[tex]v_{1} =\frac{Q_1}{A_1} =\frac{0.2}{\frac{\pi }{4} \times 0.24^2} =4.42 m/s\\v_{2} =\frac{Q_2}{A_2} =\frac{0.2}{\frac{\pi }{4} \times 0.2^2} =6.37 m/s[/tex]
[tex]h_{L}=\frac{P_1-P_2}{\gamma}+z_1-z_2+\frac{v_1^{2}-v_2^{2} }{2g}[/tex]
[tex]h_{L}=\frac{(8-7.3)\times 100 }{9.81} +0+5+\frac{4.42^2-6.37^2}{2\times 9.81}[/tex]
[tex]h_L=7.135+3.927\\h_L=11.062m[/tex]
exit velocity head = [tex]\frac{v_2^{2} }{2g}[/tex]=2.068m
head loss as a function of exit velocity head is=[tex]\frac{11.062}{2.068}[/tex]
[tex]h_L=K\times V_e[/tex]
head loss as a function of exit velocity head =5.35m
Sketch and label a simple reheat cycle along with the appropriate T-s diagram?
Answer:
A reheat cycle is used to increase the overall efficiency of the power plant.
Explanation:
The Reheat cycle is used --
1. to increase the efficiency of the turbine
2. to maintain the quality of the steam
3. to provide higher pressure ratio
The reheat cycle is used to to increase the net work output of the power plant.In reheat cycle, two turbines , one low pressure turbine and one high pressure turbine is used to increase the efficiency. The main purpose of the reheat cycle is to maintain the quality of the steam to 0.85 at the exit of the turbine.
The figure given below shows a reheat rankine cycle.
In the cycle,
Process 1-2 is high pressure turbine
Process 2-3 is the reheater
Process 3-4 is the low pressure turbine
Process 4-5 is condenser
Process 5-6 is pump
Process 6-1 is boiler
A reheat rankine cycle is 30 -40% efficient than a simple rankine cycle.
An actual vapour compression system comprises following process represents a. 1-2 Compression process b. 2-3 Condens 1 (or heat rejection from the condenser) c. 3-4 Irreversible expansion d. 4-1 Evaporation (or) heat addition to the evaporator Sketch the processes on T-S diagram.
Answer:
Explanation:
The deatailed diagram of VCRS is given below such
1-2=Isentropic compression in which temperature increases at constant entropy
2-3=Isobaric heat rejection i.e. heat rejected at constant pressure(condensation)
3-4=Irreversible expansion or throttling in which enthalpy remains constant
4-1=Isobaric heat addition(Evaporation)
Design a solid steel shaft to transmit 14 hp at a speed of 2400rpm f the allowable shearing stress is given as 3.5 ksi.
Answer:
Is required a 0.8 inches diameter steel shaft.
Explanation:
With the power P and the rotating speed n (RPM), we can find the torque applied:
T = P/N
Before calculating the torque, we convert the power and rotating speed units:
[tex]P = 14\ HP * 550\ \frac{\frac{lb.ft}{s}}{HP} *\frac{12\ in}{ft} = 92400\ \frac{lb.in}{s} [/tex]
[tex]n=2400\ RPM .\frac{2\pi/60\frac{rad}{s}}{RPM}= 251\frac{rad}{s}[/tex]
Replacing the values, the torque obtained is:
[tex]T = \frac{92400\ lb.in/s}{251\ rad/s} = 368\ \ lb.in[/tex]
Then the maximum shearing stress will be located at the edge of the shat at the Maximus radius:
[tex]Smax =\ \frac{T.R}{J}[/tex]
Here J is the moment of inertia and R a radius. For a solid shaft, it is calculated by:
[tex]J =\frac{\pi.D^4}{32}[/tex]
Where D is shaft's diameter. Replacing the expression of J in
[tex]Smax =\frac{T.R}{\frac{\pi.D^4}{32}}[/tex]
As the radius is half of the diameter:
[tex]Smax =\frac{T.D}{\frac{2*\pi.D^4}{32\\} } = \frac{16T}{\pi.D^3}[/tex]
For the maximum stress of 3.5 ksi (3500 psi = 3500\ lb/in^2) and the calculated torque:
[tex]Smax = \frac{16.368\ lb.in}{3500\ lb/in^2*\pi.D^3}[/tex]
Solving for D:
[tex]D =\sqrt[3]{16.368\ lb.in / (3500\pi\ lb/in^2)}} = 0.8\ in[/tex]
Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Coercion
Answer: d)Coercion
Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)
What do you think are the advantages and disadvantages of isothermal constant volume high extension cycle? And how efficient do you think it can be?
Answer Explanation :
ADVANTAGES OF ISOTHERMAL PROCESS
EASE OF MAINTENANCE: there is absence of direct contact between the combustible gas mixture and all moving part in this process LESS NOISE : These engines does not have valves so it is very simple in construction and produce less noiseBETTER PERFORMANCE : these engine have better performance than other enginesLOW TEMPERATURE : these engine can work on low temperature which is also an advantageDISADVANTAGES OF ISOTHERMAL PROCESS :
GREATER VOLUME AND GREATER WEIGHT: These engine require large volume and large weightSLOWER START: In this process the engine have very slower start which is a disadvantage HIGHER ECONOMIC COST : In this process we need more money this is also an disadvantage
A refrigeration cycle rejects Qn 500 Btu/s to a hot reservoir at 540 R, while receiving c200 Btu/s at 240°R. This refrigeration cycle a)- is internally reversible b)- is irreversible c)- is impossible d)- cannot be determined
Answer:
(b) Irreversible cycle.
Explanation:
Given;
[tex]T_2=540R ,Q_2= 500 Btu/s ,T_1=240 R ,Q_1= 200 Btu/s [/tex]
To find the validity of cycle
[tex]\oint _R\frac{dQ}{T}\leq0[/tex]
If it is zero then cycle will be reversible cycle and if it is less than zero then cycle will be irreversible cycle.These are possible cycle.
If it is greater than zero ,then cycle will be impossible .
Now find
[tex]\dfrac{Q_1}{T_1}-\dfrac{Q_2}{T_2}=\dfrac{200}{240}-\dfrac{500}{540}[/tex]
[tex]\dfrac{Q_1}{T_1}-\dfrac{Q_2}{T_2}[/tex]= -0.09
It means that this cycle is a irreversible cycle.
Why is a Screw Pump a quiet operating pump?
Answer:
Screw pumps like those that are used in viscous fluids transportation have lubricating properties. The fluid flows in a line axially without any disturbence. Since, the motion of fluid is free from any sort of rotation even at high speeds there is no turbulence and motion of the pump is quite. Thus screw Pumps provides smooth operation with extremely low pulsation, lower noise levels and higher efficiency.
Injector orifice patterms and size will affect propellant mixing and distribution. a)-True b)-False
Answer: True
Explanation: Injector orifice is the factor which describes the size of the opening of the injector .There are different pattern and size of the opening for the injector which affects the mixture of the chemical substance that is used for the production of the energy that is known as propellant.
The pattern and size of the orifice will define the variation in the amount of energy that could be produced.Thus the statement given is true.
What does WCS stand for? A. Western CAD System B. Worldwide Coordinate Sectors C. World Coordinate System D. Wrong CAD Settings
Answer:
The correct answer is C. World Coordinate System
Explanation:
The World Coordinate System has to do with that coordinate system which is fixed in the activities of the CADing. There is a default system in which we can refer to them as soon as we want to manipulate the objects and add new elements.
_____What is matrix in tungsten carbide cutting tool? a)- Chromium b)- Manganese c)- Cobalt d)- Aluminum
Answer:
The correct option is C) Cobalt
Explanation:
The cemented carbides are hard substances which are extensively used as cutting tools materials. In the cemented carbides, substances like tungsten-carbide are used as aggregates.
A tungsten carbide- cobalt cutting tool is a metal matrix composite, in which the tungsten carbide is the aggregate and the cobalt is the matrix.
Therefore, in the tungsten carbide cutting tool, cobalt is used as the matrix.
Burn rate can be affected by: A. Variations in chamber pressure B. Variations in initial grain temperature C. Gas flow velocity D. All of the above
Answer: D) All of the above
Explanation:
Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.
Are spheroidized steels considered as composite? If so, what is the dispersed phase a)- No b)- Yes, Chromium Carbides c)- Yes, Iron Carbides d)- Yes, Intermetallic Compounds
Answer: c)-Yes, spheroidized steel are considered as composite.the dispersed phase is iron carbide.
Explanation: Spheroidized steel are the alloy that have iron as the basic part that have been heat treated to increase their ductility and malleability property .They are considered as composite because they are made up of iron alloys. The heat treatment is usually for the carbon steel and so the dispersed phase that is obtained is iron carbide.
A composite material is a mix of two different materials such as ceramics and metals fused together to the atomic level to form another substance with more improved propertied.a)-True b)-False
Answer:
The correct option is (A) TRUE
Explanation:
A composite material is the material having characteristics that are improved and different from the constituting materials. A composite material is produced by combining two or more materials. The constituting materials have significantly different physical or chemical properties.
Some of the composite materials include-Ceramic matrix composites, Metal matrix composites etc.
A ceramic matrix composite is formed by embedding ceramic fibers in a ceramic matrix.
Also, a tungsten carbide- cobalt is a metal matrix composite, in which the tungsten carbide and cobalt metal matrix are fused together.
Therefore, the given statement: A composite material is a mix of two different materials that are fused together on the atomic level to form a new substance with improved properties is TRUE.
What is flow energy? Do fluids at rest possess any flow energy?
Flow energy is the energy possessed by a fluid due to its motion, which is quantified by kinetic energy density. Based on Bernoulli's equation, fluids at rest do not have flow energy but can have potential energy. The total mechanical energy of a fluid includes potential and kinetic energy components.
Explanation:Flow energy refers to the energy that fluid possesses due to its motion. This form of energy can be described by kinetic energy density, which is part of the total mechanical energy in a flowing fluid. According to Bernoulli's equation, the total mechanical energy of a fluid is composed of potential energy and kinetic energy. The kinetic energy of the fluid relates to its velocity and is expressed as (1/2)ρV^2, where ρ is the density of the fluid and V is the fluid velocity. On the other hand, fluids at rest do not possess flow energy or kinetic energy related to movement, but they can still have potential energy due to their position.
According to conservation of energy principles, the work done by the fluid's pressure changes the fluid's kinetic and potential energies, per unit volume. No additional flow energy is present in a fluid at rest; however, it may still exert pressure at the bottom of a container or well due to its potential energy.