Answer:
[tex]h_{1}[/tex] = 2017.25 kJ/kg
Explanation:
GIVEN DATA:
Exit velocity [tex]v_{2}[/tex] = 250 m/s
outlet enthalpy [tex]h_{2}[/tex]= 1986 kJ/kg
inlet velocity [tex]v_{1}[/tex]= 0
heat transfer Q = 0
from steady flow energy equation(SFEE) between inlet and exit point
[tex]h_{1}+\frac{v_{1}^{2}}{2}=h_{2}+\frac{v_{2}^{2}}{2}+ Q[/tex]
[tex]h_{1}=h_{2}+\frac{v_{2}^{2}}{2}[/tex]
[tex]h_{1}[/tex]=2017.25 kJ/kg
Which of the following will decrease the value of the time constant for an RL circuit? (a) increasing resistance (b) decreasing capacitance (c) decreasing resistance (d) increasing inductance
Answer:
(a) increasing resistance
Explanation:
TIME CONSTANT FOR RL CIRCUIT : In RL circuit resistor and inductor are present only time constant for RL circuit is given by
time constant =[tex]\frac{L}{R}[/tex] From this expression it is clear that is we increase the value of resistance then denominator value increases and so the overall value decreases because we know that in a fraction when denominator value is increase then overall value is decreases so option (a) will be the correct option
A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 90mm OD, a 1.65mm wall thickness, and Poisson's ratio 0.334. The purchase order specifies a minimum yield strength pf 320 MPa. What is the factor of safety if the pressure-release valve is set at 3.5 MPa?
Given:
Outer Diameter, OD = 90 mm
thickness, t = 0.1656 mm
Poisson's ratio, [tex]\mu[/tex] = 0.334
Strength = 320 MPa
Pressure, P =3.5 MPa
Formula Used:
1). [tex]Axial Stress_{max}[/tex] = [tex]P[\frac{r_{o}^{2} + r_{i}^{2}}{r_{o}^{2} - r_{i}^{2}}][/tex]
2). factor of safety, m = [tex]\frac{strength}{stress_{max}}[/tex]
Explanation:
Now, for Inner Diameter of cylinder, ID = OD - 2t
ID = 90 - 2(1.65) = 86.7 mm
Outer radius, [tex]r_{o}[/tex] = 45mm
Inner radius, [tex]r_{i}[/tex] = 43.35 mm
Now by using the given formula (1)
[tex]Axial Stress_{max}[/tex] = [tex]3.5[\frac{45^{2} + 43.35^{2}}{45^{2} - 43.35^{2}}][/tex]
[tex]Axial Stress_{max}[/tex] = [tex]3.5\times 26.78[/tex] =93.74 MPa
Now, Using formula (2)
factor of safety, m = \frac{320}{93.74} = 3.414
What is Differential Analysis in fluid mechanics?
Answer:
Differential analysis is used when it is needed to determine the detail information of the flow i.e. pressure or stress variation along any point.
Explanation:
In fluid mechanics, sometime situation arise in which we need to determine in detail about flow characteristics like stress and pressure variation.
To find these flow characteristics some relationship need to imply either at a point or at very small volume and analysis of flow at very small point is known as differential analysis.
Example: pressure and shear stress variation in a line of the wing of a plane.
A steel rod 6mm in diameter is stretched with a tensile force of 400N. Calculate the tensile stress experienced by the steel rod
Answer:
Tensile stress is 14.15 N/[tex]mm^{2}[/tex]
Explanation:
When any object is subjected to an external force, the body offers a resisting force which is equal and opposite to the external load. This resisting force is called stress. Thus stress is defined as the force acting perpendicular to the given cross sectional area of the object.
Mathematically, stress , σ = [tex]\frac{force }{area}[/tex]
Given : Tensile force, F = 400 N
Diameter of the rod, d = 6 mm
Area of the rod is given by, A = [tex]\frac{\pi }{4}\times d^{2}[/tex]
= [tex]\frac{\pi }{4}\times 6^{2}[/tex]
=28.27 [tex]mm^{2}[/tex]
Therefore, the tensile tress is, σ = [tex]\frac{force }{area}[/tex]
= [tex]\frac{400 }{28.27}[/tex]
= 14.149 N/[tex]mm^{2}[/tex]
[tex]\simeq[/tex] 14.15 N/[tex]mm^{2}[/tex]
Thus, tensile stress experieced by the rod is 14.15 N/[tex]mm^{2}[/tex]
A gas in a piston-cylinder asscmbly undcrgocs a process for which the rclationship bctwcen pressurc and volumc is pV^2=constant The initial pressurc is 1 bar, the initial volume is 0.4 m^3, and the final pressure is 9 bar. Determine the work for the process, in kJ constant.
Given:
[tex]pV^{2}[/tex] = constant (1)
⇒ [tex]p_{1}V_{1}^{2} = p_{2}V_{2}^{2}[/tex] (2)
[tex]p_{1} = 1 bar = 1\times 10^{5}[/tex]
[tex]p_{2} = 9 bar = 9\times 10^{5}[/tex]
[tex]V_{1} = 0.4 m^{3}[/tex]
[tex]V_{2} = ? m^{3}[/tex]
Solution:
Here, from eqn (1), the polytropic constant is '2' ( Since, here [tex]pV^{n}[/tex] = [tex]pV^{2}[/tex] )
Now, using eqn (2), we get
[tex]V_{2}^{2} =\frac{p_{2}}{p_{1}}\times V_{1}^{2}[/tex]
putting the values in above eqn, we get-
[tex]V_{2}^{2} =\frac{9}{1}\times 0.4^{2}[/tex]
[tex]V_{2} = 1.2 m^{3}[/tex]
Now, work for the process is given by:
[tex]W = \frac{p_{2}V_{2} - p_{1}V_{1}}{1 - n}[/tex] (3)
where,
n = potropic constant = 2
Using Eqn (3), we get:
[tex]W = \frac{9\times 10^{5}\times 1.2 - 1\times 10^{5}\times 0.4}{1 - 2}[/tex]
W = - 240 kJ
What do you understand by equilibrium of a system? What is meant by thermodynamic equilibrium? Explain in detail.
Answer:
Equilibrium of a system is defined as when the concentration of reactants and products of a chemical reaction are in equilibrium and their ratio does not vary. In another words we can say that, when the forward reaction rate is similar to the reverse reaction rate.
Thermodynamics equilibrium is defined as in an isolated system when there is no change in energy and entropy. It is determined by the its intensive properties like volume and pressure.
Which of the following describes the action of a capacitor? (a) creates a de resistance (b) converts ac into de (c) opposes changes in the flow of current (d) stores electrical energy
Answer:
From the multiple choices provided for the action of capacitor, option
(d) stores electrical energy
is correct
Explanation:
A capacitor is basically a two terminal device that stores electrical energy in the electric field in its vicinity. It is apassive element.
The property of a capacitor to store electrical energy or the effect of a capacitor is known as its capacitance. The capacitance of a capacitor is given by:
Q = CV or C = [tex]\frac{Q}{V}[/tex]
It is originally known as condensor and it can be said that a capacitor adds capacitance to the circuit.
What is an example of a linear actuator?
Answer and Explanation :
LINEAR ACTUATOR-In simple terms a linear actuator is a mechanical device that creates motion in a straight line. Linear actuators are used in machine tools and industrial machinery, in computers peripherals such as drives and printers in valve and dampers and it is used where linear motion is required.
EXAMPLE OF LINEAR ACTUATOR: There are many types of motors that are used in a linear system these include DC MOTOR (with brush or brushless) STEPPER MOTOR a linear actuator can be used to operate a large valve in refinery. Hydraulic pump is an also a good example of linear actuator.
What is an atom? a. it is a property of material b. it is specification of any material c. it the part of a substance d. it is a grain boundary
Answer:
Out of the multiple options provided,
option (c) it is the part of a substance
is correct
Explanation:
An atom is basically the smallest particle of matter or we can call it as building block of matter. It is as a brick to the wall or a cell to the body.
Matter is build up of a large number of these atoms. Atoms carbon(C), hydrogen(H), oxygen(O) combines to form molecules such as carbon dioxide([tex]CO_{2}[/tex]), water([tex]H_{2}O[/tex]), methane([tex]CH_{4}[/tex]), etc and then these molecules combines to form a substance.
It cannot be property, neither can it be specification of any material as these two characterizes the material and it can't be grain boundary for sure as grain boundary is the interface between two grain or crystal particles.
This shows that the smallest unit of matter is atom and it is the part of a substance
What is BCC unit cell and what is FCC unit cell? (10%) Sketch FCC and BCC unit cells by using a small dot to represent an atom.
Answer:
BCC is Body Centered Cubic structure and FCC is Face Centered Cubic structure.
Explanation:
BCC
In Body centered cubic structure, atoms exits at each corner of the cube and one atom is at the center of the cube. The total number of atoms that a BCC unit cell contains are 2 atoms and the coordination number of the BCC unit cell is 8, that is the number of neighbouring atoms. Packing factor which determines how loosely or closely a structure is packed by atoms is 0.68 for a BCC Unit cell. The packing efficiency is found to be 58%. Metals like tungsten, chromium, vanadium etc exhibit BCC structure.
A face Centered cubic structure consists of an atom at each corner of the cube and an atom at each face center of the cube. The total number of atoms that a BCC unit cell contains are 4 atoms per unit cell and the coordination number of FCC is 12. Packing factor of a FCC unit cell is 0.74%, thus FCC structure is closely packed than a BCC structure. Metals like aluminium, nickel, cooper, cadmium, gold exhibits FCC structure.
As temperature decreases a ductile material can become brittle - ductile-to-brittle transition
Answer:
Yes ,at low temperature ductile material behaves like brittle.
Explanation:
Yes,as temperature decreases a ductile material can become brittle.
In metals ductile to brittle transition temperature is around 0.1 to 0.2 Tm(melting point temperature) and in ceramics ductile to brittle transition temperature is around 0.5 to 0.7 Tm(melting point temperature) .
We can easily see that from graph between fracture toughness and temperature.In the graph when temperature is low then the ductile material is behaving like brittle material.But when temperature is above a particular value then material behaves like ductile.
A properly installed window quilt can provide an additional insulation of R-4 to windows. For a window with a U-factor of 0.33, determine the percentage of heat lost for a quilt-covered win- dow compared to a bare window.
Answer:
Heat loss 67%.
Explanation:
We know that heat transfer
Q=AUΔT
Where U is the overall heat transfer coefficient ,A is the area and ΔT is the temperature difference.
Now heat transfer in terms of U-factor
[tex] Q_1=AU_1ΔT[/tex]
[tex] Q_2=AU_2ΔT[/tex]
Given that temperature difference is same in both condition so
[tex]\dfrac{Q_1}{U_1}=\dfrac{Q_2}{U_2}[/tex]
[tex]\dfrac{Q_1}{Q_2}=\dfrac{U_1}{U_2}[/tex]
[tex] heat\ loss=\dfrac{Q_2-Q_1}{Q_1}[/tex]
[tex] heat\ loss=\dfrac{U_2-U_1}{U_1}[/tex]
Given that[tex]U_2=0.33U_1[/tex]
[tex] heat\ loss=\dfrac{0.33U_1-U_1}{U_1}[/tex]
Heat loss 67%.
A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box and then it is placed in a large tank containing 12 ft of water so that it will sink just to the bottom of the tank. Determine the mass of stones placed in the box, in tons.
Answer:
44.95 tonnes
Explanation:
According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces
It is given that empty weight of box = 40 tons
Let the mass of the stones to be placed be = M tonnes
Thus the combined mass of box and stones = (40+M) tonnes..........(i)
Since the box will displace water equal to it's volume V we have [tex]volume of box = 25ft*10ft*12ft= 3000ft^{3}[/tex]
[tex]Volume= 84.95m^{3}[/tex]
[tex]Since 1ft^{3} =0.028m^{3}[/tex]
Now the weight of water displaced = [tex]Weight =\rho \times Volumewhererho[/tex] is density of water = 1000kg/[tex]m^{3}[/tex]
Thus weight of liquid displaced = [tex]\frac{84.95X1000}{1000}tonnes=84.95 tonnes[/tex]..................(ii)
Equating i and ii we get
40 + M = 84.95
thus Mass of stones = 44.95 tonnes
________What is the main role of dispersed phase in ceramic matrix composites? a)- to increases yield strength b)-to increase ultimate tensile strength c)-to increase plain strain fracture toughness d)-to reduce hardness
Answer: b) to increase the ultimate tensile strength
Explanation: Ceramic matrix composites(CMC) are the materials which consist of ceramic material as well as composites .There are two pats present in it - matrix part which is present for holding the material together and reinforcement part which provides the toughness to the material. When the ceramic matrix composites have the dispersed phase,it is basically due to the increasing of the toughness of the material and hence increasing the tensile strength .
In solid motors, HTPB and PBAN are two common types of plasticizers. a) True b) False
Answer:
The given statement is True
Explanation:
Solid motor uses plasticizers out of which Poly Butadiene Acrylic Acid acrylo Nitrile (PBAN) and Hydroxy Terminator Poly Butadiene (HTPB) are the most commonly used. These are used in space shuttles.
PBAN is comparatively stick, thick and it has a stinky smell as compared to HTPB. PBAN can be mixed and cured at high temperastures few days later wheras HTPB can be mixed and cured at room termperatures within a day.
The propellant can be ready to use for flight a day later.
A_____ transducer is a device that can convert an electronic controller output signal into a standard pneumatic output. A. pneumatic-to-current (P/I) B. voltage-to-pneumatic (V/P) C. current-to-pneumatic (I/P) D. pneumatic-to-voltage (P/V)
Answer:
The correct answer is
option C. current to pneumatic (V/P)
Explanation:
A current to pneumatic controller is basically used to receive an electronic signal from a controller and converts it further into a standard pneumatic output signal which is further used to operate a positioner or control valve. These devices are reliable, robust and accurate.
Though Voltage and current to pressure transducers are collectively called as electro pneumatic tranducers and the only electronic feature to control output pressure in them is the coil.
What is a cascade refrigeration system?
Answer Explanation:
CASCADE REFRIGERATION SYSTEMS :
cascade refrigeration systems are commonly used in the liquefaction of natural gas and some other gases
cascade refrigeration system is also known as freezing system. It uses two types of refrigerants having different boiling points.this system is employed to obtain temperature of -40 to -80It allows stable ultra low temperature operationexamples of cascade refrigeration system: LNG (Liquefied natural gas ) plants mostly used cascade refrigeration system
ADVANTAGES OF CASCADE REFRIGERATION SYSTEMS:
energy is saved because the system allows the use of refrigerant gasesthe running cost is inexpensive repair is easyA cascade refrigeration system utilizes multiple linked refrigeration cycles to achieve ultra-low temperatures not possible with a single cycle. It features a cascade control mechanism for precise temperature management, using heat exchangers to interconnect and sequentially cool each stage.
Explanation:Understanding Cascade Refrigeration SystemsA cascade refrigeration system is a complex refrigeration strategy that employs multiple refrigeration cycles linked together to achieve cooling. Typically, these systems are utilized in environments requiring very low temperatures, which cannot be efficiently or cost-effectively achieved through a single refrigeration cycle. The essence of a cascade system lies in its deployment of two or more refrigeration units in series, each referred to as a "stage," with each successive stage operating at a lower temperature range than its predecessor.
The interconnection between these stages involves a heat exchanger, where a refrigerant from one cycle cools the condenser of the next lower temperature cycle. This process is crucial for achieving the desired low temperatures. One common application is in industrial refrigeration, where maintaining precise low temperatures is critical for process efficiency and product quality.
A key feature of cascade refrigeration systems is the use of cascade control, which enhances system responsiveness to temperature changes and maintains tight control over the process. This control mechanism consists of a master-slave relationship, where the output of one control loop (master) serves as the setpoint for the next (slave), creating a highly integrated control environment. Such setup enables rapid adjustment to changing operating conditions, ensuring the system swiftly responds to disturbances while maintaining the desired temperature levels.
Carbon nanotubes can be a considered toxic agent due to: (a)- Its ability to produce toxic gases (b)Its ability to penetrate skin quickly due its very small diameter (c)Its ability to mix with water (d)Difficulties in handling it properly
Answer: b)Its ability to penetrate skin quickly due to its very small diameter
Explanation: Carbon nano tubes(CNT) are the material widely used in the medical field due to the atomic structure of it ans also have small size. Toxicity in the carbon nano tubes is because their small sized atomic particles which can enter the skin by penetration or inhalation. But are still preferred in the medicine because having unique properties like mechanical property, chemical property,surface property etc.
Reliability is defined as the_______that a component or system will perform its intended function for a specific period of time.
Answer:
Probaility
Explanation:
The reliability of a component, system or unit can be defined as the probability that that component can operate for a certain period of time (mission time) without losing its function.
A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe has an inside diameter of 8.5 cm. The annulus between the cylinder ad the pipe contains stationary oil. The oil has a specific gravity of 0.92 and a kinematic viscosity of 5.57 x 10-4 m2/s. Most nearly, what is the force needed to move the cylinder along the pipe at a constant velocity of 1 m/s?
Answer :
The force needed to move the cylinder is 25.6 N
Further explanationGiven that,
Length of the cylinder, l = 0.5 m
Outer diameter of the cylinder, d = 8 cm = 0.08 m
Outer radius of the cylinder, [tex]r=0.04\ m[/tex]
Inside diameter of the pipe, d = 8.5 cm = 0.085 m
Inside radius of the pipe, [tex]r=0.0425\ m[/tex]
Specific gravity of the oil, [tex]\rho=0.92[/tex]
Density of oil, [tex]d=\rho\times \rho_w[/tex]
Kinematic viscosity of the oil, [tex]v=5.57\times 10^{-4}\ m^2/s[/tex]
Velocity of the cylinder, u = 1 m/s
We need to find the force needed to move the cylinder. Let the force is F.
Specific gravity is defined as the ratio of the density of the substance to the density of water.
Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.
Absolute viscosity is given by :
[tex]v=\dfrac{\mu}{d}[/tex]
Where, [tex]d[/tex] = density of oil
And [tex]d=\rho\times \rho_w[/tex] (density of oil = specific gravity × density of water )
[tex]d=0.92\times 10^3\ kg/m^3[/tex]
So,
[tex]\mu=v\times d[/tex]..............(1)
[tex]\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3[/tex]
[tex]\mu=0.512\ Pa-s[/tex]
The separation between the cylinder and pipe is given by :
[tex]dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m[/tex]
[tex]d_p\ and\ d_c[/tex] are diameter of pipe and cylinder respectively.
The mathematical expression for the Newton's law of viscosity can be written as:
[tex]\tau\propto\dfrac{du}{dy}[/tex]
[tex]\tau=\mu\times \dfrac{du}{dy}[/tex]..........(2)
Where
[tex]\tau[/tex] = Shear stress, [tex]\tau=\dfrac{F}{A}[/tex]............(3)
[tex]\mu[/tex] = viscosity
[tex]\dfrac{du}{dy}[/tex] = rate of shear deformation
On rearranging equation (1), (2) and (3) we get :
[tex]\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}[/tex]...............(4)
A is the area of the cylinder, [tex]A=2\pi rl[/tex]
Equation (4) becomes :
[tex]F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl[/tex]..............(5)
[tex]A=\pi d\times l[/tex]
[tex]A=\pi \times 0.08\ m\times 0.5\ m[/tex]
[tex]A=0.125\ m^2[/tex]
Now, equation (5) becomes :
[tex]F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl[/tex]
[tex]F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2[/tex]
F = 25.6 N
Learn moreKinematic viscosity : https://brainly.com/question/12947932
Keyword :Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.
(a) what is Linear equation (b) Why Laplace's equation is linear
Answer:
A) Linear Equation -
Linear equation has only one independent variable and when the linear equation plotted on a graph it forms a straight line. It is made up of two expressions equal to each other in a equation. Linear equation graph fits the Y= mx+a ( m=slope).
B) Laplace's equation is linear as it is a second order partial differential equation. So if we put dependent variable in differential equation it always show result in linear.
Most important question is "What is linear equation?". So, I will only answer that.
What is linear equation?
Linear equation is defined as one degree equation which make a straight line on a graph. Or one can say that when the highest degree of an equation is one than we call ot linear equation.
For example, 3x+4y= 8
In this equation there are three terms 3x, 4y and 8. Here in this equation "x" and "y" are variables while 8 is a constant, 3 is the coefficient of variable "x" and 4 is the coefficient of variable "y". The highest power of the variable "x" is one while the highest power of the variable "y" is also one. Hence, it is a one degree equation. Therefore, it is also known as linear equation.
Q: How to determine linear equation?
Ans: In order to determine a linear equation, look at the highest power of an equation. If the highest power of the equation is one, than it is a linear equation. If the highest power of the equation is not one or other than one like two or three, than it is not a linear equation.
For example, 2xy + 5 = 10
There are also three terms in this equation. The term "2xy" is considered as one term. The power of the variable "x" is one while the power of the variable "y" is also one hence, one plus one is equal to two. Thus the highest power of this equation is two. Therefore, it is neither one degree equation, nor a linear equation.
Q: How can we solve a linear equation in two variables?
Ans: There are different methods of solving linear equation in two variables, but we will solve a linear equation in two variables by adding equation one and equation two. After that, we will get the value of one variable and than we will put the value of that variable in the second equation also known as substitution method in order to get the value of second variable.
For example, x + y = 8 .... Equation One
X - y = 6 .... Equation Two
Now we will add equation one and equation two, as a result variable "y" will be cancelled out. We will get "2x = 14". So we will divide both sides by 2. Hence, we will get "x=7". Now we can put the value of "x" in either equation one or in equation two. For instance, we put the value of "x=7" in equation one. We get:
7 + y = 8
y = 8 - 7
y = 1
Thus, we successfully solve the linear equation in two variables where the value of variable "x=7" while the value of variable "y=1".
Air is entering a 4200-kW turbine that is operating at its steady state. The mass flow rate is 20 kg/s at 807 C, 5 bar and a velocity of 100 m/s. This air then expands adiabatically, through the turbine and exits at a velocity of 125 m/s. Afterwards the air then enters a diffuser where it decelerates isentropically to a velocity of 15 m/s and a pressure of 1 bar. Using the ideal gas model, determine, (a) pressure and temperature of the air at the turbine exit, in units of bar and Kelvin. (b) Entropy production rate in the turbine in units of kW/k, and (c) draw the process on a T-s Diagram.
Answer:
a)[tex]T_2=868.24 K[/tex] ,[tex]P_2=2.32 bar[/tex]
b) [tex]s_2-s_1=0.0206[/tex]KW/K
Explanation:
P=4200 KW ,mass flow rate=20 kg/s.
Inlet of turbine
[tex]T_1[/tex]=807°C,[tex]P_1=5 bar,V_1=100 m/s[/tex]
Exits of turbine
[tex]V_2=125 m/s[/tex]
Inlet of diffuser
[tex]P_3=1 bar,V_3=15 m/s[/tex]
Given that ,use air as ideal gas
R=0.287 KJ/kg-k,[tex]C_p[/tex]=1.005 KJ/kg-k
Now from first law of thermodynamics for open system at steady state
[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]
Here given that turbine is adiabatic so Q=0
Air treat ideal gas PV=mRT, Δh=[tex]C_p(T_2-T_1)[/tex]
[tex]w=\dfrac{P}{mass \ flow\ rate}[/tex]
[tex]w=\dfrac{4200}{20}[/tex]
w=210 KJ/kg
Now putting the values
[tex]1.005\times (273+807)+\dfrac{100^2}{2000}=1.005T_2+\dfrac{125^2}{2000}+210[/tex]
[tex]T_2=868.24 K[/tex]
Now to find pressure
We know that for adiabatic [tex]PV^\gamma =C[/tex] and for ideal gas Pv=mRT
⇒[tex]\left (\dfrac{T_2}{T_1}\right )^\gamma=\left (\dfrac{P_2}{P_1}\right )^{\gamma -1}[/tex]
[tex]\left(\dfrac{868.24}{1080}\right )^{1.4}=\left (\dfrac{P_2}{5}\right )^{1.4-1}[/tex]
[tex]P_2=2.32 bar[/tex]
For entropy generation
[tex]s_2-s_1=1.005\ln\dfrac{868.24}{1080}-0.287\ln\dfrac{2.32}{5}[/tex]
[tex]s_2-s_1=0.00103[/tex]KJ/kg_k
[tex]s_2-s_1=0.00103\times 20[/tex]KW/K
[tex]s_2-s_1=0.0206[/tex] KW/K
We can recycle glass as many time as we want. a)-True b)-False
Answer: True
Explanation: Yes , we can recycle glass endlessly, as many times as we want without the degradation of quality and purity factor because they are made from the widely available domestic components for example:- sand , limestone etc. These material are responsible for the successful recycling process of glass which does not produce any loss.
Why is it important (in the context of systems engineering management) to become familiar with analytical methods? Provides some specific examples.
Understanding analytical methods is essential in systems engineering for problem-solving and system comprehension, with real-world applications in areas like environmental monitoring.
Explanation:The importance of analytical methods in systems engineering management is crucial for problem-solving and comprehending the physics of the situation. Analytical methods help in selecting the right system and developing solutions efficiently. For instance, in the field of environmental monitoring programs, the use of analytical methods aids in assessing system health and trends accurately.
Briefly describe the ionic bonding and covalent bonding. What is the key difference between these two bonding
Answer:
Explanation:
Ionic bonding are formed when two opposite ion attract each other. Ionic bond is generally between metal and non metal. Example NaCl
Covalent bonding is the bonding which is formed by the sharing of electron pair covalent bond is generally between two non metal.Example H₂O
Ionic bond is generally electrostatic in nature which lead to attraction of positive and negative ion
ionic bond does not have definite shape where covalent has definite shape
melting point of covalent bond is low where as ionic bond have high melting point.
Two one-kilogram bars of gold are initially at 50°C and 300°C. The bars are brought in contact with each other. What is their final temperature, the entropy change of each block, and the total entropy generation in the process.
Given:
[tex]T_{1}[/tex] = 50°C = 273+50 =323 K
[tex]T_{2}[/tex] = 300°C = 273+300 =573 K
Solution:
We know that:
specific heat for gold, c = 0.129 J/g°C
Also, change in entropy, ΔS is given by:
[tex]\Delta S = cln\frac{T_{f}}{T_{i}}[/tex]
After the bars brought in contact with each other,
final temperature, [tex]T_{f}[/tex] = [tex]\frac{T_{1}+T_{2}}{2}[/tex]
final temperature, [tex]T_{f}[/tex] = [tex]\frac{323+573}{2}[/tex] = 448K
Now, entropy for first gold bar, using eqn-1
[tex]\Delta S = cln\frac{T_{f}}{T_{1}}[/tex]
[tex]\Delta S_{1} = 0.129ln\frac{448}{323}[/tex] =0.042 J/K
[tex]\Delta S_{2}[/tex] = 0.129ln [tex]\frac{448}{573}[/tex] = - 0.032 J/K
Total entropy generation,
[tex]\Delta S_{1}[/tex] = [tex]\Delta S_{1}[/tex] + [tex]\Delta S_{2}[/tex]
[tex]\Delta S[/tex] = 0.042 + (- 0.032) = 0.010 J/K
Describe the operational principle of a unitary type air conditioning equipment with a suitable sketch.
Answer:
Operational Principle of a Unitary Type Air Conditioning Equipment:
A unitary air conditioning system is basically a room type air conditioning system which comprises of an outdoor unit, a compressor for compressing
coolant, a heat exchanger (outdoor) for heat exchange, an expander attached to the heat exchanger for expansion of coolant and a duct.
It continuously removes heat and moisture from inside an occupied space and cools it with the help of heat exchanger and condensor in the condensing unit and discharges back into the same occupied indoor space that is supposed to be cooled.
The cyclic process to draw hot air, cool it down and recalculation of ther cooled air keeps the indoor occupied space at a lower temperature needed for cooling at home, for industrial processes and many other purposes.
refer to fig 1
The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature of 1623°C. The initial volume is 0.05 m^3 and the gas expands through a volume ratio of 20 according to the law PV^1.25 = constant. Calculate: a) Work transfer b) Heat transfer
Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
[tex]P_{1}[/tex] = 5 Mpa
[tex]T_{1}[/tex] = 1623°C
= 1896 K
[tex]V_{1}[/tex] = 0.05 [tex]m^{3}[/tex]
Also given [tex]\frac{V_{2}}{V_{1}} = 20[/tex]
Therefore, [tex]V_{2}[/tex] = 1 [tex]m^{3}[/tex]
R = 0.27 kJ / kg-K
[tex]C_{V}[/tex] = 0.8 kJ / kg-K
Also given : [tex]P_{1}V_{1}^{1.25}=C[/tex]
Therefore, [tex]P_{1}V_{1}^{1.25}[/tex] = [tex]P_{2}V_{2}^{1.25}[/tex]
[tex]5\times 0.05^{1.25}=P_{2}\times 1^{1.25}[/tex]
[tex]P_{2}[/tex] = 0.1182 MPa
a). Work transfer, δW = [tex]\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}[/tex]
[tex]\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}[/tex]
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
= [tex]\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}[/tex]
=[tex]\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W[/tex]
=[tex]\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200[/tex]
= 197.7 kJ
A M14 x 2 hexagonal head bolt is used to clamp together two 15 mm steel plates. Young's modulus of the bolt and the plates can be taken as 207 GPa. (i)-Determine a suitable length for the bolt. (ii)-Determine the bolt stiffness (iii)-Determine the stiffness of the members
Answer:
(i) 50 mm
(ii) 874.62 m N/m
(iii) 3116.45 m N/m
Explanation:
Given data
hexagonal head bolt = M14 x 2
steel plate = 15 mm
Young's modulus = 207 Gpa
Solution
1st part
dia of bolt (D) = 14 mm and height (H) = 12.8 mm from table
we know grip length = thickness of this plate i.e 30 mm , i.e (15mm+15mm)
so hexagonal bolt length = grip length + H
hexagonal bolt length = 30 mm + 12.8 mm = 42.80 mm i.e = 45 mm (round off)
2nd part
bolt stiffness = [tex]\frac{Ad*At*young modulus}{Ad*Lt*At*Ld}[/tex]
here Lt is length of thread = 2d +6mm
d is 14 so length of thread = 2*14 +6 = 34 mm
At from table 8-L i.e. = 115 mm2
Ad area of without thread part = [tex]\pi /4[/tex]×[tex]d^{2}[/tex]
Ad= [tex]\pi /4[/tex]×[tex]14^{2}[/tex] = 153.94 mm2
Ld is length of bolt without thread = length of bolt - Lt = 45 -34 = 11 mm
last Lt thread part length = length of bolt - Ld = 30-11 = 19 mm
put all these value in bolt stiffness i.e.
bolt stiffness = [tex]\frac{153.94*115*207}{153.94*19+115*11}[/tex] = 874.62
3rd part
stiffness of member = [tex]\frac{0.5774 \pi Ed}{2 ln (5\frac{0.5774L +0.5d}{0.5774L +2.5d})}[/tex]
here l is 30 and d is 14
so
stiffness of member = [tex]\frac{0.5774 \pi (207) 14}{2 ln (5\frac{0.5774(30) +0.5(14)}{0.5774(30) +2.5(14)})}[/tex]
stiffness of member = 3116.45 m N/m
Water, in a 150 in^3 rigid tank, initially has a temperature of 70°F and an enthalpy of 723.5 Btu/lbm. Heat is added until the water becomes a saturated vapor. Calculate: a) The initial quality of the water b) The total mass of the water c) The temperature of the water at the final state.
Answer:
a)initial quality of water 0.724.
b)mass of water =2.45 kg
c)T=70°F
Explanation:
h=732.5 Btu/lbm
h=763.33 KJ/kg (1 Btu=1.05 kj)
T=70°F⇒T=21.1°C
[tex]V=150 in^3=0.002458m^3[/tex]
(a)From steam table
Properties of saturated steam at 21.1°C
[tex]h_f= 88.56\frac{KJ}{Kg} ,h_g= 2539.49\frac{KJ}{Kg}[/tex]
To find dryness fraction
[tex]h=h_f+x(h_g-h_f)\frac{KJ}{Kg}[/tex]
[tex]763.33=88.56+x(2539.49-88.56)\frac{KJ}{Kg}[/tex]
x=0.27
So initial quality of water 0.724.
(b)
[tex]v=v_f+x(v_g-v_f)\frac{m^3}{Kg}[/tex]
where v is specific volume
From steam table at 21.1°C
[tex]v_f= 0.001\frac{m^3}{Kg} ,v_g= 54.13\frac{m^3}{Kg}[/tex]
V=[tex]m_f\times v_f[/tex]
0.002458=[tex]m_f\times 0.001[/tex]
So mass of water =2.45 kg
(c)
Actually water will take latent heat ,it means that heating of will take place at constant temperature and constant pressure.So we can say that final temperature of water will remain same (T=70°F).