State the two Carnot corollaries (principles)

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.

Answer:

(b) vpltage

Explanation:

we know the expression for voltage across the inductor V=L[tex]\frac{di}{dt}[/tex] which clearly shows voltage across the inductor is directly proportional to rate of change of current similarly current across the capacitor I=C[tex]\frac{dv}{dt}[/tex] from the expression we can see that current across the capacitor is directly proportional to rate of change of voltage. so from above discussion it is clear that response of an iductor to current is similar to response of capacitor to voltage

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]

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

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.

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.

Answer:

T = 1.17 x [tex]10^{-3}[/tex] N-m

Explanation:

Given :

Gap between the two plates , dy = 1 mm

                                                  dy = 1 x [tex]10^{-3}[/tex] m

Angular velocity of the top plate , ω = 5 rad/s

Diameter of the plate, D = 20 cm

Radius of the plate, R = 10 cm

                                    = 0.1 m

Temperature of the kerosene = 40°C

Viscosity of kerosene at 40°C = 0.0015 Pa-s

Now let us take a small elemental ring of thickness dr at a radius r.

Therefore, area of this elemental ring of dr = 2πrdr

Now linear velocity at radius r = ω x r

                                                   5r m/s

Now applying Newtons law of viscosity we get,

Shear stress, τ = μ.[tex]\frac{du}{dy}[/tex]

   [tex]\Rightarrow \frac{F_{s}}{A}=\mu .\frac{du}{dy}[/tex]

  [tex]\Rightarrow \frac{F_{s}}{A}=\mu .\frac{5r-0}{1\times 10^{-3}}[/tex]

  [tex]\Rightarrow \frac{F_{s}}{A}=\mu \times 10^{3}\times 5r[/tex]

  [tex]F_{s}=\mu \times 10^{3}\times 5r\times 2\pi rdr[/tex]

  [tex]F_{s}=5 \times 10^{3}\times \mu \times r\times 2\pi rdr[/tex]

  [tex]F_{s}=\frac{18849}{400}\times r^{2}dr[/tex]

Now we know torque due to small strip,

 dT = [tex]F_{s}[/tex] x r

 dT = [tex]\frac{18849}{400}\times r^{3}dr[/tex]

Therefore total torque for r=0 to r=R can be calculated. So by integrating,

[tex]\int dT=\int_{0}^{R}\frac{18849}{400}\times r^{3}dr[/tex]

[tex]T = \frac{18849}{400}\times \frac{R^{4}}{4}[/tex]

[tex]T = 47.1225\times \frac{0.1^{4}}{4}[/tex]

T = 1.17 x [tex]10^{-3}[/tex] N-m

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  

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

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.

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).

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.

Answer Explanation:

CASCADE REFRIGERATION SYSTEMS :

cascade refrigeration systems are commonly used in the liquefaction of natural gas and some other gases

examples of cascade refrigeration system: LNG (Liquefied natural gas ) plants mostly used cascade refrigeration system

ADVANTAGES OF CASCADE REFRIGERATION SYSTEMS:

A 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.

A 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.

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

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

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%.

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 .

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.

Answer: d) property tests

Explanation: Product service life can be referred as the life that define the service that can be provided by the product manufactured.The service life contains the testing and calculation of the product's quality, reliability, maintenance factor etc. These factors are known as the property of the product and so is calculated by the property test. Therefore option (d) is the correct option because other option does not define the factors for defining the product service life.

Understanding analytical methods is essential in systems engineering for problem-solving and system comprehension, with real-world applications in areas like environmental monitoring.

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.

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.

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

Answer :  

The force needed to move the cylinder is 25.6 N

Given 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

Kinematic viscosity : https://brainly.com/question/12947932

Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.  

Answer:  Exploration includes plethora of activities and depend upon the kind  of exploration a person is doing. But most include some of the basic activities like research , investigation, planning and execution.

Suppose we want to explore new petroleum sites then we would have to start with studying the geography of that area, then according to our research we will analyse the hot spots or the sector where probability of finding of oil field is highest, post that appropriate man power is skilled professionals, tools and machinery will be brought at the site so that execution can take place.

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.

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

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.  

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.

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.

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