What is the thermal efficiency of this reheat cycle in terms of enthalpies?
Answers
Answer:
[tex]\eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}[/tex]
Explanation:
For close gas turbine:
Gas turbine works on Brayton cycle.Gas turbine have lots of applications like ,it is use in aircraft,in land applications etc.
Reheating is the method to improve the efficiency of the gas turbine.In reheating gas is expanding in two turbine instead of one turbine alone.Two turbine like high pressure turbine and low pressure turbine are used for expansion.
In the above diagram 1-2 is a compressor,2-3 heat addition,3-4 high pressure turbine,4-5 reheating of cycle 5-6 low pressure turbine,6-1 heat rejection,
We know that [tex]\eta =\frac{W_{net}}{Q_{s}}[/tex]
Now take [tex]h_{1},h_{2},,h_{3},h_{4},h_{5},h_{6}[/tex] represent the enthalpy of point 1,2,3,4,5,6 in the cycle respectively.
So total heat supplied [tex]Q_S[/tex]=
[tex]\left (h_3-h_2\right )+\left (h_5-h_4\right )[/tex]
Net work out put
[tex]W_{net}[/tex]=[tex]\left (h_5-h_6\right )-\left (h_2-h_1\right )[/tex]
So efficiency [tex]\eta =\frac{W_{net}}{Q_{s}}[/tex]
[tex]\eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}[/tex]
Related Questions
Solid rockets can experience significant 2 phase flow. a) True b) False
Answers
Answer:
the answer is false solid rockets can experience significant 2 phase flow
The smallest crystal lattice defects is a) cracks b) point defects c) planar defects d) dislocations.
Answers
Answer:b) Point defects
Explanation: The point defect is the tiny defect that occurs in the lattice. It usually occurs when there is the atoms or ions missing in the lattice structure that creates a irregularity in the structure.The name point defect itself describes that the occurring defect is having a size of point thus is the smallest defect. Therefore option(b) is the correct option.
A negative normal strain can be considered to increase or decrease volume depending on the coordinate system used. a)True b)- False
Answers
Answer:
The given statement "A negative normal strain can be considered to increase or decrease volume depending on coordinate system used" is
b) False
Explanation:
Normal strain refers to the strain due to normal stress which is when the applied stress is perpendicular to the surface.
Negative normal strain results in compression or contraction further leading to a decrease in volume while a positive normal strain results in elongation thus giving rise to an increase in volume.
How to called crystalline lattice changes at the temperature change a) oxidation; b) primary crystallization c) allotropy d) anisotropy.
Answers
Answer: (c)Allotropy
Explanation: Allotropy is defined as the feature of the material to exist in different chemical forms or state while being in the same physical state. So, allotropy responsible for maintenance of the crystalline lattices during the temperature changes and does not alter the physical state of it. It helps elements to exist in different forms by modifying the structure of the chemical state.
Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5s
Answers
To hoist a load 6 meters in 1.5 seconds, the cable must be drawn in by the motor at a constant speed of 4 meters per second.
Explanation:The question asks to determine the constant speed at which a cable must be drawn in by a motor to hoist a load to a certain height within a given time frame. This can be solved by understanding the basic concepts of distance, speed, and time.
To find the constant speed, we use the formula:
Speed = Distance / Time. In this case, the distance is 6 meters (the height the load needs to be hoisted) and the time is 1.5 seconds.
Plugging the numbers into the formula gives:
Speed = 6m / 1.5s = 4 m/s.
Therefore, the cable must be drawn in by the motor at a constant speed of 4 meters per second to hoist the load 6 meters in 1.5 seconds.
A spherical, stainless steel (k 16 W m1 K-1) tank has a wall thickness of 0.2 cm and an inside diameter of 10 cm. The inside surface of the tank wall is held constant at 25 oC and the outside surface heat transfer coefficient is 6 W m2 K. Calculate the rate of heat loss from the tank when the outside air temperature is 15°C.
Answers
Answer:
the rate of heat loss is 2.037152 W
Explanation:
Given data
stainless steel K = 16 W [tex]m^{-1}K^{-1}[/tex]
diameter (d1) = 10 cm
so radius (r1) = 10 /2 = 5 cm = 5 × [tex]10^{-2}[/tex]
radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 × [tex]10^{-2}[/tex]
temperature = 25°C
surface heat transfer coefficient = 6 6 W [tex]m^{-2}K^{-1}[/tex]
outside air temperature = 15°C
To find out
the rate of heat loss
Solution
we know current is pass in series from temperature = 25°C to 15°C
first pass through through resistance R1 i.e.
R1 = ( r2 - r1 ) / 4[tex]\pi[/tex] × r1 × r2 × K
R1 = ( 5.2 - 5 ) [tex]10^{-2}[/tex] / 4[tex]\pi[/tex] × 5 × 5.2 × 16 × [tex]10^{-4}[/tex]
R1 = 3.825 × [tex]10^{-3}[/tex]
same like we calculate for resistance R2 we know i.e.
R2 = 1 / ( h × area )
here area = 4 [tex]\pi[/tex] r2²
area = 4 [tex]\pi[/tex] (5.2 × [tex]10^{-2}[/tex])² = 0.033979
so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )
R2 = 4.90499
now we calculate the heat flex rate by the initial and final temp and R1 and R2
i.e.
heat loss = T1 -T2 / R1 + R2
heat loss = 25 -15 / 3.825 × [tex]10^{-3}[/tex] + 4.90499
heat loss = 2.037152 W
Describe ICP/OES in detail.
Answers
Answer:
ICP -OES stand for inductively coupled plasma optical emission spectroscopy
Explanation:
It is techniques that known as trace level technique which help to identify and quantify the element present in sample by using spectra emission.
The analysis process include desolvates, ionization and excitation of the sample. The sample is identify by analyzing the emission line from it and quantify by analyzing the intensity of same emission lines.
Answer:
ICP stands for Insane Clown Posse. It is a rap group. The group's members are Violent J and Shaggy 2 Do*e
Explanation:
What is meant by thermodynamic property? How do you classify the property? Explain with specific examples.
Answers
Answer:
The condition of very system can be identified by some physical characteristics of the system known as Thermodynamic Property of the system. For example, pressure, temperature, volume, etc.
Explanation:
Thermodynamic properties are nothing but the physical characteristics of any body by which the physical condition of the body can be described. properties describes the different states of a system. They are macroscopic.
Properties are of two types ---
Intensive property
Extensive property
Intensive Property : Those properties of a system which does not depend upon mass of the system are known Intensive property.
For example, pressure, temperature, density,specific volume, etc
Extensive Property : Properties that depends on the mass oh the system is called Extensive properties.
For example, energy, volume, etc.
At any given state the specific internal energy of a gas is always greater its specific enthalpy. a)True b) False
Answers
Answer:
(b)False
Explanation:
We know that specific Internal energy of gas u=[tex]C_v[/tex]T
and specific enthalpy of gas h=[tex]C_p[/tex]T
If we take the case of air we know that
[tex]C_v[/tex]=0.707 KJ/Kg=K , [tex]C_p[/tex]=1.005 KJ/Kg=K
If we take A fixed temperature T=300 K
so u=212.1 KJ/ kg ,h=301.5 KJ/kg
So we can say that specific enthaply of gas is always greater than its specific internal energy.
An aluminium alloy used for making cans is cold rolled into a strip of thickness 0.3mm and width 1m. It is coiled round a drum of diameter 15cm, and the outer diameter of the coil is 1m. In the cold rolled condition, the dislocation density is approximately 1015 m-2. Estimate: (i) The mass of aluminium on the coil; (ii) The total length of strip on the coil; (iii) The total length of dislocation in the coiled strip.
Answers
Answer:
1. Mass = 2070 kg
2.Total length of strip = 2556 m
3. Total length of dislocation = 7.67 X[tex]10^{14}[/tex] m
Explanation:
Given:
Aluminium coil thickness, t = 0.3 mm
= 0.3 X [tex]10^{-3}[/tex] m
Width of the coil,w = 1 m
Drum diameter, d = 15 cm
= 0.15 m
Coil outer diameter, d = 1 m
Dislocation density = [tex]10^{15}[/tex] [tex]m^{2}[/tex]
1). Area of the coil, A = [tex]\frac{\pi }{4}\times[/tex] ( [tex]d_{coil} ^{2}[/tex]-[tex]d_{drum} ^{2}[/tex])
A = [tex]\frac{\pi }{4}\times (1^{2}-0.15^{2})[/tex]
A = 0.767 [tex]m^{2}[/tex]
Volume of the coil,V = A X w
= 0.767 X 1
= 0.767 [tex]m^{3}[/tex]
We know density of aluminum at STP = 2.7 X [tex]10^{3}[/tex]
Therefore, mass of the aluminum coil is,
Mass,m = Density of aluminium X Volume
= 2.7 X [tex]10^{3}[/tex] X 0.767
= 2070 kg
Mass = 2070 kg
2). Total length of trip of coil is given by
L = Volume of coil / area of strip
= [tex]\frac{0.767}{1\times 0.3\times 10^{-3}}[/tex]
= 2556 m
Total length of strip = 2556 m
3). Total length of dislocation of the coiled strip = volume X dislocation density
= 0.767 X [tex]10^{15}[/tex]
= 7.67 X [tex]10^{14}[/tex]
Total length of dislocation = 7.67 X[tex]10^{14}[/tex] m
Why factor of safety is more than 2 in the gears ? What does effect ?
Answers
Answer:
explained
Explanation:
Gear is a mechanical components designed for transfer of torque or power from one shaft to the other. Gear designing is a costly affair. The configuration and geometry of gears are such that its designing is tedious task. A lot of precision is required to design a gear. This is why factor of safety of gears are always kept higher.
The higher factor of safety accounts for durability of gears. Gears once made can used for about 200 hundred years. The wear and tear are less and the failure of gears is avoided. And hence whole machine failure can be avoided.
Convective heal transfer is defined as______
Answers
Answer: Convective heat transfer is defined as the transfer of heat from a particular place to another with the help of fluid movements.
Explanation: Convective heat also knows as convection that is transfer of heat from one place to another place by the help of fluid movements around the area having different temperatures. This process has many examples in everyday life to understand the process better.
Ex.- the melting of ice from solid to liquid form as it comes in contact of high temperature as heat surrounds the ice in form of air.
Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed Mold c)- Preforming d)- Postforming e)- None of the above f)- All the above
Answers
Answer:
it is f all of the above
Explanation:
let me know if im right
im not positive if im right but i should be right
What is Potential flow?
Answers
Answer: Potential flow states about the velocity field considering it as gradient of scalar function such as the velocity potential. It is considered same as irrotational flow.
Explanation:In the study of fluid dynamics , potential flow is stated as the term for describing about the velocity field taking it as the gradient of a scalar function. Therefore, potential flow is stated by an irrotational velocity field and there is a irrotation because of the curl of the gradient of scalar quantity is zero.Potential flow is considered same as the irrotational flow that is the particles in the fluids that do not rotate.
Not a characteristic property of ceramic material (a) high temperature stability (b) high mechanical strength (c) low elongation (d) low hardness
Answers
Low hardness is not a characteristic property of ceramic material.
Answer: Option D
Explanation:
One of the most important properties of ceramic material is the hardness that the material displays. The hardness owes to the joining of brittle fracture and plastic flow that makes the material to defend against penetration.
The hardness can be tested with the Vickers test. The hardness of ceramic makes its dominance in the use of construction purpose and manufacture of products. Other general properties are less conductivity, high melting temperature, etc.
Not a characteristic property of ceramic material (d) low hardness.
(a) High temperature stability is a defining feature of ceramics, as they can withstand extreme temperatures without significant degradation.
(b) High mechanical strength is another key characteristic, making ceramics suitable for applications where strength and durability are essential.
(c) Low elongation refers to the minimal deformation under stress, which is typical of ceramics due to their brittle nature.
(d) However, low hardness is not a characteristic property of ceramics. Ceramics are known for their high hardness, making them resistant to wear and abrasion.
A 4.4 HP electric motor spins a shaft at 2329 rpm. Find: The torque load carried by the shaft is closest to: Select one: a)-27.06 ft*lb b. 19.24 ft*lb ? c)-31.17 ftlb d) 119.07 ftlb e)-9.92 ft*lb
Answers
Answer:
Load carried by shaft=9.92 ft-lb
Explanation:
Given: Power P=4.4 HP
P=3281.08 W
Power: Rate of change of work with respect to time is called power.
We know that P=[tex]Torque\times speed[/tex]
[tex]\omega=\frac{2\pi N}{60}[/tex] rad/sec
So that P=[tex]\dfrac{2\pi NT}{60}[/tex]
So 3281.08=[tex]\dfrac{2\pi \times 2329\times T}{60}[/tex]
T=13.45 N-m (1 N-m=0.737 ft-lb)
So T=9.92 ft-lb.
Load carried by shaft=9.92 ft-lb
An indirect contact heat exchanger operating at steady state contains a shell with a single tube through it. The fluid flowing through the tube enters at 1kg/s with and enthalpy of 100kJ/kg. It exits with an enthalpy of 500kJ/kg. The fluid flowing through the shell enters with a mass flow rate of 4kg/s and an enthalpy of 1000kJ/kg. Determine the enthalpy at the exit of the shell.
Answers
Answer:900 KJ/kg
Explanation:
Given data
mass flow rate of fluid through tube is=1 kg/s
Initial enthalpy of fluid through tube=100 KJ/kg
Final enthalpy of fluid through tube=500KJ/kg
mass flow rate of fluid through shell is=4 kg/s
Initial enthalpy of fluid through shell=1000 KJ/kg
Final enthalpy of fluid through shell=[tex]h_2[/tex]
since heat lost by Shell fluid is equal to heat gain by Tube fluid
heat lost by Shell fluid=[tex]4\times c\left ( 1000-h_2\right )[/tex]
Heat gain by tube Fluid=[tex]1\times c\left ( 500-100\right )[/tex]
Equating both heats
[tex]4\times c[/tex][tex]\left ( 1000-h_2\right )[/tex]=[tex]1\times c\left ( 500-100\right )[/tex]
[tex]h_2[/tex]=900 KJ/kg
Air is compressed by a compressor from v1 = 1.0 m3/kg to v2 = 0.71 m3/kg in a reversible, isothermal process. The air temperature is maintained constant at 25 oC during this process as a result of heat transfer to the surroundings. Air moves through the compressor at a rate of m = 1 kg/s. a)- Determine the entropy change of the air per kg of air. b)- What is the power required by the compressor? c)- What is the rate at which entropy leave the compressor?
Answers
Answer:
(a)[tex]s_2-s_1[/tex]= -0.098 KJ/kg-K
(b)P= 29.8 KW
(c) [tex]S_{gen}[/tex]= -0.098 KW/K
Explanation:
[tex]V_1=1m^3/kg,V_2=0.71m^3/kg,[/tex] mass flow rate= 1 kg/s.
T=25°C
Air treating as ideal gas
(a)
We know that entropy change for ideal gas between two states
[tex]s_2-s_1=mC_v\ln \frac{T_2}{T_1}+mR\ln \frac{V_2}{V_1}[/tex]
Given that this is isothermal process so
[tex]s_2-s_1=mR\ln \frac{V_2}{V_1}[/tex]
[tex]s_2-s_1=1\times 0.287\ln \frac{0.71}{1}[/tex]
[tex]s_2-s_1[/tex]= -0.098 KJ/kg-K
(b)
Power required
[tex]P=\dot{m}T\Delta S[/tex]
[tex]P=1\times (273+25)(s_2-s_1)[/tex]
[tex]P=1\times (273+25)(-0.098)[/tex]
P= -29.8 KW (Negative sign means it is compression process.)
(c)
Rate of entropy generation [tex]S_{gen}[/tex]
[tex]S_{gen}=\dot{m}T\Delta S[/tex]
[tex]S_{gen}[/tex]=1(-0.098)
[tex]S_{gen}[/tex]= -0.098 KW/K
What are "primary alpha" and "primary beta"?
Answers
Consider a falling mass(m) under gravity(9.8m/s). Initial velocity of the mass is 5 m/s upwards. Derive expressions for the velocity and the position of the mass(m) in terms of time and initial velocity/position of the mass. a) -How long will the mass take to reach the maximum height position? b)- What would be the maximum height the mass reach relative to its initial position?
Answers
Answer:
a) 0.51 s
b) 1.275m
Explanation:
using equation of linear motion
v=u+gt...........................(1)
[tex]v^{2} -u^{2}=2gh[/tex]...........(2)
[tex]s=ut+\frac{1}{2} gt^2[/tex].......(3)
a) as the ball is thrown upward -ve 'g' will be acting on the body
as the body reaches to the maximum height the final velocity(v) becomes
zero so from equation (1)
0=5-9.8t
[tex]t=\frac{5}{9.8}[/tex]
t=0.51s
b) Now for maximum height calculation using equation (2)
[tex]v^{2} -u^{2}=2gh[/tex]
v=0
[tex]h=\frac{-u^2}{2g}[/tex]
[tex]h=\frac{-5^2}{2\times-9.8}[/tex]
h=1.275m
Consider a fully developed flow in a circular pipe (negligible entrance effects), how does the convective heat transfer coefficient vary along the flow direction? a) Gradually decrease b) Gradually increase c) Remain constant d) There is not enough information to determine
Answers
Answer: A) Gradually decrease
Explanation:
The convection value of heat transfer rate are gradually decreasing with the flow of the heat. Flow in a circular pipe, flow direction does not change in the velocity path. The average of the coefficient of heat transfer and the number of pipes are needed and the effects are get neglected so that is why the flow are fully developed.
Silicon is an intrinsic semiconductor. Adding a small amount of phosphorus provides extra electrons. As a resuit, phosphorus is an p-type dopant. (True , False )
Answers
It's an n-type dopants, as it makes the silicone n-type after the doping. so, False
Oxygen enters an insulated 14.2-cm-diameter pipe with a velocity of 60 m/s. At the pipe entrance, the oxygen is at 240 kPa and 20°C; and, at the exit, it is at 200 kPa and 18°C Calculate the rate at which entropy is generated in the pipe.
Answers
Answer:
Entropy generation==0.12 KW/K
Explanation:
[tex]s_2-s_1=C_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}[/tex]
[tex]s_2-s_1=0.891\ln \frac{291}{293}-0.2598\ln \frac{200}{240}[/tex]
[tex]s_2-s_1=0.0412\frac{KJ}{kg-K}[/tex]
Mass flow rate= [tex]\rho\times\dfrac{\pi}{4}d^2V[/tex]
[tex]\rho_1=\dfrac {P_1}{RT_1}[/tex]
[tex]\rho_1=\dfrac {240}{0.2598\times 293}[/tex]
[tex]\rho_1=3.51\frac{kg}{m^3}[/tex]
mass flow rate=[tex]\rho_1A_1V_1[/tex]
So by putting the values
Mass flow rate=2.97 kg/s
So entropy generation=(2.97)(0.0412)
=0.12 KW/K
Which of the following is/are not a cutting tool material (mark all that apply)? a. High-speed steel b. Brass c. Coated carbide d. Diamond
Answers
Answer:
The correct option is : b. Brass
Explanation:
The cutting tool materials are materials that are used to make cutting tools. The cutting tools serve a very important roles in the machinery such as milling cutters. The materials used for making the cutting tools must be tougher and harder than the material that is being cut, at all temperatures.
Some of the cutting tool materials are tool steels (carbon tool steel and high speed steel), cemented carbides and super hard materials such as diamond.
Therefore, brass is not used for making cutting tool.
Using the saturated table, the specific entropy of of subcooled liquid water at 75 C and 200 kPa is______
Answers
Answer Explanation :
We know that entropy is the measure of molecular randomness it is denoted by S It is a thermodynamic property of gas it is an extensive property means its depend on the mass of the the system and specific entropy is defined as the entropy per unit mass its unit is KJ/kg.k
by using standard table the specific entropy for the subcooled liquid is 1.057 KJ/kg.k standard table is used for knowing the specific entropy of different liquid and gaes
For water the critical temperature is 374 C, water at T- 400 C exist as vapor (____)
Answers
Answer:
The statement is False.
Explanation:
Critical temperature is defined as the temperature at and above which the liquid and the gaseous states are indistinguishable. This is the temperature at and above which the liquid cannot be liquefied.
Above critical temperature or pressure, the substance is in supercritical fluid (SCF) state. A supercritical fluid is the one in which liquid and gaseous state can not be distinguished. It can dissolve like liquids and can effuse like solids.
Water above its critical temperature exists in SCF state which is neither vapor nor liquid.
Thus, the statement is false.
A disk brake has two pads which cover 45 degrees of the disk. The outside radius is 6.0 inch and the inside radius is 4.0 inch. Assume a coefficient of friction of 0.4, and a max pressure, pa=100 psi. a) Find the force required to apply one pad. b) Find the torque capacity for both pads.
Answers
Answer:
f = 628.32 lb
t = 2513.28 lb-inc
Explanation:
given data:
θ = 45°
outside radius = 6 inch
inside radius = 4 inch
coefficient of friction = 0.4
max pressure = 100 psi
a) determine force required for applying one pad
f = [tex] \frac{\theta }{360}* 2\pi *p_{max}*r_{i}(r_{o}-r_{i})[/tex]
f = [tex] \frac{45 }{360}* 2\pi *100*4(6-4)[/tex]
f = 628.32 lb
b) torque capacity (t)
t = [tex]\mu *f*r_{average}^{}[/tex]
t = 0.4 *628.32*5
torque = 1256.64 lb-inc
for both pad = 2 * 1256.64 =2513.28 lb-inc
Before cutting coarse screw threads, the operator should lubricate: A. The leadscrew and gearbox B. The ways and cross slide C. The carriage and half-nuts D. A1l of the above
Answers
Answer:
(d) all of the above
Explanation:
before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial
so option (D) will be correct because we need lubricate in all the given parts
Give two advantages of a four-high rolling mill opposed to a two-high rolling mill for the same output diameter.
Answers
Answer:
Four- high rolling mill Two-high rolling mill
1.Small roll radius.That is why required 1.High roll radius.That is required low power. . why required high power.
2. Low roll separating force. 2.High roll separating
force
(true/false) Moment thickness, is an index that is proportional to an increment in momentum flow due to the presence of the boundary layer. if false explain why?
Answers
Answer:
True
Explanation:
Moment thickness, is an index that is proportional to an increment in momentum flow due to the presence of the boundary layer.