Answer:
a)[tex]\eta_{st}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
b)[tex]\eta_{sc}=\dfrac{\Delta h_{ideal}}{\Delta _{actual}}[/tex]
c)[tex]\eta_{sn}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
Explanation:
a)
Adiabatic turbine
Adiabatic turbine means turbine can not reject or take heat from surrounding.
Isentropic efficiency of turbine can be define as the ratio of actual work out put to the Ideal or isentropic work out put.Ideal means when turbine will give maximum work and there is no any friction losses we can say when process is isentropic.
Isentropic efficiency of turbine=(Actual work output)/(Ideal work output)
[tex]\eta_{st}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
b)
Adiabatic compressor
Isentropic efficiency of compressor can be define as the ratio of ideal or isentropic work in put to the actual work in put.
Isentropic efficiency of turbine=(Ideal work input)/(actual work input)
[tex]\eta_{sc}=\dfrac{\Delta h_{ideal}}{\Delta _{actual}}[/tex]
c)Adiabatic nozzle
We know that nozzle is device which used to accelerate the fluid.
Basically it covert pressure energy to kinetic energy.
Isentropic efficiency of nozzle can be define as the ratio of actual enthalpy drop put to the Ideal or isentropic enthalpy drop.
[tex]\eta_{sn}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
Horizontal wind turbines have same design for offshore and on shore wind farms. a)-True b)- False
Answer: False
Explanation: Horizontal axis wind turbines are usually used for generation of the electric power on the off-shore. The generation of horizontal-axis wind turbine works well when it is installed away from the shore because it supports large sized wind turbines so that they can generate high amount of electricity.They are usually not preferred for the on-shore wind farms because they can have small sized wind turbines only.Therefore the statement given is false.
What are the qualities required for a refrigerant? What is meant by the term "flash chamber?"
Answer:
A good quality refrigerant should be eco friendly.
Explanation:
A refrigerant is a substance that can extract and transfer heat from body to another body or medium.
The desirable qualities required for a refrigerant are :
1. A refrigerant should not deplete ozone layer.
2. A good quality refrigerant should have a low boiling point.
3. It should also have a low melting point.
4. Thermal conductivity of the refrigerant should be high for fast heat transfer.
5. It should have low specific heat.
6. It should have high latent heat.
7. It should have low vapour density.
8. Refrigerant should have high critical pressure and temperature.
9. It should have high enthalpy of vapourization for maximum heat absorption.
10. Refrigerants should not be toxic in nature and non flammable.
11. It should have high coefficient of performance for the working temperature range.
12. It should be easily available and cheap.
Flash Chamber :
A flash chamber in the refrigeration system is also know as the mixing chamber. It is normally used in multistage refrigeration system and is placed in between the expansion valve and the evaporator.
The flash chamber sends only the liquid refrigerant to the evaporator by seperating the liquid from the vapour refrigerant in order to increase the efficiency.
In vibration analysis, can damping always be disregarded?
In vibration analysis, damping cannot always be disregarded. This is especially the case when the system is excited near the resonance frequency.
How are fluids distributed and transported in a fluid power system?
Answer and explanation :
Fluid distribution is a new technique to produce and to transmit power from one place to other its play a major role in power distribution it is a process of using fluid (any type of fluid as oil or water ) under pressure to generate to control or to transmit
fluid power system is divided into two types
Hydraulic fluid power systempneumatic fluid power system
Thermosetting polymers are polymers that becomes soft and pliable when heated. ( True , False )
Answer:
Thermosetting polymers are polymers that becomes soft and pliable when heated is false
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
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.
What is a truss? What separates a truss from a frame and other forms of rigid bodies?
Answer:
The application of force is the main difference between truss and frame.
Explanation:
Truss:
Truss is a collection of beams,which use to handle the tensile and compression loads . That collection of beams creates rigid structure.
The load on the truss will be acting always at the at the hinge. Truss is widely used in the construction areas.
Frame:
Like truss, it is also a combination of beams and used to handle the loads. The main difference between truss and frame is the application of load. In the frame load can apply at the any point of the member of frame along with hinge.
Truss are connected by pin joint and can not transfer moment ,on the other hand frames are connected by rigid joint like welding so frame can transfer moment.
Truss and frame both forms a rigid structure and is used in the construction areas.
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
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.
To select the center of a line what option must be selected? A. Ortho B. Dynamic input C. Object snap
Answer:
The correct answer is c. object snap
Explanation:
In Autocad, the object snap is defined as a drawing aid that is used together with other different commands to help to draw accurately. It also allows snapping onto a specific object location when there is a picking point. and thus, it helps to select the center of a line.
In an adiabatic process, the temperature of the system remains constant. a)- True b)- False
Answer:
A. True
Explanation:
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
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
A Carnot cooler operates with COP = 11, whose ambient temperature is 300K. Determine the temperature at which the refrigerator absorbs heat.
Answer:
275 Kelvin
Explanation:
Coefficient of Performance=11
[tex]T_H=\text {Absolute Temperature of high temperature reservoir=300 K}[/tex]
[tex]T_L=\text {Absolute Temperature of low temperature reservoir}[/tex]
[tex]\text {Coefficient of performance for carnot cooler}\\=\frac {T_L}{T_H-T_L}\\\Rightarrow 11=\frac{T_L}{300-T_L}\\\Rightarrow 11(300-T_L)=T_L\\\Rightarrow 3300-11T_L=T_L\\\Rightarrow 3300=T_L+11T_L\\\Rightarrow 3300=12T_L\\\Rightarrow T_L=\frac {3300}{12}\\\Rightarrow T_L=275\ K\\\Therefore \text{Temperature at which the refrigerator absorbs heat=275 Kelvin}[/tex]
Fluid enters a device at 4 m/s and leaves it at 2 m/s. If there is no change in the PE of tihe flow, and there is no heat and (non-flow) work across boundaries of the device, what is the increase in specific enthalpyof the fluid (hg-hi) in kJ/kg? Assume steady state operation of the device.
Answer:
[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]
Explanation:
Now from first law for open system
[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]
Here given Q=0 ,w=0
So [tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}[/tex]
[tex]V_1=4 m/s,V_2=2 m/s[/tex]
[tex]h_1+\dfrac{4^2}{2000}=h_2+\dfrac{2^2}{2000}[/tex]
[tex]h_2-h_1=6\times 10^{-3}[/tex]
So increase in specific enthalpy
[tex]h_2-h_1=6\times 10^{-3}\frac{KJ}{Kg}[/tex]
Give two advantages of a four-high rolling mill opposed to a two-high rolling mill for the same output diameter.
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
A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twice of that of loose side, F1= 2F2. Please calculate the F1, Fe, and Fo.
Answer:
F₁ = 1500 N
F₂ = 750 N
[tex]F_{e}[/tex] = 500 N
Explanation:
Given :
Power transmission, P = 7.5 kW
= 7.5 x 1000 W
= 7500 W
Belt velocity, V = 10 m/s
F₁ = 2 F₂
Now we know from power transmission equation
P = ( F₁ - F₂ ) x V
7500 = ( F₁ - F₂ ) x 10
750 = F₁ - F₂
750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )
∴F₂ = 750 N
Now F₁ = 2 F₂
F₁ = 2 x F₂
F₁ = 2 x 750
F₁ = 1500 N , this is the maximum force.
Therefore we know,
[tex]F_{max}[/tex] = 3 x [tex]F_{e}[/tex]
where [tex]F_{e}[/tex] is centrifugal force
[tex]F_{e}[/tex] = [tex]F_{max}[/tex] / 3
= 1500 / 3
= 500 N
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
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.
The manufacturing department of a factory is the only department that needs to be practicing Lean Manufacturing philosophy. a)- True b)- False
A negative normal strain can be considered to increase or decrease volume depending on the coordinate system used. a)True b)- False
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.
The uniform wall thickness that is usually targeted for plastic injection molded parts is roughly: A. 0.5 mm B. 3 mm C. 7 mm D. 12 + mm
Answer:
B. 3 m
Explanation: For plastic injection moulded the thickness is generally between 2 mm to 3 mm
the wall is not too thick because during cooling process there should be defects so thickness of wall is no too high and there is also a problem if we use thicker wall that we need more material for moulding process so the thickness should be in between 2 to 3 mm which is in option B so option B will be the correct option
What is refrigeration capacity and what is meant by a "ton" of refrigeration?
Answer:
1 ton refrigeration =3.517 kJ/s = 3.517 kW
Explanation:
Refrigeration capacity is defined at the measure of the effective cooling capacity of a refrigerator which is expressed in Btu per hour or in tons.
1 ton capacity is a unit of air conditioning and refrigeration which measure the capacity of air conditioning and refrigeration unit.
One ton is equal to removal of 3025kcal heat per hour
1 ton refrigeration = 200 Btu/min = 3.517 kJ/s = 3.517 kW = 4.713 HP
Discuss the importance of dust and fluff removal from spinning mills and how it is realised in air conditioning plants
Answer:
Removal of dust and fluff from spinning mill is important as it has adverse and detrimental effects on the health of the workers in these industries. Tiny and microscopic particles of various substances present in the surrounding air is transferred from one place to another and these causes various respiratory diseases and pose health hazards for the workers and make work environment unhealthy and hazardous thus affecting the over all efficiency and productivity.
Cotton dust , the major pollutant, when breathed in affetcs the lungs badly and workers experience symptoms such as respiratory problems, coughing, tightness in chest, etc. Thus to ensure proper health of the workers spinning mills have been provided with powerful air conditioning to ensure purity of air, to maintain proper moisture levels and to ensure dust and fluff removal.
The dust and fluff laiden air is humidified, purified and then recirculated. Optimization of number of air changes/hour to clean air stream and prevent any health risk of the workers.
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?
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 the major limitations of melt spinning method? a)-good for limited alloys and limited shapes b)-good for limited shapes, high cost c)-good for limited thickness and limited shapes d)-high cost, high energy consumption
Answer:d
Explanation:
Melt spinning is the economical process to manufacture synthetic fiber.
Major limitations of Melt spinning is
only suitable for thermoplastic fiberHigh heat inputRequire periodic maintenance of machinePrecise control of temperature is must.Although it is very efficient method as it does not create any environment Pollution
Air enters a compressor at 100 kPa, 10°C, and 220 m/s through an inlet area of 2 m2. The air exits at 2 MPa and 240°C through an area of 0.5 m2. Including the change in kinetic energy, determine the power consumed by this compressor, in kW.
Answer:
Power consume by compressor=113,726.87 KW
Explanation:
Given:[tex]P_{1}=100KPa ,V_{1}=200 m/s,T_{1}=283 K, A_{1} =2m^2[/tex]
[tex]P_{2}=2000KPa ,T_{2}=513 K,A_{2}=0.5m^2[/tex]
Actually compressor is an open system, so here we will use first law of thermodynamics for open system .
We know that first law of thermodynamics for steady flow
[tex]h_{1}+\frac{V_{1} ^{2} }{2}+Q=h_{2}+\frac{V_{2} ^{2} }{2}+W[/tex]
We know that[tex]C_{p}=1.005\frac{Kj}{KgK}[/tex]and we take the air as ideal gas.
System is in steady state then mass flow rate in =mass flow rate out
Mass flow rate= [tex]density\times area\times velocity[/tex]
So mass flow rate =[tex]\rho _{1}V_{1}A_{1}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
=1.23×200×2 Kg/s
=541.17 Kg/s
[tex]\rho _{1}V_{1}A_{1}=\rho _{2}V_{2}A_{2}[/tex]
[tex]\rho _{2}=13.58\frac{Kg}{{m}^3}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
[tex]V_{2}[/tex]=80.07 m/s
Enthalpy of ideal gas h=[tex]C_{p}\times T[/tex]
So[tex] h_{1}=1.005\times283=284.41\frac{Kj}{Kg}[/tex]
[tex]h_{2}=1.005\times513=515.56\frac{Kj}{Kg}[/tex]
Now by putting the values
[tex]284.41+\frac{220 ^{2} }{2000}+Q=515.56+\frac{80.07 ^{2} }{2000}+W[/tex]
Here Q=0 because heat transfer is zero here.
W= -210.15 KJ/kg
So power consume by compressor=541.17×210.15
=113,726.87 KW
If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the above
Answer:
Explanation:
load = 4500lb lift height= 30 ft
time =15 s
velocity=[tex]\frac{30}{15}[/tex] ft/s
velocity=2 ft/s
power = force[tex]\times[/tex] velocity
power=[tex]{4500}\times2[/tex]
power= 9000 lb ft/s
1 hp= 550 lb ft/s
power= [tex]\frac{9000}{550} =16.36[/tex] hp
there are totally (___) slip system in an FCC metal.
Answer: 12
Explanation: In FCC metal lattice there are total four octahedral slip plane and six direction where each are common to two octahedral plane and so that gives total of three slip direction . So the total slip system is multiplication of the slip plane and the slip direction that is twelve.
Slip system=Slip plane×Slip direction
Slip system =4×3=12
Therefore there are total 12 slip system in the FCC metal lattice.
An oscillating mechanism has a maximum displacement of 3.2m and a frequency of 50Hz. At timet-0 the displacement is 150cm. Express the displacement in the general form Asin(wt + α).
Given:
max displacement, A = 3.2 m
f= 50 Hz
at t = 0, displacement, d = 150 cm = 1.5 m
Solution:
Displacement in the general form is represented by:
d = Asin(ωt ± α)
d = 3.2sin(2πft ± α)
d = 3.2sin(100πt ± α)
where,
A = 3.2 m,
ω = 2πf = 100π
Now,
at t = 0,
1.5 = 3.2sin(100π(0) ± α )
1.5 = 3.2sinα
sin α = [tex]\frac{1.5}{3.2}[/tex] = 0.4687
α = [tex]sin^{-1}(0.46875)[/tex] = 27.95° = 0.488 radian
Now, we can express displacement in the form of 'Asin(wt + α)' as:
d = 3.2sin(100πt ± 0.488 )
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) has the following dimensions: ID = 150mm, OD = 250 mm and thickness = 37 mm. What is the rotational speed in rpm that would lead to the flywheel's fracture?
Answer:
N = 38546.82 rpm
Explanation:
[tex]D_{1}[/tex] = 150 mm
[tex]A_{1}= \frac{\pi }{4}\times 150^{2}[/tex]
= 17671.45 [tex]mm^{2}[/tex]
[tex]D_{2}[/tex] = 250 mm
[tex]A_{2}= \frac{\pi }{4}\times 250^{2}[/tex]
= 49087.78 [tex]mm^{2}[/tex]
The centrifugal force acting on the flywheel is fiven by
F = M ( [tex]R_{2}[/tex] - [tex]R_{1}[/tex] ) x [tex]w^{2}[/tex] ------------(1)
Here F = ( -UTS x [tex]A_{1}[/tex] + UCS x [tex]A_{2}[/tex] )
Since density, [tex]\rho = \frac{M}{V}[/tex]
[tex]\rho = \frac{M}{A\times t}[/tex]
[tex]M = \rho \times A\times t[/tex][tex]M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t[/tex]
[tex]M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37[/tex]
[tex]M = 8252963901[/tex]
∴ [tex]R_{2}[/tex] - [tex]R_{1}[/tex] = 50 mm
∴ F = [tex]763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}[/tex]
F = 33618968.38 N --------(2)
Now comparing (1) and (2)
[tex]33618968.38 = 8252963901\times 50\times \omega ^{2}[/tex]
∴ ω = 4036.61
We know
[tex]\omega = \frac{2\pi N}{60}[/tex]
[tex]4036.61 = \frac{2\pi N}{60}[/tex]
∴ N = 38546.82 rpm
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
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
A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat from geothermal water that enters the evaporator at 500C at a rate of 0.065 kg/s and leaves at 40°C. The refrigerant enters the evaporator at 20°C with a quality of 23 percent and leaves at the inlet pressure as saturated vapor. The refrigerant loses 300 W of heat to the surroundings as it flows through the compressor and the refrigerant leaves the compressor at 1.4 MPa at the same entropy as the inlet. Determine: (a) The degrees of subcooling of the refrigerant in the condenser, b)-The mass flow rate of the refrigerant . (c) The heating load and the COP of the pump, and d)-The theoretical minimum power input to the compressor for the same heating load.
Answer:2.88
Explanation: