For a 4-bar linkage with ri =7-in, r2 =3-in, r3= 9-in, and r =8-in, determi the minimum and maximum transmission angles.
Answers
Answer:
[tex]\mu_{min}[/tex]=[tex]26.38^{\circ}[/tex]
[tex]\mu_{max}[/tex]=[tex]71.79^{\circ}[/tex]
Explanation:
[tex]r_{1}[/tex]=7 in, [tex]r_{2}[/tex]=3 in, [tex]r_{3}[/tex]=9in
,[tex]r_{4}[/tex]=8 in
Transmission angle (μ ):
It is the acute angle between coupler and the output (follower) link.
Here we consider link [tex]r_{1}[/tex] as fixed link ,[tex]r_{2}[/tex] as input link ,link [tex]r_{3}[/tex] as coupler and link [tex]r_{4}[/tex] as output link.
As we know that
[tex]\cos\mu_{max}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}-\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
[tex]\cos\mu_{min}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}+\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
When link [tex]r_{2}[/tex] will be horizontal in left side direction then transmission angle will be minimum and when link [tex]r_{2}[/tex] will be horizontal in right side direction then transmission angle will be maximum.
Now by putting the values we will find
[tex]\cos\mu_{max}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}-\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
[tex]\cos\mu_{max}=0.3125[/tex]
[tex]\mu_{max}=71.79^\circ[/tex]
[tex]\cos\mu_{min}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}+\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
[tex]\cos\mu_{min}=0.8958[/tex]
[tex]\mu_{min}=26.38^\circ[/tex]
Hence, The minimum and maximum angle of transmission angle is 26.38° and 71.79° respectively.
Related Questions
Water flows at the rate of 200 I/s upwards through a tapered vertical pipe. The diameter at Marks(3) CLO5) the bottom is 240 mm and at the top 200 mm and the length is 5m. The pressure at the bottom is 8 bar, and the pressure at the topside is 7.3 bar. Determine the head loss through the pipe. Express it as a function of exit velocity head.
Answers
Answer: 5.35m
Explanation:
By using energy equation:
[tex]\frac{P_1}{\gamma}+z_1+\frac{v_1^{2} }{2g} =\frac{P_2}{\gamma}+z_2+\frac{v_2^{2} }{2g}+h_{L}[/tex]
[tex]\gamma=specific weight[/tex]
[tex]v_{1} =\frac{Q_1}{A_1} =\frac{0.2}{\frac{\pi }{4} \times 0.24^2} =4.42 m/s\\v_{2} =\frac{Q_2}{A_2} =\frac{0.2}{\frac{\pi }{4} \times 0.2^2} =6.37 m/s[/tex]
[tex]h_{L}=\frac{P_1-P_2}{\gamma}+z_1-z_2+\frac{v_1^{2}-v_2^{2} }{2g}[/tex]
[tex]h_{L}=\frac{(8-7.3)\times 100 }{9.81} +0+5+\frac{4.42^2-6.37^2}{2\times 9.81}[/tex]
[tex]h_L=7.135+3.927\\h_L=11.062m[/tex]
exit velocity head = [tex]\frac{v_2^{2} }{2g}[/tex]=2.068m
head loss as a function of exit velocity head is=[tex]\frac{11.062}{2.068}[/tex]
[tex]h_L=K\times V_e[/tex]
head loss as a function of exit velocity head =5.35m
Why is a Screw Pump a quiet operating pump?
Answers
Answer:
Screw pumps like those that are used in viscous fluids transportation have lubricating properties. The fluid flows in a line axially without any disturbence. Since, the motion of fluid is free from any sort of rotation even at high speeds there is no turbulence and motion of the pump is quite. Thus screw Pumps provides smooth operation with extremely low pulsation, lower noise levels and higher efficiency.
An automobile having a mass of 884 kg initially moves along a level highway at 68 km/h relative to the highway. It then climbs a hill whose crest is 69 m above the level highway and parks at a rest area located there. For the automobile, determine its changes in kinetic energy, in kJ
Answers
Answer:
ΔK.E. = - 142.72 kJ
Explanation:
mass = 884 kg
initial velocity = 68 km/h = 68 \times \frac {5}{18} = 18.89 m/s
final velocity = 0 m/s
height = 69 m
change in kinetic energy :
ΔK.E. = [tex]\dfrac{1}{2}m(v_f^2-v_i^2)[/tex]
ΔK.E. =[tex]\dfrac{1}{2}\times 884 \times (0^2-18.89^2)[/tex]
ΔK.E. =-142,716.05 J
ΔK.E. =-142.72 kJ
hence change in kinetic energy of the automobile is -142.72 kJ
A 12-ft circular steel rod with a diameter of 1.5-in is in tension due to a pulling force of 70-lb. Calculate the stress in the rod
Answers
Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:
[tex] A=\pi*D^2/4 [/tex]
Replacing the diameter the area results:
[tex] A= 17.76 in^2 [/tex]
Therefore the the stress results:
[tex] σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi [/tex]
______number can be used to describe the relative growth of the hydraulic boundary layer and the thermal boundary layer. a) Reynolds b) Stanton c) Nusselt d) Prandtl e) Fourier
Answers
Answer: d) Prandtl number
Explanation: Prandtl number is basically defined as the ratio between the fluid's viscosity to the thermal conductivity.It doesn't have any sort of dimension. The fluids which are discovered with the small Prandtl numbers are considered as good fluids as they have a smooth rate of flow and as the number increases the fluid are not considered as reliable. Thus,option (d) is the correct option.
Burn rate can be affected by: A. Variations in chamber pressure B. Variations in initial grain temperature C. Gas flow velocity D. All of the above
Answers
Answer: D) All of the above
Explanation:
Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.
A refrigeration cycle rejects Qn 500 Btu/s to a hot reservoir at 540 R, while receiving c200 Btu/s at 240°R. This refrigeration cycle a)- is internally reversible b)- is irreversible c)- is impossible d)- cannot be determined
Answers
Answer:
(b) Irreversible cycle.
Explanation:
Given;
[tex]T_2=540R ,Q_2= 500 Btu/s ,T_1=240 R ,Q_1= 200 Btu/s [/tex]
To find the validity of cycle
[tex]\oint _R\frac{dQ}{T}\leq0[/tex]
If it is zero then cycle will be reversible cycle and if it is less than zero then cycle will be irreversible cycle.These are possible cycle.
If it is greater than zero ,then cycle will be impossible .
Now find
[tex]\dfrac{Q_1}{T_1}-\dfrac{Q_2}{T_2}=\dfrac{200}{240}-\dfrac{500}{540}[/tex]
[tex]\dfrac{Q_1}{T_1}-\dfrac{Q_2}{T_2}[/tex]= -0.09
It means that this cycle is a irreversible cycle.
Which of the following components of a PID controlled accumulates the error over time and responds to system error after the error has been accumulated? a)- Proportional b)- Derivative c)- Integral d)- ON/OFF.
Answers
Answer:
D Is the answer babe....
Answer:
d)- ON/OFF.
Explanation:
ON/OFF components of a PID controlled accumulates the error over time and responds to system error after the error has been accumulated.
What do you think are the advantages and disadvantages of isothermal constant volume high extension cycle? And how efficient do you think it can be?
Answers
Answer Explanation :
ADVANTAGES OF ISOTHERMAL PROCESS
EASE OF MAINTENANCE: there is absence of direct contact between the combustible gas mixture and all moving part in this process LESS NOISE : These engines does not have valves so it is very simple in construction and produce less noiseBETTER PERFORMANCE : these engine have better performance than other enginesLOW TEMPERATURE : these engine can work on low temperature which is also an advantageDISADVANTAGES OF ISOTHERMAL PROCESS :
GREATER VOLUME AND GREATER WEIGHT: These engine require large volume and large weightSLOWER START: In this process the engine have very slower start which is a disadvantage HIGHER ECONOMIC COST : In this process we need more money this is also an disadvantage
A disk-shaped part is to be cast out of aluminum. The diameter 500 mm and thickness = 20 mm. If the mold constant = 3.0 s/mm2 in Chvorinov's Rule, how long will it take the casting to solidify, in minutes?
Answers
Answer:
t =253.8s
Explanation:
Chvorinov's Rule can be written as:
[tex]t=B(\frac{V}{A} )^{n}[/tex]
where t is the solidification time,
V is the volume of the casting,
A is the surface area of the casting that contacts the mold,
n is a constant
B is the mold constant
The S.I. units of the mold constant B are s/m2.
According to Askeland, the constant n is usually 2.
[tex]V=\frac{\pi D^{2}h }{4} = 3.9*10^6[/tex]mm3
[tex]As=\pi D h+2\frac{\pi }{4} D^{2} =0.424*10^6[/tex] mm2
[tex]V/A=9.198[/tex]mm
[tex]t = 3.0*9.198^2[/tex] =253.8s
Answer:
Chvorinov's Rule with Askeland Method: t = 4.286694102 minutes
Chvorinov's Rule with Degarmo Method:
Minimum time required at constant n = 1.5 : t = 1.408751434 minutesMaximum time required at constant n = 2.0 : t = 4.286694102 minutesExplanation:
Data:
Aluminum disc
Diameter (D) = 500 mm
Thickness = Height (h) = 20 mm
Mold Constant (C) = 3.0 sec / [tex]mm^{2}[/tex]
Required:
Solidification time (t) in minutes = ?
Formula:
The solidification time can be found by using the Chvorinov's Rule:
[tex]t = C (\frac {V}{A})^{n}[/tex]
Where;
t = solidification time
C = mold constant
V = Volume of disc
A = Surface area of disc
n = constant
Note: According to Askeland n = 2.0 and According to Degarmo n varies 1.5 to 2.0 therefore , we will do for both method and by Degarmo method we can predict maximum and minimum solidification time.
Solution:
First, we will find the volume of the disc
disc = cylinder
therefore, Volume of cylinder is given by:
[tex]V = \frac{\pi }{4} * D^{2} * H[/tex]
Where:
V = Volume of Cylinder
H = Height of disc
D = Diameter of disc
Now, putting dimensional values in above equation
[tex]V = \frac{\pi }{4} * 500^{2} *20[/tex]
V = 3926990.817 [tex]mm^{3}[/tex]
Second, we will find the surface area of the disc
Therefore, surface area of cylinder is given by:
[tex]A = (\pi * D * H) + (2 * \frac{\pi }{4} * D^{2} )[/tex]
Where:
A = Surface area of disc
D = Diameter of disc
H = Height of disc
Now, putting dimensional values in above equation
[tex]A = (\pi * 500 * 20) + (2 * \frac{\pi }{4} * 500^{2} )[/tex]
A = 424115.0082 [tex]mm^{2}[/tex]
Finally, Moving towards the final solution
Chvorinov's Rule with Askeland Method n = 2:Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]
t = 257.2016461 seconds
Converting to minutes
t = 4.286694102 minutes
Chvorinov's Rule with Degarmo Method n = 1.5 (Minimum Solidification Time)Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{1.5}[/tex]
t = 84.52508604 seconds
Converting to minutes
t = 1.408751434 minutes
Chvorinov's Rule with Degarmo Method n = 2.0 (Maximum Solidification Time)
Rewriting the equation:
[tex]t = C (\frac {V}{A})^{2}[/tex]
Putting the dimensional and constants values in the equation
[tex]t = 3.0 (\frac {3926990.817}{424115.0082})^{2}[/tex]
t = 257.2016461 seconds
Converting to minutes
t = 4.286694102 minutes
For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) False
Answers
Answer:
(b)False
Explanation:
Given:
Prandtl number(Pr) =1000.
We know that [tex]Pr=\dfrac{\nu }{\alpha }[/tex]
Where [tex]\nu[/tex] is the molecular diffusivity of momentum
[tex]\alpha[/tex] is the molecular diffusivity of heat.
Prandtl number(Pr) can also be defined as
[tex]Pr=\left (\dfrac{\delta }{\delta _t}\right )^3[/tex]
Where [tex]\delta[/tex] is the hydrodynamic boundary layer thickness and [tex]\delta_t[/tex] is the thermal boundary layer thickness.
So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
So hydrodynamic layer will be thicker than the thermal boundary layer.
I need help on Problem 2.5
I would like an explanation of how you got your answer. Thank you!!
Answers
Answer:
0.424
Explanation:
The electrical energy is the electrical power times time:
E = Pt
E = (IV)t
E = (1.5 A) (110 V) (300 s)
E = 49,500 J
The heat absorbed by the cookie dough is the mass times specific heat capacity times increase in temperature:
Q = mCΔT
Q = (1 kg) (4200 J/kg/K) (5 K)
Q = 21,000 J
So the fraction of electrical energy converted to internal energy is:
Q / E = 21,000 / 49,500
Q / E = 0.424
The remainder of the electrical energy is used to do work.
Which of the following is/are not a common crystal structure in metals (mark all that apply)? a. Face-centered cubic (FCC) b. Face-centered orthorhombic (FCOR) C. Body-centered cubic (BCC) (HCP)
Answers
Answer: b) FCOR( Face-centered orthorhombic)
Explanation: Face centered orthorhombic lattice is the lattice that has eight lattice corner points and they also have each face with center lattice point.This lattice structure is usually not common in metals.Whereas the face centered lattice structure , body structure lattice and hexagonal close packed lattice are common in metal.Thus option (b) is the correct option.
A composite material is a mix of two different materials such as ceramics and metals fused together to the atomic level to form another substance with more improved propertied.a)-True b)-False
Answers
Answer:
The correct option is (A) TRUE
Explanation:
A composite material is the material having characteristics that are improved and different from the constituting materials. A composite material is produced by combining two or more materials. The constituting materials have significantly different physical or chemical properties.
Some of the composite materials include-Ceramic matrix composites, Metal matrix composites etc.
A ceramic matrix composite is formed by embedding ceramic fibers in a ceramic matrix.
Also, a tungsten carbide- cobalt is a metal matrix composite, in which the tungsten carbide and cobalt metal matrix are fused together.
Therefore, the given statement: A composite material is a mix of two different materials that are fused together on the atomic level to form a new substance with improved properties is TRUE.
Describe the steps involved in accomplishing a life-cycle cost analysis (LCCA). I have listed the steps. just describe. a)-Operational Performance. b)-Salvage externalities c)- Value vs. Risk d)- Initial expenditure e)- Maintenance implications
Answers
Explanation:
a). Operational Performance
It is defined as the parameter that describes the percentage of accuracy of performing an operation or carrying out an activity.
b). Salvage externalities
Salvage in Life cycle cost analysis is a process of estimating the value of the remaining assets in the organisation,.
c). Value Vs Risk
When we take risk in doing any activity we know the value of accomplising the activity. So value relates directly with risk. When the value of a certain task is high, the risk involve in it is also high.
d). Initial expenditure
Initial expenditure is nothing but the cost involve in starting a particular acitvity or task at the starting phase.
e). Maintenance implications
It lays emphasis in maintaining the cost of every possible parameters that are involve in the activity. It includes labour, machines, positions, energy, facilities, etc.
What is the Principle of Entropy Increase?
Answers
Answer and Explanation:
The increase in entropy principle is defined as the process in which the total change in entropy of system with its adiabatic surroundings is always positive or equal to zero. The increase in entropy mostly takes place when a solid becomes liquid because randomness is increases when solid becomes liquid so entropy is also increases.
example of increase of entropy is when solid burns and become ash, ice melting
Which atom bond in atomic interaction combines electrons, filling its valence zone a) Van der Vaals bond; b) a covalent bond; c) ionic bond; d) metallic bond.
Answers
Answer: Covalent bond
Explanation: Covalent bond is the bond that gets created when there is a sharing of electrons among atoms and hence creating atomic interaction. The bond formed is from the shared pair because they allow the atoms or ions to achieve stability by completely filling the outer shell of the electron and thus form the covalent bond .Therefore, the correct option is the option(b) .
Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Coercion
Answers
Answer: d)Coercion
Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)
What i s the value of a capacitor with 250 V applied and has 500 pC of charge? (a) 200 uF (b) 0.5 pF (c) 500 uF (d) 2 pF
Answers
Answer:
(d) 2 pF
Explanation: the charge on capacitor is given by the expression
Q=CV
where Q=charge
C=capacitance
V=voltage across the plate of the capacitor
here we have given Q=500 pF, V=250 volt
using this formula C=[tex]\frac{Q}{V}[/tex]
=500×[tex]10^{-12}[/tex]×[tex]\frac{1}{250}[/tex]
=2×[tex]10^{-12}[/tex]
=2 pF
The number-average molecular weight of a poly(styrene-butadiene alternating copolymer is 1,350,000 g/mol; determine the average number of styrene and butadiene repeat units per molecule.
Answers
To find the average number of styrene and butadiene units in a poly(styrene-butadiene) copolymer with Mn of 1,350,000 g/mol, add the molecular weights of styrene and butadiene to get the molecular weight of the repeat unit (158.24 g/mol) and divide Mn by this number to get approximately 8530 repeat units, which includes about 4265 of each monomer type.
Explanation:To calculate the average number of styrene and butadiene repeat units per molecule for a poly(styrene-butadiene) copolymer with a number-average molecular weight (Mn) of 1,350,000 g/mol, we need the molecular weights of the monomer units. The molecular weight of styrene (C8H8) is approximately 104.15 g/mol, and the molecular weight of butadiene (C4H6) is approximately 54.09 g/mol. Since the copolymer is alternating, each repeat unit consists of one styrene and one butadiene unit.
To find the total molecular weight of the repeat unit, we add the molecular weights of styrene and butadiene: 104.15 g/mol + 54.09 g/mol = 158.24 g/mol. Dividing the number-average molecular weight of the copolymer by the molecular weight of the repeat unit gives us the average number of repeat units per molecule: 1,350,000 g/mol \/ 158.24 g/mol \approximately 8530 repeat units\.
The average number of styrene units per molecule will be approximately 4265, and the same for butadiene, since there is one of each in every repeat unit in an alternating copolymer structure.
What does WCS stand for? A. Western CAD System B. Worldwide Coordinate Sectors C. World Coordinate System D. Wrong CAD Settings
Answers
Answer:
The correct answer is C. World Coordinate System
Explanation:
The World Coordinate System has to do with that coordinate system which is fixed in the activities of the CADing. There is a default system in which we can refer to them as soon as we want to manipulate the objects and add new elements.
What is the uncertainty in position of an electron of an atom if there is t 2.0 x 10' msec uncertainty in its velocity? Use the reduced Planck's constant and electron mass 9.19 x 103 kg.
Answers
Answer:
18931.4
Explanation:
Given : velocity of the electron = 2.0 [tex]\times[/tex]10
mass of the electron = 9.19[tex]\times[/tex] 103
we know that reduced planks constant, h = 6.5821[tex]\times[/tex] [tex]10^{-16}[/tex] eV s
We know from uncertainity principle,
[tex]\Delta \textup{x}.\Delta \textup{v} = \frac{h}{\dot{m}}[/tex]
[tex]\Delta \textup{x} = \frac{h}{\dot{m}\times \Delta \textup{v}}[/tex][tex]\Delta \textup{x} = \frac{6.5821\times 10^{-16}}{9.19\times 103\times 2.0\times 10}[/tex]
[tex]\Delta \textup{x}[/tex] = 18931.4 m
Hence, uncertainty in position of the electron is 18931.4
Design a solid steel shaft to transmit 14 hp at a speed of 2400rpm f the allowable shearing stress is given as 3.5 ksi.
Answers
Answer:
Is required a 0.8 inches diameter steel shaft.
Explanation:
With the power P and the rotating speed n (RPM), we can find the torque applied:
T = P/N
Before calculating the torque, we convert the power and rotating speed units:
[tex]P = 14\ HP * 550\ \frac{\frac{lb.ft}{s}}{HP} *\frac{12\ in}{ft} = 92400\ \frac{lb.in}{s} [/tex]
[tex]n=2400\ RPM .\frac{2\pi/60\frac{rad}{s}}{RPM}= 251\frac{rad}{s}[/tex]
Replacing the values, the torque obtained is:
[tex]T = \frac{92400\ lb.in/s}{251\ rad/s} = 368\ \ lb.in[/tex]
Then the maximum shearing stress will be located at the edge of the shat at the Maximus radius:
[tex]Smax =\ \frac{T.R}{J}[/tex]
Here J is the moment of inertia and R a radius. For a solid shaft, it is calculated by:
[tex]J =\frac{\pi.D^4}{32}[/tex]
Where D is shaft's diameter. Replacing the expression of J in
[tex]Smax =\frac{T.R}{\frac{\pi.D^4}{32}}[/tex]
As the radius is half of the diameter:
[tex]Smax =\frac{T.D}{\frac{2*\pi.D^4}{32\\} } = \frac{16T}{\pi.D^3}[/tex]
For the maximum stress of 3.5 ksi (3500 psi = 3500\ lb/in^2) and the calculated torque:
[tex]Smax = \frac{16.368\ lb.in}{3500\ lb/in^2*\pi.D^3}[/tex]
Solving for D:
[tex]D =\sqrt[3]{16.368\ lb.in / (3500\pi\ lb/in^2)}} = 0.8\ in[/tex]
Injector orifice patterms and size will affect propellant mixing and distribution. a)-True b)-False
Answers
Answer: True
Explanation: Injector orifice is the factor which describes the size of the opening of the injector .There are different pattern and size of the opening for the injector which affects the mixture of the chemical substance that is used for the production of the energy that is known as propellant.
The pattern and size of the orifice will define the variation in the amount of energy that could be produced.Thus the statement given is true.
For high temperature deformation, the bigger the gran sine, the higher the creep rate. a)-True b)- False
Answers
Answer:
The given statement is False.
Explanation:
This is because at high temperature the creep rate depends on grain boundary area, increasing with an increase in grain boundary area thus decreasing the grain size. Thus at higher temperatures, grain size has opposite effect , the bigger the grain size the slower the slower is the creep rate, the larger the grain boundary area.
Answer:
a
Explanation:
a
An actual vapour compression system comprises following process represents a. 1-2 Compression process b. 2-3 Condens 1 (or heat rejection from the condenser) c. 3-4 Irreversible expansion d. 4-1 Evaporation (or) heat addition to the evaporator Sketch the processes on T-S diagram.
Answers
Answer:
Explanation:
The deatailed diagram of VCRS is given below such
1-2=Isentropic compression in which temperature increases at constant entropy
2-3=Isobaric heat rejection i.e. heat rejected at constant pressure(condensation)
3-4=Irreversible expansion or throttling in which enthalpy remains constant
4-1=Isobaric heat addition(Evaporation)
The contact angle between the mercury surface and capillary tube wall is______ A) Less than 90 B) Equal to 90 C) Greater than 90 D) Varying with mercury level
Answers
Answer:
The Answer to the question is :
Explanation:
The contact angle between the mercury surface and capillary tube wall is Greater than 90.
If the surface of the solid is hydrophobic, the contact angle will be greater than 90 °. On very hydrophobic surfaces the angle can be greater than 150º and even close to 180º.
In closed forging, why is flash needed and what is the effect with the overall pressure?
Answers
Explanation:
Closed die forging:
This is also known as impression die forging.By using the two die ,impression are made to produce desired forge product.A small gap provides between these two dies ,this small gap is called flash gutter.By the help of flash gutter excess ,material can flow and forms flash.
This flash plays an important role in the closed die forging.Generally friction is high in the flash gutter due to high ratio of length to thickness.So material feels high pressure and confirm the filling of die cavity.
Sketch and label a simple reheat cycle along with the appropriate T-s diagram?
Answers
Answer:
A reheat cycle is used to increase the overall efficiency of the power plant.
Explanation:
The Reheat cycle is used --
1. to increase the efficiency of the turbine
2. to maintain the quality of the steam
3. to provide higher pressure ratio
The reheat cycle is used to to increase the net work output of the power plant.In reheat cycle, two turbines , one low pressure turbine and one high pressure turbine is used to increase the efficiency. The main purpose of the reheat cycle is to maintain the quality of the steam to 0.85 at the exit of the turbine.
The figure given below shows a reheat rankine cycle.
In the cycle,
Process 1-2 is high pressure turbine
Process 2-3 is the reheater
Process 3-4 is the low pressure turbine
Process 4-5 is condenser
Process 5-6 is pump
Process 6-1 is boiler
A reheat rankine cycle is 30 -40% efficient than a simple rankine cycle.
An important material for advanced electronic technologies is the pure silicon.a)-True b)-False
Answers
Answer: False
Explanation:
Pure silicon is a good semiconductor, which conducts electricity when is mixed with some other component and can also work as an insulator.It is found abundantly on the earth's crust area. It is widely used in mixed form in the advanced electronic technologies but not in pure silicon form . So silicon is the important material in the advanced electronic technologies but not the pure silicon form.