Given:
diameter of sphere, d = 6 inches
radius of sphere, r = [tex]\frac{d}{2}[/tex] = 3 inches
density, [tex]\rho}[/tex] = 493 lbm/ [tex]ft^{3}[/tex]
S.G = 1.0027
g = 9.8 m/ [tex]m^{2}[/tex] = 386.22 inch/ [tex]s^{2}[/tex]
Solution:
Using the formula for terminal velocity,
[tex]v_{T}[/tex] = [tex]\sqrt{\frac{2V\rho g}{A \rho C_{d}}}[/tex] (1)
[tex](Since, m = V\times \rho)[/tex]
where,
V = volume of sphere
[tex]C_{d}[/tex] = drag coefficient
Now,
Surface area of sphere, A = [tex]4\pi r^{2}[/tex]
Volume of sphere, V = [tex]\frac{4}{3} \pi r^{3}[/tex]
Using the above formulae in eqn (1):
[tex]v_{T}[/tex] = [tex]\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}[/tex]
[tex]v_{T}[/tex] = [tex]\sqrt{\frac{2gr}{3C_{d}}}[/tex]
[tex]v_{T}[/tex] = [tex]\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}[/tex]
Therefore, terminal velcity is given by:
[tex]v_{T}[/tex] = [tex]\frac{27.79}{\sqrt{C_d}}[/tex] inch/sec
I need help on Problem 2.5
I would like an explanation of how you got your answer. Thank you!!
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.
Injector orifice patterms and size will affect propellant mixing and distribution. a)-True b)-False
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
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
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
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
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
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.
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?
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
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
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]
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.
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.
Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Coercion
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)
In closed forging, why is flash needed and what is the effect with the overall pressure?
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?
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 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
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
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.
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) .
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.
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
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.
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.
For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) False
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.
Which of these is not applicable to tuning for some Overshoot on Start-up? Select one: 1)-No overshoot during normal modulating control only 2)-KC-0.2 KU 3)- KC 0.33 KU 4)- KC 0.59 KU 5)- KC-0.78 KU
Answer: 5) KC-0.78 KU
Explanation:
KC-0.78 KU is not defined for tuning of overshoot on the Start-up as, this method is only applicable for there is some turning constants and also there is no overshoot during the normally modulating control. But some overshoot at start up are applicable. For the continuous cycling method, some overshoot is same as for closed loop tuning.
______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
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.
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
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 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.
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
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
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.
Primary Creep: slope (creep rate) decreases with time
Answer:
true
Explanation:
Creep is known as the time dependent deformation of structure due to constant load acting on the body.
Creep is generally seen at high temperature.
Due to creep the length of the structure increases which is not fit for serviceability purpose.
When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.
When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.
A wooden block (SG = 0.6) floats in oil (GS = 0.8). What fraction of the volume of the block is submerged in oil?
Answer:
3/4 th fraction of the volume of block is submerged in oil.
Explanation:
We know that
density of the block, ρ[tex]_{b}[/tex] =SG[tex]\times[/tex]density of water
= 0.6[tex]\times[/tex] 1000
= 600 kg/[tex]m^{3}[/tex]
density of the oil, ρ[tex]_{o}[/tex] =SG [tex]\times[/tex] denity of water
= 0.8[tex]\times[/tex] 1000
= 800 kg/[tex]m^{3}[/tex]
Let acceleration due to gravity, g = 9.81 m/[tex]s^{2}[/tex]
Volume of block submerged in oil is [tex]V_{1}[/tex]
Volume of block above the oil surface is [tex]V_{2}[/tex]
The total volume of the block is V = [tex]V_{1}[/tex]+[tex]V_{2}[/tex]
Therefore for a partially submerged body, we know that
Buoyant force = total weight of the block
and
total weight of the block = Weight of the fluid displaced by the block
ρ[tex]_{b}[/tex]\timesV\timesg = ρ[tex]_{o}[/tex]\times[tex]V_{1}[/tex]\times g
600([tex]V_{1}[/tex]+[tex]V_{2}[/tex]) = 800 [tex]V_{1}[/tex]
600[tex]V_{1}[/tex] + 600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex]
600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex] - 600[tex]V_{1}[/tex]
600[tex]V_{2}[/tex] = 200[tex]V_{1}[/tex]
[tex]\frac{V_{1}}{V_{2}}[/tex] = [tex]\frac{600}{200}[/tex]
Thus [tex]V_{1}[/tex] = 600
[tex]V_{2}[/tex] = 200
V = 600+200
= 800
Therefore fraction of volume of the block submerged in oil is,
[tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{600}{800}[/tex]
[tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{3}{4}[/tex]
The heat rate is essentially the reciprocal of the thermal efficiency. a)- True b)- False
Answer:
a). TRUE
Explanation:
Thermal efficiency of a system is the defined as the ratio of the net work done to the total heat input to the system. It is a dimensionless quantity.
Mathematically, thermal efficiency is
η = net work done / heat input
While heat rate is the reciprocal of efficiency. It is defined as the ratio of heat supplied to the system to the useful work done.
Mathematically, heat rate is
Heat rate = heat input / net work done
Thus from above we can see that heat rate is the reciprocal of thermal efficiency.
Thus, Heat rate is reciprocal of thermal efficiency.
An important material for advanced electronic technologies is the pure silicon.a)-True b)-False
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.
The Molybdenum with an atomic radius 0.1363 nm and atomic weight 95.95 has a BCC unit cell structure. Calculate its theoretical density (g/cm^3).
Solution:
Given :
atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m
atomic wieght, M = 95.96
Cell structure is BCC (Body Centred Cubic)
For BCC, we know that no. of atoms per unit cell, z = 2
and atomic radius, r =[tex]\frac{a\sqrt{3} }{4}[/tex]
so, a = [tex]\frac{4r}{\sqrt{3}}[/tex]
m = mass of each atom in a unit cell
mass of an atom = [tex]\frac{M}{N_{A} }[/tex],
where, [tex]N_{A}[/tex] is Avagadro Number = 6.02×10^{23}
volume of unit cell = a^{3}
density, ρ = [tex]\frac{mass of unit cell}{volume of unit cell}[/tex]
density, ρ = [tex]\frac{z\times M}{a^{3}\times N_{A}}[/tex]
ρ = [tex]\frac{2\times 95.95}{(\frac{4\times 0.1363\times 10^{-9}}{\sqrt{3}})^{3}\times (6.23\times 10^{23})}[/tex]
ρ = 10.215gm/[tex]cm^{3}[/tex]
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)
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.
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
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.
Why is a Screw Pump a quiet operating pump?
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.
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.
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]