Answer:[tex]\mu =9\times 10^{-3}Pa-s[/tex]
Explanation:
Distance between Plates(dy)=4.5 mm
Relative Velocity(du)=5 m/s
We know shear stress is given by [tex]\tau =10 Pa[/tex]
[tex]\tau =\frac{\mu du}{dy}[/tex]
where du=relative Velocity
dy=Distance between Plates
[tex]10=\frac{\mu \times 5}{4.5\times 10^{-3}}[/tex]
[tex]\mu =9\times 10^{-3}Pa-s[/tex]
The horizontal component of acceleration, ay during a projectile motion is usually assumed to be_________ a)-9.81 m/s^2 b)-Zero c)- Constant d)- 32.2 m/s^2
Answer:
b) zero
Explanation:
The horizontal component of acceleration during projectile motion is usually assumed to be zero.Because in projectile motion horizontal component of velocity will remain the constant and we know that rate of change of velocity with time is called acceleration.So when velocity is constant then acceleration will be zero.
In projectile motion ,gravitational acceleration will be in only vertical direction.
The barrel of a bicycle tire pump becomes quite warm during use. Explain the mechanisms responsible for the temperature increase.
Answer:
The air heats up when being compressed and transefers heat to the barrel.
Explanation:
When a gas is compressed it raises in temperature. Assuming that the compression happens fast and is done before a significant amount of heat can be transferred to the barrel, we could say it is an adiabatic compression. This isn't exactly true, it is an approximation.
In an adiabatic transformation:
[tex]P^{1-k} * T^k = constant[/tex]
For air k = 1.4
SO
[tex]P0^{-0.4} * T0^{1.4} = P1^{-0.4} * T1^{1.4}[/tex]
[tex]T1^{1.4} = \frac{P1^{0.4} * T0^{1.4}}{P0^{0.4}}[/tex]
[tex]T1^{1.4} = \frac{P1}{P0}^{0.4} * T0^{1.4}[/tex]
[tex]T1 = T0 * \frac{P1}{P0}^{0.4/1.4}[/tex]
[tex]T1 = T0 * \frac{P1}{P0}^{0.28}[/tex]
SInce it is compressing, the fraction P1/P0 will always be greater than one, and raised to a positive fraction it will always yield a number greater than one, so the final temperature will be greater than the initial temperature.
After it was compressed the hot air will exchange heat with the barrel heating it up.
What is meant by the thickness to chord ratio of an aerofoil?
Answer:
Chord ratio:
It is the ratio of thickness to the chord.It is also knows as thickness ratio.
Chord ratio measure the performance of wing when it is operating at the transonic speed.
When the speed cross the speed of sound wave then that wave creates the shock wave and these shock leads to produce drag force on the aerofoil profile.When Mach number is one then it means that if Mach increase then it will leads increase then drag force.
When a 2-kg block is suspended from a spring, the spring is stretched a distance of 20 mm. Determine the natural frequency and the period of vibration for a 0.5-kg block attached to the same spring
Answer:
7.05 Hz
Explanation:
The natural frequency of a mass-spring system is:
[tex]f = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}[/tex]
To determine the constant k of the spring we use Hooke's law:
Δl = F / k
k = F / Δl
In the first case the force was the weight of the 20 kg mass and the Δl was 20 mm.
F = m * a
F = 2 * 9.81 = 19.6 N
Then:
k = 19.6 / 0.02 = 980 N/m
Therefore:[tex]f = \frac{1}{2 \pi}\sqrt{\frac{980}{0.5}} = 7.05 Hz[/tex]
An object at a vertical elevation of 20 m and a speed of 5 m/s decreases in elevation to an elevation of 1 m. At this location, the object has a velocity of 15 m/s. The mass of the object is 68 kg. Assuming the object is the system, determine if there is any work transfer associated with the object (there is no heat transfer). The object is solid, incompressible and its temperature does not change during the process. If there is work transfer, is work done on or by the object? Assume the acceleration of gravity g = 9.81 m/s2.
Answer with Explanation:
We know that from the principle of work and energy we have
Work done on/by a body =ΔEnergy of the body.
Now as we know that energy of a body is the sum of it's kinetic and potential energy hence we can find out the magnitude of the final and initial energies as explained under
[tex]E_{initial}=P.E+K.E\\\\E_{initail}=mgh_1+\frac{1}{2}mv_1^{2}\\\\68\times 9.81\times 20+\frac{1}{2}\times 68\times (5)^{2}=14191.6Joules[/tex]
Similarly the final energy is calculated to be
[tex]E_{final}=P.E+K.E\\\\E_{final}=mgh_2+\frac{1}{2}mv_2^{2}\\\\68\times 9.81\times 1+\frac{1}{2}\times 68\times (15)^{2}=8317.08Joules[/tex]
As we can see that the energy of the object has changed thus by work energy theorem we conclude that work transfer is associated with the object.
Part 2)
The change in the energy of the body equals [tex]8317.08-14191.6=-5874.52Joules[/tex]
Since the energy is lost by the system hence we conclude that work is done by the object.
The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k= 0.029 W/ m K. The measured temperature difference across a 25-mm-thick sheet of the material is T1 - T2 = 12°C. a. What is the heat flux through a 3 mx 3 m sheet of the insulation? b. What is the rate of heat transfer through the sheet of insulation? c. What is the thermal resistance of the sheet due to conduction?
Answer:
a. The heat flux through the sheet of insulation is 19.92 W/m^2
b. The rate of heat transfer through the sheet of insulation is 125.28 W
c. The thermal resistance of the sheet due to the conduction is 0.86 Km^2/W.
Explanation:
From the heat conduction Fourier's law it can be state for a wall of width e and area A :
q = Q/ΔT = k*A* (T2-T1)/e
Where q is the rate of heat transfer, k the conductivity constant, and T2 and T1 the temperatures on the sides of the wall. Replacing the values in the correct units, we obtained the rate of heat transfer:
q = 0.029 W/*mK * (3m*3m) * (12°K) / (0.025m)
(The difference in temperatures in Kelvin is the same than in Celcius degres).
q = 0.029 W/*mK * (9 m^2) * (12°K) / (0.025m) = 125.28 W
The heat flux is calculated by dividing q by the area of the wall:
q/A = k* (T2-T1)/e = 0.029 W/*mK * (12°K) / (0.025m) = 19.92 W/m^2
The thermal resistance of the sheet is defined as:
R = e / k
Replacing the values in the proper units:
R = 0.025 m / 0.029 W/*mK = 0.86 Km^2/W
What is the specific volume of H2O at 1000F and 2000 psia?
Answer:
The specific volume of H2O at 1000F and 2000 psia is 0.3945 ft^3/lb
Explanation:
In figure you can see that for 2000 psia saturation temperature of water is 636 F, so at 1000 F water is at vapor phase. Then, we have to use superheated steam table. From the table the specific volume of H2O at 1000F and 2000 psia is 0.3945 ft^3/lb
Answer:
[tex]0.3945\frac{ft^{3}}{lb}[/tex]
Explanation:
The specific volume is the inverse of the density, so
[tex]v=\frac{1}{d}[/tex]
[tex]d=\frac{m}{V}[/tex]
[tex]v=\frac{V}{m}[/tex]
For superheated steam you can use the table and locate the temperature an pressure given by te problem, in its correspondent value that is [tex]0.3945\frac{ft^{3}}{lb}[/tex]
The Danner process is used to make glass_____
Answer:
The Danner process is used to make glass tubes
Explanation:
In this process the molten glass passes from the feeder to a hollow ceramic cylinder that is tilted downward, so that the molten glass can "run"; This ceramic cylinder is spinning constantly.
As the molten glass passes through the Danner tube, compressed air is blown to prevent the glass from being damaged. On the opposite side of the feeder, i.e, on the other side of the tube a bubble is also formed so-called drawing onion, the glass tube is removed in a free buckling in a traction line from the onion.
When the stretching speed remains constant and an increase in the blowing pressure occurs, larger tube diameters are created and the wall thickness decreases.
This process allows tube diameters between 2 and 60 mm.
This is how glass bottles are made.
What are primary and secondary super-heaters?
Explanation:
Superheater has two types of parts which are:
The primary super-heater The secondary super-heaterPrimary super-heater is first heater which is passed by the steam after steam comes out of steam drum.
After steam is heated on super primary heater, then the steam is passed on secondary super-heater so to be heated again. Thus, on secondary super-heater, the steam formed is hottest steam among others.
Steam from secondary super-heater which becomes the superheated steam, flow to rotate the High-Pressure Turbine.
A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a conical section at the bottom, right circular cylindrical mid-section and a hemispherical dome to cover the top. The radius of the tank is 1.5 m, the cylindrical side-walls will be 4.0 m in height, and the apex of the conic section at the bottom has an included angle of 60°. If the tank is filled to the top of the cylindrical side-walls, what is the tank capacity in liters?
Answer:
The volume up to cylindrical portion is approx 32355 liters.
Explanation:
The tank is shown in the attached figure below
The volume of the whole tank is is sum of the following volumes
1) Hemisphere top
Volume of hemispherical top of radius 'r' is
[tex]V_{hem}=\frac{2}{3}\pi r^3[/tex]
2) Cylindrical Middle section
Volume of cylindrical middle portion of radius 'r' and height 'h'
[tex]V_{cyl}=\pi r^2\cdot h[/tex]
3) Conical bottom
Volume of conical bottom of radius'r' and angle [tex]\theta [/tex] is
[tex]V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}[/tex]
Applying the given values we obtain the volume of the container up to cylinder is
[tex]V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}[/tex]
Hence the capacity in liters is [tex]V=32.355\times 1000=32355Liters[/tex]
The Reynolds number is a dimensionless group defined for a fluid flowing in a pipe as Re Durho/μ whereD is pipe diameter, u is fluid velocity, rho is fluid density, and μ is fluid viscosity.When the value of the Reynolds number is less than about 2100, the flow is laminar—that is, the fluid flows in smooth streamlines. For Reynolds numbers above 2100, the flow is turbulent, characterized by a great deal of agitation. Liquid methyl ethyl ketone (MEK) flows through a pipe with an inner diameter of 2.067 inches at an average velocity of 0.48 ft/s. At the fluid temperature of 20°C the density of liquid MEK is 0.805 g/cm3 and the viscosity is 0.43 centipoise [1 cP 1:00 103 kg/ m s]. Without using a calculator, determine whether the flow is laminar or turbulent. Show your calculations.
The Reynolds number is approximately 14,067, indicating turbulent flow of MEK through the pipe, as Re > 2100 denotes turbulence.
let's calculate the Reynolds number for the flow of liquid methyl ethyl ketone (MEK) through the pipe.
Given:
- Pipe diameter (D) = 2.067 inches = 0.0512 meters (converted from inches to meters)
- Fluid velocity (u) = 0.48 ft/s = 0.1463 meters/s (converted from feet per second to meters per second)
- Fluid density (ρ) = 0.805 g/cm³ = 805 kg/m³ (converted from grams per cubic centimeter to kilograms per cubic meter)
- Fluid viscosity (μ) = 0.43 centipoise = 0.43 x 10² kg/(m*s)(converted from centipoise to kg/(m*s))
The Reynolds number (Re) is calculated as:
[tex]\[ Re = \frac{(D \cdot u \cdot \rho)}{\mu} \][/tex]
Plugging in the values:
[tex]\[ Re = \frac{(0.0512 \, \text{m} \cdot 0.1463 \, \text{m/s} \cdot 805 \, \text{kg/m³})}{0.43 \times 10^{-3} \, \text{kg/(m*s)}} \][/tex]
[tex]\[\text{Re} = \frac{(0.0074991 \, \text{m}^2/\text{s}^2 \cdot 805 \, \text{kg/m}^3)}{0.43 \times 10^{-3} \, \text{kg/(m*s)}}\]\[\text{Re} = \frac{6.0498755 \, \text{kg/(m*s)}}{0.43 \times 10^{-3} \, \text{kg/(m*s)}}\]\[\text{Re} \approx \frac{6.0498755}{0.00043}\]\[\text{Re} \approx 14,067.35\][/tex]
Since the calculated Reynolds number is greater than 2100 (Re > 2100), the flow of MEK through the pipe is turbulent.
The energy conversion that takes place in an evaporative cooler is: a. thermal energy to chemical energy b. thermal energy to heat c. cooling of thermal energy d. heat energy to mechanical energy
Answer:
C.Cooling of thermal energy
Explanation:
In evaporative cooling liquid water convert into vapor by using thermal energy.In this out side air is used to evaporate the water.
Hot air enters from outside and this hot air gets contact with wet matrix .In wet matrix temperature of air will reduce and temperature of water will increases .After wet matrix heat transfer take place between air and water .After that by using force convection air pushed out and produce cooling.
The condition of irrotationality for a two-dimensional flow is satisfied when rotation w everywhere is (less than — equal to — more than) zero.
Answer:
Zero.
Explanation:
Lets take velocity V is given as in 2 dimensional flow
V=u i +v j
V=f(x,y)
u=f(x,y)
v=f(x,y)
Rotational ability ω given as
[tex]\omega =\dfrac{1}{2}\left (\dfrac{\partial v}{\partial x}-\dfrac{\partial u} {\partial x}\right)[/tex]
If
ω=0 ,then flow will be irrotational.
ω ≠0 ,then flow will be rotational.
So we can say that , ω will be zero every where for irrotational two dimensional flow .
A 600 MW power plant has an efficiency of 36 percent with 15
percent of the waste heat being released to the atmosphere as stack
heat and the other 85 percent taken away in the cooling water.
Instead of drawing water from a river, heating it, and returning it
to the river, this plant uses an evaporative cooling tower wherein
heat is released to the atmoshphere as cooling water is
vaporized.
At what rate must 15C makeup water be provided from the river to
offset the water lost in the cooling tower?
Answer:
401.3 kg/s
Explanation:
The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).
qw = 0.85 * q2
q2 = 0.64 * q1
p = 0.36 * q1
q1 = p /0.36
q2 = 0.64/0.36 * p
qw = 0.85 *0.64/0.36 * p
qw = 0.85 *0.64/0.36 * 600 = 907 MW
In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)
The latent heat for the vaporization of water is:
SLH = 2.26 MJ/kg
So, to dissipate 907 MW
G = qw * SLH = 907 / 2.26 = 401.3 kg/s
Answer:
m = 367.753 kg.s
Explanation:
GIVEN DATA:
Power plant capacity W=600 MW
Plant efficiency is [tex]\eta = 36\%[/tex]
[tex]efficiency = \frac{W}{QH}[/tex]
[tex]0.36 = \frac{600}{QH}[/tex]
QH = 1666.6 MW
from first law of thermodynamics we hvae
QH -QR = W
Amount of heat rejection is QR = 1066.66 MW
As per given information we have 15% heat released to atmosphere
[tex]QR = 0.15 \tiimes 1066.66 = 159.99 MW[/tex]
AND 85% to cooling water
[tex]Q = 0.85 \times 1066.66 = 906.66 MW[/tex]
from saturated water table
at temp 150 degree c we have Hfg = 2465.4 kJ/kg
rate of cooiling water is given as = mhfg
[tex]906.66 \times 1000 KW = m \times 2465.4[/tex]
m = 367.753 kg.s
where m is rate of makeup water that is added to offset
A steam piston system is heated by 72 Btu of heat. 7 Btu of heat is lost to the environment. The piston does 12 Btu of work as it rises. What is the change in the energy of the water steam in the system?
Answer:
ΔU = 53 BTU
Explanation:
In thermodynamics we consider heat entering a system as positive and heat leaving as negative.
Also, work performed by the system is positive and work applied to the system is negative.
The total heat applied to the system is:
Q = 72 - 7 = 65 BTU
The work is:
L = 12 BTU
The first law of thermodynamics states that:
Q = L + ΔU
Where ΔU is the change in internal energy.
Rearranging:
ΔU = Q - L
ΔU = 65 - 12 = 53 BTU
What is the movement of the piston from top dead center (TDC) to bottom dead center (BDC) called?
Answer:
Piston movement of TDC to BDC is called stroke.
Explanation:
Step1
In any engine fuel is burned inside the cylinder and forces the piston to move from top dead center to bottom dead center. First piston is moved from top dead center to bottom dead center to allow the suction of fuel inside the cylinder. Then piston move toward top dead center and compresses the fuel. This process is called compression of fuel. Then fuel is burned with the spark inside the cylinder and the piston moves toward bottom dead center. This process is called expansion process. Last process is the expansion process that allows the exhaust out of the cylinder.
Step2
The piston reciprocates from bottom position of the cylinder to top position of the cylinder and vice versa. So, the motion of piston from top center to bottom center is called stroke. One movement of piston from TDC to BDC is called one stroke. There are total four strokes in petrol engine and diesel engine. So, two strokes are equal to one cycle or one rotation of crank and four strokes are equal to two rotation of crank.
Thus, piston movement of TDC to BDC is called stroke.
Answer:
The return stroke
Explanation:
Motion from top dead center to bottom dead center is called the return stroke because motion from the bottom to the top is the forward stroke.
Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) 354 mg (45 km) / (0.0356 kN), (b) (0.00453 Mg) (201 ms), and (c) 435 MN/23.2 mm.
The question requires calculating expressions using metric units and prefixes, converting them into SI units with appropriate prefixes, and ensuring these calculations are accurate to three significant figures.
Explanation:The question involves evaluating expressions with different units and converting them to SI units with appropriate prefixes, which requires a good understanding of metric prefixes and significant figures. Furthermore, examples of such conversions are provided to aid in this process.
Evaluations with SI Units and PrefixesFor part (a), you convert milligrams to kilograms, kilometers to meters, and kilonewtons to newtons, then perform the division ensuring that the answer is to three significant figures.For part (b), convert megagrams to kilograms and multiply by the given number of seconds, again reporting the answer to three significant figures.For part (c), convert meganewtons to newtons and millimeters to meters, divide, and round to three significant figures.Each conversion is an exercise in carefully applying metric prefixes and understanding how to manipulate them within calculations.
Determine the specific volume of superheated water vapor at 15 Mpa and 1000C, using (a) the ideal-gas equation, and (b) the steam tables.
Answer:
a)[tex]v=25.56\ kg/m^3[/tex]
b) [tex]v=38.8\ kg/m^3[/tex]
Explanation:
Given that
Pressure P = 15 MPa
Temperature = 1000 C = 1273 K
a)If assume as ideal gas
The gas constant for super heated steam R is 0.461 KJ/kg.K.
We know that ideal gas equation
P v =RT
15 x 1000 x v =0.461 x 1273
[tex]v=25.56\ kg/m^3[/tex]
b) By using steam table
From steam table we can see that volume of super heated vapot at 15 MPa and 1273 K .
[tex]v=38.8\ kg/m^3[/tex]
An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inch and a standard deviation of 0.07 inch. Assume that the shaft diameter follows a Gaussian distribution. What percentage of the diameter of the total shipment of shafts will fall between 2.39inch and 2.60 inch?
Answer: 78.89%
Explanation:
Given : Sample size : n= 1200
Sample mean : [tex]\overline{x}=2.45 [/tex]
Standard deviation : [tex]\sigma=0.07[/tex]
We assume that it follows Gaussian distribution (Normal distribution).
Let x be a random variable that represents the shaft diameter.
Using formula, [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponds to 2.39 will be :-
[tex]z=\dfrac{2.39-2.45}{0.07}\approx-0.86[/tex]
z-value corresponds to 2.60 will be :-
[tex]z=\dfrac{2.60-2.45}{0.07}\approx2.14[/tex]
Using the standard normal table for z, we have
P-value = [tex]P(-0.86<z<2.14)=P(z<2.14)-P(z<-0.86)[/tex]
[tex]=P(z<2.14)-(1-P(z<0.86))=P(z<2.14)-1+P(z<0.86)\\\\=0.9838226-1+0.8051054\\\\=0.788928\approx0.7889=78.89\%[/tex]
Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%
When a 20-lb weight is suspended from a spring, the spring is stretched a distance of 1 in. Determine the natural frequency and the period of vibration for a 30-lb weight attached to the same spring
Answer:
natural frequency = 2.55 Hz
period of vibration = 0.3915 s
Explanation:
given data
weight = 20 lb
distance = 1 in = [tex]\frac{1}{12}[/tex] ft
weight = 30 lb
to find out
Determine the natural frequency and the period of vibration
solution
we first calculate here stiffness k by given formula that is
k = [tex]\frac{weight}{diatnace}[/tex] ..........1
k = [tex]\frac{20}{1/12}[/tex]
k = 240 lb/ft
so
frequency = [tex]\sqrt{\frac{k}{m} }[/tex] ..................2
put here value k and mass m = [tex]\frac{weight}{g}[/tex]
frequency = [tex]\sqrt{\frac{240}{30/32.2} }[/tex]
frequency = 16.05 rad/s
and
period of vibration = [tex]\frac{2* \pi }{frequency}[/tex]
period of vibration = [tex]\frac{2* \pi }{16.05}[/tex]
period of vibration = 0.3915 s
and
natural frequency = [tex]\frac{1 }{period of vibration}[/tex]
natural frequency = [tex]\frac{1 }{0.3915}[/tex]
natural frequency = 2.55 Hz
A power company desires to use groundwater from a hot spring to power a heat engine. If the groundwater is at 95 deg C and the atmosphere is at 20 deg C, what is the maximum power output for a mass flow of 0.2 kg/s? Assume the water is cooled to atmospheric temperature.
Answer:
W = 12.8 KW
Explanation:
given data:
mass flow rate = 0.2 kg/s
Engine recieve heat from ground water at 95 degree ( 368 K) and reject that heat to atmosphere at 20 degree (293K)
we know that maximum possible efficiency is given as
[tex]\eta = 1- \frac{T_L}{T_H}[/tex]
[tex]\eta = 1 - \frac{ 293}{368}[/tex]
[tex]\eta = 0.2038[/tex]
rate of heat transfer is given as
[tex]Q_H = \dot m C_p \Delta T[/tex]
[tex]Q_H = 0.2 * 4.18 8(95 - 20)[/tex]
[tex]Q_H = 62.7 kW[/tex]
Maximuim power is given as
[tex]W = \eta Q_H[/tex]
W = 0.2038 * 62.7
W = 12.8 KW
A galvanometer has a coil with a resistance of 24.0 Ω, and a current of 180 μA causes it to deflect full scale. If this galvanometer is to be used to construct an ammeter that can read up to 10.0 A, what shunt resistor is required? A galvanometer has a coil with a resistance of 24.0 , and a current of 180 μA causes it to deflect full scale. If this galvanometer is to be used to construct an ammeter that can read up to 10.0 A, what shunt resistor is required? 234 µΩ 123 µΩ 342 µΩ 423 µΩ 432 µΩ
Answer:
shunt resistor is 432µΩ
correct option is 432µΩ
Explanation:
given data
resistance R = 24.0 Ω
full scale deflection current Ig = 180 μA
galvanometer current I = 10.0 A
to find out
what shunt resistor is required
solution
we will apply here full scale deflection current formula that is
Ig = I × [tex]\frac{r}{r + R}[/tex] ...............1
here r is shunt current and R is resistance and I is galvanometer
put here all value
180 × [tex]10^{-6}[/tex] = 10 × [tex]\frac{r}{r + 24}[/tex]
r = 18 × [tex]10^{-6}[/tex] × 24
r = 432 × [tex]10^{-6}[/tex] Ω
so shunt resistor is 432µΩ
correct option is 432µΩ
Shunt resistor required for 10.0 A ammeter: approximately 432 μΩ.
To construct an ammeter using a galvanometer, a shunt resistor is connected in parallel to the galvanometer. The shunt resistor diverts most of the current, allowing only a fraction of the current to pass through the galvanometer, thus enabling it to measure higher currents.
The relationship between the current passing through the galvanometer and the shunt resistor can be expressed using the formula:
[tex]\[I_{\text{total}} = I_{\text{g}} + I_{\text{s}}\][/tex]
Where:
[tex]- \(I_{\text{total}}\)[/tex] is the total current (10.0 A)
[tex]- \(I_{\text{g}}\)[/tex] is the current passing through the galvanometer (180 μA)
[tex]- \(I_{\text{s}}\)[/tex] is the current passing through the shunt resistor
Given that the galvanometer resistance [tex]\(R_{\text{g}} = 24.0 Ω\)[/tex] and full-scale deflection current \[tex](I_{\text{g}} = 180 μA\),[/tex] we can calculate the shunt resistance required using Ohm's Law:
[tex]\[I_{\text{g}} = \frac{V_{\text{g}}}{R_{\text{g}}}\][/tex]
Where:
- [tex]\(V_{\text{g}}\)[/tex] is the voltage across the galvanometer
Since the galvanometer is designed to deflect full-scale at 180 μA, the voltage across it is:
[tex]\[V_{\text{g}} = I_{\text{g}} \times R_{\text{g}}\]\[V_{\text{g}} = (180 \times 10^{-6}) \times 24\]\[V_{\text{g}} = 0.00432 \, \text{V}\][/tex]
The current passing through the shunt resistor can be calculated using:
[tex]\[I_{\text{s}} = \frac{V_{\text{total}}}{R_{\text{s}}}\][/tex]
Since the total current is 10.0 A and the voltage across the shunt resistor is the same as the voltage across the galvanometer (as they are in parallel), we have:
[tex]\[I_{\text{s}} = \frac{10.0}{R_{\text{s}}}\][/tex]
Now, the total current equals the sum of the currents through the galvanometer and shunt resistor:
[tex]\[10.0 = 180 \times 10^{-6} + \frac{0.00432}{R_{\text{s}}}\][/tex]
Solving for [tex]\(R_{\text{s}}\):[/tex]
[tex]\[R_{\text{s}} = \frac{0.00432}{10.0 - 180 \times 10^{-6}}\]\[R_{\text{s}} ≈ \frac{0.00432}{10.0}\]\[R_{\text{s}} ≈ 432 \, \text{μΩ}\][/tex]
So, the required shunt resistor is approximately 432 μΩ. Therefore, the closest answer is 432 μΩ.
Please define the coefficient of thermal expansion?
Answer:
The coefficient of thermal expansion tells us how much a material can expand due to heat.
Explanation:
Thermal expansion occurs when a material is subjected to heat and changes it's shape, area and volume as a result of that heat. How much that material changes is dependent on it's coefficient of thermal expansion.
Different materials have different coefficients of thermal expansion (i.e. It is a material property and differs from one material to the next). It is important to understand how materials behave when heated, especially for engineering applications when a change in dimension might pose a problem or risk (eg. building large structures).
A settling tank with 50 foot diameter and a SWD of 9 feet treats
aflow of 15, 000 gpd. What is the detention time?
Answer:
Detention time will be 9 days
Explanation:
We have given diameter of setting tank d = 50 feet
Radius of setting tank [tex]r=\frac{d}{2}=\frac{9}{2}=25 feet[/tex]
SWD = 9 FEET
Area [tex]A=\pi r^2=3.14\times 25^2=1962.5ft^2[/tex]
So volume [tex]V = area\times SWD=1962.5\times 9=17671.5ft^3[/tex]
We have given flow Q = 15000 gpd
So Q= 15000×0.13368 =2005.22 [tex]ft^3[/tex] per day
Detention time is given by [tex]=\frac{volume}{Q}=\frac{17671.5}{2005.22}=9days[/tex]
So detention time will be 9 days
A second inventor was driving down the highway in her Prius one day with her hand out the window. She happened to be driving through the middle of a wind farm when an idea snapped into her head. She thought to herself, what if she mounted a small windmill generator on the roof of her Prius and wired it into the battery? She thought, with a little head wind to get her started she could just drive off under wind power! The faster she drove, the more power she would make, and the faster she could go… What’s wrong with her idea? a. As a side question, Priuses make use of regenerative braking. Why is it called regenerative braking? Why don’t cars advertise regenerative accelerators?
Answer:
Explanation:
It wouldn't work because the wind energy she would be collecting would actually come from the car engine.
The relative wind velocity observed from a moving vehicle is the sum of the actual wind velocity and the velovity of the vehicle.
u' = u + v
While running a car will generate a rather high wind velocity, and increase the power generated by a wind turbine, the turbine would only be able to convert part of the wind energy into electricity while adding a lot of drag. In the end, it would generate less energy that what the drag casuses the car to waste to move the turbine.
Regenerative braking uses an electric generator connected to the wheel axle to recover part of the kinetic energy eliminated when one brakes the vehicle. Normal brakes dissipate this energy as heat, a regenerative brake uses it to recharge a batttery. Note that is is a fraction of the energy that is recovered, not all of it.
A "regenerative accelerator" makes no sense. Braking is taking kinetic energy out of the vehicle, while accelerating is adding kinetic energy to it. Cars accelerate using the power from their engines.
For a motor purchased for a tank project, it claims that at 12V it produces 6000 RPM. This motor is attached to a 50:1 gear reduction gearbox. The output of this gearbox is attached to a 2.5" wheel. Determine the speed of the tank under these conditions in miles/hour (mph).
Answer:
0.889 mph
Explanation:
Given:
Voltage = 12 V
RPM produced, N₁ = 6000
Gear reduction gearbox, G = 50:1
Diameter of the wheel = 2.5" = 2.5 × 0.0254 = 0.0635 m
Now,
The output speed = [tex]\frac{N_1}{\textup{G}}=\frac{6000}{50}[/tex] = 120 RPM
Thus,
The speed, v = [tex]\frac{\pi DN_2}{60}[/tex]
on substituting the respective values, we get
v = [tex]\frac{\pi\times0.0635\times120}{60}[/tex]
or
v = 0.39878 m/s
Also,
1 m/s = 2.23 mph
thus,
v = 0.39878 × 2.23 = 0.889 mph
Write the equation to estimate heat flow q due to convection.
Answer:
q= h ΔT
Explanation:
Heat flow due to convection given as follows
Q= h A ΔT
Heat flux per unit area given as
q= h ΔT
Where h is the heat transfer coefficient,A is the surface area and ΔT is the temperature difference.
Lets take one plate having temperature [tex]T_1[/tex] and air is flowing on the plate with temperature [tex]T_2[/tex] and air having heat transfer coefficient h.Dimensions of plate is l x b.
Area of plate
[tex]A=lbm^2[/tex]
So heat transfer
[tex]Q=h\times l\times b(T_1-T_2)[/tex]
In an oscillating LC circuit, L ! 25.0 mH and C ! 7.80 mF. At time t 0 the current is 9.20 mA, the charge on the capacitor is 3.80 mC, and the capacitor is charging.What are (a) the total energy in the circuit, (b) the maximum charge on the capacitor, and (c) the maximum current? (d) If the charge on the capacitor is given by q ! Q cos(vt $ f), what is the phase angle f? (e) Suppose the data are the same, except that the capacitor is discharging at t ! 0.What then is f?
Answer:
a) 926 μJ
b) 3.802 mC
c) 8.61 A
d) 0.0721
e) 3.2137
Explanation:
The energy in the inductor is
[tex]El = \frac{1}{2}*L*I^2[/tex]
[tex]El = \frac{1}{2}*25*10^{-3}*(9.2*10^{-3})^2 = 1.06*10^{-6} J = 1.06 \mu J[/tex]
The energy store in a capacitor is
[tex]Ec = \frac{1}{2}*C*V^2[/tex]
The voltage in a capacitor is
V = Q/C
[tex]V = \frac{3.8*10^{-3}}{7.8*10^{-3}} = 0.487 V[/tex]
Therefore:
[tex]Ec = \frac{1}{2}*7.8*10^{-3}*0.487^2 = 9.256*10^{-4} J = 925.6 \mu J[/tex]
The total energy is Et = 925.6 + 1.1 = 926.7 μJ
At a certain point all the energy of the circuit will be in the capacitor, at this point it will have maximum charge
[tex]Ec = \frac{1}{2}*C*V^2[/tex]
V = Q/C
[tex]Ec = \frac{1}{2}*C*(\frac{Q}{C})^2[/tex]
[tex]Ec = \frac{1}{2}*\frac{Q^2}{C}[/tex]
[tex]Q^2 = 2*Ec*C[/tex]
[tex]Q = \sqrt{2*Ec*C}[/tex]
[tex]Q = \sqrt{2*926*10{-6}*7.8*10^{-3}} = 3.802 * 10{-3} C = 3.802 mC[/tex]
When the capacitor is completely empty all the energy will be in the inductor and current will be maximum
[tex]El = \frac{1}{2}*L*I^2[/tex]
[tex]I^2 = 2*\frac{El}{L}[/tex]
[tex]I = \sqrt{2*\frac{El}{L}}[/tex]
[tex]I = \sqrt{2*\frac{926.7*10^{-3}}{25*10^{-3}}} = 8.61 A[/tex]
At t = 0 the capacitor has a charge of 3.8 mC, the maximum charge is 3.81 mC
q = Q * cos(vt + f)
q(0) = Q * cos(v*0 + f)
3.8 = 3.81 * cos(f)
cos(f) = 3.8/3.81
f = arccos(3.8/3.81) = 0.0721
If the capacitor is discharging it is a half cycle away, so f' = f + π = 3.2137
The pilot of an airplane reads the altitude 6400 m and the absolute pressure 45 kPa when flying over the city. Calculate the local atmospheric pressure in that city in kPa and in mmHg. Take the densities of air and mercury to be 0.828 kg/m3 and 13600 kg/m3, respectively. Show this using the systematic solution method.
Answer:
1) The absolute pressure equals = 96.98 kPa
2) Absolute pressure in terms of column of mercury equals 727 mmHg.
Explanation:
using the equation of pressure statics we have
[tex]P(h)=P_{surface}-\rho gh...............(i)[/tex]
Now since it is given that at 6400 meters pressure equals 45 kPa absolute hence from equation 'i' we obtain
[tex]P_{surface}=P(h)+\rho gh[/tex]
Applying values we get
[tex]P_{surface}=45\times 10^{3}+0.828\times 9.81\times 6400[/tex]
[tex]P_{surface}=96.98\times 10^{3}Pa=96.98kPa[/tex]
Now the pressure in terms of head of mercury is given by
[tex]h_{Hg}=\frac{P}{\rho _{Hg}\times g}[/tex]
Applying values we get
[tex]h=\frac{96.98\times 10^{3}}{13600\times 9.81}=0.727m=727mmHg[/tex]
Sketch the velocity profile for laminar and turbulent flow.
Answer:
The laminar flow is generally given in high viscosity fluids such as honey or oil, it has the characteristic of flowing in an orderly manner, the walls of the tube have a zero speed while in the center it has a maximum speed.
turbulent flow is characterized by fluid velocity vectors presenting themselves in a disorderly manner and in all directions.
I attached the drawings for the velocity profile in laminar and turbulent flow.