Answer: the thermal conductivity of the sample is 22.4 W/m . °C
Explanation:
We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
ASSUMPTIONS
1. Steady operating conditions exist since the temperature readings do not change with time.
2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.
3. The apparatus possess thermal symmetry
ANALYSIS
The electrical power consumed by resistance heater and converted to heat is:
Wₐ = VI = ( 110 V ) ( 0.4 A ) = 44 W
Q = 1/2Wₐ = 1/2 ( 44 A )
Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so
A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²
Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be
Q = kA ( ΔT/L ) → k = QL / AΔT
k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )
k = 22.4 W/m . °C
Consider a solid round elastic bar with constant shear modulus, G, and cross-sectional area, A. The bar is built-in at both ends and subject to a spatially varying distributed torsional load t(x) = p sin( 2π L x) , where p is a constant with units of torque per unit length. Determine the location and magnitude of the maximum internal torque in the bar.
Answer:
[tex]\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]
Explanation:
Given that
Shear modulus= G
Sectional area = A
Torsional load,
[tex]t(x) = p sin( \frac{2\pi}{ L} x)[/tex]
For the maximum value of internal torque
[tex]\dfrac{dt(x)}{dx}=0[/tex]
Therefore
[tex]\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}[/tex]
Thus the maximum internal torque will be at x= 0.25 L
[tex]t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]
Describe the meaning of the different symbols and abbreviations found on the drawings/documents that they use (such as BS8888, surface finish to be achieved, linear and geometric tolerances, electronic components, weld symbols and profiles, pressure and flow characteristics, torque values, imperial and metric systems of measurement, tolerancing and fixed reference points)
Answer:
Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.
There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.
Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.
Explanation:
what are the conditions for sheet generator to build up its voltage?
Answer:
There are six conditions
1. Poles should contain some residual flux.
2. Field and armature winding must be correctly connected so that initial mmm adds residual flux.
3. Resistance of field winding must be less than critical resistance.
4. Speed of prime mover of generator must be above critical speed.
5. Generator must be on load.
6. Brushes must have proper contact with commutators.
Explanation:
The speed above which an airplane will experience structural damage when a load is applied, instead of stalling, is called the ______________ speed and varies with weight
Answer:
Maneuvering speed.
Explanation:
The speed above which an airplane will experience structural damage when a load is applied, instead of stalling, is called the maneuvering speed and varies with weight.
In aeronautical engineering, the maneuvering speed (Va) of an aircraft such as an aeroplane, helicopter, or jet is an airspeed limitation which is mainly selected by an aircraft designer.
Generally, at speeds higher or greater than the manoeuvring speed, aircraft pilots are advised not to attempt a full deflection of any flight control surface because it's capable of resulting in a damage to the structure of an aircraft.
If you're a pilot, to find the maneuvering speed of an aircraft, you should look at the flight manual of the aircraft or on the cockpit placard in the aircraft. The maneuvering speed of an aircraft is a calibrated speed and should not be exceeded by any pilot.
A wall 0.12 m thick having a thermal diffusivity of 1.5 × 10-6 m2/s is initially at a uniform temperature of 97°C. Suddenly one face is lowered to a temperature of 20°C, while the other face is perfectly insulated. Use the explicit finite-difference technique with space and time increments of 30 mm and 300 s to determine the temperature distribution at at 45 minutes.
Answer:
at t = 45 s :
To = 61.7⁰c, T1 = 55.6⁰c, T2 = 49.5⁰c, T3 = 34.8⁰C
Explanation:
Wall thickness = 0.12 m
thermal diffusivity = 1.5 * 10^-6 m^2/s
Δt ( time increment ) = 300 s
Δ x = 0.03 m ( dividing wall thickness into 4 parts assuming the system to be one dimensional )
using the explicit finite-difference technique
Detailed solution is attached below
What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Answer:
E = 2940 J
Explanation:
It is given that,
Mass, m = 12 kg
Position at which the object is placed, h = 25 m
We need to find the potential energy of the mass. It is given by the formula as follows :
E = mgh
g is acceleration due to gravity
[tex]E=12\times 9.8\times 25\\\\E=2940\ J[/tex]
So, the potential energy of the mass is 2940 J.
In the LC-3 data path, the output of the address adder goes to both the MARMUX and the PCMUX, potentially causing two very different register transfers to take place. Why does this not happen
Answer:
no need for that
Explanation:
they are not the same at all
A 60-Hz 220-V-rms source supplies power to a load consisting of a resistance in series with an inductance. The real power is 1500 W, and the apparent power is 4600 VA.
a. Determine the value of the resistance.
b. Determine the value of the inductance.
Answer:
(a) The value of the resistance is 3.431 Ω
(b) The value of the inductance is 0.0264 H
Explanation:
Given;
frequency of the source, f = 60 Hz
rms voltage, V-rms = 220 V
real power, Pr = 1500 W
apparent power, Pa = 4600 VA
(a). Determine the value of the resistance
[tex]P_r = I_{rms}^2R[/tex]
where;
R is resistance
[tex]I_{rms} = \frac{Apparent \ Power}{V_{rms}} \\\\I_{rms} = \frac{P_a}{V_{rms}}\\\\I_{rms}= \frac{4600}{220} \\\\I_{rms}= 20.91 \ A[/tex]
Resistance is calculated as;
[tex]R = \frac{P_r}{I_{rms}^2} \\\\R = \frac{1500}{(20.91)^2} \\\\R = 3.431 \ ohms[/tex]
(b). Determine the value of the inductance.
[tex]Q_L = I_{rms}^2 X_L[/tex]
where;
[tex]Q_L[/tex] is reactive power
[tex]X_L[/tex] is inductive reactance
[tex]Apparent \ power = \sqrt{Q_L^2 + P_r^2} \\\\P_a^2 = Q_L^2 + P_r^2\\\\Q_L^2 = P_a^2 - P_r^2\\\\Q_L^2 = 4600^2 - 1500^2\\\\Q_L^2 = 18910000\\\\Q_L = \sqrt{18910000}\\\\Q_L = 4348.56 \ VA[/tex]
inductive reactance is calculated as;
[tex]X_L = \frac{Q_L}{I_{rms}^2} \\\\X_L = \frac{4348.56}{(20.91)^2} \\\\X_L = 9.95 \ ohms[/tex]
inductance is calculated as;
[tex]X_L = \omega L\\\\X_L = 2\pi f L\\\\L = \frac{X_L}{2\pi f} \\\\L = \frac{9.95}{2\pi *60} \\\\L = 0.0264 \ H\\\\L = 26.4 \ mH[/tex]
Technician A says that proper footwear may include both leather and steel-toed shoes. Technician B says that leather-soled shoes provide slip resistance. Who is correct
Given:
We have given two statements.
Statement 1: Proper footwear may include both leather and steel-toed shoes.
Statement 2: Leather-soled shoes provide slip resistance.
Find:
Which statement is true.
Solution:
A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.
Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.
Leather-soled shoes don't provide slop resistance.
Therefore, both the Technicians are wrong.
From the statements made by both technician A and Technician B, we can say that; both technicians are wrong.
We are given the statements made by both technicians;
Technician A: Proper footwear may include both leather and steel-toed shoes.
Technician B: Leather-soled shoes provide slip resistance.
Now, they are talking about safety shoes to be worn in workshops.
A shoe that is Slip resistant will have rubber soles and tread patterns that can help to have better grip of wet or greasy floors.
This is the type of shoe that should be worn by technicians in the workshop.
Thus, Technician A is wrong because proper footwear does not include leather shoes.
Similarly, technician B is also wrong because leather shoes are not safety shoes.
Read more about slip resistant shoes at; https://brainly.com/question/17411739
A series circuit contains four resistors. In the circuit, R1 is 80 , R2 is 60 , R3 is 90 , and R4 is 100 . What is the total resistance? A. 330 B. 250 C. 460 D. 70.3
When you shift your focus, everything you
see is still in perfect focus.
True or false
Answer:
true
Explanation:
true
Answer:
I believe this is true
Explanation:
If your looking at something and you look at something else everything is still in perfect view and clear, in focus.
hope this helps :)
Given the unity feedback system
G(s)= K(s+4)/s(s+1.2)(s+2)
Find:
a. The range of K that keeps the system stable
b. The value of K that makes the system oscillate
c. The frequency of oscillation when K is set to the value that makes the system oscillate
Answer:
A.) 0 > K > 9.6
B.) K = 9.6
C.) w = +/- 2 sqrt (3)
Explanation:
G(s)= K(s+4)/s(s+1.2)(s+2)
For a closed loop stability, we can analyse by using Routh - Horwitz analysis.
To make the pole completely imaginary, K must be equal to 9.6 Because for oscillations. Whereas, one pair of pole must lie at the imaginary axis.
Please find the attached files for the solution
The following liquids are stored in a storage vessel at 1 atm and 25°C. The vessels are vented with air. Determine whether the equilibrium vapor above the liquid will be flammable. The liquids are:________.
a. Acetone
b. Benzene
c. Cyclohexane
d. Toluene Problem
Answer:
The liquids are TOLUENE because the equilibrum vapor above it will be flammable ( D )
Explanation:
Liquids stored at : 1 atm , 25⁰c and they are vented with air
Determining whether the equilibrum vapor above the liquid will be flammable
We can determine this by using Antoine equation to calculate saturation vapor pressure also apply Dalton's law to determine the volume % concentration of air and finally we compare answer to flammable limits to determine which liquid will be flammable
A) For acetone
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A = 16.6513
B = 2940.46
C = -35.93
T = 298 k input values into Antoine equation
therefore ; [tex]p^{out}[/tex] = 228.4 mg
calculate volume percentage using Dalton's law
= V% = (saturation vapor pressure / pressure ) *100
= (228.4 mmHg / 760 mmHg) * 100 = 30.1%
The liquid is not flammable because its UFL = 12.8%
B) For Benzene
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A= 15.9008
B = 2788.52
C = -52.36
T = 298 k input values into the above equation
[tex]p^{out}[/tex] = 94.5 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= (94.5 / 760 ) * 100 = 12.4%
Benzene is not flammable under the given conditions because its UFL =7.1%
C) For cyclohexane
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten appendix E ( chemical process safety (3rd edition) )
A = 15.7527
B = 2766.63
c = -50.50
T = 298 k
solving the above equation using the given values
[tex]p^{out}[/tex] = 96.9 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= ( 96.9 mmHg /760 mmHg) * 100 = 12.7%
cyclohexane not flammable under the given conditions because its UFL= 8%
D) For Toluene
using the Antoine equation to calculate saturation vapor pressure
[tex]In(P^{out} ) = A - \frac{B}{C + T}[/tex]
values gotten from appendix E ( chemical process safety (3rd edition) )
A = 16.0137
B = 3096.52
C = -53.67
T = 298 k
solving the above equation using the given values
[tex]p^{out}[/tex] = 28.2 mmHg
calculate volume percentage using Dalton's law
V% = (saturation vapor pressure / pressure ) *100
= (28.2 mmHg / 760 mmHg) * 100 = 3.7%
Toluene is flammable under the given conditions because its UFL= 7.1%
Determine the normal stress in a ball, which has an outside diameter of 160 mm and a wall thickness of 3.8 mm, when the ball is inflated to a gage pressure of 78 kPa.
Answer:
The normal stress is 0.7821 MPa
Explanation:
The external diameter D = 160 mm
The thickness t = 3.8 mm = 3.8 x 10^-3 m
gauge pressure P = 78 kPa = 78 x 10^3 Pa
The maximum shear stress τmax = ?
The external radius of the shell from the external surface R = D/2 = 160/2 = 80 mm
The internal radius of the shell r = R - t
==> 80 - 3.8 = 76.2 mm
Therefore the internal diameter d = 2r = 2 x 76.2 = 152.4 mm
==> d = 152.4 x 10^-3 m
The normal stress σ = [tex]\frac{Pd}{4t}[/tex] = [tex]\frac{78*10^{3}*152.4*10^{-3} }{4*3.8*10^{-3} }[/tex] = 782052.63 Pa
==> σ = 0.7821 MPa
Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa m0.5. It has been determined that fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
Answer: 164.2253 MPa
Explanation:
First we find the half internal crack which is = length of surface crack /2
so α = 8.6 /2 = 4.3mm ( 4.3×10⁻³m )
Now we find the dimensionless parameter using the critical stress crack propagation equation
∝ = K / Y√πα
stress level ∝ = 112Mpa
fracture toughness K = 26Mpa
dimensionless parameter Y = ?
SO working the formula
Y = K / ∝√πα
Y = 26 / 112 (√π × 4.3× 10⁻³)
Y = 1.9973
We are asked to find stress level for internal crack length of 4m
so half internal crack is = length of surface crack /2
4/2 = 2mm ( 2 × 10⁻³)
from the previous formula critical stress crack propagation equation
∝ = K / Y√πα
∝ = 26 / 1.9973 √(π × 2 × 10⁻³)
∝ = 164.2253 Mpa
Who plays a role in the financial activities of a company?
O A. Just employees
O B. Just managers
O C. Only members of the finance and accounting department
O D. Everyone at the company, including managers and employees
Hey,
Who plays a role in the financial activities of a company?
O D. Everyone at the company, including managers and employees
Answer:
Everyone at the company, including managers and employees
Explanation:
. A belt drive is desired to couple the motor with a mixer for processing corn syrup. The 25-hp electric motor is rated at 950 rpm and the mixer must operate as close to 250 rpm as possible. Select an appropriate belt size, commercially available sheaves, and a belt for this application. Also calculate the actual belt speed and the center distance.
Answer:
Hello the table which is part of the question is missing and below are the table values
For a 5V belt the available diameters are : 5.5, 5.8, 5.9, 6.2, 6.3, 6.6, 12.5, 13.9, 15.5, 16.1, 18.5, 20.1
Answers:
belt size = 140 in with diameter of 20.1n
actual speed of belt = 288.49 in/s
actual center distance = 49.345 in
Explanation:
Given data :
Electric motor (driver sheave) speed (w1) = 950 rpm
Driven sheave speed (w2) = 250 rpm
pick D1 ( diameter of driver sheave) = 5.8 in ( from table )
To select an appropriate belt size we apply the equation for the velocity ratio to get the diameter first
VR = [tex]\frac{w1}{w2}[/tex] = 950 / 250
also since the speed of belt would be constant then ;
Vb = w1r1 = w2r2 ------- equation 1
r = d/2
substituting the value of r into equation 1
equation 2 becomes : [tex]\frac{w1}{w2} = \frac{d2}{d1}[/tex] = VR
Appropriate belt size ( d2) can be calculated as
d2 = [tex]\frac{w1d1}{w2}[/tex] = [tex]\frac{950 * 5.8}{250}[/tex] = 22.04
From the given table the appropriate belt size would be : 20.1 because it is the closest to the calculated value
next we have to determine the belt length /size
[tex]L = 2C + \frac{\pi }{2} ( d1+d2) + \frac{(d2-d1)^2}{4C}[/tex]
inputting all the values into the above equation including the value of C as calculated below
L ≈ 140 in
Calculating the center distance
we use this equation to get the ideal center distance
[tex]d2< C_{ideal} < 3( d1 +d2)[/tex]
22.04 < c < 3 ( 5.8 + 20.1 )
22.04 < c < 77.7
the center distance is between 22.04 and 77.7 but taking an average value
ideal center distance would be ≈ 48 in
To calculate the actual center distance we use
[tex]C = \frac{B+\sqrt{B^2 - 32(d2-d1)^2} }{16}[/tex] -------- equation 3
B = [tex]4L -2\pi (d2 + d1 )[/tex]
inputting all the values into (B)
B = 140(4) - 2[tex]\pi[/tex]( 20.01 + 5.8 )
B ≈ 399.15 in
inputting all the values gotten Back to equation 3 to get the actual center distance
C = 49.345 in ( actual center distance )
Calculating the actual belt speed
w1 = 950 rpm = 99.48 rad/s
belt speed ( Vb) = w1r1 = w1 * [tex]\frac{d1}{2}[/tex]
= 99.48 * 5.8 / 2 = 288.49 in/s
An inventor claims to have developed a heat pump that produces a 200-kW heating effect for a 293 K heatedzone while only using 75 kW of power and a heat source at 273 K. Justify the validity of this claim.
Answer:
From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.
Explanation:
Heat generated Q = 200 kW
power input W = 75 kW
Temperature of heated region [tex]T_{h}[/tex] = 293 K
Temperature of heat source [tex]T_{c}[/tex] = 273 K
For this engine,
coefficient of performance COP = Q/W = 200/75 = 2.67
The maximum theoretical COP obtainable for a heat pump is given as
COP = [tex]\frac{T_{h} }{T_{h} - T_{c} }[/tex] = [tex]\frac{293 }{293 - 273 }[/tex] = 14.65
From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.
A concentric tube heat exchanger is used to cool a solution of ethyl alcohol flowing at 6.93 kg/s (Cp = 3810 J/kg-K) from 65.6 degrees C to 39.4 degrees C using water flowing at 6.30 kg/s at a temperature of 10 degrees C. Assume that the overall heat transfer coefficient is 568 W/m2-K. Use Cp = 4187 J/kg-K for water.
a. What is the exit temperature of the water?
b. Can you use a parallel flow or counterflow heat exchanger here? Explain.
c. Calculate the rate of heat flow from the alcohol solution to the water.
d. Calculate the required heat exchanger area for a parallel flow configuration
e. Calculate the required heat exchanger area for a counter flow configuration. What happens when you try to do this? What is the solution?
A car travels from A, due north to a town B 4 km away. It then travels due east until it arrives town C 5 km from B. determine the distance of town C from A
Answer:
A to C = 6.4 km
Explanation:
A to B = 4 km
B to C = 5 km
A to C = using pythagorean theorem
a² + b² = c²
a = A to B = 4
b = B to C = 5
c = A to C
c² = 4² + 5²
c = 6.4 km (A to C)
According to the scenario, the distance between town C from town A is found to be 6.40 Km.
Which background does this question depend on?The background that this question depends on is known as the direction-based question. These types of questions completely depend on the distance of moving bodies like cars, persons, or any other objects as well with respect to the initial position.
According to the question,
The distance between town A to town B = 4 km.
The distance between town B to town C = 5 km.
Now, according to the Pythagoras theorem, the distance between town C to town A is as follows:
[tex]AC^2[/tex] = [tex]AB^2 +BC^2[/tex].
[tex]AC^2[/tex] = [tex]4^2+5^2[/tex]
[tex]AC^2[/tex] = 16 + 25 = 41.
AC = √41 = 6.40 km.
Therefore, the distance between town C from town A is found to be 6.40 Km.
To learn more about Pythagoras' theorem, refer to the link:
https://brainly.com/question/343682
#SPJ5
Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa m0.5. It has been determined that fracture results at a stress of 112 MPa when the maximum internal crack length is 8.6 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
Answer:
the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
Explanation:
From the given information; the objective is to compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
The Critical Stress for a maximum internal crack can be expressed by the formula:
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
where;
[tex]\sigma_c[/tex] = critical stress required for initiating crack propagation
[tex]K_{lc}[/tex] = plain stress fracture toughness = 26 Mpa
Y = dimensionless parameter
a = length of the internal crack
given that ;
the maximum internal crack length is 8.6 mm
half length of the internal crack will be 8.6 mm/2 = 4.3mm
half length of the internal crack a = 4.3 × 10⁻³ m
From :
[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]
[tex]Y= \dfrac{26}{112 \times \sqrt{\pi \times 4.3 \times 10 ^{-3}}}[/tex]
[tex]Y= \dfrac{26}{112 \times0.1162275716}[/tex]
[tex]Y= \dfrac{26}{13.01748802}[/tex]
[tex]Y=1.99731315[/tex]
[tex]Y \approx 1.997[/tex]
For this same component and alloy, we are to also compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.
when the length of the internal crack a = 3mm
half length of the internal crack will be 3.0 mm / 2 = 1.5 mm
half length of the internal crack a =1.5 × 10⁻³ m
From;
[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]
[tex]\sigma_c = \dfrac{26}{1.997 \sqrt{\pi \times 1.5 \times 10^{-3}}}[/tex]
[tex]\sigma_c = \dfrac{26}{0.1370877444}[/tex]
[tex]\sigma_c =189.6595506[/tex]
[tex]\sigma_c =[/tex] 189.66 MPa
Thus; the required stress level at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa
A single-phase transformer has 480 turns on the primary and 90 turns on the secondary. The mean length of the flux path in the core is 1.8 m and the joints are equivalent to an airgap of 0.1 mm. The value of the magnetic field strength for 1.1 T in the core is 400 A/m, the corresponding core loss is 1.7 W/kg at 50 Hz and the density of the core is 7800 kg/m3. If the maximum value of the flux density is to be 1.1 T when a p.D. Of 2200 V at 50 Hz is applied to the primary, calculate: a. The cross-sectional area of the core; b. The secondary voltage on no load; c. The primary current and power factor on no load.
Answer:
a) cross sectional area of the core = 0.0187 m²
b) The secondary voltage on no-load = 413 V
c) The primary currency and power factor on no load is 1.21 A and 0.168 lagging respectively.
Explanation:
See attached solution.
Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent to the maximum shear stress.
In this exercise we have to calculate the formula that will be able to determine the length of the cantilevered, like this:
[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]
So to determinated the maximum tensile and compreensive stress due to bending we can describe the formula as:
[tex]\sigma_b = \frac{MC}{I}[/tex]
Where,
[tex]\sigma_b[/tex] is the compressive stress or tensile stress[tex]M[/tex] is the B.M [tex]C[/tex] is the N.A distance[tex]I[/tex] is the moment of interiorSo making this formula for the max, we have:
[tex]\sigma_c=\frac{MC}{I} \\\sigma_T=-\sigma_c=-\frac{MC}{I}\\\sigma_{max}=M_{max}\\[/tex]
With all this information we can put the formula as:
[tex]\sigma_{max}C=\frac{M_{max}C}{I}[/tex]
See more about stress in the beam at brainly.com/question/23637191
Cold water at 20 degrees C and 5000 kg/hr is to be heated by hot water supplied at 80 degrees C and 10,000 kg/hr. You select from a manufacturer's catalog a shell-and-tube heat exchanger (one shell with two tube passes) having a UA value of 11,600 W/K. Determine the hot water outlet temperature.
Answer:
59°C
Explanation:
Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)
and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)
Therefore the minimum and maximum heat capacities are:
Cmin = Cc = 5803.2(W/K)
Cmax = Ch = 11634.3(W/K)
The capacity ratio is:
Cr = Cmin / Cmax = 0.499 = 0.5
The maximum possible heat transfer rate is:
Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)
And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99
Given that from the appropriate graph in the handouts we can read = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)
Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C
If you see a red, a green, and a white light on another boat, what does this tell you?
A boat is approaching you head on.
The red and green lights are sidelights that are positioned on the port side (red) (left as facing the bow) and starboard (green) (right as facing the bow) side of the boat. Various white lights are required depending on the size of the boat, but generally, a white masthead light and stern light are required. See the US Coast Guard site in the link below for more specific information.
Hope this helps
what's the maximum shear on a 3.0 m beam carrying 10 kN/m?
Answer:
max shear = R = V = 15 kN
Explanation:
given:
load = 10 kn/m
span = 3m
max shear = R = V = wL / 2
max shear = R = V = (10 * 3) / 2
max shear = R = V = 15 kN
4. ""ABC constriction Inc."" company becomes the lowest in the bed process to get a $21 million construction project for ""Northern Inc."". Now ""ABC construction Inc."" planning to make a formal contract agreement
Answer:
hello your question is incomplete here is the complete question
. “ABC construction Inc.” company becomes the lowest in the bed process to get a $21 million construction project for “Northern Inc.”. Now “ABC construction Inc.” planning to make a formal contract agreement with the “Northern Inc.”. What are the main elements of this agreement to consider it as a legal contract?
Answer : elements of the agreement
offeracceptancecapacity certaintyconsiderationintention to create legal relationExplanation:
Offer : an offer is the beginning element for any valid agreement to be started or reached between two or more bodies. ABC construction would have to make an offer first for the agreement to be valid
Acceptance: This is part where by the company "Northern Inc" after receiving the offer from ABC construction Inc would have to consent to the approval of the offer made.
capacity : This the element of the agreement that helps to ensure that both parties have the legal and financial backings to embark on the contract agreement .
certainty : This element ensures that both parties understands the terms and conditions attached to the agreement and this to ensure that there are no bogus conditions
Consideration : This is a very vital element because the both parties have to give something in return while going into a valid agreement
Intention to create legal relation : Legal relations are applied to contract agreements whereby both parties want the contract agreement to b legally enforced and this is important in order to prevent contract breach by any party involved in the agreement
Some characteristics of clay products such as (a) density, (b) firing distortion, (c) strength, (d) corrosion resistance, and (e) thermal conductivity are affected by the extent of vitrification. Will they increase or decrease with increasing degree of vitrification?
1. (a) increase (b) decrease (c) increase (d) decrease (e) increase
2. (a) decrease (b) increase (c) increase (d) increase (e) decrease
3. (a) decrease (b) decrease (c) increase (d) decrease (e) decrease
4. (a) increase (b) increase (c) increase (d) increase (e) increase
5. (a) increase (b) decrease (c) decrease (d) increase (e) decrease
Explanation:
1. increase This due to increase in the pore volume.
2.increase . This is due to the fact that more liquid phase will be present at the firing.
3. Increase. This increase is because of the fact that clay on cooling forms glass.Thus, gaining more strength as the liquid phase formed fills in pore volume.
4. Increase, Rate of corrosion depends upon the surface area exposed.Since, upon vitrification surface area would increase, therefore corrosion increases.
5. Increase , glass has higher thermal conductivity than the pores it fills.
A transformer winding contains 900 turns of wire which creates 400 ohms of primary importance how many terms of the same size wire required to create a secondary impedance of 25 ohms
Answer: 255
255 turns are required to create 25 ohms of secondary impedance.
Explanation:
Given that,
Number of turns in primary wire N₁ = 900
impedance on Primary wire Z₁ = 400 ohms
Number of turns in Secondary wire N₂ = ?
impedance on Secondary wire Z₂ = 25 ohms
we know that, the relationship between turn and impedance is
Zp / Zs = ( Np / Ns )²
(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²
there fore
Z₁ / Z₂ = ( N₁ / N₂ )²
Now we substitute
( 400 / 25 ) = ( 900 / N₂ )²
400 / 25 = 900² / N₂²
we cross multiple to get our N₂
400 × N₂² = 900² × 25
N₂² = ( 900² × 25 ) / 400
N₂² = ( 810000 × 25 ) / 400
N₂² = 20250000 / 400
N₂² = 50625
N₂ = √50625
N₂ = 225
Therefore 255 turns are required to create 25 ohms of secondary impedance.
steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a 1000 ft ramps at a high speed vo and travels 600ft in 7 s at constant deceleration before its speed is reduced to uo/2. Assuming the same constant deceleration.
Determine:
a. The additional time required for the truck to stop.
b. The additional distance traveled by the truck.
Answer:
a. 6 seconds
b. 180 feet
Explanation:
Images attached to show working.
a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.
b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.