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.
The re truck is moving forward at a speed of 35 mi /hr and is decelerating at the rate of 10 ft/sec2. Simultaneously, the ladder is being raised and extended. At the instant considered the angle is 30° and is increasing at the constant rate of 10 deg /sec. Also at this instant the extension b of the ladder is 5 ft, with b ˙ = 2 ft/sec and b ¨ =−1 ft/sec2. For this instant determine the acceleration of the end A of the ladder (a) with respect to the truck and (b) with respect to the ground.
Answer:
(a). -1.76i + 0.76j ft/s^2.
(b). -10.42i + 5.7j ft/s^2.
Explanation:
So, we are given the following data or parameters in the question/problem above;
=> "The fire truck is moving forward at a speed = 35 mi /hr and is decelerating at the rate = 10 ft/sec2."
=> "At the instant considered the angle = 30° and is increasing at the constant rate = 10 deg /sec. Also at this instant the extension b of the ladder = 5 ft, with b ˙ = 2 ft/sec and b ¨ =−1 ft/sec2"
STEP ONE: Convert the angular speed to rad/s and speed to ft/s.
Angular speed => π/18 rad/s
Speed => 35 × 1.46667 = 52.33 ft/s
STEP TWO: determine the coordinate form of the acceleration of the truck.
= - 10(cos 30°)i - sin(30°)j [π/18 × K] × 25i[π/18 × K] × 2i × [π/9 × K] - 1i.
= -10.42i + 5.7j ft/s^2.
PLEASE NOTE: The step TWO above is the solution to the second part (b) of this question which is the with respect to the ground.
STEP THREE:
This, from step two above;
( -10.42i + 5.7j ) - ( acceleration of the truck).
= - 10.42i + 5.7j - - 10(cos 30°)i - sin(30°)j = -1.76i + 0.76j ft/s^2.
An AISI/SAE 4340-A steel rod with the yield strength of 450 MPa, 2.0 m long will be subjected to a tensile force, must have the minimum weight possible, and must behave elastically for this load. The elastic modulus of steel is 207 GPa. What is the engineering strain of the rod
Answer: 0.002174
Explanation:
Given that the
Yield strength rho = 450 MPa
Length = 2 m
Elastic modulus E= 207 GPa
According to Hook's law, if the elastic limits is not reached, the elastic modulus is the ratio of elastic strength to the elastic strain ə
E = rho/ə
Make ə the subject of formula
ə = rho/ E
ə = (450 × 10^6) / (207 × 10^9)
ə = 2.174 × 10^-3
Therefore, the engineering strain which depends on engineering stress and elastic modulus is 2.174 × 10^-3
Elastic Strain has no S.I Units.
A solid shaft is subjected to an axial load P = 200 kN and a torque T = 1.5 kN.m. a) Determine the diameter of the shaft if the maximum shear stress should not exceed 100 Mpa. b) Using the diameter, determine the maximum normal stress.
Answer:
a) 42 mm
b) 144.4 MPa
Explanation:
Load P = 200 kN = 200 x 10^3 N
Torque T = 1.5 kN-m = 1.5 x 10^3 N-m
maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa
diameter of shaft d = ?
From T = τ * [tex]\frac{\pi }{16}[/tex] * [tex]d^{3}[/tex]
substituting values, we have
1.5 x 10^3 = 100 x 10^6 x [tex]\frac{3.142 }{16}[/tex] x [tex]d^{3}[/tex]
[tex]d^{3}[/tex] = 7.638 x 10^-5
d = [tex]\sqrt[3]{7.638 * 10^-5}[/tex] = 0.042 m = 42 mm
b) Normal stress = P/A
where A is the area
A = [tex]\frac{\pi d^{2} }{4}[/tex] = [tex]\frac{3.142*0.042^{2} }{4}[/tex] = 1.385 x 10^-3
Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = 144.4 MPa
The data listed below are claimed for power cycles operating between hot and cold reservoirs at 1000 K and 400 K, respectively. For each case determine whether such a cycle is in keeping with the first and second laws of thermodynamics.
a) Qh=300 kJ, W(cycle)=160kJ, Qc=140 kJ
b) Qh=300 kJ, W(cycle)=180kJ, Qc=120 kJ
c) Qh=300 kJ, W(cycle)=170kJ, Qc=140 kJ
d) Qh=300 kJ, W(cycle)=200kJ, Qc=100 kJ
Answer:
a) This cycle observes the First and Second Laws of Thermodynamics, b) This cycle observes the First and Second Laws of Thermodynamics, c) This cycle does not observe the First Law of Thermodynamics, d) This cycle does not observe the Second Law of Thermodynamics.
Explanation:
The Carnot cycle offers a reliable criterion to determine the maximum theoretical efficiency ([tex]\eta_{th,max}[/tex]) for a power cycle in term of the temperatures of hot and cold reservoirs and expressed in percentage:
[tex]\eta_{th, max} = \left(1-\frac{T_{L}}{T_{H}} \right)\times 100\,\%[/tex]
Where:
[tex]T_{L}[/tex], [tex]T_{H}[/tex] - Temperatures of cold and hot reservoirs, measured in kelvins.
If [tex]T_{L} = 400\,K[/tex] and [tex]T_{H} = 1000\,K[/tex], the maximum theoretical thermal efficiency is:
[tex]\eta_{th, max} = \left(1-\frac{400\,K}{1000\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{th,max} = 60\,\%[/tex]
In addition, the real efficiency of the heat engine is described by the following formula:
[tex]\eta_{th} = \left(1-\frac{Q_{L}}{Q_{H}} \right)\times 100\,\%[/tex]
Where:
[tex]Q_{H}[/tex] - Heat absorbed by the heat engine from hot reservoir, measured in kilojoules.
[tex]Q_{L}[/tex] - Heat released by the heat engine to the cold reservoir, measured in kilojoules.
The following conditions must be observed by all heat engines:
First Law of Thermodynamics
[tex]W = Q_{H}-Q_{L}[/tex]
Second Law of Thermodynamics
[tex]\eta_{th} \leq \eta_{th,max}[/tex]
Now, each cycle is checked:
a) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 140\,kJ[/tex] and [tex]W = 160\,kJ[/tex]
First Law of Thermodynamics
[tex]W = 300\,kJ-140\,kJ[/tex]
[tex]W = 160\,kJ[/tex]
This cycle observes the First Law of Thermodynamics.
Second Law of Thermodynamics
[tex]\eta_{th} = \left(1-\frac{140\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 53.333\,\%[/tex]
This cycle observes the Second Law of Thermodynamics.
b) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 120\,kJ[/tex] and [tex]W = 180\,kJ[/tex]
First Law of Thermodynamics
[tex]W = 300\,kJ-120\,kJ[/tex]
[tex]W = 180\,kJ[/tex]
This cycle observes the First Law of Thermodynamics.
Second Law of Thermodynamics
[tex]\eta_{th} = \left(1-\frac{120\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 60\,\%[/tex]
This cycle observes the Second Law of Thermodynamics.
c) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 140\,kJ[/tex] and [tex]W = 170\,kJ[/tex]
First Law of Thermodynamics
[tex]W = 300\,kJ-140\,kJ[/tex]
[tex]W = 160\,kJ[/tex]
This cycle does not observe the First Law of Thermodynamics.
Second Law of Thermodynamics
[tex]\eta_{th} = \left(1-\frac{140\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 53.333\,\%[/tex]
This cycle observes the Second Law of Thermodynamics.
d) [tex]Q_{H} = 300\,kJ[/tex], [tex]Q_{L} = 200\,kJ[/tex] and [tex]W = 100\,kJ[/tex]
First Law of Thermodynamics
[tex]W = 300\,kJ-100\,kJ[/tex]
[tex]W = 200\,kJ[/tex]
This cycle observes the First Law of Thermodynamics.
Second Law of Thermodynamics
[tex]\eta_{th} = \left(1-\frac{100\,kJ}{300\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 66.667\,\%[/tex]
This cycle does not observe the Second Law of Thermodynamics.
Based on the first parameters, the power cycle obeys both the First and Second Law of Thermodynamics.
Given the following data:
Temperature of hot reservoir = 1000 K.Temperature of cold reservoir = 400 K.How to verify which law a power cycle obeys.In order to verify a power cycle obeys the first and second laws of thermodynamics, we would use the Carnot cycle.
Mathematically, the maximum theoretical efficiency for a power cycle in terms of the temperature is given by this formula:
[tex]\eta_{th, max}=(1-\frac{T_c}{T_h} ) \times 100[/tex]
Where:
[tex]T_c[/tex] is the temperature of cold reservoir.[tex]T_h[/tex] is the temperature of hot reservoir.Substituting the given parameters into the formula, we have;
[tex]\eta_{th, max}=(1-\frac{400}{1000} ) \times 100\\\\\eta_{th, max}=60\%[/tex]
Similarly, the real efficiency of a power cycle, we have:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100[/tex]
The conditions for a power cycle.According to the First Law of Thermodynamics, the following condition must be met:
[tex]W_{cycle}=Q_h-Q_c[/tex]
According to the Second Law of Thermodynamics, the following condition must be met:
[tex]\eta_{th }\leq \eta_{th, max }[/tex]
Next, we would determine whether or not each obeys the first and second laws of thermodynamics:
When [tex]Q_h=300\; kJ, \;W_{(cycle)}=160kJ, \;and\;Q_c=140\; kJ[/tex]
For the First Law:
[tex]W_{cycle}=Q_h-Q_c\\\\160=300-140\\\\160=160[/tex]
Therefore, the power cycle obeys the First Law of Thermodynamics.
For the Second Law:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{140}{300} ) \times 100\\\\\eta_{th}=53.33\%\\\\\eta_{th }\leq \eta_{th, max } =53.33\%\leq 60\%[/tex]
Therefore, the power cycle obeys the Second Law of Thermodynamics.
Using the second parameters.When [tex]Q_h=300\; kJ, \;W_{(cycle)}=180kJ, \;and\;Q_c=120\; kJ[/tex]
For the First Law:
[tex]W_{cycle}=Q_h-Q_c\\\\180=300-120\\\\180=180[/tex]
Therefore, the power cycle obeys the First Law of Thermodynamics.
For the Second Law:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{120}{300} ) \times 100\\\\\eta_{th}=60\%\\\\\eta_{th }\leq \eta_{th, max } =60\%\leq 60\%[/tex]
Therefore, the power cycle obeys the Second Law of Thermodynamics.
Using the third parameters.When [tex]Q_h=300\; kJ, \;W_{(cycle)}=170kJ, \;and\;Q_c=140\; kJ[/tex]
For the First Law:
[tex]W_{cycle}=Q_h-Q_c\\\\170=300-140\\\\170\neq 160[/tex]
Therefore, the power cycle doesn't obey the First Law of Thermodynamics.
For the Second Law:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{140}{300} ) \times 100\\\\\eta_{th}=53.33\%\\\\\eta_{th }\leq \eta_{th, max } =53.33\%\leq 60\%[/tex]
Therefore, the power cycle obeys the Second Law of Thermodynamics.
Using the fourth parameters.When [tex]Q_h=300\; kJ, \;W_{(cycle)}=200kJ, \;and\;Q_c=100\; kJ[/tex]
For the First Law:
[tex]W_{cycle}=Q_h-Q_c\\\\200=300-100\\\\200=200[/tex]
Therefore, the power cycle obeys the First Law of Thermodynamics.
For the Second Law:
[tex]\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{100}{300} ) \times 100\\\\\eta_{th}=66.67\%\\\\\eta_{th }\leq \eta_{th, max } \neq 63.33\%\geq 60\%[/tex]
Therefore, the power cycle doesn't obey the Second Law of Thermodynamics.
Read more on thermodynamics here: brainly.com/question/11628413
You subjected a rod under the cyclic stress with the maximum stress of 200 MPa and minimum stress of 20 MPa. The fatigue limit was determined to be ~100 MPa. How many cycles can this materials sustain before failure?
Answer:
The material will not fail
Explanation:
A rod subjected under cyclic stress will fail if the cyclic stress it is subjected to is a constant maximum value that is above the fatigue limit of the rod. but in this problem the Rod is subjected to a cyclic stress that ranges from 200 MPa(maximum stress) and 20 MPa ( minimum stress). this simply means that at all times the Rod will not experience maximum stress of 200 MPa and its Fatigue limit is also set at ~100 MPa
attached is the diagram showing the cyclic stress the rod is subjected to
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace
Answer:
T = 858.25 s
Explanation:
Given data:
Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),
initial uniform temperature ( Ti ) = 200 c
Final temperature = 550 c
convection coefficient = 250 w/m^2 k
products combustion temp = 800 c
calculate how long the plate should be left in the furnace ( to attain 550 c )
first calculate/determine the Fourier series Number ( Fo )
[tex]\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}[/tex]
= 0.4167 = [tex]1.0396e^{-0.4888*Fo}[/tex]
therefore Fo = 3.8264
Now determine how long the plate should be left in the furnace
Fo = [tex](\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )[/tex]
k = 48
p = 7830
L = 0.1
Input the values into the relation and make t subject of the formula
hence t = 858.25 s
True or false : In improper integrals infinte intervals mean that both of the integration limits are should be infinity
Answer:
An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration
Explanation:
A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where it is traveling is 60°F. Determine the drag on the fin when the submarine is traveling at 2.5 ft/s.
Answer:
[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]
Explanation:
Given that:
The height of a triangular stabilizing fin on its stern is 1 ft tall
and it length is 2 ft long.
Temperature = 60 °F
The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.
From these information given; we can have a diagrammatic representation describing how the triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.
The diagram can be found in the attached file below.
If we recall ,we know that;
Kinematic viscosity v = [tex]1.2075 \times 10^{-5} \ ft^2/s[/tex]
the density of water ρ = 62.36 lb /ft³
[tex]Re_{max} = \dfrac{Ux}{v}[/tex]
[tex]Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}[/tex]
[tex]Re_{max} = 414078.6749[/tex]
[tex]Re_{max} = 4.14 \times 10^5[/tex] which is less than < 5.0 × 10⁵
Now; For laminar flow; the drag on the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:
[tex]dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2[/tex]
where;
[tex](2-x) dy[/tex] = strip area
[tex]Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}[/tex]
Therefore;
[tex]dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})[/tex]
[tex]dF_D = 1.136 \times(2-x)^{1/2} \ dy[/tex]
Let note that y = 0.5x from what we have in the diagram,
so , x = y/0.5
By applying the rule of integration on both sides, we have:
[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy[/tex]
[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy[/tex]
Let U = (2-2y)
-2dy = du
dy = -du/2
[tex]F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}[/tex]
[tex]F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du[/tex]
[tex]F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2[/tex]
[tex]F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2[/tex]
[tex]F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ][/tex]
[tex]F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ][/tex]
[tex]F_D = 1.071031071 \ lbf[/tex]
[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]
On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. How do you explain this behavior?
Answer:
Enthalpy is a function of pressure hence normalized enthalpy departure values will approach zero with reduced pressure approaching zero
Explanation:
On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. this is because enthalpy is a function of pressure therefore as the Pressure is reducing towards the zero value, the gas associated with the pressure tends to behave more like an Ideal gas.
For an Ideal gas the Normalized enthalpy departure value will be approaching the zero value.
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
Which of these properties generally applies to transfer lines? A. It is difficult to introduce product modifications. B. It is difficult to store products between individual machines. C. The process can run even if one of the machines fails. D. Products can take different paths through the system.
Answer:
It is difficult to introduce products modification
While having a discussion, Technician A says that you should never install undersized tires on a vehicle. The vehicle will be lower, and the speedometer will no longer be accurate. Technician B says that the increase in engine rpm for a given speed will result in a decrease in fuel economy. Who is correct
Answer:
Both technician A and technician B are correct.
Explanation: Vehicle manufacturers always specify the size of the tires required for a given vehiclefor optimal efficiency,this will ensure that the speedometer is accurate and the level of the vehicle is good enough to ensure the vehicle works efficiently.
It is also a known fact that an increase in a vehicle's rpm(revolution per minute) will eventually lead to increased fuel consumption which means the fuel economy of the vehicle will be reduced making the vehicle less efficient in its fuel consumption.
. 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 adiabatic compressor receives 1.5 meter cube per second of air at 30 degrees celsius and 101 kpa. The discharge pressure is 505 kpa and the power supplied is 325 kW, what is the discharge temperature
Answer:
The discharge temperature is 259.82 K
Explanation:
In this question, we are concerned with calculating the discharge temperature
Please check attachment for complete solution
g 940 The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass of 50 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P
Answer:
μb = 0.096
μc = 0.073
Explanation:
member AB:
-800( 4/3 ) + Nb (2) = 0
Nb (2) = 3200/3
Nb = 533.3N
Post BC:
summation of force along the y axis=0
Nc + Nb + 150(3/5 ) -50(9.81)=0
Nc + 533.3 + 150(3/5 ) -50(9.81)=0
Nc = 933.83 N
Also (-4/5)(150)(3) + Fb(0.7)= 0
Fb = (4/5)(150)(3)/0.7 = 51.429 N
Likewise alog the x axis,
4/5(150) - Fc -Fb = 0
4/5(150) - Fc -51.429 = 0
Fc = 4/5(150) -51.429 =68.571 N
μb = Fb/Nb = 51.429/533.3 = 0.096
μc = Fc/Nc = 68.571 / 933.83 = 0.073
Describe the similarities and differences between circuits with resistors combined in series and circuits with resistors combined in parallel
Answer:
from the below explanation... we can say that, in the series circuit, flowing current remains the same at each part of the circuit. While in parallel circuits, the voltage across two endpoints of the branches is the same as the supplied voltage.
Explanation:
1.
The components in a series circuit are arranged in a single path from one end of supply to another end. However, the multiple components in a parallel circuit are arranged in multiple paths wrt the two end terminals of the battery.
2.
In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit.
3.
In the series circuit, different voltage exists across each component in the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.
4.
A fault in one of the components of the series circuit causes hindrance in the operation of a complete circuit. As against fault in a single component in a parallel network do not hinder the functioning of another part of the circuit.
5.
The detection of a fault in case of a series circuit is difficult, but it is quite easy in parallel circuits.
6.
The equivalent resistance in case of a series circuit is always more than the highest value of resistance in the series connection. While the equivalent resistance in the parallel circuit is always less than any of the individual resistances in parallel combination.
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
An elastic cable is to be designed for bungee jumping from a tower 130 ft high. The specifications call for the cable to be 85 ft long when unstretched, and to stretch to a total length of 100 ft when a 750-lb weight is attached to it and dropped from the tower.
Determine:
a. The required spring constant k of the cable.
b. How close to the ground a 185-lb man will come if he uses this cable to jump from the tower?
Answer:
a) The spring constant is 50 lb/ft
b) The man is 26.3 ft close to the ground.
Explanation:
Height of tower is 130 ft
Specification calls for a cable of length 85 ft
the maximum this length stretches is 100 ft when subjected to a load of 750 lb
The extension of the cable is calculated from the formula from Hooke's law
F = kx
where F is the load or force on the cable
k is the spring constant of the cable
x is the extension on the cable
a) The extension on the cable is
x = 100 ft - 85 ft = 15 ft
substituting into the formula above, we'll have
750 = k*15
k = 750/15 = 50 lb/ft
b) for a 185 lb man, jumping down will give an extension gotten as
F = kx
185 = 50*x
x = 185/50 = 3.7 ft
The total length of the cable will be extended to 100 ft + 3.7 ft = 103.7 ft
closeness to the ground = 130 ft - 103.7 ft = 26.3 ft
Fences four feet high can be used to:____________
A. Deter casual trespassers
B. Provide reasonable defense from external intrusion
C. Discourage a determined intruder
D. Slow down a determined intruder
Answer:
A. Deter casual trespassers
Explanation:
In safeguarding valued property, fences are used to serve as a psychological and physical barrier to intending intruders. Depending on its height, the fence could serve several purposes. A fence that is three to four foot high would only serve to deter casual trespassers. This is because any intruder would make an effort to scale a four-foot-high fence.
However, when an intruder is determined he would make an effort to scale a fence that is even up to eight-foot-high. So to provide optimum security, a very high fence supported with barb wire and a closed-circuit television camera would help to ensure that the environment is monitored against any determined intruder.
One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid surface from a height of 120 in., the height of the first bounce of the ball must be in the range 55 in. <= h <= 60 in. Determine the range of the coefficients of restitution of the tennis balls satisfying this requirement. Any ideas on this?
Answer:
At temperature is and relative humidity is 86% therefore, the humidity ratio is 0.0223 and the specific volume is 14.289
At temperature is and Relative humidity is 40% therefore, the humidity ratio is 0.0066 and the specific volume is 13.535.
To calculate the mass of air can be calculated as follows:
Now , we going to calculate the volume,
The time which is required to fill the cistern can be calculated as follows:
Now, putting the value in above formula we get,
Therefore, the hours required to fill the cistern is 4.65 hours.
Explanation:
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
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
A cantilever beam is 4000 mm long span and has a u.d.l. of 0.30 kN/m. The flexural stiffness is 60 MNm². Calculate: 1. Slope 2. Deflection.
Answer:
1. Slope = 53.3 x 10⁻⁶
2. Deflection = -0.00016m
Explanation:
given:
let L = 4 m (span of cantilever beam)
let w = 300 N/m (distributed load)
let EI =60 MNm² (flexural stiffness)
dy w * L³ 300 x 4³
1. slope = ------- = --------- = ------------------- = 53.3 x 10⁻⁶
dx 6EI 6 x 60x10⁶
wL⁴ 300 x 4⁴
2. Deflection = y = - ----------- = - ------------------ = -0.00016m
8EI 8 x 60x10⁶
therefore the deflection is 0.16mm downwards.
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.
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:
"A three-phase, 60-Hz, 4-pole, 440-V (line-line, rms) induction-motor drive has a full-load (rated) speed of 1750 rpm. The rated torque is 40 Nm. Keeping the air gap flux-density peak constant at its rated value, (a) plot the torque-speed characteristics (the linear portion) for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz. (b) This motor is supplying a load whose torque demand increases linearly with speed, such that it equals the rated torque of the motor at the rated motor speed. Calculate the speeds of operation at the four values of frequency in part (a)."
Answer:
A) slip(s1) = (1800 - 1750) / 1800 = 0.0277
B) 1800 rpm, 1350 rpm, 900 rpm, 450 rpm
Explanation:
Given data
frequency = 60 Hz,
Line - line rms = 440 - V
3 phase induction-motor drive
number of poles = 4
Full-load rated speed = 1750
rated torque = 40 Nm
A) The plot of torque-speed characteristics for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz is attached below
first we calculate the rated speed:
Ns = [tex]\frac{120f}{p}[/tex]
f = 60 Hz . p (number of poles) = 4
Ns[tex]_{1}[/tex] = [tex]\frac{120(60)}{4}[/tex] = 1800
full loaded rated value = 1750.
slip(s1) = (1800 - 1750) / 1800 = 0.0277
considering a linearly condition the slip is low
[tex]\frac{T1}{T2} = \frac{S1}{S2} * \frac{F2}{F1}[/tex]
S1 = 0.0277
f1 = 60 Hz
hence s2 = 0.018 therefore Ns2 = 1500
B) The speeds of operation at : 60 Hz, 45 Hz, 30 Hz, 15 Hz
for 60 Hz :
Ns = [tex]\frac{120f}{p}[/tex] = (120*60) / 4 = 1800 rpm
for 45 Hz:
Ns = 120(f) / p = (120*45) / 4 = 5400 /4 = 1350 rpm
for 30 Hz:
Ns = 120(f) / p = (120*30) / 4 = 3600 / 4 = 900 rpm
for 15 Hz:
Ns = 120(f) / p = (120*15) / 4 = 1800 / 4 = 450 rpm
what is the rated power output in ( kw) of a 8 pole motor designed to an IEC 180L motor frame ?
Answer:
P=11 kW
Explanation:
Given that
Number of poles= 8
I.E.C. 180L motor frame
From data book , for 8 poles motor at 50 Hz
Speed = 730 rpm
Power factor = 0.75
Efficiency at 100 % load= 89.3 %
Efficiency at 50 % load= 89.1 %
Output power = 11 kW
Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.
P=11 kW
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:
1). Mention any four operations that requires airlines. 2). Explain how airflow is applicable to the above mentioned operations.
Answer:
Following are the answer to this question:
Explanation:
1)
Following the four operations in the airlines:
Landside operations:
In Airlines, the airports are divided into areas on the countryside and on the airside, in which landside region is available to the public, although strictly controlled access to the airside zone. Its area covers all areas of the airport across the aircraft, including parts of the buildings which can only be reached by customers and employees.
Airside Operations:
It's also committed to ensuring which air operations military exercises Ballarat airfields are safe and secure. It includes the provision of parking and flight escort services to itinerant and automates. Organizing operational response to incidents, accidents, or emergencies at the airport.
Billing and invoicing Operations:
This requires several steps, each of which must be performed with absolute accuracy to ensure that perhaps the airport operator is adequately paid for supplying passengers with all the services and infrastructure. After this, the receipts want to be produced and sent to customers on the airline.
Information management:
In this, it collects all the data about the customers and employees and it also helps in finding new routes and after collecting the data it processes on them.
2)
It involves many activities in the airline, including dispatch, flight preparation, flight watch, weather information source, activities control, ground-to-air communications, and staff coordination, scheduling, and maintenance planning. Computing and expert programs are constantly being used to handle unpredictable activities.
An Otto cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and pressure ratio (rp) for constant volume heat addition, determine the efficiency and other values listed below. The gas constant for air is R = 0.287 kJ/ kg.K
T1= 310K
P1(kpa)= 100
r=11.5
rp =1.95
Required:
a. Determine the specific internal energy (kJ/kg) at state 1.
b. Determine the relative specific volume at state 1.
c. Determine the relative specific volume at state 2.
d. Determine the temperature (K) at state 2.
e. Determine the specific internal energy (kJ/kg) at state 2.
Answer:
A) 222.58 kJ / kg
B) 0.8897 M^3/ kg
c) 0.7737 m^3/kg
D) 746.542 k
E) 536.017 kj/kg
efficiency = 58% ( approximately )
Explanation:
Given Data :
Gas constant (R) = 0.287 kJ/ kg.K
T1 = 310 k
P1 ( Kpa ) = 100
r = 11.5 ( compression ratio )
rp = 1.95 ( pressure ratio )
A ) specific internal energy at state 1
= Cv*T1 = 0.718 * 310 = 222.58 kJ / kg
B) Relative specific volume at state 1
= P1*V1 = R*T1 ( ideal gas equation )
V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3
V1 = 88.97 / 100 = 0.8897 M^3/ kg
C ) relative specific volume at state 2
Applying r ( compression ratio) = V1 / V2
11.5 = 0.8897 / V2
V2 = 0.8897 / 11.5 = 0.7737 m^3/kg
D) The temperature (k) at state 2
since the process is an Isentropic process we will apply the p-v-t relation
[tex]\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }[/tex]
hence T2 = [tex]9^{1.4-1} * 310[/tex] = 2.4082 * 310 = 746.542 k
e) specific internal energy at state 2
= Cv*T2 = 0.718 * 746.542 = 536.017 kj/kg
efficiency = output /input = 390.3511 / 667.5448 ≈ 58%
attached is a free hand diagram of an Otto cycle is attached below