Answer:
minimum required diameter of the steel linkage is 3.57 mm
Explanation:
original length of linkage l = 10 m
force to be transmitted f = 2 kN = 2000 N
extension e = 5 mm= 0.005 m
maximum stress σ = 200 N/mm^2 = [tex]2*10^{8} N/m^{2}[/tex]
maximum stress allowed on material σ = force/area
imputing values,
200 = 2000/area
area = 2000/([tex]2*10^{8}[/tex]) = [tex]10^{-5}[/tex] m^2
recall that area = [tex]\pi d^{2} /4[/tex]
[tex]10^{-5}[/tex] = [tex]\frac{3.142*d^{2} }{4}[/tex] = [tex]0.7855d^{2}[/tex]
[tex]d^{2} = \frac{10^{-5} }{0.7855}[/tex] = [tex]1.273*10^{-5}[/tex]
[tex]d = \sqrt{1.273*10^{-5} }[/tex] = [tex]3.57*10^{-3}[/tex] m = 3.57 mm
maximum diameter of the steel linkage d = 3.57 mm
If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I send
Answer:
32000 bits/seconds
Explanation:
Given that :
there are 16 signal combinations (states) = 2⁴
bits n = 4
and a baud rate (number of signals/second) = 8000/second
Therefore; the number of bits per seconds can be calculated as follows:
Number of bits per seconds = bits n × number of signal per seconds
Number of bits per seconds = 4 × 8000/second
Number of bits per seconds = 32000 bits/seconds
Find the heat flow from the composite wall as shown in figure. Assume one dimensional flow KA=150 W/m°C , KB=25 W/m°C, KC=60 W/m°C , KD=60 W/m°C
Answer:
The heat flow from the composite wall is 1283.263 watts.
Explanation:
The conductive heat flow through a material, measured in watts, is represented by the following expression:
[tex]\dot Q = \frac{\Delta T}{R_{T}}[/tex]
Where:
[tex]R_{T}[/tex] - Equivalent thermal resistance, measured in Celsius degrees per watt.
[tex]\Delta T[/tex] - Temperature gradient, measured in Celsius degress.
First, the equivalent thermal resistance needs to be determined after considering the characteristics described below:
1) B and C are configurated in parallel and in series with A and D. (Section II)
2) A and D are configurated in series. (Sections I and III)
Section II
[tex]\frac{1}{R_{II}} = \frac{1}{R_{B}} + \frac{1}{R_{C}}[/tex]
[tex]\frac{1}{R_{II}} = \frac{R_{B}+R_{C}}{R_{B}\cdot R_{C}}[/tex]
[tex]R_{II} = \frac{R_{B}\cdot R_{C}}{R_{B}+R_{C}}[/tex]
Section I
[tex]R_{I} = R_{A}[/tex]
Section III
[tex]R_{III} = R_{D}[/tex]
The equivalent thermal resistance is:
[tex]R_{T} = R_{I} + R_{II}+R_{III}[/tex]
The thermal of each component is modelled by this:
[tex]R = \frac{L}{k\cdot A}[/tex]
Where:
[tex]L[/tex] - Thickness of the brick, measured in meters.
[tex]A[/tex] - Cross-section area, measured in square meters.
[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Celsius degree.
If [tex]L_{A} = 0.03\,m[/tex], [tex]L_{B} = 0.08\,m[/tex], [tex]L_{C} = 0.08\,m[/tex], [tex]L_{D} = 0.05\,m[/tex], [tex]A_{A} = 0.01\,m^{2}[/tex], [tex]A_{B} = 3\times 10^{-3}\,m^{2}[/tex], [tex]A_{C} = 7\times 10^{-3}\,m^{2}[/tex], [tex]A_{D} = 0.01\,m^{2}[/tex], [tex]k_{A} = 150\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]k_{B} = 25\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]k_{C} = 60\,\frac{W}{m\cdot ^{\circ}C}[/tex] and [tex]k_{D} = 60\,\frac{W}{m\cdot ^{\circ}C}[/tex], then:
[tex]R_{A} = \frac{0.03\,m}{\left(150\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}[/tex]
[tex]R_{A} = \frac{1}{50}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{B} = \frac{0.08\,m}{\left(25\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (3\times 10^{-3}\,m^{2})}[/tex]
[tex]R_{B} = \frac{16}{15}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{C} = \frac{0.08\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (7\times 10^{-3}\,m^{2})}[/tex]
[tex]R_{C} = \frac{4}{21}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{D} = \frac{0.05\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}[/tex]
[tex]R_{D} = \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{I} = \frac{1}{50} \,\frac{^{\circ}C}{W}[/tex]
[tex]R_{III} = \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{II} = \frac{\left(\frac{16}{15}\,\frac{^{\circ}C}{W} \right)\cdot \left(\frac{4}{21}\,\frac{^{\circ}C}{W}\right)}{\frac{16}{15}\,\frac{^{\circ}C}{W} + \frac{4}{21}\,\frac{^{\circ}C}{W}}[/tex]
[tex]R_{II} = \frac{16}{99}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{T} = \frac{1}{50}\,\frac{^{\circ}C}{W} + \frac{16}{99}\,\frac{^{\circ}C}{W} + \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}[/tex]
Now, if [tex]\Delta T = 400\,^{\circ}C - 60\,^{\circ}C = 340\,^{\circ}C[/tex] and [tex]R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}[/tex], the heat flow is:
[tex]\dot Q = \frac{340\,^{\circ}C}{\frac{2623}{9900}\,\frac{^{\circ}C}{W} }[/tex]
[tex]\dot Q = 1283.263\,W[/tex]
The heat flow from the composite wall is 1283.263 watts.
Describe experimental factors that could be modified, and unalterable properties of materials used.
Answer:
a. mechanical properties
b. thermal properties
c. chemical properties
d. electical properties
e. magnetic properties
Explanation:
a. The mechanical properties of a material are those properties that involve a reaction to an applied load.The most common properties considered are strength, ductility, hardness, impact resistance, and fracture toughness, elasticity, malleability, youngs' modulus etc.
b. Thermal properties such as boiling point , coefficient of thermal expansion , critical temperature , flammability , heat of vaporization , melting point ,thermal conductivity , thermal expansion ,triple point , specific heat capacity
c. Chemical properties such as corrosion resistance , hygroscopy , pH , reactivity , specific internal surface area , surface energy , surface tension
d. electrical properties such as capacitance , dielectric constant , dielectric strength , electrical resistivity and conductivity , electric susceptibility , nernst coefficient (thermoelectric effect) , permittivity etc.
e. magnetic properties such as diamagnetism, hysteresis, magnetostriction , magnetocaloric coefficient , magnetoresistance , permeability , piezomagnetism , pyromagnetic coefficient
A vehicle experiences hard shifting. Technician A says that the bell housing may be misaligned. Technician B says that incorrect oil may have been put in the transmission. Who is correct? Group of answer choices
Answer:
Technician B
Explanation:
Vehicle hard shifting is a situation whereby the vehicle faces difficulty or shakes when changing gears/speed.
Actually technician B is correct, the primary reason for hard shifting is low level of transmission fluid, hard shifts can also be caused by excessive line pressure due to a clog or malfunctioning shift solenoid.
A closed, 5-m-tall tank is filled with water to a depth of 4 m. The top portion of the tank is filled with air which, as indicated by a pressure gage at the top of the tank, is at a pressure of 20 kPa. Determine the pressure that the water exerts on the bottom of the tank.
Answer:
The pressure that the water exerts on the bottom of the tank is 59.2 kPa
Explanation:
Given;
height of tank, h = 5m
height of water in the tank, [tex]h_w[/tex] = 4m
pressure at the top of the tank, [tex]P_{top}[/tex] = 20 kPa
The pressure exerted by water at the bottom of the tank is the sum of pressure on water surface and pressure due to water column.
[tex]P_{bottom} = \gamma h + P_{top}\\\\P_{bottom} = (9.8*10^3*4 \ \ + \ 20*10^3)Pa\\\\P_{bottom} = 59200 \ Pa\\\\P_{bottom} = 59.2 \ kPa[/tex]
Therefore, the pressure that the water exerts on the bottom of the tank is 59.2 kPa
For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _______ and using the Maximum Shear Stress Theory the material will __________
a. fail / not fail
b. fail /fail
c. not fail/fail
d. not fail/not fail
Answer:
Option A - fail/ not fail
Explanation:
For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16% of the cylinder volume at bottom dead center and the crankshaft rotates at 2400 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.5 lbf/in. 2 and a temperature of 60 8 F at the beginning of compression. The maximum temperature in the cycle is 5200 8 R.
Based on this model,
1- Write possible Assumptions no less than three assumptions
2- Draw clear schematic for this problem
3- Determine possible Assumptions no less than three assumptions
4- Draw clear schematic for this problem.
5- calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.
Answer:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The net work per cycle is 845.88 kJ/kg
The power developed in horsepower ≈ 45374 hP
Explanation:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m
Stroke length = 3.4 in. = 0.08636 m.
The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³
The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³
The clearance volume, v₂ = 9.59 × 10⁻⁵ m³
p₁ = 14.5 lbf/in.² = 99973.981 Pa
T₁ = 60 F = 288.706 K
[tex]\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}[/tex]
Otto cycle T-S diagram
T₂ = 288.706*[tex]6.25^{0.393}[/tex] = 592.984 K
The maximum temperature = T₃ = 5200 R = 2888.89 K
[tex]\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}[/tex]
T₄ = 2888.89 / [tex]6.25^{0.393}[/tex] = 1406.5 K
Work done, W = [tex]c_v[/tex]×(T₃ - T₂) - [tex]c_v[/tex]×(T₄ - T₁)
0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg
The power developed in an Otto cycle = W×Cycle per second
= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.
An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression when the lintel carries a total uniformly distributed load of 10000 lb on a simple span of 6 ft. Also, calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange
Answer:
hello your question lacks some information attached is the complete question
A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in
(ii) maximum bending stress in compression = 0.7413*10^6 Ib-in
B) (i) The average shear stress at the neutral axis = 0.7904 *10 ^5 psi
(ii) Average shear stress at the web = 18.289 * 10^5 psi
(iii) Average shear stress at the Flange = 1.143 *10^5 psi
Explanation:
First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)
A) Calculate the maximum bending stress in tension and compression
lintel load = 10000 Ib
simple span = 6 ft
( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS
I = 53.54
i) The maximum bending stress (fb) in tension=
= [tex]\frac{M_{mm}Y }{I}[/tex] = [tex]\frac{6.48 * 10^6 * 2.375}{53.54}[/tex] = 0.287 * 10^6 Ib-in
ii) The maximum bending stress (fb) in compression
= [tex]\frac{M_{mm}Y }{I}[/tex] = [tex]\frac{6.48 *10^6*(8.5-2.375)}{53.54}[/tex] = 0.7413*10^6 Ib-in
B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange
i) The average shear stress at the neutral axis
V = [tex]\frac{wL}{2}[/tex] = [tex]\frac{1000*6*12}{2}[/tex] = 3.6*10^5 Ib
Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - [tex]\frac{0.5}{2}[/tex] ) * [tex]\frac{(2.375 - (\frac{0.5}{2} ))}{2}[/tex]
= 5.878 in^3
t = VQ / Ib = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi
ii) Average shear stress at the web ( value gotten from the shear stress at the flange )
t = 1.143 * 10^5 * (8 / 0.5 ) psi
= 18.289 * 10^5 psi
iii) Average shear stress at the Flange
t = VQ / Ib = [tex]\frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}[/tex]
= 1.143 *10^5
A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. If the inside surface temperature is 1450 °C and outer surface temperature is 90°C, estimate the temperature between layers and calculate the heat loss in KJ/h-m2. Assume k for fire brick 4 KJ/m-h°C, k for common brick= 2.8 KJ/m-h°C, k for 85% magnesia = 0.25 KJ/m-h°C and k for steel 240 KJ/m-h°C, k
Answer:
fire brick / common brick : 1218 °Ccommon brick / magnesia : 1019 °Cmagnesia / steel : 90.06 °Cheat loss: 4644 kJ/m^2/hExplanation:
The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...
R = d/k
so the thermal resistances of the layers of furnace wall are ...
R₁ = 0.200/4 = 0.05 °C·m²·h/kJ
R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ
R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ
R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ
So, the total thermal resistance is ...
R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ
__
The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)
__
The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:
fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C
so, the fire brick interface temperature at the common brick is ...
1450 -232 = 1218 °C
For the next layers, the interface temperatures are ...
common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C
magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C
_____
Comment on temperatures
Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.
Consider two double-pipe counter-flow heat exchangers that are identical except that one is twice as long as the other. Is the longer heat exchanger more likely to have a higher effectiveness
Answer:
Both pipes will have the same effectiveness
Explanation:
Heat exchanger effectiveness is defined as the ratio of the actual amount of heat transferred to the maximum possible amount of heat that could be transferred with an infinite area.
The actual amount of heat transferred for counter-flow heat exchangers is given as;
[tex]q = C_h(T_h,_i -T_h,_o)[/tex]
where;
Ch is the specific heat capacity
Th,i is the inlet temperature of the hot fluid
Th,o is the outlet temperature of the hot fluid
The maximum possible amount of heat that could be transferred with an infinite area, is given as;
[tex]q_{max} = C_{min}(T_h,_i-T_c,_i)[/tex]
where;
Cmin is the minimum heat transfer coefficient for hot and cold fluid
Heat exchanger effectiveness is calculated as;
[tex]\epsilon = \frac{q}{q_{max}}[/tex]
From the formula above, Heat exchanger effectiveness is independent of the pipe length.
Therefore, both pipes will have the same effectiveness
Anytime scaffolds are assembled or __________, a competent person must oversee the operation.
a. Drawn
b. Disassembled
c. Thought
d. Made
When scaffolds are now being construct or deconstruct, a competent person must supervise the work and train everybody who'll be assisting, and the further discussion can be defined as follows:
The competent person is also responsible for proposing whether fall protection is required for each scaffold erected. In constructing a scaffold, there are specific criteria for the ground the scaffold is constructed. On the products and components used to build the scaffold, its height in relation to the foundation. It's platform's design, and whether or not high efficiency is needed to supervise the installation.Therefore, the final answer is "Option B".
Learn more:
brainly.com/question/16049673
A nail gun operates using pressurized air, which is supplied through a 1/4-in diameter hose. The gun requires 75 psi to operate with a 2.6 ft3 /min airflow. If the air compressor develops 90 psi, determine the maximum allowable length of hose that can be used for its operation. Assume incompressible flow and a smooth pipe. Take =0.000238 slug/ft3 and = 1.6×10-4 ft2 /s.
Answer:
If we assumed flow was laminar, L = 1840 m
But in reality, this flow is in the turbulent region and L = 0.00000304 m = (1.197 × 10⁻⁴) inch
Explanation:
We first check the region of flow of the fluid by computing the Reynolds number
Re = (ρvD/μ)
Listing all the parameters and converting to SI units
ρ = density of the fluid = 0.000238 slug/ft³ = 0.123 kg/m³
v = velocity of flow = (Q/A)
Q = volumetric flowrate of the air = 2.6 ft³/min = 0.0012271 m³/s
A = Cross sectional Area of the pipe = (πD²/4)
D = Pipe diameter = (1/4) inch = 0.00635 m
A = (π×0.00635²/4) = 0.00003167 m²
v = (0.0012271/0.00003167) = 38.746 m/s
μ = dynamic viscosity of the fluid = (Kinematic viscosity) × (density)
Kinematic viscosity = (1.6 × 10⁻⁴) ft²/s = (1.486 × 10⁻⁵) m²/s
μ = (1.486 × 10⁻⁵) × (0.123) = 0.0000018278 = (1.83 × 10⁻⁶) Pa.s
Re = (0.123×38.746×0.00635)/(1.83 × 10⁻⁶) = 16,556.824214903
This Reynolds number is in the turbulent flow region.
From Hagen-Poiseulle equation, the volumetric flowrate for laminar and turbulent flow is given as
Q = (πD⁴ΔP)/(128μL) for laminar flow
ΔP = (0.241 × ρ⁰•⁷⁵ × μ⁰•²⁵ × L × Q¹•⁷⁵)/D⁴•⁷⁵ for turbulent flow
Since our flow is turbulent
ΔP = (0.241 × ρ⁰•⁷⁵ × μ⁰•²⁵ × L × Q¹•⁷⁵)/D⁴•⁷⁵
L = (ΔP.D⁴•⁷⁵)/(0.241 ρ⁰•⁷⁵ μ⁰•²⁵ Q¹•⁷⁵)
Listing all the parameters and converting to SI units
L = Length of the pipe = ?
ΔP = Pressure drop across the flow channel = 90 - 75 = 15 psi = 103421.4 Pa
D = Pipe diameter = (1/4) inch = 0.00635 m
ρ = density of the fluid = 0.000238 slug/ft³ = 0.123 kg/m³
μ = dynamic viscosity of the fluid = (1.83 × 10⁻⁶) Pa.s
Q = volumetric flowrate of the air = 2.6 ft³/min = 0.0012271 m³/s
L = (0.123 × 0.00635⁴•⁷⁵) ÷ (0.241 × 0.123⁰•⁷⁵ × 0.0000018278⁰•²⁵ × 0.0012271¹•⁷⁵)
L = (4.4986 × 10⁻¹²) ÷ (1.481 × 10⁻⁶)
L = 0.0000030378 m = 0.00000304 m = (1.197 × 10⁻⁴) inch
If we assumed that flow was laminar
Q = (πD⁴ΔP)/(128μL)
L = (πD⁴ΔP)/(128μQ)
L = (π × 0.00635⁴ × 103421.4) ÷ (128 × 0.0000018278 × 0.0012271)
L = (0.0005282691) ÷ (0.0000002871)
L = 1840 m
Hope this Helps!!!
A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz. Determine the tube thickness that should be used if the shearing stress is not to exceed 60 MPa.
Answer:
25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
Explanation:
Given data
steel tube : outer diameter = 50-mm
power transmitted = 100 KW
frequency(f) = 34 Hz
shearing stress ≤ 60 MPa
Determine tube thickness
firstly we calculate the ; power, angular velocity and torque of the tube
power = T(torque) * w (angular velocity)
angular velocity ( w ) = 2[tex]\pi[/tex]f = 2 * [tex]\pi[/tex] * 34 = 213.71
Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s
next we calculate the inner diameter using the relation
[tex]\frac{J}{c_{2} } = \frac{T}{t_{max} }[/tex] = 467.92 / (60 * 10^6) = 7.8 * 10^-6 m^3
also
c2 = (50/2) = 25 mm
[tex]\frac{J}{c_{2} }[/tex] = [tex]\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )[/tex] = [tex]\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ][/tex]
therefore; 0.025^4 - [tex]c^{4} _{1}[/tex] = 0.050 / [tex]\pi[/tex] (7.8 *10^-6)
[tex]c^{4} _{1}[/tex] = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)
39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8
[tex]c_{1} = \sqrt[4]{26.66 * 10^{-8} }[/tex] =
THE TUBE THICKNESS
[tex]c_{2} - c_{1}[/tex] = 25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
An example of a transient analysis involving the 1st law of thermodynamics and conservation of mass is the filling of a compressed air tank. Assume that an air tank is being filled using a compressor to a pressure of 5 atm, and that it is being fed with air at a temperature of 25°C and 1 atm pressure. The compression process is adiabatic. Will the temperature of the air in the tank when it is done being filled i.e. once the pressure in the tank reaches 5 atm), be greater than, equal to, or less that the temperature of the 25°C air feeding the compressor?
A. Greater than 25°C
B. Unable to determine
C. Same as 25°C
D. Less than 25°C
Answer:
The temperature will be greater than 25°C
Explanation:
In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.
mathematically
Change in the internal energy of a system ΔU = ΔQ + ΔW
in an adiabatic process, ΔQ = 0
therefore
ΔU = ΔW
where ΔQ is the change in heat into the system
ΔW is the work done by or done on the system
when work is done on the system, it is conventionally negative, and vice versa.
also W = pΔv
where p is the pressure, and
Δv = change in volume of the system.
In this case, work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C
A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 ∘ with respect to the horizontal. The spring is then released.
Required:
a. If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?
b. If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?
c. Now the incline has a coefficient of kinetic friction μk. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction μk?
Answer:
a) The mass moves a distance of 0.625 m up the slope before coming to rest
b) The distance moved by the mass when it is connected to the spring is 0.6 m
c) [tex]\mu = 0.206[/tex]
Explanation:
Spring constant, k = 70 N/m
Compression, x = 0.50 m
Mass placed at the free end, m = 2.2 kg
angle, θ = 41°
Potential Energy stored in the spring, [tex]PE= 0.5 kx^2[/tex]
[tex]PE = 0.5 * 70 * 0.5^2\\PE = 8.75 J[/tex]
According to the principle of energy conservation
PE = mgh
8.75 = 2.2 * 9.8 * h
h = 0.41
If the mass moves a distance d from the spring
sin 41 = h/d
sin 41 = 0.41/d
d = 0.41/(sin 41)
d = 0.625 m
The mass moves a distance of 0.625 m up the slope before coming to rest
b) If the mass is attached to the spring
According to energy conservation principle:
Initial PE of spring = Final PE of spring + PE of block
[tex]0.5kx_1^2 = 0.5kx_2^2 + mgh\\x_2 = d - x_1 = d - 0.5\\h = d sin 41\\0.5*70*0.5^2 = 0.5*70*(d-0.5)^2 + 2.2*9.8*d*sin41\\8.75 = 35(d^2 - d + 0.25) + 14.15d\\8.75 = 35d^2 - 35d + 8.75 + 14.15d\\35d^2 = 20.85d\\d = 0.6 m[/tex]
The distance moved by the mass when it is connected to the spring is 0.6 m
3) The spring potential is converted to increased PE and work within the system.
mgh = Fd + 0.5kx²...........(1)
d = x , h = dsinθ
kinetic friction force , F = μmgcosθ
mgdsinθ + μmg(cosθ)d = 0.5kd²
mgsinθ + μmgcosθ = 0.5kd
sinθ + μcosθ = kd/(2mg)
[tex]\mu = \frac{\frac{kd}{2mg} - sin\theta}{cos\theta} \\\\\mu = \frac{\frac{70*0.5}{2*2.2*9.8} - sin41}{cos41} \\\\\mu = 0.206[/tex]
Answer:
A) l = 0.619m
B) l = 0.596m
C) μ = 0.314
Explanation:
The data given is:
k = 70 N/m
x = 0.5 m
m = 2.2 kg
θ = 41°
(FIGURES FOR EACH PART ARE ATTACHED AT THE BOTTOM. CONSULT THEM FOR BETTER UNDERSTANDING)
Part A
Gain in Gravitational Potential Energy = Loss in Elastic Potential Energy
mgh = (1/2)kx²
(2.2)(9.8)h = (1/2)(70)(0.5)²
h = 0.406 m
sinθ = h/l
l = h / sinθ
l = 0.406/sin41
l = 0.619m
Part B
Loss in Elastic Potential Energy in compressed spring = Gain in Gravitational Potential Energy + Gain in Elastic Potential Energy in stretched spring
(1/2)kx² = mgh + (1/2)k(l - 0.5)²
(1/2)(70)(0.5)² = (2.2)(9.8)(l·sin41)) + (1/2)(70)(l² - l + 1/4)
8.75 = 14.15(l) + 35(l²) - 35(l) + 8.75
35(l²) -20.85(l) = 0
l = 0.596m
Part C
Loss in Elastic Potential Energy = Gain in Gravitational Potential Energy + Work done against friction
(1/2)kx² = mgh + Fd
(1/2)kx² = mg(dsinθ) + μRd
(1/2)kx² = mg(dsinθ) + μ(mg · cosθ)d
(1/2)kx² = mgd (sinθ + μ(cosθ))
(1/2)(70)(0.5)² = (2)(9.8)(0.5) (sin41 + μcos41)
8.75 = 6.43 + 7.4μ
μ = 0.314
Calculate the resistance using Voltage and current, again using voltage and power, again using current and power, and again using R1 and R2 recording the calculations for Run 3 rows 41-56
Answer:
R = V / I , R = V² / P, R = P / I²
Explanation:
For this exercise let's use ohm's law
V = I R
R = V / I
Electric power is defined by
P = V I
ohm's law
I = V / R
we substitute
P = V (V / R)
P = V² / R
R = V² / P
the third way of calculation
P = (i R) I
P = R I²
R = P / I²
help mhee why are you u an enigner
Answer:
help me why are you an enginer
Explanation:
because lives
If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially, the block is at rest
Answer:
115.2 W
Explanation:
The computation is shown below:
As we know that
Power = F . v
[tex]F_H = F cos \theta[/tex]
[tex]F_H = 30 \frac{4}{5}[/tex]
[tex]F_H = 24N[/tex]
Now we solve for V
[tex]V = V_0 + at[/tex] a = 24N ÷ 20Kg
But V_0 = 0 a = 1.2 m/s^2
F_H = ma V = 0 + (1.2) (4)
a = F_H ÷ m V = 4.8 m/s
Therefore
Power = F_Hv
= (24) (4.8)
= 115.2 W
By applying the above formuals we can get the power
Select True/False for each of the following statements regarding aluminum / aluminum alloys: (a) Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor. (b) Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum. (c) Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic. (d) Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation. (e) The relatively low melting point of aluminum is often considered a significant limitation for high-temperature structural applications.
Explanation:
(a) Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor.
False, aluminium is not susceptible to any corrosion by the presence of water vapor.
(b) Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their micro structures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.
True.
(c) Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.
False, aluminium is stable at high temperatures and does not oxidizes.
(d) Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.
False,pure aluminium is not resistant to the creep deformation.
(e) The relatively low melting point of aluminum is often considered a significant limitation for high-temperature structural applications.
False.
In this exercise, we have to analyze the statements that deal with aluminum and its properties, thus classifying it as true or false:
A) False
B) True
C) False
D) False
E) True
Analyzing the statements we can classify them as:
(a) For this statement we can say that it is False, aluminium is not susceptible to any corrosion by the presence of water vapor.
(b) For this statement we can say that it is True.
(c) For this statement we can say that it is False, aluminium is stable at high temperatures and does not oxidizes.
(d) For this statement we can say that it is False, pure aluminium is not resistant to the creep deformation.
(e) For this statement we can say that it is True.
See more about aluminum properties at brainly.com/question/12867973
Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the inlet temperature for each turbine stage is 1400 K. The pressure ratios across each turbine stage are equal. The turbine stages and compressor each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k= 1.4.
Calculate:
a. the thermal efficiency of the cycle.
b. the back work ratio.
c. the net power developed, in kW.
Answer:
a. 47.48%
b. 35.58%
c. 2957.715 KW
Explanation:
[tex]T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}[/tex]
T₁ = 300 K
[tex]\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }[/tex]
[tex]T_{2s}[/tex] = 579.21 K
T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K
T₃ = T₂ + [tex]\epsilon _{regen}[/tex](T₅ - T₂)
T₄ = 1400 K
Given that the pressure ratios across each turbine stage are equal, we have;
[tex]\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{5s}[/tex] = 1400×[tex]\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }[/tex] = 1007.6 K
T₅ = T₄ + ([tex]T_{5s}[/tex] - T₄)/[tex]\eta _t[/tex] = 1400 + (1007.6- 1400)/0.8 = 909.5 K
T₃ = T₂ + [tex]\epsilon _{regen}[/tex](T₅ - T₂)
T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K
T₆ = 1400 K
[tex]\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{7s}[/tex] = 1400×[tex]\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }[/tex] = 1007.6 K
T₇ = T₆ + ([tex]T_{7s}[/tex] - T₆)/[tex]\eta _t[/tex] = 1400 + (1007.6 - 1400)/0.8 = 909.5 K
a. [tex]W_{net \ out}[/tex] = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg
Heat supplied is given by the relation
cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg
Thermal efficiency of the cycle = (Net work output)/(Heat supplied)
Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%
b. [tex]bwr = \dfrac{W_{c,in}}{W_{t,out}}[/tex]
bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%
c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW
A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-vapor mixture with a quality of 0.94 at a pressure of 325 kPa and a velocity of 15 m/s. In addition, there is heat transfer from the turbine to the surroundings for 560 kW. Find the power produced by the turbine and express it in kW?
Answer:
The power produced by the turbine is 23309.1856 kW
Explanation:
h₁ = 3755.39
s₁ = 7.0955
s₂ = sf + x₂sfg =
Interpolating fot the pressure at 3.25 bar gives;
570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175
2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345
h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg
Power output of the turbine formula =
[tex]Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ][/tex]
Which gives;
[tex]560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ][/tex]
= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856
[tex]- \dot{W }[/tex] = -22749.1856 - 560 = -23309.1856 kJ
[tex]\dot{W }[/tex] = 23309.1856 kJ
Power produced by the turbine = Work done per second = 23309.1856 kW.
Hot coffee is contained in a cylindrical thermos bottle that is of length L-0.3m and is lying on its side (horizontally). The coffee container consists of a glass flask of diameter DI-7 cm sperated from an aluminum housing of diameter D2-0.08 m by air at atmospheric pressure. If the inner and outer temperatures are 75oC and 35 oC respectively, what is the heat loss from the coffee due to convection?
Note: The diagram attached below is the completion of the given question.
Also calculate the heat loss from the coffee due to radiation and the total heat loss
Answer:
[tex]Q_{conv} = 16.04 W\\Q_{rad} = 3.20 W\\H_{loss} = 19.24 W[/tex]
Explanation:
Glass diameter of the coffee container, [tex]D_1 = 7 cm = 0.07 m[/tex]
Glass radius of the coffee container, r₁ = 0.07/2 = 0.035 m
Diameter of the aluminium housing, [tex]D_2 = 0.08 m[/tex]
Radius of the aluminium housing, r₂ = 0.08/2 = 0.04 m
Inner temperature, T₁ = 75°C = 348 K
Outer temperature, T₂ = 35°C = 308 K
[tex]\epsilon_1 = 0.25\\\epsilon_2 = 0.25[/tex]
[tex]T_f = \frac{T_1 + T_2}{2} \\T_f = \frac{348+308}{2} \\T_f = 328 K[/tex]
At [tex]T_f = 328 K[/tex], P = 1 atm
k = 0.0284 w/m-k, v = 23.74 * 10⁻⁶, [tex]\alpha = 26.6 * 10^{-6}[/tex], pr = 0.703, [tex]\beta = 3.05 * 10^{-3} K^{-1}[/tex]
[tex]H_{loss} = Q_{rad} + Q_{conv}[/tex]
[tex]Q_{conv} = \frac{2 \pi k L (T_1 - T_2)}{ln(\frac{r_2}{r_1} )} \\\\Q_{conv} = \frac{2 \pi* 0.0284 * 0.3 (75 - 35)}{ln(\frac{0.04}{0.035} )} \\Q_{conv} = 16.04 W[/tex]
[tex]Q_{rad} = \frac{\sigma (\pi D_1 L) (T_1^4 - T_2^4)}{\frac{1}{\epsilon_1 } + \frac{1 - \epsilon_2}{\epsilon_2} (\frac{\gamma_1}{\gamma_2}) }[/tex]
[tex]Q_{rad} = \frac{5.67*10^{-8}(\pi * 0.07*0.3) (348^4 - 308^4)}{\frac{1}{0.25 } + \frac{1 - 0.25}{0.25} (\frac{0.035}{0.04}) } \\\\Q_{rad} = 3.20 W[/tex]
[tex]H_{loss} = 3.20 + 16.04\\H_{loss} = 19.24 W[/tex]
Identify the correct statements in the context of friction factors of laminar and turbulent flows
a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces
b) In turbulent flow, the tubes with rough surfaces have much lower friction factors than the tubes with smooth surfaces.
c) In laminar flow, the friction factor is dependent on the surface roughness
d) In laminar flow, the friction factor is independent of the surface roughness.
Answer:
a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces.
Explanation:
Turbulent flow is a type of fluid flow in which fluid will undergo irregular fluctuations. The tubes with rough surfaces have higher friction factors than the tubes with smooth surfaces. In laminar flow the effect of effect of surface roughness is negligible on friction factors.
A long corridor has a single light bulb and two doors with light switch at each door.
Design logic circuit for the light; assume that the light is off when both switches are
in the same position.
Answer:
Light = A xor B
Explanation:
If switches A and B produce True or False, then Light will be True for ...
Light = A xor B
Which of the following are the main psychological domains?
Answer:
Domain 1: Biological (includes neuroscience, consciousness, and sensation) Domain 2: Cognitive (includes the study of perception, cognition, memory, and intelligence) Domain 3: Development (includes learning and conditioning, lifespan development, and language) i hope this helps you.
If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially, the block is at rest
Answer:
The power of force F is 115.2 W
Explanation:
Use following formula
Power = F x V
[tex]F_{H}[/tex] = F cos0
[tex]F_{H}[/tex] = (30) x 4/5
[tex]F_{H}[/tex] = 24N
Now Calculate V using following formula
V = [tex]V_{0}[/tex] + at
[tex]V_{0}[/tex] = 0
a = [tex]F_{H}[/tex] / m
a = 24N / 20 kg
a = 1.2m / [tex]S^{2}[/tex]
no place value in the formula of V
V = 0 + (1.2)(4)
V = 4.8 m/s
So,
Power = [tex]F_{H}[/tex] x V
Power = 24 x 4.8
Power = 115.2 W
Question 1 The first choice for how to reduce or eliminate a hazard is: a) Engineering controls b) Workplace controls c) Personal protective equipment d) Administrative controls
Answer:
a) Engineering controls.
Explanation:
Hazard can be defined as any agent or source that has the potential to constitute danger and cause harm, damage or adverse bodily injuries, health effects on a vulnerable individual, property or group of people.
Generally, can be classified into various categories such as mechanical, biological, chemical, physical, psychosocial and ergonomic hazard. These hazards are either human induced or natural.
Some examples of hazard are radiation, fire, flood, chemicals, drought, vapor or steam, exposed live wire, dust particles, electrical circuits and equipments etc.
The first choice for how to reduce or eliminate a hazard is engineering controls. Engineering controls of hazards involves the process of protecting, shielding or guarding individuals by eliminating the agent of hazards or through the use of barriers between the hazard and the vulnerable individual or group of people.
Basically, engineering controls when properly designed, maintained and used effectively would help to mitigate hazards and keep the work environment relatively safe for workers.
Examples of engineering controls are;
1. Ventilation systems.
2. Machine or equipment guards.
3. Radiation shields.
4. Safety interlocks.
5. Sound dampening equipments.
What information is needed to set up sales tax in QuickBooks Online for a client who only does business in their home state?
Answer:
Their company address
When their last tax period started
How often they have to file a tax return
When they started collecting sales tax for the agency
Explanation:
For setup the sales tax information in Quickbooks online for a client who only does business in their home state, we need these information which are given below:
1. Their company addresses
2. Last Tax Time period started
3. How frequently they filed the tax return
3. when they begin to received sales tax
Therefore all the other options are not valid. Hence, ignored it
An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectively, and the respective atomic weights are 22.7 and 91.4 g/mol, what is the density (in g/cm3) of this material?
A. 0.438g/cm3
B. 0. 571g/cm3
C. 1.75g/cm3
D. 3.50g/cm3
Answer:
c) 1.75 g/cm³
Explanation:
Given that
Radii of the A ion, r(c) = 0.137 nm
Radii of the X ion, r(a) = 0.241 nm
Atomic weight of the A ion, A(c) = 22.7 g/mol
Atomic weight of the X ion, A(a) = 91.4 g/mol
Avogadro's number, N = 6.02*10^23 per mol
Solution is attached below
Fluid flow in a pipe, the pressure decrease due to throttle valve from 10 bar to 100 kpa if the specific volume of fluid increase from 0.3 m3 /kg to 1.8 m3/kg. Find the change in internal energy during the process
Answer:
Fluid flow in a pipe, the pressure decrease due to throttle valve from 10 bar to 100 kpa if the specific volume of fluid increase from 0.3 m3 /kg to 1.8 m3/kg. Find the change in internal energy during the process
Explanation:
hope it will helps you