Explanation:
The two types of furnaces used in steel production are:
Basic oxygen furnace
In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.
Electric arc furnace
Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.
Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep the pipe bend in place? Is it: a) 0.76 N b) 1.08 N c) 1.52 N d) 0.25 N e) 2.56 N
Answer:
b)1.08 N
Explanation:
Given that
velocity of air V= 45 m/s
Diameter of pipe = 2 cm
Force exerted by fluid F
[tex]F=\rho AV^2[/tex]
So force exerted in x-direction
[tex]F_x=\rho AV^2[/tex]
[tex]F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2[/tex]
F=0.763 N
So force exerted in y-direction
[tex]F_y=\rho AV^2[/tex]
[tex]F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2[/tex]
F=0.763 N
So the resultant force R
[tex]R=\sqrt{F_x^2+F_y^2}[/tex]
[tex]R=\sqrt{0.763^2+0.763^2}[/tex]
R=1.079
So the force required to hold the pipe is 1.08 N.
A thin film of oil (v=0.001 m^2/s) 2mmthick flows down to
a surface inclined at 30 degrees to thehorizontal. What is the
maximum and mean velocity offlow?
Answer:
Maximum and mean velocity will be equal i.e 0.5 m/s
Explanation:
given data:
viscosity = 0.001 m^2/s
Thickness of thin film = 2 mm = 0.002 m
Neglecting the body weight hence no shear stress
Mean velocity is given V
[tex]V = \frac{\nu}{D}[/tex]
[tex]V = \frac{0.001}{2*10^{-3}}[/tex]
V = 0.5 m/s
Maximum and mean velocity will be equal i.e 0.5 m/s
A person is attempting to start a 20hp diesel engine with a hand crank. The distance from the handle to the center of rotation is 12 inches. The person is able to apply 25 lbf tangent to the circular path of the handle. The crank rotates at 2 rounds per second. How much power (Btu/min and hp) is the person providing?
Answer: 0.57 hp, 24.2 Btu/min
Explanation:
Hi!
The power P delivered when rotating a crank at angular speed ω, applying a torque τ is given by:
P = τ*ω
In this case torque is τ = 12 in * 25 lbf = 300 lbf*in
Angular speed is 2 round per seconds, which is 120 RPM. Then power is:
P = 300*120 lbf*in*RPM
1 hp = (lbf*in*RPM) / 63,025
P = 0.57 hp
1 hp = 42,4 Btu/min
Then P = 24.2 Btu/min
Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat transfer rate. The air exits the compressor at 20 m/s through a tube of 10 cm diameter.
Answer:
The heat is transferred is at the rate of 752.33 kW
Solution:
As per the question:
Temperature at inlet, [tex]T_{i} = 20^{\circ}C[/tex] = 273 + 20 = 293 K
Temperature at the outlet, [tex]T_{o} = 200{\circ}C[/tex] = 273 + 200 = 473 K
Pressure at inlet, [tex]P_{i} = 80 kPa = 80\times 10^{3} Pa[/tex]
Pressure at outlet, [tex]P_{o} = 800 kPa = 800\times 10^{3} Pa[/tex]
Speed at the outlet, [tex]v_{o} = 20 m/s[/tex]
Diameter of the tube, [tex]D = 10 cm = 10\times 10^{- 2} m = 0.1 m[/tex]
Input power, [tex]P_{i} = 400 kW = 400\times 10^{3} W[/tex]
Now,
To calculate the heat transfer, [tex]Q[/tex], we make use of the steady flow eqn:
[tex]h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}[/tex]
where
[tex]h_{i}[/tex] = specific enthalpy at inlet
[tex]h_{o}[/tex] = specific enthalpy at outlet
[tex]v_{i}[/tex] = air speed at inlet
[tex]p_{s}[/tex] = specific power input
H and H' = Elevation of inlet and outlet
Now, if
[tex]v_{i} = 0[/tex] and H = H'
Then the above eqn reduces to:
[tex]h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}[/tex]
[tex]Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}[/tex] (1)
Also,
[tex]p_{s} = \frac{P_{i}}{ mass, m}[/tex]
Area of cross-section, A = [tex]\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}[/tex]
Specific Volume at outlet, [tex]V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s[/tex]
From the eqn:
[tex]P_{o}V_{o} = mRT_{o}[/tex]
[tex]m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s[/tex]
Now,
[tex]p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg[/tex]
Also,
[tex]\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg[/tex]
Now, using these values in eqn (1):
[tex]Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW[/tex]
Now, rate of heat transfer, q:
q = mQ = [tex]0.925\times 813.33 = 752.33 kW[/tex]
The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated with an external surface finish. c)- both answers 1 and 2 d)- all of the above.
Answer:
a)Are generally associated with factor.
Explanation:
We know that losses are two types
1.Major loss :Due to friction of pipe surface
2.Minor loss :Due to change in the direction of flow
As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.
Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as
[tex]Losses=K\dfrac{V^2}{2g}[/tex]
Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−1. The exhaust from the turbine is carried through a 15-cm-diameter pipe and is at 50 kPa and 100°C [state 2]. What is the power output of the turbine?
H1 = 3150.7 kJ/kg V1 = 0.2004 m3/kg
H2 = 2682.6 kJ/kg V2 = 3.4181 m3/kg
Answer:
Power output, [tex]P_{out} = 178.56 kW[/tex]
Given:
Pressure of steam, P = 1400 kPa
Temperature of steam, [tex]T = 350^{\circ}C[/tex]
Diameter of pipe, d = 8 cm = 0.08 m
Mass flow rate, [tex]\dot{m} = 0.1 kg.s^{- 1}[/tex]
Diameter of exhaust pipe, [tex]d_{h} = 15 cm = 0.15 m[/tex]
Pressure at exhaust, P' = 50 kPa
temperature, T' = [tex]100^{\circ}C[/tex]
Solution:
Now, calculation of the velocity of fluid at state 1 inlet:
[tex]\dot{m} = \frac{Av_{i}}{V_{1}}[/tex]
[tex]0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}[/tex]
[tex]0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}[/tex]
[tex]v_{i} = 3.986 m/s[/tex]
Now, eqn for compressible fluid:
[tex]\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}[/tex]
Now,
[tex]\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}[/tex]
[tex]\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}[/tex]
[tex]\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}[/tex]
[tex]v_{e} = 19.33 m/s[/tex]
Now, the power output can be calculated from the energy balance eqn:
[tex]P_{out} = -\dot{m}W_{s}[/tex]
[tex]P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}[/tex]
[tex]P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW[/tex]
The power output of the Turbine is; 225.69 kW
What is the Power Output?We are given
Pressure of steam; P = 1400 kPa
Temperature of steam at state 1; T = 350°C
Diameter of pipe; d₁ = 8 cm = 0.08 m
Mass flow rate; m' = 0.1 kg/s
Diameter of exhaust pipe; d₂ = 15 cm = 0.15 m
Pressure at exhaust; P' = 50 kPa
Temperature at state 2; T' = 100°C
Area; A = πd²/4
A = π * 0.08²/4
A = 0.0016π m²
We can find the find initial velocity from the formula;
v₁ = m' * V₁/A
v₁ = (0.1 * 0.2004/(0.0016π))
v₁ = 3.986 m/s
From equation of compressible fluid, we know that;
(A₁ * v₁)/V₁ = (A₂ * v₂)/V₂
A₂ = πd₂²/4
A₂ = π * 0.15²/4
A₂ = 0.005625π m²
(0.0016π * 3.986)/0.2004 = (0.005625π * v₂)/3.4181
Solving for v₂ gives;
v₂ = 19.33 m/s
Finally power output is gotten from the expression;
P_out = -m'(H₂ - H₁) + (v₂² - v₁²)/2
P_out = -0.1(2682.6 - 3150.7) + (19.33² - 3.986²)/2
P_out = 225.69 kW
Read more about Power Output at; https://brainly.com/question/25573309
For a gearbox power and efficiency test apparatus that accommodates interchangeable gearboxes, drive up to 0.5kW. How to select the motor for the test apparatus?
Answer:
First we must know the efficiency of the gearbox in which energy is lost by friction between its components, we divide this efficiency by the nominal power and find the real power.
With this value knowing the speed of rotation of the gearbox we select an engine that has a power just above.
Also keep in mind that the motor torque must be greater than the resistance offered by the gearbox and the 0.5Kw machine
It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine whether his new crown was pure gold (SG = 19.3). Archimedes measured the weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N. Was it pure gold?
Answer with Explanation:
The crown will be pure if it's specific gravity is 19.3
Now by definition of specific gravity it is the ratio between the weight of an object to the weight of water of equal volume
Since it is given that the weight of the crown is 11.8 N we need to find it's volume
Now According to Archimedes principle when the crown is immersed into water the water shall exert a force in upwards direction on the crown with a magnitude equaling to weight of the water displaced by the crown
Mathematically this is the difference between the weight of the crown in air and weight when immersed in water
Thus Buoyant force is [tex]F_{B}=11.8-10.9=0.9N[/tex]
Now by Archimedes principle This force equals in magnitude to the weight of water of same volume as of the crown
Thus the specific gravity of the crown equals
[tex]S.G=\frac{11.8}{0.9}=13.11[/tex]
As we see that the specific gravity of the crown material is less than that of pure gold hence we conclude that it is impure.
Given:
[tex]\to SG_{gold} = 19.3 \\\\[/tex]
weight of air [tex]W_{air} = 11.8\ N \\\\[/tex]
water weight [tex]W_{water} = 10.9 \ N \\\\[/tex]
To Find:
using the buoyancy B that calculates the difference of weight=?
Solution:
Using formula:
[tex]\to B = W_{air} - W_{water} \\\\[/tex]
[tex]= 11.8\ N - 10.9\ N\\\\ = 0.9 \ N\\\\[/tex]
Using a formula for calculating weight:
[tex]\to W_{air} = SG_{\gamma water}\times V_{crown}\\\\ \to W_{water} = B(SG - 1)[/tex]
calculating the [tex]\bold{SG_{crown}}[/tex]:
[tex]\to SG_{crown}=1+\frac{W_{water}}{B} \\\\[/tex]
[tex]= 1 + \frac{10.9\ N}{0.9\ N}\\\\= 1 + 12.11\\\\= 13.11[/tex]
Knowing that the [tex]SG_{gold}= 19.3[/tex] indicates that the crown is not made of pure gold.
[tex]\to \bold{SG_{crown}= 13.1}[/tex] The crown is not composed entirely of gold.
Find out more information about the gold here:
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What is the range of a 32-bit signed integer?
Answer:
From -2147483647 to 2147483648.
Explanation:
A 32 bit integer has 2^32 = 4294967296 possible values. A signed integer has positive and negative values as well as the zero.
Of these values (2^32)/2 will be positive, (2^32)/2 - 1 will be negative and one will be the zero.
Therefore the range is from -(2^32)/2 + 1 to (2^32)/2.
This can also be expressed as from -2147483647 to 2147483648.
20 gallons of an incompressible liquid exert a force of 50 lbf at the earth’s surface. What force in lbf would 3 gallons of this liquid exert on the surface of the moon? The gravitational acceleration on the surface of the moon is 5.51 ft/s^2.
Answer:
1.29 lbf
Explanation:
Weight is a force, it is the product of a mass by the acceleration of gravity.
f = m * a
Gallons are a unit of volume. The relationship of volume with mass is:
m = V * δ
δ is the density (mass per unit of volume)
Then the weight of a liquid is:
w = V * δ * g
Rearranging:
δ = w / (V * g)
The acceleration of gravity on Earth is 32.2 ft/s^2
δ = 50 / (20 * 32.2) = 0.078 lbm/gallon
Knowing this and the gravity on the Moon we can calculate how much would 3 gallons of this liquid weight on the Moon.
w = V * δ * g
w = 3 * 0.078 * 5.51 = 1.29 lbf
Use Newton's law of universal gravitation to calculate the weight of a 90 kg person standing on the surface of the earth to the nearest 1 N.
To find the weight of a 90 kg person on Earth, use the formula w = mg, where g is 9.80 m/s², resulting in a weight of 882 N to the nearest 1 N.
Explanation:To calculate the weight of a person on Earth using Newton's law of universal gravitation, you use the formula:
w = mg
Where w is the weight in newtons (N), m is the mass in kilograms (kg), and g is the acceleration due to gravity on Earth, which is 9.80 m/s².
Given a person has a mass of 90 kg, their weight can be calculated as follows:
w = (90 kg) × (9.80 m/s²) = 882 N
Therefore, the weight of a 90 kg person standing on the surface of the Earth is 882 N to the nearest 1 N.
A slight breeze is blowing over the hot tub above and yields a heat transfer coefficient h of 20 W/m2 -K. The air temperature is 75 F. If the surface area of the hot tub is 7.5 m2 , what is the heat loss (heat rate) due to convection? The temp of Hot Tube is 102F.
Answer:4050 W
Explanation:
Given
Heat transfer Coefficient(h)=[tex]20 W/m^2-K[/tex]
Air temperature =75 F
surface area(A)=[tex]7.5 m^2[/tex]
Temperature of hot tube is 102 F
We know heat transfer due to convection is given by
[tex]Q=hA\left ( \Delta T\right )[/tex]
[tex]Q=20\times 7.5\left ( 102-75\right )=4050 W[/tex]
A rigid tank holds 10 lbm of 160 °F water. If the quality of the water is 0.5 then what is the pressure in the tank and volume of the tank?
Answer:
The pressure and volume of the tank are 4.74703 psi and 385.968 ft³ respectively.
Explanation:
Volume is constant as the tank is rigid. Take the saturation condition of water from the steam table for pressure at 160°F.
Given:
Mass of the water is 10 lb.
Dryness fraction is 0.5.
Temperature of water is 160°F.
From steam table at 160°F:
The pressure in the tank is 4.74703 psi.
Specific volume of saturated water is 0.0163918 ft³/lb.
Specific volume of saturated steam is 77.1773 ft³/lb.
Calculation:
Step1
From steam table at 160°F:
The pressure in the tank is 4.74703 psi.
Step2
Specific volume of tank is calculated as follows:
[tex]v=v_{f}+x(v_{g}-v_{f})[/tex]
[tex]v=0.0163918 +0.5(77.1773 -0.0163918)[/tex]
[tex]v=0.0163918 +38.58045[/tex]
v=38.5968 ft³/lb.
Step4
Volume is calculated as follows:
[tex]V=v\times m_{t}[/tex]
[tex]V=38.5968\times10[/tex]
V=385.968 ft³.
Thus, the pressure and volume of the tank are 4.74703 psi and 385.968 ft³ respectively.
Write a matrix, that is a lower triangular matrix.
Answer:
[tex]\left[\begin{array}{ccc}10&0&0\\14&25&0\\57&18&39\end{array}\right][/tex]
Explanation:
A lower triangular matrix is one whose elements above the main diagonal are zero meanwhile all the main diagonals elements and below are nonzero elements. This is one of the two existing types of triangular matrixes. Attached you will find a image referring more about triangular matrixes.
If there is any question, just let me know.
Air at 20' C in the cylinder of a diesel engine is compressed from an initial pressure of 1 atm and volume of 800 cm3 to 60 cm3. Assume that air behaves as an ideal gas with γ-1.40 and that the compression is adiabatic. Find the final pressure and temperature of the air.
Answer:
The pressure is 37.56 atm.
Temperature at the end of adiabatic compression is 825.73 K.
Explanation:
Given that
[tex]T_1=20C[/tex]
[tex]P_1=1\ atm[/tex]
[tex]V_1=800\ cm^3[/tex]
[tex]V_2=60\ cm^3[/tex]
So r=800/60=13.33
γ=1.4
For process 1-2
[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]
[tex]\dfrac{T_2}{273+20}=13.33^{1.4-1}[/tex]
[tex]T_2=825.73\ K[/tex]
[tex]\dfrac{P_2}{P_1}=r^\gamma[/tex]
[tex]\dfrac{P_2}{1}=13.33^{1.4}[/tex]
[tex]P_2=37.56\ atm[/tex]
So
The pressure is 37.56 atm.
Temperature at the end of adiabatic compression is 825.73 K.
In fluid mechanics the term manifold generally applies to: a)- a quantity b)- static head pressure c)- surface leakage d)- a fluid distribution system
Answer:
The correct answer is option 'd':Fluid distribution system
Explanation:
In fluid mechanics 'manifold' is a term used to define a pipe into which many other pipes drain or a pipe which branches into many different smaller pipes.
Manifolds are of many types such as exhaust manifold,hydraulic manifold, inlet manifold ,e.t.c.
Manifolds serve an important purpose as they are used to exhaust the gas from different piston chambers into a single exhaust.
You are wearing skates and standing on a frictionless ice rink. Strapped to your back is jet pack that can produce a constant horizontal force of 75 lbf = 334 N. If you switch on the jet pack, what is your speed after 5 seconds?
Answer:
23.2 m/s
Explanation:
A jetpack appliang a force on a body would result in an acceleration following Newton's equation:
f = m * a
Rearranging:
a = f / m
The problem doesn't state a mass, so I'll use mine, which is
m = 72 kg
Then:
a = 334 / 72 = 4.64 m/s^2
The equation for speed under constant acceleration is:
V(t) = V0 + a * t
Since I was standing before turning the jetpack on
V0 = 0
So:
V(5) = 0 + 4.64 * 5 = 23.2 m/s
How much calcium chloride will react with100mg of soda ash?
Answer:
105mg of calcium chloride [tex]CaCl_{2}[/tex]
Explanation:
The molecular formula of calcium chloride is [tex]CaCl_{2}[/tex] and the molecular formula of soda ash is [tex]Na_{2}CO_{3}[/tex]
First of all you should write the balanced reaction between both compounds, so:
[tex]CaCl_{2}+Na_{2}CO_{3}=2NaCl+CaCO_{3}[/tex]
Then you should have the molar mass of the two compounds:
Molar mass of [tex]CaCl_{2}=110.98\frac{g}{mol}[/tex]
Molar mass of [tex]Na_{2}CO_{3}=105.98\frac{g}{mol}[/tex]
Now with stoichiometry you can find the mass of calcium chloride that reacts with 100mg of soda ash, so:
[tex]100mgNa_{2}CO_{3}*\frac{1molNa_{2}CO_{3}}{105.98gNa_{2}CO_{3}}*\frac{1molCaCl_{2}}{1molNa_{2}CO_{3}}*\frac{110.98gCaCl_{2}}{1molCaCl_{2}}=105mgCaCl_{2}[/tex]
Determine the activation energy for diffusing copper into gold if you know that the diffusion coefficient is 3.98 x 10-13 m2/s at 980 degreeC and the diffusion coefficient is 3.55 x 10-16 m2/s at 650degreeC.
Answer:
[tex]Q_d = 0.0166 J/mol[/tex]
Explanation:
given data:
diffusion coefficient [tex]= 3.98 \times 10^{-13} m^2/s[/tex]
[tex]T_1 = 980 Degree\ celcius = 1253 K[/tex]
[tex]T_2 = 650 Degree\ celcius = 923 K[/tex]
[tex]D_2 = 3.55\times 10^{-16} m^2/s[/tex]
Activation energy is given as[tex] = -2.3R \frac{ \Delta log D}{\Delta\frac{1}{T}}[/tex]
[tex] Q_d = -2.3 R [\frac{logD_1 - logD_2}{ \frac{1}{T_1} - \frac{1}{T_2}}][/tex]
[tex]Q_d = -2.3 \times 8.31 \frac{log(3.98*10^{-13} - log(3.55*10^{-16}}{ \frac{1}{1253} - \frac{1}{923}}[/tex]
[tex]Q_d = 0.0166 J/mol[/tex]
For an orthogonal cutting operation, tool material is HSS, rake angle is 15° , chip thickness is 0.5 mm, speed is 55 m/min and feed is 0.3 mm/rev. The shear plane angle (in degrees) is
Answer:
The shear plane angle is 34.45°.
Explanation:
Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the edge of the surface. For orthogonal cutting operation feed is the chip thickness.
Given:
Rake angle is 15°.
Uncut chip thickness is 0.5 mm.
Speed is 55 m/min.
Feed or the chip thickness is 0.3 mm.
Calculation:
Step1
Chip reduction ratio is calculated as follows:
[tex]r=\frac{t_{c}}{t}[/tex]
[tex]r=\frac{0.3}{0.5}[/tex]
r = 0.6
Step2
Shear angle is calculated as follows:
[tex]tan\phi=\frac{rcos\alpha}{1-rsin\alpha}[/tex]
Here, [tex]\phi[/tex] is shear plane angle, r is chip reduction ratio and [tex]\alpha[/tex] is rake angle.
Substitute all the values in the above equation as follows:
[tex]tan\phi=\frac{rcos\alpha}{1-rsin\alpha}[/tex]
[tex]tan\phi=\frac{0.6cos15^{\circ}}{1-0.6sin15^{\circ}}[/tex]
[tex]tan\phi=\frac{0.57955}{0.8447}[/tex]
[tex]\phi=34.45^{\circ}[/tex]
Thus, the shear plane angle is 34.45°.
Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one end of the beam is free and one end is fixed. The cross section has a base of 100 mm and height of 300 mm, length of the beam is 5.9 m, E=20.5x10^10 N/m2 and density of 7830 kg/m3. Write your answer in rad/sec with 2 decimal points.
Answer:
The natural angular frequency of the rod is 53.56 rad/sec
Explanation:
Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure
We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by
[tex]\Delta x=\frac{PL^3}{3EI}[/tex]
Hence we can write
[tex]P=\frac{3EI\cdot \Delta x}{L^3}[/tex]
Comparing with the standard spring equation [tex]F=kx[/tex] we find the cantilever analogous to spring with [tex]k=\frac{3EI}{L^3}[/tex]
Now the angular frequency of a spring is given by
[tex]\omega =\sqrt{\frac{k}{m}}[/tex]
where
'm' is the mass of the load
Thus applying values we get
[tex]\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}[/tex]
[tex]\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec[/tex]
Answer:
its a lil confusing
Explanation:
A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied weight is 500 N. Rotational speed is 1774. How many kW of power will be developed?
Answer:
P = 80.922 KW
Explanation:
Given data;
Length of load arm is 900 mm = 0.9 m
Spring balanced read 16 N
Applied weight is 500 N
Rotational speed is 1774 rpm
we know that power is given as
[tex] P = T\times \omega[/tex]
T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm
[tex]\omega[/tex] angular speed [tex]=\frac{2 \pi N}{60} [/tex]
Therefore Power is
[tex]P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65 watt[/tex]
P = 80.922 KW
Explain in less than 100 words, how you can obtain the young's modulus of a material using a tensile machine.
Answer with Explanation:
In a tensile test in an tensile machine the following steps followed to obtain the young's modulus:
1) A specimen of material with known cross sectional area and gauge length is loaded axially.
2) The load is increased and the corresponding change in the gauge length of the material is noted by using a deflectometer or a dial gauge.
3) The ratio between Force applied and the nominal cross section known as stress is calculated for each applied value of force until the material fails.
4) The ratio between the change in gauge length and the original gauge length is found this ratio is known as strain.
5) A graph is plotted between the values calculated in step 3 and 4 above.
6) The slope of the line at the origin of the graph gives the young's modulus of the material.
A 2-kg plastic tank that has a volume of 0.2m^3 is filled with water. Assuming the density of water is 999 kg/m 3 ; determine the weight of the combined system.
Answer:
Mass of the combined system = 201.8 Kg
Weight of the combined system = 1977.64 N
Explanation:
Given:
Mass of the plastic tank = 2 kg
Volume of the plastic tank = 0.2 m³
Density of water = 999 Kg/m³
Now, the volume of water in the tank = volume of tank = 0.2 m³
Also,
Mass = Density × Volume
therefore,
Mass of water = 999 × 0.2 = 199.8 Kg
Thus, the combined mass = Mass of tank + Mass of water in the tank
or
the combined mass = 2 + 199.8 = 201.8 Kg
Also,
Weight = Mass × Acceleration due to gravity
or
Weight = 201.8 × 9.8 = 1977.64 N
A(n)____ topology is the most reliable.
Answer:
Star topology
Explanation:
Topology:
Topology is the connection of networks .These network connect by nodes.
Type of topology:
1.Bus topology
2.Star topology
3.Ring topology
4.Mesh topology
Star topology is the most reliable topology .Because failure of node node does not affect the other nodes.In star topology network are connect in star form.
Ring topology is the most undesirable topology.In ring topology network connects in the ring form.In ring topology failure of one node affect the whole network.
What is the difference between absolute and gage pressure?
Explanation:
Step1
Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.
The expression for absolute pressure is given as follows:
[tex]P_{ab}=P_{g}+P_{atm}[/tex]
Here, [tex]P_{ab}[/tex] is absolute pressure, [tex]P_{g}[/tex] is gauge pressure and[tex]P_{atm}[/tex] is atmospheric pressure.
Step2
Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.
Determine the maximum weight of the flowerpot that can besupported without exceeding a cable tension of 50 lb in eithercable AB or AC.
Answer:
Maximum weight = 76.64 lb
Explanation:
In the second figure attached it can be seen the free body diagram of the problem.
The equation of equilibrium respect x-axis is
[tex] \sum F_x = 0 [/tex]
[tex] F_{AC} \times sin 30 - F_{AB} \times \frac{3}{5} = 0 [/tex]
[tex] F_{AC} = 1.2 \times F_{AB} [/tex]
So, cable segmet AC supports bigger tension than segment AB. If [tex] F_{AC} = 50 lb [/tex] then
[tex] F_{AB} = \frac{F_{AC}}{1.2} [/tex]
[tex] F_{AB} = 41.67 lb [/tex]
The equation of equilibrium respect y-axis is
[tex] \sum F_y = 0 [/tex]
[tex] F_{AB} \times \frac{4}{5} + F_{AC} \times cos 30 - W = 0 [/tex]
[tex] 41.67 lb \times \frac{4}{5} + 50 lb \times cos 30 - W = 0 [/tex]
[tex] W = 76.64 lb [/tex]
A flat plate that is 0.3m wide and 1m long is pulled over a film of water that is 0.2mm thick. The water is at 20 C. If the pulling force is 20N, what is the velocity of the plate?
Answer:
The velocity of the pulling plate is 14.98 m/s.
Explanation:
The shear stress caused by the 20 Newton force on the plate is given by
[tex]\tau =\frac{20}{Area}=\frac{20}{0.3\times 1.0}=66.66N/m^2[/tex]
Now by newton's law of viscosity we have
[tex]\tau =\mu\cdot \frac{\Delta u}{\Delta y}[/tex]
where
[tex]\mu [/tex] is the dynamic viscosity of the water at 20 degree
[tex]\frac{\Delta u}{\Delta y}[/tex] is the velocity gradient
Applying values we get
[tex]66.66=8.90\times 10^{-4}\times \frac{U}{0.2\times 10^-3}\\\\\therefore u=14.98m/s[/tex]
Under a construction management (CM) contract, should the CM firm's responsibilities normally begin at the construction phase, the design phase, the planning phase, or the conceptual phase?
The CM firm's responsibilities under a construction management contract should begin at the planning phase. This early engagement allows for a cohesive approach to project management, involving definition of scope, budgeting, and risk assessment. Construction managers work closely with designers and contractors throughout the project, ensuring it meets design specifications and safety standards.
Explanation:Under a construction management (CM) contract, the CM firm's responsibilities should begin at the planning phase, which precedes the design phase, construction phase, and conceptual phase. During the planning phase, the CM firm is involved in defining project objectives, scope, and feasibility, providing valuable input that shapes the overall project. The expertise of construction managers is utilized to help with site selection, project scheduling, budgeting, and risk management. Their early involvement ensures that the construction phase flows more smoothly, as potential issues may have been identified and addressed beforehand. Construction managers play a crucial role in maintaining the integrity of the design and adherence to specifications throughout the construction process.
Designers, on the other hand, continue their involvement even after generating detailed technical drawings and digital models by acting as on-site supervisors during construction. Their responsibility remains to ensure that contractors follow the design's specifications, indicating direct involvement in the contract construction phase. The construction site can present various risks, including the exposure of workers to dangerous materials which designers need to consider while specifying project materials. The collaborative effort between the CM firm, designers, and contractors is fundamental to achieving a successful and safe project completion, fulfilling the client's expectations, and managing any equity issues related to contract allocations.
The three major network category are_____,_____ and_____
Answer:
The three major network categories are LAN (Local Area Network), MAN (Metropolitan Area Network) and WAN (Wide Area Network)
Explanation:
A Local Area Network (LAN) is the simplest form of a network. It is usually limited to a geographical area and consists of a group of computers in the same organisation linked together. In most cases, the same technology is used for all computers in this network (eg. ethernet).
A Metropolitan Area Network (MAN) can connect multiple LANs as long as they are close to each other (geographically). MANs allow high speed remote connection as if the devices were on the same LAN.
A Wide Area Network (WAN) can connect multiple LANs across large geographical distances. Unlike MANs, the speed might differ based on factors such as distance. The internet is the most popular example of a WAN.