Answer:
a) The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.
b) a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)
Explanation:
The stack of plywood has a certain mass. The weight will depend on that mass.
w = m * g
There will be a normal reaction between the stack and the bed of the truck, this will be:
nr = -m * g * cos(θ)
Being θ the tilt angle of the bed.
The static friction force will be:
ffs = - m * g * cos(θ) * fJK
The dynamic friction force will be:
ffd = - m * g * cos(θ) * fik
These forces would produce accelerations
affs = -g * cos(θ) * fJK
affd = -g * cos(θ) * fik
affs = -9.81 * cos(θ) * 0.4 = -3.92 * cos(θ)
affd = -9.81 * cos(θ) * 0.3 = -2.94 * cos(θ)
These accelerations oppose to movement and must be overcome by another acceleration to move the stack.
The acceleration of the truck is horizontal, the horizontal component of these friction forces is:
affs = -3.92 * cos(θ)^2
affd = -2.94 * cos(θ)^2
The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.
Assuming the bed has a lenght L.
The horizontal movement will be over a distance cos(θ) * L because L is tilted.
Movement under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
In this case X0 = 0, V0 = 0, and a will be the sum of the friction force minus the acceleration of the truck.
If we set the frame of reference with the origin on the initial position of the stack and the positive X axis pointing backwards, the acceleration of the truck will now be negative and the dynamic friction acceleration positive.
L * cos(θ) = 1/2 * (2.94 * cos(θ)^2 - a) * 0.4^2
2.94 * cos(θ)^2 - a = 2 * 0.16 * L * cos(θ)
a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)
You are riding in an elevator that is going up at 10 ft/s. You are holding your cell phone 5 ft above the floor when it suddenly slips out of your hand and falls to the floor. Will it hit the elevator floor in more time, equal time, or less time than it would take if the elevator were standing still? Show work justifying your answer.
Answer:
It falls at the same speed in both cases.
Explanation:
If I were standing still the phone would be in free fall after slipping out of my hand.
I set a frame of reference with origin on the ground and the positive Y axis pointing up.
It would slip at t0 = 0, from a position Y0 = 5 ft, with a speed of Vy0 = 0.
It would be subject to an gravitational acceleration of -32.2 ft/s^2.
Since acceleration is constant:
Y(t) = Y0 + Vy0 * t + 1/2 * 4 * t^2
When it hits the floor at t1 it will be at Y(t1) = 0
0 = 5 + 0 * t1 - 16.1 * t1^2
16.1 * t1^2 = 5
t1^2 = 5 / 16.1
[tex]t1 = \sqrt{0.31} = 0.55 s[/tex]
If the elevator is standing still it would take 0.55 s to hit the ground.
Now, if the elevator is moving up at 10 ft/s.
The frame of reference will have its origin at the place the floor of the elevator is at t = 0, and stay there as the elevetor moves. The floor of trhe elevator will have a position of Ye = 10 * t
Vy0 = 10 ft/s because it will be moving initially at the same speed as the elevator.
And it will hit the floor of the elevator not at 0, but at
Ye = 10 * t2
So:
10 * t2 = 5 + 10 * t2 - 16.1 * t2^2
0 = 5 - 16.1 * t2^2
16.1 * t1^2 = 5
t1^2 = 5 / 16.1
[tex]t1 = \sqrt{0.31} = 0.55 s[/tex]
It falls at the same speed in both cases.
What is specific gravity? How is it related to density?
Answer:
Specific gravity is defined as the ratio of the Densities of the two substances.
Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{1000}}[/tex]
Explanation:
Specific gravity is defined as the ratio of the Densities of the two substances.
For the standardization, the density ration of densities is calculated with respect to the density of water i.e the denominator is the density of water.
Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{Density of water}}[/tex]
Also,
At STP density of water is 1000 Kg/m³
Therefore the relation between the specific gravity and density is,
The Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{Density of water at STP}}[/tex]
or
Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{1000}}[/tex]
What is 220 C in degrees Fahrenheit (F)?
Answer:
428°F
Explanation:
The equation to convert degrees Celsius to degrees fahrenheit is
°F (degrees fahrenheit) = (9/5 * °C (degrees celsius) ) + 32
°F = (9/5 * 220 °C (degrees celsius)) +32 = 428 °F (degrees fahrenheit)
If 3.7 grams of a gas contains 3.7 × 10^22 molecules, what is the molar mass of this gas in units of g/mol?
Answer:
Molar mass of the gas will be 60.65 gram/mole
Explanation:
We have given mass of gas = 3.7 gram
Gas contains [tex]3.7\times 10^{22}[/tex]
We know that any gas contain [tex]6.022\times 10^{23}[/tex] molecules in 1 mole
So number of moles [tex]=\frac{3.7\times 10^{22}}{6.022\times 10^{23}}=0.061[/tex]
We know that number of moles [tex]n=\frac{mass\ in\ gram}{molar\ mass}[/tex]
So [tex]0.061=\frac{3.7}{molar\ mass}[/tex]
Molar mass = 60.65 gram/mole
In crash tests, a shock absorber is used to slow the test car. The shock absorber consists of a piston with small holes that moves in a cylinder containing water. Viscous dissipation in the water transforms work into heat. How much heat will be transferred from the water after a 2000 kg car crashes into the shock absorber at a speed of 40 m/s?
Answer:
The heat transferred to water equals 1600 kJ
Explanation:
By the conservation of energy we have
All the kinetic energy of the moving vehicle is converted into thermal energy
We know that kinetic energy of a object of mass 'm' moving with a speed of 'v' is given by
[tex]K.E=\frac{1}{2}mv^{2}[/tex]
Thus
[tex]K.E_{car}=\frac{1}{2}\times 2000\times 40^{2}=1600\times 10^{3}Joules[/tex]
Thus the heat transferred to water equals [tex]1600kJ[/tex]
How much heat (Btu) is prod uced by a 150-W light bulb that is on for 20-hours?
Answer:
heat produced by 150 watt bulb is 510 Btu
Explanation:
Given data:
Power of bulb is 150 W
Duration for light is 20 hr
We know that 1 Watt = 3.4 Btu
hence using above conversion value we can calculate the power in Btu unit
so for [tex]150 W\ bulb = 150\times 3.4 = 510 Btu[/tex]
therefore, 150 W bulb consist of 510 Btu power for 20 hr
output heat by 150 watt bulb is 510 Btu
A reciprocating compressor takes a compresses it to 5 bar. Assuming that the compression is reversible and has an index, k, of 1.3, find the final temperature. charge of air at 1 bar & 20°C and a) T2= 1093 K b) T2=151.8 K c) T2=983.6 K d) T2 =710.9 K e) T2= 424.8 K
Answer:
final temperature is 424.8 K
so correct option is e 424.8 K
Explanation:
given data
pressure p1 = 1 bar
pressure p2 = 5 bar
index k = 1.3
temperature t1 = 20°C = 293 k
to find out
final temperature t2
solution
we have given compression is reversible and has an index k
so we can say temperature is
[tex]\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }[/tex] ...........1
put here all these value and we get t2
[tex]\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }[/tex]
t2 = 424.8
final temperature is 424.8 K
so correct option is e
At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of gravity, in ft/s2, at that elevation? If the balloon drifts to another elevation where g = 32.05 ft/s2, what is her weight, in lbf, and mass, in lb?
Answer:
1) g=31.87ft/s^2
2)m=120
W=119.64lbf
Explanation:
first part
the weight of a body with mass is calculated by the following equation
W=mg
we convert 120lb to slug
m=120lbx1slug/32.147lb=3.733slug
solving for g
g=W/m
g=119/3.733=31.87ft/s^2
second part
the mass is the same
m=120lb=3.733slug
Weight
W=3.733slug*32.05ft/s^2=119.64lbf
calculate the viscosity(dynamic) and kinematic viscosity of airwhen
the temperature is 288.15K and the density is 1.23kg/m3.
Answer:
(a) dynamic viscosity = [tex]1.812\times 10^{-5}Pa-sec[/tex]
(b) kinematic viscosity = [tex]1.4732\times 10^{-5}m^2/sec[/tex]
Explanation:
We have given temperature T = 288.15 K
Density [tex]d=1.23kg/m^3[/tex]
According to Sutherland's Formula dynamic viscosity is given by
[tex]{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}[/tex], here
μ = dynamic viscosity in (Pa·s) at input temperature T,
[tex]\mu _0[/tex]= reference viscosity in(Pa·s) at reference temperature T0,
T = input temperature in kelvin,
[tex]T_0[/tex] = reference temperature in kelvin,
C = Sutherland's constant for the gaseous material in question here C =120
[tex]\mu _0=4\pi \times 10^{-7}[/tex]
[tex]T_0[/tex] = 291.15
[tex]\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s[/tex]when T = 288.15 K
For kinematic viscosity :
[tex]\nu = \frac {\mu} {\rho}[/tex]
[tex]kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec[/tex]
For two 0.2 m long rotating concentric cylinders, the velocity distribution is given by u(r) = 0.4/r - 1000r m/s. If the diameters are 2 cm and 4 cm, respectively, calculate the fluid viscosity if the torque on the inner cylinder is measured to be 0.0026 N*m.
Answer:
5.9*10^-3 Pa*s
Explanation:
The fluid will create a tangential force on the surface of the cylinder depending on the first derivative of the speed respect of the radius.
τ = μ * du/dr
u(r) = 0.4/r - 1000*r
The derivative is:
du/dr = -1/r^2 - 1000
On the radius of the inner cylinder this would be
u'(0.02) = -1/0.02^2 - 1000 = -3500
So:
τ = -3500 * μ
We don't care about the sign
τ = 3500 * μ
That is a tangential force per unit of area.
The area of the inner cylinder is:
A = h * π * D
And the torque is
T = F * r
T = τ * A * D/2
T = τ * h * π/2 * D^2
T = 3500 * μ * h * π/2 * D^2
Then:
μ = T / (3500 * h * π/2 * D^2)
μ = 0.0026 / (3500 * 0.2 * π/2 * 0.02^2) = 5.9*10^-3 Pa*s
A walker’s cassette tape player uses four AA batteries in series to provide 6V to the player circuit. The four alkaline battery cells store a total of 200 watt-seconds of energy. If the cassette player is drawing a constant 10mA from the battery pack, how long will the cassette operate at normal power?
Answer:
The cassette player will operate at normal power for 3333.33 seconds.
Explanation:
The first step is to identify the operating voltage and the operating current with the purpose to determine the power that the cassette player consumes. Remember that power equals the voltage multiplied by the current [tex]P=V\times I[/tex], where P is in Watts (W), V is in Volts (V) and I is in Amperes (A).
The problem says that four batteries are connected in series to provide a voltage of 6V to the player circuit. So, the operating voltage is 6V, [tex]V=6V[/tex]
Then, the problem says that cassette player draws a constant current of 10mA from the battery pack. So, the operating current is 10mA or 0.01A, [tex]I=0.01A[/tex]
From previous, it could be said that the cassette player consumes 0.06W.
[tex]P=V\times I=(6V)\times (0.01A)=0.06W[/tex]
Now, the idea is to calculate how long the cassette will operate at 0.06W.
The problem says that the battery pack stores [tex]200\, W\cdot s[/tex], it means that the battery pack could provide 200W in a second; after a second, the battery pack will not work properly. However, the battery pack just have to provide 0.06W so, it will last more time. For calculating that, you must divide the total power per time the cell can provide by the power that the cassette player needs to work.
[tex]t=\frac{200W\cdot s}{0.06W}=3333.33s[/tex]
As you can see, the W units are canceled and second remains.
Thus, the cassette player will operate at normal power for 3333.33 seconds.
A piston-cylinder device contains 0.1 m3 of liquid water and 0.9 m² of water vapor in equilibrium at 800 kPa. Heat is transferred at constant pressure until the temperature reaches 350°C. Determine the initial temperature, total mass and final volume of the water. Show the process on a P-v diagram with respect to saturation lines.
Answer:
Initial temperature = 170. 414 °C
Total mass = 94.478 Kg
Final volumen = 33.1181 m^3
Diagram = see picture.
Explanation:
We can consider this system as a close system, because there is not information about any output or input of water, so the mass in the system is constant.
The information tells us that the system is in equilibrium with two phases: liquid and steam. When a system is a two phases region (equilibrium) the temperature and pressure keep constant until the change is completed (either condensation or evaporation). Since we know that we are in a two-phase region and we know the pressure of the system, we can check the thermodynamics tables to know the temperature, because there is a unique temperature in which with this pressure (800 kPa) the system can be in two-phases region (reach the equilibrium condition).
For water in equilibrium at 800 kPa the temperature of saturation is 170.414 °C which is the initial temperature of the system.
to calculate the total mass of the system, we need to estimate the mass of steam and liquid water and add them. To get these values we use the specific volume for both, liquid and steam for the initial condition. We can get them from the thermodynamics tables.
For the condition of 800 kPa and 170.414 °C using the thermodynamics tables we get:
Vg (Specific Volume of Saturated Steam) = 0.240328 m^3/kg
Vf (Specific Volume of Saturated Liquid) = 0.00111479 m^3/kg
if you divide the volume of liquid and steam provided in the statement by the specific volume of saturated liquid and steam, we can obtain the value of mass of vapor and liquid in the system.
Steam mass = *0.9 m^3 / 0.240328 m^3/kg = 3.74488 Kg
Liquid mass = 0.1 m^3 /0.00111479 m^3/kg = 89.70299 Kg
Total mass of the system = 3.74488 Kg + 89.70299 Kg = 93,4478 Kg
If we keep the pressure constant increasing the temperature the system will experience a phase-change (see the diagram) going from two-phase region to superheated steam. When we check for properties for the condition of P= 800 kPa and T= 350°C we see that is in the region of superheated steam, so we don’t have liquid water in this condition.
If we want to get the final volume of the water (steam) in the system, we need to get the specific volume for this condition from the thermodynamics tables.
Specific Volume of Superheated Steam at 800 kPa and 350°C = 0.354411 m^3/kg
We already know that this a close system so the mass in it keeps constant during the process.
If we multiply the mass of the system by the specific volume in the final condition, we can get the final volume for the system.
Final volume = 93.4478 Kg * 0.354411 m^3/kg = 33.1189 m^3
You can the P-v diagram for this system in the picture.
For the initial condition you can calculate the quality of the steam (measure of the proportion of steam on the mixture) to see how far the point is from for the condition on all the mix is steam. Is a value between 0 and 1, where 0 is saturated liquid and 1 is saturated steam.
Quality of steam = mass of steam / total mass of the system
Quality of steam = 3.74488 Kg /93.4478 Kg = 0,040 this value is usually present as a percentage so is 4%.
Since this a low value we can say that we are very close the saturated liquid point in the diagram.
The initial temperature of the water is 170.41 °C.
The initial volume of the water is 1 cubic meter.
The mass of the water is 4.620 kilograms.
The final volume of the water is 1.62828 cubic meters.
The process is described in the [tex]P-\nu[/tex] diagram attached below.
How to determine the water properties in a piston-cylinder device
By steam tables we find the missing properties of water:
State 1[tex]p = 800\,kPa[/tex], [tex]T = 170.41\,^{\circ}C[/tex], [tex]\nu = 0.21643\,\frac{m^{3}}{kg}[/tex], [tex]h = 2563.62\,\frac{kJ}{kg}[/tex] (Liquid-Vapor Mix) ([tex]x = 90\,\%[/tex])
State 2[tex]p = 800\,kPa[/tex], [tex]T = 350\,^{\circ}C[/tex], [tex]\nu = 0.35442\,\frac{m^{3}}{kg}[/tex], [tex]h = 3162.2\,\frac{kJ}{kg}[/tex] (Superheated Vapor)
Now we proceed to calculate the initial temperature, initial volume, mass and the final volume of the water:
Initial temperature[tex]T_{1} = 170.41\,^{\circ}C[/tex]
The initial temperature of the water is 170.41 °C. [tex]\blacksquare[/tex]
Initial volume[tex]V_{1} = 0.9\,m^{3}+0.1\,m^{3}[/tex]
[tex]V_{1} = 1\,m^{3}[/tex]
The initial volume of the water is 1 cubic meter. [tex]\blacksquare[/tex]
Mass[tex]m = \frac{V_{1}}{\nu_{1}}[/tex] (1)
[tex]m = 4.620\,kg[/tex]
The mass of the water is 4.620 kilograms. [tex]\blacksquare[/tex]
Final volume[tex]V_{2} = m\cdot \nu_{2}[/tex] (2)
[tex]V_{2} = 1.62828\,m^{3}[/tex]
The final volume of the water is 1.62828 cubic meters. [tex]\blacksquare[/tex]
The process is described in the [tex]P-\nu[/tex] diagram attached below. [tex]\blacksquare[/tex]
To learn more on piston-cylinder devices, we kindly invite to check this verified question: https://brainly.com/question/6334891
A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to convert the ice to water at 200°F?
Answer:
Heat required =7126.58 Btu.
Explanation:
Given that
Mass m=20 lb
We know that
1 lb =0.45 kg
So 20 lb=9 kg
m=9 kg
Ice at -15° F and we have to covert it at 200° F.
First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.
We know that
Specific heat for ice [tex]C_p=2.03\ KJ/kg.K[/tex]
Latent heat for ice H=336 KJ/kg
Specific heat for ice [tex]C_p=4.187\ KJ/kg.K[/tex]
We know that sensible heat given as
[tex]Q=mC_p\Delta T[/tex]
Heat for -15F to 32 F:
[tex]Q=mC_p\Delta T[/tex]
[tex]Q=9\times 2.03(32+15) KJ[/tex]
Q=858.69 KJ
Heat for 32 Fto 200 F:
[tex]Q=mC_p\Delta T[/tex]
[tex]Q=9\times 4.187(200-32) KJ[/tex]
Q=6330.74 KJ
Total heat=858.69 + 336 +6330.74 KJ
Total heat=7525.43 KJ
We know that 1 KJ=0.947 Btu
So 7525.43 KJ=7126.58 Btu
So heat required to covert ice into water is 7126.58 Btu.
A disk is rotating around an axis located at its center. The angular velocity is 0.5 rad/s. The radius of the disk is 0.4 m. What is the magnitude of the velocity at a point located on the outer edge of the disk, in units of m/s?
Answer:
0.2 m/s
Explanation:
The velocity of a point on the edge of a disk rotating disk can be calculated as:
[tex]v=\omega*r[/tex]
Where [tex]\omega[/tex] is the angular velocity and r the radius of the disk. This leads to:
[tex]v=0.5\,rad/s\,*\,0.4\,m=0.2\,m/s[/tex]
Answer:
0.2 m/s
Explanation:
Step 1: identify the given parameters
angular velocity, ω = 0.5 rad/s
radius of the disk, r = 0.4m
Note: the inner part of the disk and outer edge spin at the same rate. This means that the velocity at inner part of the disk is the same as the outer part.
Step 2: calculate the velocity of the disk at the outer edge in m/s
Velocity = Angular velocity (rad/s) X radius (m)
Velocity = 0.5 rad/s X 0.4 m
Velocity = 0.2 m/s
The input shaft to a gearbox rotates at 2300 rpm and transmits a power of 42.6 kW. The output shaft power is 34.84 kW at a rotational speed of 620 rpm. Determine the torque of the input shaft shaft, in N-m.
Answer:
Torque at input shaft will be 176.8695 N-m
Explanation:
We have given input power [tex]P_{IN}=42.6KW=42.6\times 10^3W[/tex]
Angular speed = 2300 rpm
For converting rpm to rad/sec we have multiply with [tex]\frac{2\pi }{60}[/tex]
So [tex]2300rpm=\frac{2300\times 2\pi }{60}=240.855rad/sec[/tex]
We have to find torque
We know that power is given by [tex]P=\tau \omega[/tex], here [tex]\tau[/tex] is torque and [tex]\omega[/tex] is angular speed
So [tex]42.6\times 10^3=\tau \times 240.855[/tex]
[tex]\tau =176.8695N-m[/tex]
So torque at input shaft will be 176.8695 N-m
A rectangular tank is filled with water to a depth of 1.5 m. Its longest side measures 2.5 m. What is the moment of the force about the base on this side? Is it. a) 5.5 kN.m b) 9.2 kN.m c) 13.8 kN.m d) 27.6 kN.m e) 41.4 kN.m
Answer:
The correct answer is option 'c': 13.8 kNm
Explanation:
We know that moment of a force equals
[tex]Moment=Force\times Arm[/tex]
The hydro static force is given by [tex]Force=Pressure\times Area[/tex]
We know that the hydrostatic pressure on a rectangular surface in vertical position is given by [tex]Pressure=\rho \times g\times h_{c.g}[/tex]
For the given rectangular surface we have [tex]h_{c.g}=\frac{h}{2}=\frac{1.5}{2}=0.75m[/tex]
Thus applying the values we get force as
[tex]Force=1000\times 9.81\times 0.75\times 1.5\times 2.5=27.59kN[/tex]
This pressure will act at center of pressure of the rectangular plate whose co-ordinates is given by h/3 from base
Thus applying the calculated values we get
[tex]Moment=27.59\times \frac{1.5}{3}=13.8kN.m[/tex]
A mass of air occupying a volume of 0.15m^3 at 3.5 bar and 150 °C is allowed [13] to expand isentropically to 1.05 bar. Its enthalpy is then raised by 52kJ (note the unit) by heating at constant pressure. Assuming that all processes occur reversibly, sketch them on a p-v chart and calculate the total work done and the total heat transfer.
Answer:
Total work: -5.25 kJ
Total Heat: 52 kJ
Explanation:
V0 = 0.15
P0 = 350 kPa
t0 = 150 C = 423 K
P1 = 105 kPa (isentropical transformation)
Δh1-2 = 52 kJ (at constant pressure)
Ideal gas equation:
P * V = m * R * T
m = (R * T) / (P * V)
R is 0.287 kJ/kg for air
m = (0.287 * 423) / (350 * 0.15) = 2.25 kg
The specifiv volume is
v0 = V0/m = 0.15 / 2.25 = 0.067 m^3/kg
Now we calculate the parameters at point 1
T1/T0 = (P1/P0)^((k-1)/k)
k for air is 1.4
T1 = T0 * (P1/P0)^((k-1)/k)
T1 = 423 * (105/350)^((1.4-1)/1.4) = 300 K
The ideal gas equation:
P0 * v0 / T0 = P1 * v1 / T1
v1 = P0 * v0 * T1 / (T0 * P1)
v1 = 350 * 0.067 * 300 / (423 * 105) = 0.16 m^3/kg
V1 = m * v1 = 2.25 * 0.16 = 0.36 m^3
The work of this transformation is:
L1 = P1*V1 - P0*V0
L1 = 105*0.36 - 350*0.15 = -14.7 kJ/kg
Q1 = 0 because it is an isentropic process.
Then the second transformation. It is at constant pressure.
P2 = P1 = 105 kPa
The enthalpy is raised in 52 kJ
Cv * T1 + P1*v1 = Cv * T2 + P2*v2 + Δh
And the idal gas equation is:
P1 * v1 / T1 = P2 * v2 / T2
T2 = T1 * P2 * v2 / (P1 * v1)
Replacing:
Cv * T1 + P1*v1 + Δh = Cv * T1 * P2 * v2 / (P1 * v1) + P2*v2
Cv * T1 + P1*v1 + Δh = v2 * (Cv * T1 * P2 / (P1 * v1) + P2)
v2 = (Cv * T1 + P1*v1 + Δh) / (Cv * T1 * P2 / (P1 * v1) + P2)
The Cv of air is 0.7 kJ/kg
v2 = (0.7 * 300 + 105*0.16 + 52) / (0.7 * 300 * 105 / (105 * 0.16) + 105) = 0.2 m^3/kg
V2 = 2.25 * 0.2 = 0.45 m^3
T2 = 300 * 105 * 0.2 / (105 * 0.16) = 375 K
The heat exchanged is Q = Δh = 52 kJ
The work is:
L2 = P2*V2 - P1*V1
L2 = 105 * 0.45 - 105 * 0.36 = 9.45 kJ
The total work is
L = L1 + L2
L = -14.7 + 9.45 = -5.25 kJ
A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. Determine the minimum required wire diameter assuming a factor of safety 5 and that the yield strength of the steel is 900MPa.
Answer:
11.6 mm
Explanation:
With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:
σadm = strength / fos
σadm = 900 / 5 = 180 MPa
The stress is the load divided by the section:
σ = P / A
σ = 4*P / (π*d^2)
Rearranging:
d^2 = 4*P / (π*σ)
[tex]d = \sqrt{4*P / (\pi*\sigma)}[/tex]
[tex]d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm[/tex]
Define the hydraulic diameter for a rectangular duct
Answer with Explanation:
Hydraulic diameter is a term analogous to the diameter of the circular sectional pipe but used for the cases when the cross sectional shape of the pipe is non circular.
It serves as an equivalent diameter that is used to calculate the Reynolds number for the flow.
The hydraulic diameter is 4 times the hydraulic radius of any section.
For a rectangular duct as shown in the attached figure
[tex]R_{h}=\frac{Wetted_{Area}}{Wetted_{perimeter}}\\\\R_h=\frac{d\times b}{2(d+b)}\\\\\therefore D_{h}=4\times R_{h}=4\times \frac{db}{2(d+b)}=\frac{2db}{(d+b)} [/tex]
Where
[tex]D_{h}[/tex] is the hydraulic diameter of the duct with depth 'd' and width 'b'
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An inventor proposes an engine that operates between the 27 deg C warm surface layer of the ocean and a 10 deg C layer a few meters down. The claim is that this engine can produce 100 kW at a flow of 20 kg/s. Is this possible?
Answer:
Engine not possible
Explanation:
source temperature T1 = 300 K
sink temperature T2= 283 K
therefore, carnot efficiency of the heat engine
η= 1- T_2/T_1
[tex]\eta= 1-\frac{T_1}{T_2}[/tex]
[tex]\eta= 1-\frac{283}{300}[/tex]
= 0.0566
= 5.66%
claims of work produce W = 100 kW, mass flow rate = 20 kg/s
[tex]Q=mc_p(T_1-T_2)[/tex]
[tex]Q=20\times4.18(300-283)[/tex]
= 1421.2 kW
now [tex]\eta= \frac{W}Q}[/tex]
now [tex]\eta= \frac{100}{1421.2}[/tex]
=7%
clearly, efficiency is greater than carnot efficiency hence the engine is not possible.
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is 120 MPa?
Answer:
119.35 mm
Explanation:
Given:
Inside diameter, d = 100 mm
Tensile load, P = 400 kN
Stress = 120 MPa
let the outside diameter be 'D'
Now,
Stress is given as:
stress = Load × Area
also,
Area of hollow pipe = [tex]\frac{\pi}{4}(D^2-d^2)[/tex]
or
Area of hollow pipe = [tex]\frac{\pi}{4}(D^2-100^2)[/tex]
thus,
400 × 10³ N = 120 × [tex]\frac{\pi}{4}(D^2-100^2)[/tex]
or
D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]
or
D = 119.35 mm
Answer:
D =119.35 mm
Explanation:
given data:
inside diameter = 100 mm
load = 400 kN
stress = 120MPa
we know that load is given as
[tex]P = \sigma A[/tex]
where:
P=400kN = 400000N
[tex]\sigma = 120MPa[/tex]
[tex]A =(\frac{1}{4} \pi D^2 - \frac{1}{4}\pi (100^2)[/tex]
[tex]A=\frac{1}{4} \pi (D^2 - 10000)[/tex]
putting all value in the above equation to get the required diameter value
[tex]400 = 120*\frac{1}{4} \pi (D^2 - 10000)[/tex]
solving for
D =119.35 mm
The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 mA for 40 continuous hours. During that time, the voltage will drop from 1.5 V to 1.0 V. Assume the drop in voltage is linear with time. (2 points) How many seconds is 40 hrs? (5 points) Plot the battery voltage as a function of time. Each axis needs a label (what is being plotted), scale (the values along the axis), and units. (7 points) Plot the battery power as a function of time. Write an equation for the power from 0 hours to 40 hours. (6 points) Remember that power is the derivative of energy with respect to time so energy is the integral of power over a given time period. There are two ways to find the energy. One is to calculate the area under the power curve from 0 hours to 40 hours. The second is to perform the integration of the power function from 0 to 40 hours. Find how much energy does the battery delivers in this 40 hour interval using both methods. The numerical answer is 1620 J. You must show the correct method to get credit.
Answer:
a) 144.000 s
b) and c)Battery voltage and power plots in attached image.
[tex]V=-\frac{0.5}{144000} t + 1.5 V[tex]
[tex]P(t)=-(31.25X10^{-9}) t+0.0135[/tex] where D:{0<t<40} h
d) 1620 J
Explanation:
a) The first answer is a rule of three
[tex]s=\frac{3600s * 40h}{1h} = 144000s[/tex]
b) Using the line equation with initial point (0 seconds, 1.5 V)
[tex]m=\frac{1-1.5}{144000-0} = \frac{-0.5}{144000}[/tex]
where m is the slope.
[tex]V-V_{1}=m(x-x_{1})[/tex]
where V is voltage in V, and t is time in seconds
[tex]V=m(t-t_{1}) + V_{1}[/tex] and using P and m.
[tex]V=-\frac{0.5}{144000} t + 1.5 V[tex]
c) Using the equation V
POWER IS DEFINED AS:
[tex] P(t) = v(t) * i(t) [tex]
so.
[tex] P(t) = 9mA * (-\frac{0.5}{144000} t + 1.5) [tex]
[tex]P(t) = - (31.25X10^{-9}) t + 0.0135[/tex]
d) Having a count that.
[tex]E = \int\limits^{144000}_{0} {P(t)} \, dt = \int\limits^{144000}_{0} {v(t)*i(t)} \, dt[/tex]
[tex]E = \int\limits^{144000}_{0} {-\frac{0.5}{144000} t + 1.5*0.009} \, dt = 1620 J[/tex]
If a steel cable is rated to take 800-lb and the steel has a yield strength of 90,000psi, what is the diameter of the cable?
Answer:
d = 2.69 mm
Explanation:
Assuming the cable is rated with a factor of safety of 1.
The stress on the cable is:
σ = P/A
Where
σ = normal stress
P: load
A: cross section
The section area of a circle is:
A = π/4 * d^2
Then:
σ = 4*P / (π*d^2)
Rearranging:
d^2 = 4*P / (π*σ)
[tex]d = \sqrt{4*P / (\pi*\sigma)}[/tex]
Replacing:
[tex]d = \sqrt{4*800 / (\pi*\90000)} = 0.106 inches[/tex]
0.106 inches = 2.69 mm
Refrigerant R-12 is used in a Carnot refrigerator
operatingbetween saturated liquid and vapor during the heat
rejectionprocess. If the cycle has a high temperature of 50 deg C
and a lowtemperature of -20 deg C, find the heat transferred from
therefrigerated space, the work required, the coefficient
ofperformance and the quality at the beginning of the heat
additioncycle.
Answer:
Heat transferred from the refrigerated space = 95.93 kJ/kg
Work required = 18.45 kJ/kg
Coefficient of performance = 3.61
Quality at the beginning of the heat addition cycle = 0.37
Explanation:
From figure
[tex] Q_H [/tex] is heat rejection process
[tex] Q_L [/tex] is heat transferred from the refrigerated space
[tex] T_H [/tex] is high temperature = 50 °C + 273 = 323 K
[tex] T_L [/tex] is low temperature = -20 °C + 273 = 253 K
[tex] W_{net} [/tex] is net work of the cycle (the difference between compressor's work and turbine's work)
Coefficient of performance of a Carnot refrigerator [tex] (COP_{ref}) [/tex] is calculated as
[tex] COP_{ref} = \frac{T_L}{T_H - T_L} [/tex]
[tex] COP_{ref} = \frac{253 K}{323 K - 253 K} [/tex]
[tex] COP_{ref} = 3.61 [/tex]
From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table
[tex] Q_H = 122.5 kJ/kg [/tex]
From coefficient of performance definition
[tex] COP_{ref} = \frac{Q_L}{Q_H - Q_L} [/tex]
[tex] Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L[/tex]
[tex] Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)} [/tex]
[tex] Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)} [/tex]
[tex] Q_L = 95.93 kJ/kg [/tex]
Energy balance gives
[tex] W_{net} = Q_H - Q_L [/tex]
[tex] W_{net} = 122.5 kJ/kg - 95.93 kJ/kg [/tex]
[tex] W_{net} = 26.57 kJ/kg [/tex]
Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)
[tex] x = \frac{s_1 - s_f}{s_g - s_f} [/tex]
From figure
[tex] s_1 = s_4 = 1.165 kJ/(K kg) [/tex]
Replacing with table values
[tex] x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)} [/tex]
[tex] x = 0.37 [/tex]
Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get
[tex] h_1 = h_f + x \times (h_g - h_f) [/tex]
[tex] h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg [/tex]
[tex] h_1 = 241.58 kJ/kg [/tex]
By energy balance, [tex] W_{t} [/tex] turbine's work is
[tex] W_{t} = |h_1 - h_4| [/tex]
[tex] W_{t} = |241.58 kJ/kg - 249.7 kJ/kg| [/tex]
[tex] W_{t} = 8.12 kJ/kg [/tex]
Finally, [tex] W_{c} [/tex] compressor's work is
[tex] W_{c} = W_{net} + W_{t}[/tex]
[tex] W_{c} = 26.57 kJ/kg + 8.12 kJ/kg[/tex]
[tex] W_{c} = 34.69 kJ/kg [/tex]
A heat pump with a 2 kW motor is used to heat a building at 30 deg C. The building loses heat at a rate of 0.5 kW per degree difference to the colder ambient at T amb. The heat pump has a coefficient of performance that is 50 % of a carnot heat pump. What is the maximum ambient temperature for which the heat pump is sufficient?
Answer:
T=5.3° C
Explanation:
Given that
Power input to the pump = 2 KW
Building loses heat rate = 0.5 KW/C
So rate of heat transfer = 0.5(273+30-T)
rate of heat transfer = 0.5(303-T)
T=Ambient temperature
Building temperature = 30° C
We know that ,heat pump is used to heat the building.
COP of pump = 0.5 COP of Carnot heat pump
[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{273+30}{303-T}[/tex]
[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{303}{303-T}[/tex]
[tex]COP\ of\ pump=\dfrac{303-T}{Power}[/tex]
[tex]COP\ of\ pump=0.5\times \dfrac{303-T}{2}[/tex] -----1
And also
[tex]COP\ of\ pump=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex] ----2
So from now equation 1 and 2
[tex]\dfrac{303-T}{4}=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex]
So T= 278.38 K=5.3° C
T=5.3° C
Ambient temperature =5.3° C.
Discuss the differences between conduction and convective heat transfer.
Answer:
Basically there are two principal differences between the convection and conduction heat transfer
Explanation:
The conduction heat transfer is referred to the transfer between two solids due a temperature difference, while for, the convective heat transfer is referred to the transfer between a fluid (liquid or gas) and a solid. Also, they used different coefficients for its calculation.
We can include on the explanation that conduction thermal transfer is due to temperature difference, while convection thermal transfer is due to density difference.
A loan for $85,000 is to be paid in 10 yearly payments. Each payment is larger by $100.00. Compute the first, the fourth, and the last payment, if the time value of money is 10% per year.
Answer:
1st payment = $9350
4th payment = $9650
last 10th payment = $10250
Explanation:
given data
loan = $85000
time = 10 year
each payment larger = $100
time value of money = 10% per year
to find out
first, the fourth, and the last payment
solution
we find here actual value at end of 1st year that is
actual value = loan amount + 10% of loan
actual value N = 85000 + (10% ×85000 )
actual value N = $93500
so
1st payment is = [tex]\frac{actual value}{total time}[/tex]
1st payment is = [tex]\frac{93500}{10}[/tex]
1st payment = $9350
and
4th payment = 1st payment + 3× each payment larger
4th payment = 9350 + 3×100
4th payment = $9650
and
last 10th payment = 4th payment + 6× each payment larger
last 10th payment = 9650 + 6× 100
last 10th payment = $10250
NASA SPACE SHUTTLE QUESTION:
What lessons have we learned from the shuttle program and space travel?
-Be detailed
-full sentences
-2-3 sentence response
Explanation:
The Challenger accident made them aware of the risks. They had been a little naive, since it was never believed that such a thing could happen. This, together with Space Shuttle Columbia disaster revealed the risks of the space shuttle program, which also had very high maintenance costs.
An open tank contains ethylene glycol at 25°C. Compute the pressure at a depth of 3.0m.
Answer:
pressure at depth 3 m is 33.255 kPa
Explanation:
given data
temperature = 25°C
depth = 3 m
to find out
pressure
solution
we know pressure formula that is
pressure = ρ g h ................1
and we know specific gravity of Ethylene glycol is 1.13
so
1.13 = [tex]\frac{\rho (e)}{\rho (water)}[/tex]
ρ (Ethylene glyco) = 1130 Kg/m³
so
pressure will be by equation 1
pressure = 1130 × 9.81 × 3
pressure = 33255.9 Pa
so pressure at depth 3 m is 33.255 kPa
A 1200-kg car moving at 20 km/h is accelerated
ata constant rate of 4 m/s2 up to a speed of 75km/h.
What are the force and total time required?
Answer:
Force on the car will be 4800 N and time required to cover this distance 13.75 sec
Explanation:
We have given mass of the car = 1200 kg
Initial velocity u = 20 km/h
Final velocity v = 75 km/h
Acceleration [tex]a=4m/sec^2[/tex]
From the first equation of motion we know that
v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time
So [tex]75=20+4\times t[/tex]
t = 13.75 sec
From second law of motion we know that [tex]F=ma[/tex]
So force [tex]F=1200\times 4=4800N[/tex]