Answer:
true
Explanation:
True, there are several types of polymers, thermoplastics, thermosets and elastomers.
Thermosets are characterized by having a reticulated structure, so they have low elasticity and cannot be stretched when heated.
Because of the above, thermosetting polymers burn when heated.
The typical area of a commercial airplane's passenger window is 80.0 in^2 . At an altitude of 3.00 × 104 ft above the sea level, the atmospheric pressure is 0.350 atm. Determine the net force on the passenger window during flight at that altitude for both the English Engineering (EE) and SI unit systems. Use appropriate units and unit conversions in all steps of your calculations.
Answer:
The force over the plane windows are 764 lbf in the EE unit system and 3398 N in the international unit system.
Explanation:
The net force over the window is calculated by multiplying the difference in pressure by the area of the window:
F = Δp*A
The pressure inside the plane is around 1 atm, hence the difference in pressure is:
Δp = 1atm - 0.35 atm = 0.65 atm
Expressing in the EE unit system:
Δp = 0.65 atm * 14.69 lbf/in^2 = 9.55 lbf/in^2
Replacing in the force:
F = 9.55 lbf/in^2 * 80 in^2 = 764 lbf
For the international unit system, we re-calculate the window's area and the difference in pressure:
A = 80 in^2 * (0.0254 m/in)^2 = 0.0516 m^2
Δp = 0.65 atm * 101325 Pa = 65861 Pa = 65861 N/m^2
Replacing in the force:
F = 65861 N/m^2 *0.0516 m^2 = 3398 N
The primary heat transfer mechanism that quickly warms my hand if I hold it directly above a campfire is: a)-Radiation b)-Inductance c)- Convection d)- Conduction
Answer:
The correct answer is option 'c':Convection.
Explanation:
When we ignite a campfire the heat produced by combustion heats the air above the fire. As we know that if a gases gains heat it expands thus it's density decreases and hence it rises, if we hold our hands directly above the fire this rising hot air comes in contact with our hands thus warming them.
The situation is different if we are at some distance from the campfire laterally. Since the rising air cannot move laterally the only means the heat of the fire reaches our body is radiation.
But in the given situation the correct answer is convection.
Find the Hooke's law of orthotropic and transverse isotropic material in matrix form
Answer:
Hooke's law is a principle of physics that states that the force F needed to extend or compress a spring by some distance X is proportional to that distance. That is: F = kX, where k is a constant factor characteristic of the spring: its stiffness, and X is small compared to the total possible deformation of the spring.
Explanation:
The Phoenix with a mass of 390 kg was a spacecraft used for exploration of Mars. Determine the weight of the Phoenix, in N, (a) on the surface of Mars where the acceleration of gravity is 3.73 m/s2 and (b) on Earth where the acceleration of gravity is 9.81 m/s2.
Answer:
a) on mars W=1454.7N
b)on earth W=3825.9N
Explanation:
The weight of any body with mass is given by the following equation
W=mg
where
m=mass
g=gravity
W=weight
Remember that the weight is expresed in Newton and the units are kgm/s^2
A)weight on the surface of mars
W=(390kg)(3.73m/s^2)=1454.7N
b) on earth
W=(390kg)(9.81m/s^2)=3825.9N
Pressurized water ( 10 bar, 110°C) enters the bottom of an 10-m-long vertical tube of diameter 63 mm at a mass flow rate of 1.5 kg/s. The tube is located inside a combustion chamber, resulting in heat transfer to the tube. Superheated steam exits the top of the tube at 7 bar, 600°C. Determine the change in the rate at which the following quantities enter and exit the tube: (1) the combined thermal and flow work, (2) the mechanical energy, and (3) the total energy of the water. Also, (4) determine the heat transfer rate, . Hint: Relevant properties may be obtained from a thermodynamics text.
Answer:
(1) [tex]\Delta E = 4845.43 kW[/tex]
(2) [tex]\Delta E_{m} = 5.7319 kW[/tex]
(3) [tex]\Delta E_{t} = 4839.69 kW[/tex]
(4) q = 4839.69 kW[/tex]
Solution:
Using Saturated water-pressure table corresponding to pressure, P = 10 bar:
At saturated temperature, Specific enthalpy of water, [tex]h_{ws} = h_{f} = 762.5 kJ/kg[/tex]
At inlet:
Saturated temperature of water, [tex]T_{sw} = 179.88^{\circ}C[/tex]
Specific volume of water, [tex]V_{wi} = V_{f} = 0.00127 m^{3}/kg[/tex]
Using super heated water table corresponding to a temperature of [tex]600^{\circ}C[/tex] and at 7 bar:
At outlet:
Specific volume of water, [tex]V_{wso} = 0.5738 m^{3}/kg[/tex]
Specific enthalpy of water, [tex]h_{wo} = 3700.2 kJ/kg[/tex]
Now, at inlet, water's specific enthalpy is given by:
[tex]h_{i} = C_{p}(T - T_{sw}) + h_{ws}[/tex]
[tex]h_{i} = 4.187(110^{\circ} - 179.88^{\circ}) + 762.5[/tex]
[tex]h_{i} = -292.587 + 762.5= 469.912 kJ/kg[/tex]
(1) Now, the change in combined thermal energy and work flow is given by:
[tex]\Delta E = E_{o} - E_{i}[/tex]
[tex]\Delta E = m(h_{wo} - h_{i})[/tex]
[tex]\Delta E = 1.5(3700.2 - 469.912) = 4845.43 kW[/tex]
(2) The mechanical energy can be calculated as:
velocity at inlet, [tex]v_{i} = \rho A V_{wi}[/tex]
[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]
[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]
[tex]v_{i} = \frac{1.5\times 0.00127}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]
[tex]v_{i} = 0.542 m/s[/tex]
Similarly,, the velocity at the outlet,
[tex]v_{o} = \frac{1.5\times 0.57378}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]
[tex]v_{o} = 276.099 m/s[/tex]
Now, change in mechanical energy:
[tex]\Delta E_{m} = E_{mo} - E_{mi}[/tex]
[tex]\Delta E_{m} = m[(\frac{v_{o}^{2}}{2} + gz_{o}) - (\frac{v_{i}^{2}}{2} + gz_{i})][/tex]
[tex]\Delta E_{m} = 1.5[(\frac{276.099^{2}}{2} + 9.8(z_{o} - z_{i}) - (\frac{0.542^{2}}{2}][/tex]
[tex]\Delta E_{m} = 57319 J = 5.7319 kW[/tex]
(3) The total energy of water is given by:
[tex]\Delta E_{t} = E - E_{m} = 4845.43 - 5.7319 = 4839.69 kW[/tex]
(4) The rate of heat transfer:
q = [tex]\Delta E_{t} = 4839.69 kW[/tex]
A fluidis flowing through a capillary
tube having diameter of 1.0 mm andReynolds number of 1000.
Calculate the velocity of the fluid, ifkinematic viscosity of fluid
is 1.1012 x
10-6m2/s.
Answer:
The velocity of the fluid is 1.1012 m/s
Solution:
As per the question, for the fluid:
Diameter of the capillary tube, d = 1.0 mm = [tex]1.0\times 10^{- 3} m[/tex]
Reynolds No., R = 1000
Kinematic viscosity, [tex]\mu_{k} = 1.1012\times 10^{- 6} m^{2}/s[/tex]
Now, for the fluid velocity, we use the relation:
[tex]R = \frac{v_{f}\times d}{\mu_{k}}[/tex]
where
[tex]v_{f}[/tex] = velocity of fluid
[tex]v_{f} = \frac{R\times \mu_{k}}{d}[/tex]
[tex]v_{f} = \frac{1000\times 1.1012\times 10^{- 6}}{1.0\times 10^{- 3}} = 1.1012 m/s[/tex]
The shaft of a vacuum cleaner motor rotates with an angular acceleration of four times the shaft’s angular velocity raised to the ¾ power. The vacuum beater bar is attached to the motor shaft with pulley through a drive belt. The radii of the motor pulley and the beater bar are 0.25 in and 1.0 in respectively. Determine the angular velocity of the beater bar when t = 4 s, given that omega_0 is 1 rad/s when theta = 0.
Answer:
470 rad/s
Explanation:
The acceleration of the motor shaft is:
γ1 = 4*w1^(3/4)
When connected by a belt the pulleys have the same tangential speed
vt = w * r
vt1 = vt2
w1 * r1 = w2 * r2
w2 = w1 * r1/r2
Therefore:
γ2 = 4 * (w1 * r1/r2)^(3/4)
d(w1 * r1/r2)/dt = 4 * (w1 * r1/r2)^(3/4)
(r1/r2) * dw1/dt = 4 * (r1/r2)^(3/4) * (w1 * r1/r2)^(3/4)
dw1/dt = 4 * (r1/r2)^(-1/4) * (w1)^(3/4)
This is a differential equation.
Solving it through Wolfram Alpha:
w1(t) = (1 / 256) * (4 * (r1/r2)^(-1/4) * t - 4)^4
w1(4) = (1 / 256) * (4 * (0.25 / 1)^(-1/4) * 4 - 4)^4 = 470 rad/s
The angular velocity of the beater bar at t=4s is approximately 6.37 rad/s, based on the given angular acceleration equation and initial angular velocity.
Explanation:The angular velocity of the beater bar can be found using the relationship between angular acceleration and angular velocity. The given equation states that the angular acceleration is four times the angular velocity raised to the 3/4 power. Therefore, we can write:
α = 4 * ω^(3/4)
To find the angular velocity at t=4s, we can integrate the equation to get:
ω = 4/7 * t^7/4 + C
When t = 0, ω = ω_0 = 1 rad/s. Substituting these values, we can solve for C:
1 = 0 + C
Therefore, C = 1. Finally, we can substitute t = 4s into the equation to get the angular velocity:
ω = 4/7 * 4^7/4 + 1
Calculating this expression, we find that the angular velocity of the beater bar at t=4s is approximately 6.37 rad/s.
A worker's hammer is accidentally dropped from the 20th floor of a building under construction. With what velocity does it strike the pavement 304 ft below, and what time t is required?
Answer:
Final Velocity (Vf)= 139.864 ft/s
Time (t)= 4,34 s
Explanation:
This is a free fall problem, to solve it we will apply free fall concepts:
In a free fall the acceletarion is gravity (g) = 9,81 m/s2, if we convert it to ft/s^2 = g= 32.174 ft/s^2
Final velocity is Vf= Vo+ g*t[tex]Vf^{2} = Vo^{2} +2*g*hwhere h is height (304 ft in this case).
Vo =0 since the hammer wasn't moving when it stared to fall
Then Vf^2= 0 + 2* 32.174 ft/s^2 *304 ft
Vf^2= 19,561.8224 ft^2/s^2
Vf=[sqrt{19561.8224 ft^2/s^2}
Vf=139.864 ft/s
Time t= (Vf-Vo)/g => (139.864 ft/s-0)/32.174 ft/s^2 = 4.34 sec
Good luck!
A train which is traveling at 70 mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 52 mi/hr as it passes a point 1/2 mi beyond A. A car moving at 52 mi/hr passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver steps on the gas.
(a) Calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 4.3 sec.
(b) find the velocity v of the car as it reaches the crossing.
Answer:
a) 0 mi/s^2
b) 52 mi/s
Explanation:
Assuming the crossing is 1/2 mile past point A and that point B is near point A (it isn't clear in the problem)
The train was running at 70 mi/h at point A and with constant deceleration reachesn the crossing 1/2 mile away with a speed of 52 mi/h
The equation for position under constant acceleration is:
X(t) = X0 + V0 * t + 1/2 * a * t^2
I set my reference system so that the train passes point A at t=0 and point A is X = 0, so X0 = 0.
Also the equation for speed under constant acceleration is:
V(t) = V0 + a * t
Replacing
52 = 70 + a * t
Rearranging
a * t = 52 - 70
a = -18/t
I can then calculate the time it will take it to reach the crossing
1/2 * a * t^2 + V0 * t - X(t) = 0
Replacing
1/2 (-18/t) * t^ + 70 * t - 1/2 = 0
-9 * t + 70 * t = 1/2
61 * t = 1/2
t = (1/2)/61 = 0.0082 h = 29.5 s
And the acceleration is:
a = -18/0.0082 = -2195 mi/(h^2)
To beath the train the car must reach the crossing in 29.5 - 4.3 = 25.2 s
X(t) = X0 + V0 * t + 1/2 * a * t^2
52 mi/h = 0.0144 mi/s
1/2 = 0 + 0.0144 * 25.2 + 1/2 * a * 25.2^2
1/2 = 0.363 + 317.5 * a
317.5 * a = 0.5 - 0.363
a = 0.137/317.5 = 0.00043 mi/s^2 (its almost zero)
The car should remain at about constant speed.
It will be running at the same speed.
What colour is best for radiative heat transfer? a. Black b. Brown c. Blue d. White
Answer:
The correct answer is option 'a': Black
Explanation:
As we know that for an object which is black in color it absorbs all the electromagnetic radiation's that are incident on it. Thus if we need to transfer energy to an object by radiation the most suitable color for the process is black.
In contrast to black color white color is an excellent reflector, reflecting all the incident radiation that may be incident on it hence is the least suitable material for radiative heat transfer.
Find the specific volume and internal energy of compressed liquid water at 100 C and 10 MPa, using both the compressed liquid tables and the saturated liquid approximation. What percent error do you get using the saturated liquid approximation?
Answer:
specific volume v = 0.0010385 m³/kg
and internal energy u = 416.23 KJ/kg
specific volume vf = 0.001043 m³/kg
and internal energy uf = 419.06 KJ/kg
% error in specific volume = 0.43 %
% error in internal energy = 0.679 %
Explanation:
given data
pressure = 10MPa
temperature = 100 C
to find out
specific volume and internal energy and What percent error do you get
solution
we know pressure 10 MPa and temperature 100 C
so from compressed liquid water table we get
specific volume v = 0.0010385 m³/kg
and internal energy u = 416.23 KJ/kg
and
by using saturated liquid approx
from saturated water table of 100 C
specific volume vf = 0.001043 m³/kg
and internal energy uf = 419.06 KJ/kg
so
% error in specific volume is
% error in specific volume = [tex]\frac{0.001043-0.0010385}{0.0010385}[/tex] × 100
% error in specific volume = 0.43 %
and
% error in internal energy is
% error in internal energy = [tex]\frac{419.06-416.23}{416.23}[/tex] × 100
% error in internal energy = 0.679 %
A 1/4th scale car is to be tested in a wind tunnel. If the full scale speed of the car is 30m/s, what should be the wind tunnel speed for Reynolds number similarity a) 30m/s b) 6m/s c) 7.5m/s d) 150m/s e) 120m/s
Answer:
e)v=120 m/s
Explanation:
Given that
Scale ratio = 1/4
Speed of car =30 m/s
lets wind tunnel speed is v
We know that Reynolds number given as
[tex]Re=\dfrac{\rho\ L\ V}{\mu }[/tex]
If all conditions taken as similar then
[tex](L\ V)_c=(L\ V)_w[/tex]
Given that
[tex]\dfrac{L_w}{L_c}=\dfrac{1}{4}[/tex]
So we can say that
4 x 30 = v x 1
v=120 m/s
A reversible cyclic device does work while exchanging heat with three constant temperature reservoirs. The three reservoirs (1,2, and 3) are at temperatures of 1000K, 300K, and 500K. Four hundred kilojoules of heat are transferred from reservoir 1 to the device, and the total work done is 100 kJ. Find the magnitude and direction of the heat transfer from the other two reservoirs.
Answer:
Lets take [tex]Q_2[/tex] heat transfer take place from 500 K reservoir to device and [tex]Q_3[/tex] from device to 300 K reservoir.
From the energy conservation we can say that
[tex]400+Q_2=100+Q_3[/tex]
[tex]300=Q_3-Q_2[/tex] -----1
For reversible process
[tex]\dfrac{400}{1000}+\dfrac{Q_2}{500}-\dfrac{Q_3}{300}=0[/tex]
[tex]5Q_3-3Q_2=600[/tex] ----2
By solving above two equation
[tex]Q_3=-150 KJ,Q_2=-450KJ[/tex]
But here sign come negative it means that
[tex]Q_2[/tex] heat transfer take place from device to 500 K reservoir and [tex]Q_3[/tex] from 300 K reservoir to device.
In this exercise we have to use the knowledge of heat transfer to calculate the heat transferred to the other two reservoirs will be:
The magnitude of the Q2 is -450 in the out direction while the Q3 is -150 in the inward direction.
From the information given in the statement, we have that:
Temperatures of 1000K, 300K, and 500K. Four hundred kilojoules of heat are transferred Total work done is 100 kJ.knowing that from conservation we can say:
[tex]400+Q_2=100+Q_3\\300=Q_3-Q_2\\\frac{400}{1000}+\frac{Q_2}{500}-\frac{Q_3}{300}=0\\[/tex]
So solving we have:
[tex]Q_3=-150KJ\\Q_2=-450KJ[/tex]
See more about heat transfer at brainly.com/question/12107378
A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at T[infinity] = 15°C. From reliability considerations, the chip temperature must not exceed T = 85°C.f the coolant is air and the corresponding convection 200 W/m2 K, what is the maximum coefficient is h allowable chip power? If the coolant is a dielectric liquid for which h 3000 W/m2 K, what is the maxi- mum allowable power?
Answer:
Q(h=200)=0.35W
Q(h=3000)=5.25W
Explanation:
first part h=200W/Km^2
we must use the convection heat transfer equation for the chip
Q=hA(Ts-T∞)
h= convective coefficient=200W/m2 K
A=Base*Leght=5mmx5mm=25mm^2
Ts=temperature of the chip=85C
T∞=temperature of coolant=15C
Q=200x2.5x10^-5(85-15)=0.35W
Second part h=3000W/Km^2
Q=3000x2.5x10^-5(85-15)=5.25W
• Suppose that a particular algorithm has time complexity T(n) = 10 ∗ 2n, and that execution of the algorithm on a particular machine takes T seconds for n inputs. Now, suppose you are presented with a machine that is 64 times as fast as your current machine. How many inputs can you process on you new machine in T seconds?
Answer:
The number of inputs processed by the new machine is 64
Solution:
As per the question:
The time complexity is given by:
[tex]T(n) = 10\times 2n[/tex]
where
n = number of inputs
T = Time taken by the machine for 'n' inputs
Also
The new machine is 65 times faster than the one currently in use.
Let us assume that the new machine takes the same time to solve k operations.
Then
T(k) = 64 T(n)
[tex]\frac{T(k)}{T(n)} = 64[/tex]
[tex]\frac{20k}{20n} = 64[/tex]
k = 64n
Thus the new machine will process 64 inputs in the time duration T
A 200L tank is evacuated and then filled through a valve connected to an air reservoir at 1 MPa and 20 °C. The valve is shut off when the pressure in the tank reaches 0.5 MPa. What is the mass of air in the tank?
Answer:
The air mass in the tank is 23.78 kg
Solution:
As per the question:
Volume of the tank, [tex]V_{t} = 200 l = 2\times 10^{- 3} m^{3}[/tex]
Pressure, P = 1 MPa = [tex]1\times 10^{6} Pa[/tex]
Temperature, T = [tex]20^{\circ}C[/tex] = 273 + 20 = 293 K
Pressure, P' = 0.5 MPa = [tex]0.5\times 10^{6} Pa[/tex]
Now,
To calculate the air mass, [tex]m_{a}[/tex] we use:
[tex]PV_{t} = m_{a}RT[/tex]
where
R = Rydberg constant = 0.287 J/kg.K
[tex]1\times 10^{6}\times 2\times 10^{- 3} = m_{a}0.287\times 293[/tex]
[tex]m_{a} = 23.78 kg[/tex]
The A-36 steel pipe has a 6061-T6 aluminum core. It issubjected to a tensile force of 200 kN. Determine the averagenormal stress in the aluminum and the steel due to thisloading.The pipe has an outer diameter of 80 mm and aninner diameter of 70mm.
Answer:
In the steel: 815 kPa
In the aluminum: 270 kPa
Explanation:
The steel pipe will have a section of:
A1 = π/4 * (D^2 - d^2)
A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2
The aluminum core:
A2 = π/4 * d^2
A2 = π/4 * 0.7^2 = 0.3848 m^2
The parts will have a certain stiffness:
k = E * A/l
We don't know their length, so we can consider this as stiffness per unit of length
k = E * A
For the steel pipe:
E = 210 GPa (for steel)
k1 = 210*10^9 * 0.1178 = 2.47*10^10 N
For the aluminum:
E = 70 GPa
k2 = 70*10^9 * 0.3848 = 2.69*10^10 N
Hooke's law:
Δd = f / k
Since we are using stiffness per unit of length we use stretching per unit of length:
ε = f / k
When the force is distributed between both materials will stretch the same length:
f = f1 + f2
f1 / k1 = f2/ k2
Replacing:
f1 = f - f2
(f - f2) / k1 = f2 / k2
f/k1 - f2/k1 = f2/k2
f/k1 = f2 * (1/k2 + 1/k1)
f2 = (f/k1) / (1/k2 + 1/k1)
f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN
f1 = 200 - 104 = 96 kN
Then we calculate the stresses:
σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa
σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa
The answer is: within the steel: 815 kPa
When within the aluminum: 270 kPa
The aluminumWhen The steel pipe will have a piece of:Then A1 = π/4 * (D^2 - d^2)After that A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2
The aluminum core is: Now A2 = π/4 * d^2Then A2 = π/4 * 0.7^2 = 0.3848 m^2
After that The parts will have a particular stiffness:k = E * A/l
We don't know their length, so we are able to consider this as stiffness per unit of length k = E * A
For the steel pipe: E = 210 GPa (for steel) k1 = 210*10^9 * 0.1178 = 2.47*10^10 N
For the aluminum: E = 70 GPak2 = 70*10^9 * 0.3848 = 2.69*10^10 NHooke's law:Δd = f / k
Since we are using stiffness per unit of length we use stretching per unit of length:ε = f / k
When the force is distributed between both materials will stretch the identical length:f = f1 + f2f1 / k1 = f2/ k2
Replacing: f1 = f - f2(f - f2) / k1 = f2 / k2f/k1 - f2/k1 = f2/k2f/k1 = f2 * (1/k2 + 1/k1)f2 = (f/k1) / (1/k2 + 1/k1)f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KNf1 = 200 - 104 = 96 kN
Then we calculate the stresses:σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPaσ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa
Find out more information about Aluminum here:
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A rigid tank holds 22 kg of 127 °C water. If 9 kg of that is liquid water what is the pressure in the tank and volume of the tank?
Answer:
The pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.
Explanation:
Volume is constant as the tank is rigid. Take the saturation condition of water from the steam table for pressure at 127°C.
Given:
Total mass of water is 22 kg.
Mass of liquid water is 9 kg.
Temperature of water is 127°C.
From steam table at 127°C:
The pressure in the tank is 246.878 Kpa.
Specific volume of saturated water is 0.00106683 m³/kg.
Specific volume of saturated steam is 0.72721 m³/kg.
Calculation:
Step1
From steam table at 127°C:
The pressure in the tank is 246.878 Kpa.
Step2
Dryness fraction is calculated as follows:
[tex]x=\frac{m_{v}}{m_{t}}[/tex]
Here, dyness fraction is x, mass of vapor is [tex]m_{v}[/tex]and total mass is [tex]m_{t}[/tex].
Substitute the values in the above equation as follows:
[tex]x=\frac{m_{v}}{m_{t}}[/tex]
[tex]x=\frac{22-9}{22}[/tex]
x = 0.59
Step3
Specific volume of tank is calculated as follows:
[tex]v=v_{f}+x(v_{g}-v_{f})[/tex]
[tex]v=0.00106683+0.59(0.72721-0.00106683)[/tex]
[tex]v=0.00106683+0.42842447[/tex]
v=0.4295 m³/kg.
Step4
Volume is calculated as follows:
[tex]V=v\times m_{t}[/tex]
[tex]V=0.4295 \times22[/tex]
V=9.449 m³.
Thus, the pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.
The temperature of a system rises by 10 °C during a heating process. Express the rise in temperature of K, R, and °F.
Explanation:
Given T = 10 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (10 + 273.15) K = 283.15 K
T = 283.15 K
The conversion of T( °C) to T(F) is shown below:
T (°F) = (T (°C) × 9/5) + 32
So,
T (°F) = (10 × 9/5) + 32 = 50 °F
T = 50 °F
The conversion of T( °C) to T(R) is shown below:
T (R) = (T (°C) × 9/5) + 491.67
So,
T (R) = (10 × 9/5) + 491.67 = 509.67 R
T = 509.67 R
A 10°C rise in temperature is equivalent to a 10 K increase in the Kelvin scale and an 18°F increase in the Fahrenheit scale. For Rankine, it's an 18 R increase.
Explanation:The temperature of a system rising by 10 °C can be expressed in different units of measurement. Firstly, since the Celsius and Kelvin scales are offset by 273.15, an increase by 10 °C is equivalent to an increase by 10 K. For the Fahrenheit scale, we use the conversion formula (°C × 9/5) + 32, which means an increase by 10 °C corresponds to an increase by 18 °F. Therefore, a rise by 10 °C equates to a rise by 10 K in the Kelvin scale and by 18 °F in the Fahrenheit scale. In the Rankine scale, which is Fahrenheit-based, the increase would be 18 R since it also uses the 9/5 scaling factor.
a bar of solid circular cross section is loaded in tension by forces p. the bar has length l=16in and diameter 0.50in. the material is a magnessium alloy having modulus of elasticity E=6.4*10^6 psi. the allowable stress in tension is epsillon allow=17000psi and the elongation of the bar must not exceed 0.04 in. what is the allowable value of the forces p?
Answer:3142 Pound force
Explanation:
Given
Length(L)=16 in.
diameter=0.5 in.
Modulus of elasticity (E)[tex]=6.4\times 10^6 psi.[/tex]
Allowable stress [tex]\left ( \sigma _{allowable}\right )=17000 psi[/tex]
Max elongation of the bar =0.04 in.
Also we Know
Elongation is given
[tex]\Delta L=\frac{PL}{AE}[/tex]
where
P=Force applied
L=length of bar
A=Cross-section
E=Modulus of Elasticity
[tex]A=\frac{\pi d^2}{4}=\frac{\pi}{16} in.^2[/tex]
[tex]0.04=\frac{P\times 16\times 16}{\pi \times 6.4\times 10^6}[/tex]
[tex]P=3142 pound\ force\approx 13.976 kN[/tex]
corresponding stress is [tex]\frac{3142}{\frac{\pi }{16}}=16,000 psi[/tex]
Which is less than Allowable stress
thus allowable value of Force P is 3142 Pound force or 13.976 kN
Engineering controls are the physical changes that employers make to the work environment or to equipment that make it safer to use.
A) True B) False
Answer:
True
Explanation:
Engineering controls are those techniques used to reduce or eliminate hazards of any condition, thereby protecting the workers.
These are mostly products that act as barriers between the worker and the hazard. This may include machinery or equipment. The common engineering controls used are glovebox, biosafety cabinet, fume hood, vented balance safety enclosure, HVAC system, lockout-tagout, sticky mat and rupture disc.
A 0.5 m^3 container is filled with a mixture of 10% by volume ethanol and 90% by volume water at 25 °C. Find the weight of the liquid.
Answer:
total weight of liquid = 4788.25 N or 488.09 kg
Explanation:
given data
total volume = 0.5 m³
volume of ethanol = 10 % of volume = 0.10 × 0.5 = 0.05 m³
volume of water = 90 % at 25 °C of volume = 0.90 × 0.5 = 0.45 m³
to find out
weight of the liquid
solution
we know that density of water at 25 is 997 kg/m³
and density of ethanol is 789 kg/m³
so weight of water is = density × volume × g
put here value and we take g = 9.81
weight of water is = 997 × 0.45 × 9.81
weight of water = 4401.25 N ......................1
weight of ethanol is = density × volume × g
put here value and we take g = 9.81
weight of ethanol is = 789 × 0.05 × 9.81
weight of ethanol = 387.00 N ...............2
so total weight of liquid = sum of equation 1 add 2
total weight of liquid = 4401.25 + 387
total weight of liquid = 4788.25 N or 488.09 kg
A 3 ft x 2 ft block moves down a 15 degree inclined slope at a speed of V = 0.2 ft/s over a thin layer (h = 0.0125 ft) of oil with a dynamic viscosity of 8.2 x 10^-2 lbf-s/ft^2. What is the weight of the block? Draw FBD.
Answer:
mg = 30.415 lbf
Explanation:
from figure body of size 3ft*2ft is tend to move down side
weight is divided into two component
vertical component = mgcos15 and
horizontal component = mg sin15
considering horizontal component equal to shear force
mgsin15 = \tau A
mgsin15 =\mu \frac{dv}{dh} A
mg =\frac{ \mu v A}{h*sin15}
=\frac{8.2*10^{-2}*0.2*3*2}{0.0125*sin15}
mg = 30.415 lbf
Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 7, and the turbine inlet temperature is 1200 K. For constant specific heats with k = 1.4 and Cp = 1.005 kJ/kg, calculate the percent thermal efficiency (enter a number only)
Answer:
The thermal efficiency of cycle is 42.6%.
Explanation:
Given that
[tex]T_1=300 K[/tex]
[tex]P_1=100KPa[/tex]
mass flow rate = 6 kg/s
Compression ratio = 7
Turbine inlet temperature = 1200 K
γ=1.4
We know that thermal efficiency of Brayton cycle given as
[tex]\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}[/tex]
Now by putting the values
[tex]\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}[/tex]
[tex]\eta=1-\dfrac{1}{7^{\frac{1.4-1}{1.4}}}[/tex]
η=0.426
So the thermal efficiency of cycle is 42.6%.
For jet aircraft engine, what is not used for the system? A- nozzle B- Turbine C- compressor D- none
Answer:
option D
Explanation:
The correct answer is option D.
For jet aircraft engine, all the system that is nozzle, turbine and Compressor are used.
Nozzle is used to provide thrust to the aircraft. It is attached at the downstream of the turbine.
Turbine is used to release hot gas which is produced in air due to burning of gas.
Engine suck air from the fan attached in front and compressor increase the pressure of the air.
If you know the measured electrical voltage and the cross-sectional area of the pipe/tube you can determine the volume? a) True b) False
Answer:
b)false
Explanation:
false, missing two variables, the current and the material.
To find the volume of the tube we must know the transverse area and the length.
It is possible to calculate the length of the tube knowing the resistance, since this is a value that depends on the material, the area and the length.
It is possible to calculate the electrical resistance using the ohm equation.
Resitence = Voltage / current.
Air enters an insulated turbine operating at steady state at 8 bar, 500K, and 150 m/s. At the exit the conditions are 1 bar, 320 K, and 10 m/s. There is no in elevation. Determine the work developed and the exergy destruction, each in kJ/kg of air flowing. Let To=300K and po=1bar significant change
Answer
given,
P₁ = 8 bar T₁ = 500 K V₁ = 150 m/s
P₂ = 1 bar T₂ = 320 K V₂ = 10 m/s
writing energy equation
h₁ + (KE)₁ + (PE)₁ + Q m = h₂ + (KE)₂ + (PE)₂ + W
[tex]W = (h_1 - h_2 ) + \dfrac{V_1^2-V_2^2}{2000}[/tex]
ideal gas property of air
T₁ = 500 K h₁ = 503.02 KJ/kg S₁ = 2.21952 kJ/kgK
T₂ = 320 K h₂ = 320.29 KJ/kg S₂ = 1.7679 kJ/kgK
[tex]W = (503.02-320.29) + \dfrac{150^2-10^2}{2000}[/tex]
W = 193.93 KJ/Kg
calculation of energy destruction
= [tex]T_0(S_2-S_1-Rln(\dfrac{P_2}{P_1}))[/tex]
= [tex]T_0(S_2-S_1+Rln(\dfrac{P_1}{P_2}))[/tex]
= [tex]300(1.7679-2.21952-\dfrac{8.314}{28.97}ln(\dfrac{8}{1}))[/tex]
=[tex]300 \times 0.145152[/tex]
=43.54 KJ/Kg
The viscosity of a fluid is 10 be measured by a viscometer constructed of two 75-cm-long concentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the two cylinder is 0.12 cm. The inner cylinder is rotated at 200 rpm, and the torque is measured to be 0.8 N m. Determine the viscosity of the fluid.
Answer:
0.023 Pa*s
Explanation:
The surface area of the side of the inner cylinder is:
A = π*d*l
A = π*0.15*0.75 = 0.35 m^2
At 200 rpm the inner cylinder has a tangential speed of:
u = w * r
u = w * d/2
w = 200 rpm * 2π / 60 = 20.9 rad/s
u = 20.9 * 0.15 / 2 = 1.57 m/s
The torque is of 0.8 N*m, this means that the force is:
T = F * r
F = T / r
F = 2*T / d
For Newtoninan fluids with two plates moving respect of each other with a fluid between the viscous friction force would be:
F = μ*A*u / y
Where
μ: viscocity
y: separation between pates
A: surface area of the plates
Then:
2*T / d = μ*A*u/y
Rearranging:
μ = 2*T*y / (d*A*u)
μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s
A 29-mm-diameter copper rod is 1.1 m long with a yield strength of 73 MPa. Determine the axial force necessary to cause the diameter of the rod to reduce by 0.01 percent, assuming elastic deformation. Check that the elastic deformation assumption is valid by comparing the axial stress to the yield strength. The axial force necessary to cause the diameter of the rod to reduce by 0.01 percent is ____kN.
Answer:
32.96 MPa
Explanation:
The Poisson ratio of copper is:
μ = 0.355
The Young's modulus of copper is:
E = 117 GPa
The equation for reduction of diameter of a rod is:
D = D0 * (1 - μ*σ/E)
Rearranging:
D = D0 - D0*μ*σ/E
D0*μ*σ/E = D0 - D
D0*μ*σ = E*(D0 - D)
σ = E*(D0 - D) / (D0*μ)
If the diameter is reduced by 0.01 percent
D = 0.9999*D0
σ = E*(D0 - 0.9999*D0) / (D0*μ)
σ = E*(0.0001*D0) / (D0*μ)
σ = 0.0001*E / μ
σ = 0.0001*117*10^9 / 0.355 = 32.96 MPa
This value is below the yield strength, therefore it is valid.
You are reading a "pressure head" of 14 feet. If water Has a specific weight of 62.4 pound/cubic feet, what is The pressure of the "pressure head"?
Answer:
Pressure of pressure head will be [tex]863.6pound/ft^2[/tex]
Explanation:
We have given pressure head = 14 feet
Specific weight of water [tex](\rho g)_{water}=62.4pound/ft^3[/tex]
Pressure is given by [tex]P=(\rho g)_{water}h[/tex], here h is pressure head.
So pressure [tex]P=(\rho g)_{water}h=62.4pound/ft^3\times 14ft=863.6pound/ft^2[/tex]
So the pressure of pressure head will be [tex]863.6pound/ft^2[/tex]