Answer: (a). Ec(μ) = 165.6 GPa
(b). Ec(∝) = 83.09 GPa
Explanation:
this is quite straightforward, so we will go step by step.
from the data we have that,
Moduli of elasticity of the metal -(Em) is 60 Gpa
Moduli of elasticity of oxide is (Ep) is 380 Gpa
volume Vp = 33% = 0.33
(a). To solve the upper bound-modulus of the elasticity is calculate thus;
Ec (μ) = EmVm + EpVp ----------------(1)
where E rep the modulus of elasticity
v rep the volume fraction
c rep the composite
Vm = 100% - Vp
Vm = 100% - 33% = 67%
Vm = 0.67
substituting the valus of Em, Vm, Ep, Vp from equation (1) we have;
Ec(μ) = (60×0.67) + (380×0.33)
Ec(μ) = 40.2 + 125.4 = 165.6 GPa
Ec(μ) = 165.6 GPa
(b). The lower bound modulus of elasticity can be calculated thus;
Ec(∝) = EmVp / EpVm + EmVp -------------- (2)
substituting values Em,Vm,Ep,Vp.
Ec(∝) = 60×30 / (380×0.67) + (60 ×0.33)
Ec(∝) = 22800 / 254.6 + 19.8 = 83.09 GPa
Ec(∝) = 83.09 GPa
cheers i hope this helps!!!!
A 10 wt % aqueous solution of sodium chloride is fed to an evaporative crystallizer which operates at 80o C. The product is a slurry of solute crystals suspended in a saturated solution. The unit is to produce 1000 kg/crystals per hour. If the pump that handles the slurry cannot handle more than 40 wt % solids, find the feed rate to the crystallizer and the evaporation rate of the water. (20 pts)
Answer: The feed rate is
17,020kg/he and the rate is 13,520kg/h
Check the attachment for step by step explanation
The clock period of a ripple counter must be longer than the total propagation
delays through all of the flip-flops. For proper operation, the period / frequency
should be:
Tmin = N x tpd where tpd is the propagation delay fmax = 1/ Tmin = 1/ (N x tpd).
If the propagation delay is 15ns for a J-K FF, what is the largest
MOD counter that can be constructed to ensure the counter will operate for
frequencies > 12 MHz?
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
Insulated Gas Turbine Air enters an adiabatic gas turbine at 1050 K and 1 MPa and leaves at 400 kPa. Kinetic and potential energy changes can be ignored. Treat the air as an ideal gas with constant specific heats. Let k = 1.4 a. Determine the theoretical exit temperature - corresponding to part "b" below, the maximum theoretical power output. b. Determine the maximum theoretical work output for the gas turbine in kJ/kg. c. If the isentropic turbine efficiency is 0.8, what is the actual work output of the turbine in kJ/kg?
Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
Explanation:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
(a) Determine the temperature of the insulated walls. (b) Determine the net radiation heat rate from surface 2 per unit conduit length. 13.48 A long conduit is constructed with diffuse, gray walls 0.5 m wide. The top and bottom of the conduit are insulated. The emissivities of the walls are ε 1 = 0.45, ε 2 = 0.65, and ε 3 = 0.15, respectively, while the temperatures of walls 1 and 2 are 500 K and 700 K, respectively.
Answer:
Explanation:
the solution to the problem is given in the pictures attached. (b) is answered first then (a). I hope the explanation helps you.Thank you
"Design a sequential circuit with two T flip-flops A and B, and one input x. When x = 0, the circuit remains in the same state. When x = 1, the circuit goes through the state transitions from 00 to 01, to 10, to 11, back to 00, and repeats. What is the boolean equation of the input of flip-flop A (TA)? Type variables and operations without blanks."
Answer:
See attached images for the diagrams and tables
A 3-phase stepping motor is to be used to drive a linear axis for a robot. The motor output shaft will be connected to a screw thread with a screw pitch of 1 mm. We want to be able to have a spatial control of at least 0.05 mm. a. How many poles should the motor have? b.How many pulses are needed from the controller every second to move the linear axis at a rate of 90 mm/sec?
The number of poles required for a 3-phase stepping motor to achieve 0.05 mm spatial control depends on the increments per revolution, which was not provided. To move the axis at 90 mm/sec, the pulses per second from the controller will be calculated based on the screw pitch and the number of increments per revolution. Additional information on the stepper motor is needed for precise calculations.
Explanation:To answer the student's questions regarding a 3-phase stepping motor for spatial control in a robotics application:
Poles of the motor: The number of poles in a stepper motor determines the resolution of movement. To achieve a spatial control of at least 0.05 mm with a screw pitch of 1 mm, the motor needs to have enough poles to allow for a fractional turn equal to that precision. Given that stepper motors can have a range of 5,000 to 10,000 increments in a 90-degree rotation, the exact number of poles required can be calculated based on the needed increments per mm.Pulses needed for motion: To move the axis at a speed of 90 mm/sec, the number of pulses required from the controller per second will depend on the revolutions per minute (rpm) the screw thread needs to turn, which is determined by the number of poles or increments per revolution of the motor. Generally, to calculate this, we would use the formula (speed in mm/sec) / (screw pitch in mm) * (number of increments per revolution), giving us the pulses per second.Without the specific number of increments per revolution of the stepping motor, it's impossible to provide an exact answer. However, typically stepper motors with more poles can provide finer control, and thus would be preferred in this application to meet the 0.05 mm spatial control requirement.
Air enters the first compressor stage of a cold air-standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The temperature at the inlet to the turbine is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%.
For k = 1.4, calculate:
(A) the thermal efficiency of the cycle.
(B) the back work ratio.
(C) the net power developed, in kW.
(D) the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T0 = 300 K.
Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 961 K, while the wire mesh on top of the grill is covered with steaks initially at 274 K. The distance between the coal bricks and the steaks is 0.20 m. Treating both the steaks and the coal bricks as blackbodies, determine the initial rate of radiation heat transfer from the coal bricks to the steaks. Also, determine the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface.
Answer:
Step 1
Given
Diameter of circular grill, D = 0.3m
Distance between the coal bricks and the steaks, L = 0.2m
Temperatures of the hot coal bricks, T₁ = 950k
Temperatures of the steaks, T₂ = 5°c
Explanation:
See attached images for steps 2, 3, 4 and 5
An airplane starts from rest, 6050 ft down a runway at uniform accelerationthen takes off with a speed of 150mi / h . It then climbs in a straight line with a uniform acceleration of 2 ft/s^ 2 until it reaches a constant speed of 195mi / h . How far has the plane traveled when it reaches this constant speed?
Given Information:
distance = s₁ = 6050 ft
velocity = v₁ = 0 mi/hr
velocity = v₂ = 150 mi/hr
velocity = v₃ = 195 mi/hr
Acceleration = a = 2 ft/s²
Required Information:
distance = s₂ = ?
Answer:
distance = s₂ = 14,399 ft
Explanation:
We know from the equations of motion,
v₃² = v₂² + 2a(s₂ - s₁)
We want to find out the distance s₂
2a(s₂ - s₁) = v₃² - v₂²
s₂ - s₁ = (v₃² - v₂²)/2a
s₂ = (v₃² - v₂²)/2a + s₁
First convert given velocities from mi/hr to ft/s
1 mile has 5280 feet and 1 hour has 3600 seconds
velocity = v₂ = 150*(5280/3600) = 220 ft/s
velocity = v₃ = 195*(5280/3600) = 286 ft/s
s₂ = (v₃² - v₂²)/2a + s₁
s₂ = (286² - 220²)/2*2 + 6050
s₂ = 33396/4 + 6050
s₂ = 8349 + 6050
s₂ = 14,399 ft
Therefore, the plane would have traveled a distance of 14,399 ft when it reaches a constant speed of 286 ft/s
Consider the problem of oxygen transfer from the interior lung cavity, across the lung tissue, to the network of blood vessels on the opposite side. The lung tissue (species B) may be approximated as a plane wall of thickness L. The inhalation process may be assumed to maintain a constant molar concentration CA(0) of oxygen (species A) in the tissue at its inner surface (x = 0), and assimilation of oxygen by the blood may be assumed to maintain a constant molar concentration CA(L) of oxygen in the tissue at its outer surface (x = L). There is oxygen consumption in the tissue due to metabolic processes, and the reaction is zero order, with N_A=-k_0.
Obtain expressions for the distribution of the oxygen concentration in the tissue and for the rate of assimilation of oxygen by the blood per unit tissue surface area.
Answer:
See attached images
Derive an expression for the axial thrust exerted by a propeller if the thrust depends only on forward speed, angular speed, size, and viscosity and density ofthe fluid. How would the expression change if gravity were a relevant variable in the case of a ship propeller?
Answer:
Find the given attachment
Logical variables: Running late? Complete the tunction RunningLate such that the logical variable on Time is true if no Traffic is true and gasEmpty is false. Ex >> n°Traffic = true ; >> gasEmpty-false; >onTime - Runninglate(noTraffic,gasEmpty) logical Your Function B Save C Reset EE MATLAB Documentation 1 function onTime - RunningLate (noTraffic,gasEmpty) 2 % complete the logical expression to the right of . using the variables n°Traffic and gasEmpty 4 onTime - 6 end Code to call your function C Reset 1 noTraffic true; gasEmpty true; 2 onTime-Runninglate(noTraffic, gasEmpty)
The function RunningLate returns true for the variable onTime when noTraffic is true and gasEmpty is false, which is achieved by the logical expression noTraffic && ~gasEmpty in MATLAB.
Explanation:The task is to complete the function RunningLate in MATLAB, which returns a logical variable onTime that is true if noTraffic is true and gasEmpty is false. The finished MATLAB code inside the function should be:
onTime = noTraffic && ~gasEmpty;This code uses the logical AND operator (&&) to check that there is no traffic, and the NOT operator (~) to check that the gas tank is not empty.
Convert 580,000 kW to [GW] with correct number of significant digits.
Answer:
[tex]x = 0.58\,GW[/tex]
Explanation:
The conversion is:
[tex]x = (580000\,kW)\cdot \left(\frac{1\,GW}{1000000\,kW} \right)[/tex]
[tex]x = 0.58\,GW[/tex]
Consider schedules S3, S4, and S5 below. Determine whether each schedule is
strict, cascadeless, recoverable, or nonrecoverable. (Determine the strictest
recoverability condition that each schedule satisfies.)
S3: r1 (X); r2 (Z); r1 (Z); r3 (X); r3 (Y); w1 (X); c1; w3 (Y); c3; r2 (Y); w2 (Z);
w2 (Y); c2;
S4: r1 (X); r2 (Z); r1 (Z); r3 (X); r3 (Y); w1 (X); w3 (Y); r2 (Y); w2 (Z); w2 (Y); c1;
c2; c3;
S5: r1 (X); r2 (Z); r3 (X); r1 (Z); r2 (Y); r3 (Y); w1 (X); c1; w2 (Z); w3 (Y); w2 (Y);
c3; c2;
Answer:
A schedule is strict if it satisfies the following conditions:
Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.
Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.
Explanation:
See attached image
Determine the convention heat transfer coefficient inside the for the flow of (a) air and (b) water at a velocity of 2 m/s in an 8-cm- diameter and 7-m-long tube when the tube is subjected to uniform heat flux from all surfaces. Use fluid properties at 25oC.
Answer:
Find the attachments sequence wise for complete solution.
A very long pipe of 0.05 m (r1) radius and 0.03 m thickness (r2 - r1) is buried at a depth of 2m (z) to transport liquid nitrogen. The pipe wall has a very low thermal conductivity of 0.0035 Wm- 1K-1 and receives a cooling power of 10 W/m to keep the liquid nitrogen at 77 K. If the thermal conductivity of the ground is 1 Wm-1K-1, what is the surface temperature of the ground
Answer:
See explaination
Explanation:
thermal conductivity is A measure of the ability of a material to transfer heat. Given two surfaces on either side of the material with a temperature difference between them, the thermal conductivity is the heat energy transferred per unit time and per unit surface area, divided by the temperature difference.
Please kindly check attachment for the step by step solution of the given problem.
A flexible pavement has a SN of 3.8 (all drainage coefficients are equal to 1.0). The initial PSI is 4.7 and the terminal serviceability is 2.5. The soil has a CBR of 9. The overall standard deviation is 0.40 and the reliability is 95%. The pavement is currently designed for 1800 equivalent 18-kip single-axle loads per day. If the number of 18-kip single-axle loads were to increase by 30%, by how many years would the pavement design life be reduced
Answer:
2.83 years
Explanation:
Please kindly see the attached file at thw attachment area for a detailed and step by step solution to the given problem.
Consider a 50 x 106 m3 lake fed by a polluted stream with a flow rate of 75 m3 /s and a pollutant concentration of 25.5 mg/L. There is also a sewage outfall that discharges 3.0 m3 /s of wastewater with a pollutant concentration of 125 mg/L. Stream and sewage wastes have a reaction rate coefficient of 5% per day. Find the steady-state (i.e., effluent) pollutant concentration and flow rate
Answer:
24.78 mg/L
Explanation:
The step-to-step explanation is written legibly with clear explanation in the diagram attached below.
Answer:
Answer: Input rate = 2.288 x 10⁶mg/s
Output rate =78 x 10³Cmg/s
Decay rate = 28.94 x 10 ⁵C mg/s
Explanation:
Assuming that complete and instantaneous mixing occurs in the lake, this implies that the concentration in the lake C is the same as the concentration of the mix leaving the lake Cm
Input rate = Output rate + KCV
Input rate = Q₁C₁ + QwCw
= (75.0m³/s x 25.5mg/L + 3.0m³/s x 125.0mg/L) x 10³L/m³
= 2.288 x 10⁶mg/s
Output rate = QmCm = (Q₁ +Qw)C
=(75 + 3.0)m³/s x Cmg/L x 10³L/m³ = 78 x 10³Cmg/s
Decay rate = KCV = 5/d x Cmg/L x 50 x 10⁶ x 10³L/m³
24 hr/d x 3600s/hr
= 28.94 x 10 ⁵C mg/s
So,
=2.288 x 10⁶ = 78 x 10³C + 28.94 x 10 ⁵C = 29.72 X 10⁵C
= 2.288 x 10⁶
29.72 X 10⁵ = 0.77 mg/l
C =
The mean of hours that the average person watches television each day is 4.18 hours with a standard deviation of 1.19 hours. Find probability that someone watches between 3 and 5 hours a day
Answer:
[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]
[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]
And we can find this probability with this difference:
[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]
And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution
Explanation:
For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:
[tex]\mu = 4.18,\sigma =1.19[/tex]
We can assume that the distribution for X is normal
[tex] X \sim N(\mu = 4.18 , \sigma =1.19)[/tex]
And we want to find this probability:
[tex] P(3 <X<5)[/tex]
And we can use the z score formula given by:
[tex] z=\frac[X- \mu}{\sigma}[/tex]
And we can find the z score for each limit and we got:
[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]
[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]
And we can find this probability with this difference:
[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]
And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution
Compare the heat transfer coefficients for laminar forced and free convection over vertical flat plates. Develop an approximate relation between the Reynolds and Rayleigh numbers such that the heat transfer coefficients for pure forced convection and pure free convection are equal.
Answer:
Check the attached images below.
Explanation:
It Is required to develop an approximate relation between . Reynolds and Grashor numbers such that the heat-transfer coefficients for pure forced convection and pure .e convection are equal, assuming laminar flow, by comparing the heat-transfer coefficients for forced or free convection over vertical hat plates.
write the equation or heat transfer coefficient Mr (arced comedian.
Kindly check the attached images below.
"From the earth to the moon". In Jules Verne’s 1865 story with this title, three men went to the moon in a shell fired from a giant cannon sunk in the earth in Florida.
(a) Find the minimum muzzle speed needed to shoot a shell straight up to an altitude equal to the 2 times earth’s radius RE.
(b) Find the minimum muzzle speed that would allow a shell to reach the height of the moon, 385,000 km, center of the earth to center of the moon
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to get the step by step explanation to the above question.
Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 1.825 m3/s and expands adiabatically to an exit state of 1 bar, 200°C. Kinetic and potential energy effects are negligible. Determine for the turbine: (a) the power developed, in kW. (b) the rate of entropy production, in kW/K. (c) the percent isentropic turbine efficiency.
Answer: (a).power developed = 776.1 kW
(b). Rate of entropy production = 1.023 kW/K
(c). efficiency = 63%
Explanation:
Let us carry a step by step process to solve this problem;
from the question we have that
P₁ = 5 bar
T₁ = 320°C
where V₁ = 0.5416 m³/Kg, S₁ = 7.5308 KJ/Kg-K and R₁ = 0.3105.6 KJ/Kg
the volumetric flow rate is given as (φ) = 1.825 m³/s
Remember that φ = ṁ V
where ṁ is the mass flowrate, and V is the volume
ṁ = φ/V = 1.825/0.5416 = 3.37 Kg/s
Also given for the Exit state;
P₂ = 1 bar
T₂ = 200°C
where V₂ = 0.5416 m³/Kg, S₂ = 7.5308 KJ/Kg-K and R₂ = 0.3105.6 KJ/Kg
(a). we are asked to determine the power developed in the Kw.
using the Flow energy equation to turbine we have;
ṁ(R₁ + V₁²/2 + gZ₁) + φ = ṁ(R₂ + V₂²/2 + gZ₂₂) + ш
canceling out terms from both steps we have that
ш = 3.37 (3105-2815.3) = 776.1 kW
Therefore the Power output is 776.1 kW
(b). The rate of entropy production in Kw/K.
Rate(en) = ṁ (S₂-S₁) = 3.37 (7.8343 - 7.5308)
Rate(en) = 1.023 kW/K
(c). The percent isentropic turbine efficiency.
Πt = (R₁-R₂) / (h₁ - h₂s)
Πt = (3105.6 - 2875.3) / (3105.6 - 2740) = 63%
Πt = 63%
cheers i hope this helped!!!!!
3-For the problems in this exercise, assume that there are no pipeline stalls and that the breakdown of executed instructions is as follows: a. In what fraction of all cycles is the data memory used b. In what fraction of all cycles is the input of the sign-extend circuit needed? c. What is this circuit doing in cycles in which its input is not needed?
Answer:
(a) 35% (b) 80% (c) Control signal is sent to a resource for activation of it's usage.
The operations are performed even in areas where it's not needed, it is ignored, because it is not used in cycles.
Explanation:
Solution
The computation of fraction is defined below:
(a) The data memory used in lw and sw instructions
Now,
The fraction value is = sw + lw
=10 + 25
= 35
therefore the fraction value is 35%
(b) The needed sign is extended for all other instructions other than ADD
which is shown below
The Fraction value = addi + beq + lw + sw
=20 +25+25+ 10 =80 %
The fraction value here is 80%
(c) Now, a control signal is been sent to resource for activation of it's usage.
The operations are carried out even in case where it's not needed, it is ignored, because it is not used in cycles.
Note: The complete question to this exercise is attached below.
The fraction of all cycles that is the data memory used is 35%.
The fraction of all cycles that is the input of the sign-extend circuit needed is 80%.The circuit is known to be sending a resource for activation of it's usage and it is also done in areas of the circuit where it's not needed and so it is left that way.What is a circuit ?A circuit is known to be a kind of closed loop via which electricity often pass through.
To solve for (a) which is the data memory that has been used in lw and sw instructions, you then add the value together:
That is: fraction value is = sw + lw
=10 + 25
So question A fraction value is 35%
To solve for me the required sign-extend circuit is depicted as:
The Fraction value = addi + beq + lw + sw
=20 +25+25+ 10 =80 %
The question (b) fraction value here is 80%
So therefore, the fraction value are 35 percent and 80 percent respectively
Learn more about circuit from
https://brainly.com/question/2969220
A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 17 in.2 and that approximately 2.1 kg had corroded away during the submersion. Assuming a corrosion penetration rate of 200 mpy for this alloy in seawater, estimate the time of submersion in years. The density of steel is 7.9 g/cm3.
Answer:
4.8 years
Explanation:
The rate of corrosion (CPR) is defined as the rate at which a metal corrodes in a specific environment. It depends on the environmental condition and the type of metal. It is expressed in inches per year or melts per year. CPR is given by:
[tex]CPR=\frac{KW}{\rho At}[/tex]
Where K is a constant = 534, W is the weight corroded = 2.1 kg = 2.1 × 10⁶mg, A is the area = 17 in², ρ is the density = 7.9 g/cm³
From the CPR equation:
[tex]t=\frac{KW}{\rho A(CPR)}=\frac{534*2.1*10^6}{7.9*17*200}= 4.17*10^4 hrs=4.8years[/tex]
An extruder barrel has a diameter of 4.22 inches and a length of 75 inches. The screw rotates at 65 revolutions per minute. The screw channel depth = 0.23 in, and the flight angle = 21.4 degrees. The head pressure at the die end of the barrel is 705 lb/in2. The viscosity of the polymer melt is given as 145 x 10-4 lb·sec/in2. Calculate the volume flow rate in in3/sec of the plastic through the barrel.
Answer:
volume flow rate Q = 53.23 in³/s
Explanation:
given data
diameter = 4.22 inches
length = 75 inches
screw rotates = 65 revolutions per minute
depth = 0.23 in
flight angle = 21.4 degrees
head pressure = 705 lb/in²
viscosity = 145 x [tex]10^{-4}[/tex] lb·sec/in²
solution
we get here volume flow rate of platstic in barrel that is express as
volume flow rate Q = volume flow rate of die - volume flow of extruder barrel ................1
here
volume flow rate extruder barrel is
flow rate = [tex]\frac{\pi \times 705 \times 4.22 \times 0.23\times sin21.4 }{12\times 145 \times 10^{-4}\times 75 }[/tex]
flow rate = 60.10
and
volume flow rate of die is express as
volume flow rate = 0.5 × π² × D² × Ndc × sinA × cosA .............2
put here value and w eget
volume flow rate = 0.5 × π² × 4.22² × 0.23 × 1 × sin21.4 × cos21.4
volume flow rate = 6.866
so put value in equation 1 we get
volume flow rate Q = 60.10 - 6.866
volume flow rate Q = 53.23
2.(10 pts)A proposed engine cycle employs an ideal gas and consists of the following sequence of transformations; a) Isothermal compression at 300 o K from a pressure of 1bar to a pressure of 30bar b) Constant pressure heating to a temperature of 1600 o K. c) Isothermal expansion at 1600 o K to the original pressure of 1 bar. d) Constant pressure cooling to a temperature of 300 o K to complete the cycle An ideal regenerator connects d) to b) so that the heat given up in d) is used for the heating in b). For an engine using a kilamole of gas find the net work in kJ and the thermal efficiency. You may assume C p
Answer:
Check the explanation
Explanation:
For ideal regeneration heat loss in cooling aqual to heat gain in compression so temperature Tb=Td as can be seen in the step by step solution in the attached images below.
Select the statement that is false.
A. If two graphs G and H are isomorphic, then they have the same total degree.
B. If two graphs G and H have the same degree sequence, then G and H are isomorphic.
C. If two graphs G and H have the same degree sequence, then G and H must have the same number of edges.
D. If two graphs G and H have the same number of edges then G and H must have the same total degree.
Answer:
D
Explanation:
the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.
The journals in a high speed oil engine are 80 mm in diameter. and 40 mm long. The radial clearance is 0.060mm. Each supports a load of 9 kN when the shaft is rotating at 3600 rpm. The bearing is lubricated with SAE 40 oil supplied at atmospheric pressure and average operating temperature is about 65oC. Using Raimondi- Boyd charts analyze the bearing under steady state operation.
Answer: S = 0.284
Explanation:
Data: d = 80 mm;
l =40 mm;
c = 0.06 mm;
F = 9kN;
n = 3600rpm = 60 rps
SAE 40 oil
T= 65°C
Therefore:
p = F / ld
= 9 x1000 /40 x 80
= 2.813 MPa
μ = 30 cp at 65°C for SAE 40 oil
S = r^2 x μ x n / c^2 x p
S= ( 40 )^2 x 30*10^-3 x 60 / (0.06)^2 x 2.813*10^6
S = 2880 / 10,126.6
S = 0.284
l/d = ½,
h o /c = 0.38
ε = e /c = 0.62
h o = 0.38 x C
= 0.382 x 0.06
=0.023mm
= 23µm
e = 0.62 x C
= 0.62 x 0.06
= 0.037 mm
Viscosity temperature curves of SAE graded oils
(r /c) f = 7.5,
S = 0.284
l /d = ½
f = 7.5 x (c / r)
= 7.5x (0.06/40)
= 0.0113
Problem 32.3 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a msss of 250 kg and have a combined center of gravity located directly above C. For the position when θ = 25°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A.
The question involves applying principles of static equilibrium and mechanical advantage in physics to calculate forces exerted by a hydraulic cylinder and at a support point in a construction telescoping arm scenario.
Explanation:The question relates to the application of static equilibrium and mechanical advantage concepts in physics to solve for forces in a telescoping arm used in construction. Specifically, it involves calculating force exerted by a hydraulic cylinder and the force at a support point when a mass is placed at a specific location on the arm.
In order to solve for the force exerted at point B by the hydraulic cylinder (part a) and the force exerted on the supporting carriage at A (part b) when the arm makes an angle heta = 25 degrees with the horizontal, one would typically use the principles of torques and static equilibrium. Summation of torques around a point, often the pivot, and summation of forces in horizontal and vertical directions are standard procedures. This requires knowledge of the geometry of the construction, the location of the center of gravity, and the mass of the workers and platform.
The information given about the forces in a wheelbarrow and cranes are analogous examples that illustrate similar principles of physics applied to different scenarios. These examples demonstrate how to calculate the mechanical advantage and forces based on lever arms and mass distribution.
The force exerted at B by the hydraulic cylinder BD and the force exerted on the supporting carriage at A can be calculated using principles of moments and equilibrium of forces, respectively.
For part (a): The force exerted at B by the hydraulic cylinder BD can be calculated using the principle of moments. By summing the moments about point A, you can find the force at B.
For part (b): The force exerted on the supporting carriage at A can be determined by considering.
(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced with tension steel As=5 in2 . The beam supports a uniformly distributed dead load (including its own weight) DL=2.2 kips/ft and a uniformly distributed live load LL=1.8 kips/ft. The properties of the materials are as follows: f’c=4000 psi, steel fy=60,000 psi. Calculate the long-term deflections in the beam after five years.
Answer:
Zx = 176In³
Explanation:
See attached image file