An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickness. Knowing that the maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight, determine the largest offset that can be used
Answers
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The largest offset that can be used is [tex]h = 0.455 \ in[/tex]
Explanation:
From the question we are told that
The diameter of the metal tube is [tex]d_m = 0.75 \ in[/tex]
The thickness of the wall is [tex]D = 0.08 \ in[/tex]
Generally the inner diameter is mathematically evaluated as
[tex]d_i = d_m -2D[/tex]
[tex]= 0.75 - 2(0.08)[/tex]
[tex]= 0.59 \ in[/tex]
Generally the tube's cross-sectional area can be evaluated as
[tex]a = \frac{\pi}{4} (d_m^2 - d_i^2)[/tex]
[tex]= \frac{\pi}{4} (0.75^2 - 0.59^2)[/tex]
[tex]= 0.1684 \ in^2[/tex]
Generally the maximum stress of the metal is mathematically evaluated as
[tex]\sigma = \frac{P}{A}[/tex]
[tex]\sigma = \frac{P}{ 0.1684}[/tex]
The diagram showing when the stress is been applied is shown on the second uploaded image
Since the internal forces in the cross section are the same with the force P and the bending couple M then
[tex]M = P * h[/tex]
Where h is the offset
The maximum stress becomes
[tex]\sigma_n = \frac{P}{A} + \frac{M r_m }{I}[/tex]
Where [tex]r_m[/tex] is the radius of the outer diameter which is evaluated as
[tex]r_m = \frac{0.75}{2}[/tex]
[tex]r_m = 0.375 \ in[/tex]
and I is the moment of inertia which is evaluated as
[tex]I = \frac{\pi}{64} (d_m^4 - d_i^4 )[/tex]
[tex]= \frac{\pi}{64}(0.75^4 - 0.59^4)[/tex]
[tex]= 0.009583 \ in^4[/tex]
So the maximum stress becomes
[tex]\sigma' = \frac{P}{0.1684} + \frac{Phr}{0.009583}[/tex]
Now the question made us to understand that the maximum stress when the offset was introduced must not exceed the 4 times the original stress
So
[tex]\sigma ' = 4 \sigma[/tex]
=> [tex]\frac{P}{0.1684} + \frac{Phr_m }{0.009583} = 4 [\frac{P}{0.1684} ][/tex]
The P would cancel out
[tex]\frac{1}{0.1684} + \frac{h(0.375)}{0.009583} = \frac{4}{0.1684}[/tex]
[tex]5.94 + 39.13h = 23.753[/tex]
[tex]39.13h = 17. 813[/tex]
[tex]h = 0.455 \ in[/tex]
Related Questions
A small vehicle is powered by a pulsejet. The available net thrust is 6000 N and the traveling speed is 200 km/hr. The gases leave the engine with an average velocity (Ve) of 360 m/s Assume pressure equilibrium exists at the outlet plane and the fuel to air ratio is 0.06.
a. Compute the mass flow rate required
b. Calculate the inlet area (assume To is 16 °C and Po is one atmosphere)
C. Calculate the thrust power
d. Calculate the propulsive efficiency
Answers
Answer:
a) The mass flow rate is 19.71 kg/s
b) The inlet area is 0.41 m²
c) The thrust power is 333.31 kW
d) The propulsive efficiency is 26.7%
Explanation:
Please look at the solution in the attached Word file.
The arrival rate at a parking lot is 6 veh.min. Vehicles start arriving at 6:00PM and when the queue reaches 36 vehicles, service begins. If company policy is that total vehicle delay should be equal to 500 veh-min, what is the departure rate?
Answers
Answer:
Departure rate = 7.65 vehicle/min
Explanation:
See the attached file for the calculation.
A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calculate the magnitude and direction of the vertical force on this enlargement when 10 f t3/s of water flow upward through the line and the pressure at the smaller end of the enlargement is 30 psi.
Answers
Answer:
F_y = 151319.01N = 15.132 KN
Explanation:
From the linear momentum equation theory, since flow is steady, the y components would be;
-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y
We are given;
Length; L = 5ft = 1.52.
Initial diameter;d1 = 12in = 0.3m
Exit diameter; d2 = 24 in = 0.6m
Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s
Initial pressure;p1 = 30 psi = 206843 pa
Thus,
initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²
Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²
Now, we know that volume flow rate of water is given by; Q = A•V
Thus,
At exit, Q2 = A2•V2
So, 0.28 = 0.28•V2
So,V2 = 1 m/s
When flow is incompressible, we often say that ;
Initial mass flow rate = exit mass flow rate.
Thus,
ρ1 = ρ2 = 1000 kg/m³
Density of water is 1000 kg/m³
And A1•V1 = A2•V2
So, V1 = A2•V2/A1
So, V1 = 0.28 x 1/0.07
V1 = 4 m/s
So, from initial equation of y components;
-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y
Where F_y is vertical force of enlargement pressure and P2 = 0
Thus, making F_y the subject;
F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2
Plugging in the relevant values to get;
F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)
F_y = 151319.01N = 15.132 KN
Air at a pressure of 1 atm and a temperature of 50 °C is in parallel flow over the top surface of a flat plate that is heated to a uniform temperature of 100 °C. The plate has a length of 0.20 m (in the flow direction) and a width of 0.10 m. The Reynolds number based on the plate length is 40,000.
a. What is the rate of heat transfer from the plate to the air?
b. If the free stream velocity of the air is doubled and the pressure is increased to 10 atm, what is the rate of heat transfer?
Answers
It is given that :
Let the mean bulk temperature [tex]$=\frac{50+100}{2}$[/tex]
[tex]$=75^\circ C$[/tex]
From the property table at 1 bar and [tex]$75^\circ C$[/tex],
[tex]$K=0.02917 \ W/\mu K, \ Pr = 0.71055 $[/tex]
Flow is laminar as Re = 4000 for laminar.
Flow Nusselt Number is given by :
[tex]$\overline{Nu} = 0.664 (Re)^{0.5} Pr^{1/3} = \frac{hd}{K}$[/tex]
[tex]$\theta = 4 \times 0.2 \times 0.1 \times (100-50)$[/tex]
[tex]$=17.32$[/tex]
At 10 bar and [tex]$75^\circ C$[/tex],
[tex]$\rho = 9.999 \ kg/m^3 , \ \mu =20.91 \times 10^{-6}$[/tex]
[tex]$K=30.05 \times 10^{-7} \ W/\mu K, \ Pr = 0.7092, \ C_p=1.019 \ kJ/kg K$[/tex]
[tex]$Re_2 = \frac{9.999 \times 2 \times V}{1 \times 20.9 \times 10^{-6}}$[/tex]
Initial, [tex]$Re_i = \frac{1 \times V}{1 \times 20.82 \times 10^{-6}}$[/tex]
[tex]$=40000$[/tex]
[tex]$V=40000 \times 0.2 \times 20.82 \times 10^{-6}$[/tex]
[tex]$Re_2 = \frac{9.999 \times 2 \times 40000}{1 \times 20.9 \times 10^{-6}}$[/tex]
[tex]$Re_2=796477.01$[/tex]
Flow is turbulent.
This Nusselt number is given by :
[tex]$Nu=(0.037)(Re)^{0.8}- 8\pi Pr^{1/3}=958.75$[/tex]
[tex]$h=\frac{958.75 \times k}{0.2}$[/tex]
[tex]$=144.05 \ W /\mu^2C$[/tex]
[tex]$\theta =144.05 \times 0.2 \times 0.1 \times (100.5)$[/tex]
[tex]$=144.05 \ \omega$[/tex]
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The entire population of a given community is examined, and all who are judged to be free from bowel cancer are questioned extensively about their diets. These people then are followed for several years to see whether or not their eating habits will predict their risk of developing bowel cancer.
a. Cross-sectional study
b. Case-control study
c. Prospective cohort study
d. Historical prospective cohort study
e. Clinical trial
f. Community trial
Answers
Answer:
type of study design is Prospective cohort study
Explanation:
This study follows overtime in a group of similar people who differ from some of the factors in the study to determine how factors may affect the outcome rate. Tests may vary depending on the hypothesis given by Cretin patients who are cigarette patients, who are most likely to be smokers, then most likely to be over 20 years of age, with a high rate of lung cancer. The effective cause of the disease is determined by the method of screening of the individuals below.Determine the angle φ at which the applied force P should act on the pipe so that the magnitude of P is as small as possible for pulling the pipe up the incline. What is the corresponding value of P? The pipe weighs W and the slope α is known. Express the answer in terms of the angle of static friction, θ = tan-1 μs.
Answers
Answer:
∅=Ф
P = W sin([tex]\alpha[/tex] + Ф)
Explanation:
First, we'll isolate and draw the free-body diagram of the pipe
Note that since the pipe is moving, the friction force is equal to the product of normal reaction force and the kinetic coefficient of friction
F = F_max = u_kN
Also note that the weight makes with the y-axis angle a because the x-axis makes the same angle with the horizontal
The expression for angle of friction is:
B = tan-1 (u_k)
From here we can express the coefficient of friction as:
u_k = tan(Ф)
Replace u_s by tan(Ф) in the expression for the friction force
F = N tan(Ф)
diagram is attached
By equating sum of forces in y-direction to zero, we can write the expression for the normal reaction force
ΣF_y = 0
N — W cos[tex]\alpha[/tex]- P sin Ф= 0
From here we can express N as:
N = W cos[tex]\alpha[/tex] -— P sin Ф
Replace N by the expression above in the expression for friction force F(written in step 1)
F = (W cos[tex]\alpha[/tex] — P sin Ф) tan( Ф) (1)
Now, we'll equate sum of forces in x-direction to zero
ΣF_x = 0
-F - W cos[tex]\alpha[/tex] + P sin Ф =0
Replace F by expression (1)
— (W cos[tex]\alpha[/tex] — P sin Ф) tan(Ф) — W sin[tex]\alpha[/tex]+pcosФ=0
-W cos [tex]\alpha[/tex] tan(Ф) + P sin Ф tan(Ф) — W sin[tex]\alpha[/tex] +pcosФ=0
P(sin Ф tan(Ф) + cosФ) — W(cos [tex]\alpha[/tex] tan(Ф) + sin [tex]\alpha[/tex])
From here we can express the force P needed to pull the pipe as:
P = W(cos[tex]\alpha[/tex] tan(Ф) + sin[tex]\alpha[/tex])/sinФ*tansФ+cosФ (2)
All we have to do now is to simplify the expression (2). We'll start by sin replacing tan(Ф) with sinФ/cosФ
P = W(cos *sinФ/cosФ + sin)/sinФ*sinФ/cosФ+cosФ *cosФ/cosФ
We can multiply the right side of equation by cosФ/cosФ
P = W(cos[tex]\alpha[/tex] *sinФ + sin[tex]\alpha[/tex]cosФ)/sin∅*sinФ+cos∅cosФ *cosФ/cosФ
Finally, we'll replace (cos[tex]\alpha[/tex] *sinФ + sin[tex]\alpha[/tex]cosФ) by sin([tex]\alpha[/tex] + Ф) and (sin∅ sinФ + cos∅ cos Ф) by cos( ∅— Ф)
P wsin([tex]\alpha[/tex] + Ф) /cos(∅ — Ф) (3)
Since the first derivative of the function is actually tangens of the angle which tangent makes with the x-axis, we'll find it by equating the first derivative by zero(this means that the tangent of the function is horizontal, i.e. that the function is at its maximum or minimum)
Note that the variable in the expression (3) is 0, since both B and a are known
dP/d∅ =d/d∅ [sin(Ф+)/cos(∅-Ф) ]
Note that sin(Ф+[tex]\alpha[/tex]) is constant since both Ф and a are known
dP/d∅ = sin(Ф+[tex]\alpha[/tex]) d/dФ [1/cos(∅-Ф) ]
Next, we'll apply the reciprocal rule
= -dP/d∅[cos(∅-Ф)]/cos^2(∅-Ф)*sin(Ф+[tex]\alpha[/tex])
Next, we'll apply the differentiation rule
=(-sin(∅-Ф))*d/d∅[∅-Ф]*sin(Ф+[tex]\alpha[/tex])/cos^2(∅-Ф)
=(d/d∅[∅]+d/d∅[-∅])*sin(Ф+[tex]\alpha[/tex])sin(∅-Ф)/cos^2(∅-Ф)
dP/d∅ =sin(Ф+[tex]\alpha[/tex])*sin(∅-Ф)/cos^2(∅-Ф) (4)
Next step will be to equate the expression (4) to zero, to determine the value of # when the function is minimum
sin(Ф+[tex]\alpha[/tex])*sin(∅-Ф)/cos^2(∅-Ф) =0
Note that sin(Ф+[tex]\alpha[/tex]) is constant, so in order for the equation above to be correct, sin(∅-Ф) needs to be equal to zero
sin(∅-Ф) = 0
Since sin 0° = sin 180° = 0, two possible solutions for ∅ are:
∅-Ф=0 Ф=∅
or
∅-Ф = 180° ∅ = 180° + Ф
Since the function for P is only good over the range 0 < ∅ < 90°, since when > 90° the friction force will change its direction, we can conclude that the minimum force P is required to move the pipe at angle:
∅=Ф
Finally, replace # by 8 in expression (3) to determine the minimum force P required to move the pipe
P = W sin([tex]\alpha[/tex] + Ф ) / cos ∅ — ∅)
P = W sin([tex]\alpha[/tex] + Ф)
2.) For a 20‐mm‐diameter tube with either water or unused engine oil flowing through it, find: a.)The mean velocity, hydrodynamic entry length, and thermal entry length for each of the fluids with a temperature of 300 K if the mass flow rate is 0.01 kg/s. b.)The mass flow rate, hydrodynamic entry length, and thermal entry length for each fluid at 400 K and a mean velocity of 0.02 m/s.
Answers
Answer:
a.) The mean velocity = 0.0318 m/s
The hydrodynamic entry length = 0.636 m
The thermal entry length = 0.004 m
(b) The mass flow rate = 0.0051 kg/s
The hydrodynamic entry length = 0.028 m
The thermal entry length = 1.419 m
Explanation:
See the attached files for the calculation.
Five hundred gallons of 89-octane gasoline is obtained by mixing 87-octane gasoline with 92-octane gasoline. (a) Write a system of equations in which one equation represents the total amount of final mixture required and the other represents the amounts of 87- and 92-octane gasoline in the final mixture. Let x and y represent the numbers of gallons of 87- and 92-octane gasoline, respectively. (b) Use a graphing utility to graph the two equations in part (a) in the same viewing window. As the amount of 87-octane gasoline increases, how does the amount of 92-octane gasoline change? (c) How much of each type of gasoline is required to obtain the 500 gallons of 89-octane gasoline?
Answers
Explanation:
a) The total volume equals the sum of the volumes.
500 = x + y
The total octane amount equals the sum of the octane amounts.
89(500) = 87x + 92y
44500 = 87x + 92y
b) desmos.com/calculator/ekegkzllqx
As x increases, y decreases.
c) Use substitution or elimination to solve the system of equations.
44500 = 87x + 92(500−x)
44500 = 87x + 46000 − 92x
5x = 1500
x = 300
y = 200
The required volumes are 300 gallons of 87 gasoline and 200 gallons of 92 gasoline.
In your new role at Wayne Industries, you have been given the freedom to propose and develop your own project ideas. You have an idea for a frictionless piston-cylinder assembly. Your design requires the expansion of 0.25 m3 of air at 3.3 MPa and 280 oC to 180 kPa in a slow, isothermal process. In order to be approved for further development, you need to submit calculations for the values of the heat transferred in the process.
Answers
Answer:
461.65 KJ/Kg
Explanation:
In this question, we are asked to calculate the values of heat transferred in the process.
Please check attachment for complete solution and step by step explanation
Neglecting the presence of friction, air drag, and other inefficiencies, how much gasoline is consumed when a 1300 kg automobile accelerates from rest to 80 km/h. Assume the density and enthalpy of gasoline are 680 kg/m3 and 45 MJ/kg respectively. Express your answer in the units of mL. Show all work and use SI units for full credit. Box your final answer(s)
Answers
Answer:
Explanation:
Given that, .
Mass of car is
M = 1300kg
Velocity of car
V = 80km/h = 80 × 1000/3600
V = 22.22m/s
Calculate the kinetic energy of the vehicle as follows:
K.E = ½ MV²
K.E = ½ × 1300 × 22.22²
K.E = 320,987.65 J
Given that,
Enthalpy is 45MJ / kg
h = 45MJ / kg
Then, enthalpy is given as.
Enthalpy = Energy / mass
h = E / m
45 × 10^6 = 320,987.65 / m
m = 320,987.65 / 45 × 10^6
m = 7.133 × 10^-3 kg
m = 7.133 mg
Also, given that, density is 680kg/m³
Density is given as
Density = mass / Volume
ρ = m / v
Then, v = m / ρ
v = 7.133 × 10^-3 / 680
v = 1.049 × 10^-5 m³
We know that
1mL = 10^-6 m³
Therefore,
v = 1.049 × 10^-5 m³ × 1mL / 10^-6m³
v = 10.49 mL
Jasper and Gemma are going to play on a teeter totter. Gemma gets on first. When Jasper gets on, Gemma moves into the air, but she does not move to the top. Which statement could correctly explain the forces acting on the teeter totter? assume that Jasper and Gemma are the same distance from one another.
Answers
Answer:
A) the forces are balanced because Jasper weighs the same as Gemma
Explanation:
Answer:
A. The forces are balanced because Jasper weighs the same as Gemma.
Explanation:
Took the test
You are an engineer at company XYZ, and you are dealing with the need to determine the maximum load you can apply to a set of bolted/clamped plates, such that you get infinite life. a) (1.5 pt) If you have an coarse thread M12 x 1.75 bolt of SAE Class 5.8, what would you recommend as the initial tightening force in the bolt? Is this initial force in the bolt the same as the clamped members?Hint: Use Fi = 0.9AtSp. b) (2 pt) If the clamped members have an effective stiffness twice that of the bolt and an external separating load varies between 0 and 5 kN, what are the alternating and mean forces on the bolt? What are the mean and alternating stresses in the bulk of the bolt? c) (1.5 pt) If the fatigue stress concentration factor is Kf=2.2 for the threads and we account for a yield strength of 400 MPa, what are the ef
Answers
Answer:
a)
[tex]F_i = 28.8 kN[/tex]
b)
Alternating and mean forces on the bolt are 0.85 kN and 29.65 kN respectively.
Mean and alternating stresses in the bulk of the bolt are 378 MPa and 10 MPa.
c)
Safety factor = 5.5
Explanation:
Basic Dimension of Isometric Screw thread" is considered for analysis.
At Nominal diameter, d = 12 mm , [tex]A_t = 84.3mm^{2}[/tex] , [tex]F_i = 0.9A_tS_p[/tex] , [tex]S_p = 380 MPa[/tex]
Rolled threads; [tex]K_f = 2.2[/tex]
Ethanol is produced in a one-liter batch fermenter by Saccharomyces cerevisiae (yeast). To begin the batch process, glucose and ammonia are added to the reactor, which has been inoculated by the yeast. The reaction produces glycerol and cell mass, in addition to ethanol, carbon dioxide, and water according to the (unbalanced) reaction below. Note that that the mass ratio of glycerol produced to glucose consumed is 0.2556. Also note that the mass ratio of H.O produced to NH, consumed is 1.058 Glucose + Ammonia → Glycerol + Ethanol + Yeast Cell Mass + Carbon Dioxide + Water a CH (8) + b NH(g)}p CH 0 (1) + (CHO) +ỊCH K04 Nụ (8) + $ 0 g) + t H (1) Balance the equation and fill out the table below. (1 point cach, 6 total) a- 1 b = 115 g of NH, are added to the reactor. At the end of the batch, 640 g of ethanol has been produced. How much glucose was added (assume complete conversion of the limiting reactant). mlar B 16 points) BONUS QUESTION USING CHAPTER 4 CONCEPTS: The standard heat of combustion (AHC) for the yeast is-22.1 kJ/mol. In order to maintain a constant temperature, how much heat must be added or removed from the reactor? Both the sign and magnitude of your answer will be graded J 3 points)
Answers
Answer:
Check the explanation
Explanation:
The process of Alcoholic fermentation involves the converting a single mole of glucose into two moles of carbon dioxide and two moles of ethanol, and in the process producing two moles of ATP. The total chemical formula for alcoholic fermentation is: C6H12O6 → 2 C2H5OH + 2 CO. Sucrose is a dimer of fructose and glucose molecules.
Kindly check the attached image below to see the full step by step explanation to the question above.
Specifically, the following methods must be implemented in the LinkedList class: (You should utilize listIterator() method already defined in the LinkedList class to obtain its LinkedListIterator object, and use the methods in the LinkedListIterator class to traverse from the first element to the last element of the linked list to define the following methods.)
Answers
Answer:
Attached to this solution is a Seventeen pages of code. Cheers!
Explanation:
Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a cold reservoir at 250 K and rejects 950 KJ to a hot reservoir at 300 K while receiving 70 kJ of work to operate. Draw a schematic of the cycle clearly indicating the hot and cold reservoir and the direction of heat and work transfers. (10 pts.)
Answers
Answer:
Impossible.
Explanation:
The ideal Coefficient of Performance is:
[tex]COP_{i} = \frac{250\,K}{300\,K-250\,K}[/tex]
[tex]COP_{i} = 5[/tex]
The real Coefficient of Performance is:
[tex]COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}[/tex]
[tex]COP_{r} = 12.571[/tex]
Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.
A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of 24 kpsi. After some calculation, the designer has selected a standard 1/2 in diameter (d) bar. The factor of safety (n) is ____. Round the answers to three significant digits.
Answers
Answer:
[tex]n = 2.36[/tex]
Explanation:
The stress experimented by the circular bar is:
[tex]\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)[/tex]
[tex]\sigma = 10.186\,kpsi[/tex]
The safety factor is:
[tex]n = \frac{24\,kpsi}{10.186\,kpsi}[/tex]
[tex]n = 2.36[/tex]
The ice on the rear window of an automobile is defrosted by attaching a thin, transparent, film type heating element to its inner surface. By electrically heating this element, a uniform heat flux may be established at the inner surface. The inside air temperature and convection heat transfer coefficient are T,i = 25°C and hi = 25 W/(m2 -K), while the outside air temperature is T,o = -10°C. For 4-mm thick window glass, determine (a) the electric power required per unit window area to maintain an inner surface temperature of 15°C and (b) the thermal conductivity of the windshield.
Answers
Answer:
A)Q = 1208.33 W/m²
B)K = 0.138 W/m.K
Explanation:
We are given;
inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K
outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K
Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K
Thickness, L = 4mm = 0.004m
convection heat transfer coefficient ; hi = 25 W/(m².K)
A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;
Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]
Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;
Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]
Q = 583.33 + 625
Q = 1208.33 W/m²
B) The formula for thermal conductivity is;
K = (QL)/(AΔT)
Where;
K is the thermal conductivity in W/m.K
Q is the amount of heat transferred through the material
L is the distance between the two isothermal planes
A is the area of the surface in square meters
ΔT is the difference in temperature in Kelvin
ΔT = 298K - 263K = 35K
Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;
k = (Q/A) x (L/ΔT)
K = 1208.33 x (0.004/35)
K = 0.138 W/m.K
(Using Python)Part 2aNumerology is the "study of the purported mystical or special relationship between a number and observed or perceived events." It has been used throughout human history as a way to attach meaning to a name, object or event using mathematics. It is considered a "pseudoscience" by modern scientists since it has no basis in observable phenomena. With that said, it makes a great programming challenge so we're going to go with it! :)What you want to do for this project is to ask the user to type in their name. Next, you will need to use a technique called "theosophical reduction" to convert their name into a number. With this technique we assign each letter of the alphabet its own number. For example, the letter "a" is equal to the number 1. "b" = 2, "c" = 3, "z" = 26, etc. You should ignore non-alphabetic characters (i.e. numbers, spaces and special characters)Once you've gotten all of the letters converted into numbers you can add them up into one single number. This is the "numerology number" for the name that the user entered.So for the name "craig" the numerology number would be:c = 3r = 18a = 1i = 9g = 73 + 18 + 1 + 9 + 7 = 38Here's are a few sample runnings of this program:Name: craigYour 'cleaned up' name is: craigReduction: 38Name: craig kappYour 'cleaned up' name is: craigkappReduction: 82Name: rumple stil skinYour 'cleaned up' name is: rumplestilskinReduction: 198Name: !rumple!stil!skinYour 'cleaned up' name is: rumplestilskinReduction: 198Name: pikachu!pikapika!Your 'cleaned up' name is: pikachupikapikaReduction: 143Name: PIKACHUpikapikaYour 'cleaned up' name is: pikachupikapikaReduction: 143Some hints:Convert the user's name to all uppercase or all lowercase before you do anything elseRemove any spaces, numbers or special characters from the name to ensure that you are only working with the letters A-ZThe ord() function may be userful to convert each character into an ASCII index
Answers
Answer:
See explaination for python programming code
Explanation:
Python programming code below
import re
s = "abc" # enter string here
#s = "hello world! HELLOW INDIA how are you? 01234"
# Short version
print filter(lambda c: c.isalpha(), s)
# Faster version for long ASCII strings:
id_tab = "".join(map(chr, xrange(256)))
tostrip = "".join(c for c in id_tab if c.isalpha())
print s.translate(id_tab, tostrip)
# Using regular expressions
s1 = re.sub("[^A-Za-z]", "", s)
s2 = s1.lower()
print s2
import string
values = dict()
for index, letter in enumerate(string.ascii_lowercase):
values[letter] = index + 1
sum = 0
for ch2 in s2:
for ch1 in values:
if(ch2 == ch1):
sum = sum + values[ch1]
print sum
Water flows through a Xylan tube at 300 K temperature and 0.5 kg/s flow rate. The inner and outer radii of the Xylan tube is 20 and 30 mm, respectively. A thin electrical heating tape wrapped around the outer surface of the Xylan tube delivers a uniform surface heat flux of 1500 W/m², while a convection coefficient of 20 W/ m K is maintained on the outer surface of the tape by ambient air at 310 K. (a) What is the outer surface temperature of the Xylan tube? (b) What is the fraction of the power dissipated by the tape, which is transferred to the water? Please draw the thermal circuit. Assume that the thermal properties of water at 300 K are as follows: u = 1100x10kg/s.m; k = 0.555 W/m.K; Pr = 7.45. Thermal conductivity of Xylan tube is k = 0.25 W/ mK.
Answers
Answer:
Find the attachments for the complete solution
Note
In the above Question heat flux is balanced with ambient heat loss and tube heating of water.
In Tube, a hollow cylinder heat flow equation has been used.
Two forces P can be applied separately or at the same time to a plate that is welded to a solid circular bar of radius r. Determine the largest compressive stress in the circular bar, (a) when both forces are applied, (b) when only one of the forces is applied.
Answers
This answer discusses the calculation of the largest compressive stress in a circular bar when two forces are applied either together or individually, highlighting the difference in stress levels due to the application of forces.
Explanation:The question involves determining the largest compressive stress in a circular bar when two forces, P, are applied either simultaneously or separately. This problem is grounded in the principles of mechanics of materials, particularly in analyzing stress and strain in materials under tension or compression.
For part (a), when both forces are applied, the compressive stress would likely be higher, as the forces cumulatively contribute to the stress. The compressive stress (σ) in a circular bar can be calculated using the formula σ = P/A, where P is the total force applied, and A is the cross-sectional area of the bar. In the case of two forces applied simultaneously, P would be the summation of both forces.
For part (b), when only one of the forces is applied, the compressive stress in the bar would be lower, as it comes from a single source. The calculation still follows σ = P/A, but with P representing only one of the forces.
- if `check_1` and `check_2` variables are both True, it should set the value of a variable `outcome` to the string 'BOTH' - elif `check_1` is True and `check_2` is False, it should set the value of a variable `outcome` to the string 'ONE' - elif `check_1` is False and `check_2` is True, it should set the value of a variable `outcome` to the string 'TWO' - else (meaning both must be False), it should set the value of a variable `outcome` to the string 'NEITHER'
Answers
Answer:
See Explaination
Explanation:
if(check1 and check2):
outcome = "BOTH"
elif(check1):
outcome = "ONE"
elif(check2):
outcome = "TWO"
else:
outcome = "NEITHER"
To save steel-handling costs, an alternative design is proposed for the beam in Problem 1 using two No. 9 Grade 75 bars to provide approximately the same steel strength as the originally proposed four No. 7
Grade 60 bars. Check to determine if the redesigned beam is satisfactory with respect to cracking according to the ACI Code. What modification could you suggest that would minimize the number of bars to reduce cost, yet satisfy requirements of crack control?
Answers
Answer:
See the explanation for the answer
Explanation:
Check to determine the redesigned beam is satisfactory for cracking:
crack width is controlled by establishing a minimum spacing.
Steps followed to check for cracking in the beams
i) According to aci code 10.6.7 if depth of beam isgreater than 36 in then skin reinforcement has to provide.the skin reinforcement to be provided should be such that it should not be greater than the actual main tension reinforcement.
ii) In second step steel stress is determined:
steel stressfs=Ms/(As*(d-hf/2)) or fs= 0.60fy
where Ms=service load moment
As*(d-hf/2)=area of reinforcement *moment area
iii) S=540/fs-2.5Cc is less than or equal to 12*(36/fs)
here Cc = clear spacing
if S=center to center spacing is with in the limit as specified above then the cracking is with in the control if not then redesign has to done.
In the given problem the data given is :
grade of steel in desigened beam is 75 and in redesigned beam is 60 so the stress in steel is 75*0.6=45ksi and 0.6*60=36ksi respectively
now the spacing is calculated for the two design and redesigned beams
the center to center spacing is given by S=540/fs-2.5Cc
For designed beam
S=540/45-(2.5*2.25)=6.375in which is less than 12*36/fs=12*36/45=9.6in hence it is safe
For redesigned beam
S=540/36-(2.5*2.25)=9.375in and it is less than the maximum spacing which is given by 12*36/fs=12*36/36=12in
Hence, the beam is within the limits and the beam is safe against the cracking.
Modifications to reduce the number of reinforcing bars
The addition of steel does not prevent cracking due to restrained shrinkage but it limits the width of crack by causing the formation of the number of narrow cracks rather than single wide crack.
Larger size bars leads to fewer cracks but wider cracks while smaller size bars leads to number of narrow cracks hence it is advisable to provide number of smaller diameter bars of equal strength of designed bars rather than larger bars.
Two concentric helical compression springs made of steel and having the same length when loaded and when unloaded are used to support a static load of 3 kN. The outer spring has D = 50 mm, d = 9 mm, and N = 5; the inner spring D = 30 mm, d = 5 mm, and N = 10. Determine the deflection and the maximum stress in each spring.
Answers
Answer:
see explaination for all the answers and full working.
Explanation:
deflection=8P*DN/Gd^4
G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2
for outer spring,
deflection=8*3*50^3*5/(70*9^4)=32.66mm
for inner spring
deflection=8*3*30^3*10/(70*5^4)=148.11mm
max stress=k*8*P*C/(3.14*d^2)
for outer spring
c=50/9=5.55
k=(4c-1/4c-4)+.615/c=1.2768
max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2
for inner spring
c=6
k=1.2525
max stress=2.29KN/mm^2
Explanation:
Data
Load = 3kn = 3000N
Modulus of rigidity = 80Gpa= 80000mpa
Outer spring diameter = 50mm
d. = 9mm
N = 5
Inner spring diameter = 30mm
d = 5mm
N = 10
Fo = outer force
Fi = inner force
Ki = stiffness of inner spring
Ko = stiffness of outer spring
Ks = stress factor
The Atbash Cipher encrypts messages by reversing lowercase letters, so ‘a’ becomes ‘z’, ‘b’ becomes ‘y’, ‘c’ becomes ‘x’, etc... Also, any space or punctuation mark gets repeated. For example, hello human! encrypts to svool sfnzm!! Encrypt msg and save the answer to a variable called encrypted (you don’t have to display anything). Note: msg will only have lowercase letters, punctuation and spaces. msg = input('Enter secret message: ', 's');
Answers
Answer:
See Explanation Below
Explanation:
// Program is written in C++ Programming Language
// Comments are used for explanatory purpose
// Program starts here
#include<iostream>
#include <bits/stdc++.h>
using namespace std;
int main()
{
// Declare 2 string variables to store the secret message and to store the encrypted text
string message, result;
// Prompt user to enter a secret message
cout<<"Enter a secret message: ";
cin>message;
// Convert the input string to char array
int n = message.length();
char char_array[n + 1];
strcpy(char_array, message.c_str());
// Initialise result
result = "";
// Declare an array of all possible alphabets a-z
char possible[26] = { 'a','b','c','d','e','f','g,','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v',w','x','y','z'};
// Generate output string
// Start by getting string position
int count = 0;
while(count<n)
{
// If current character is blank or !
if(char_array[count] = '!' || char_array[count] = ' ')
{
result+=char_array[count];
}
else
{
for(int I = 0; I<26; I++)
{
if(char_array[count] = possible[I])
{
result+=possible[25-I];
}
}
}
count++;
}
// No output required; the program stops here
return 0;
}
// End of program
Consider airflow over a plate surface maintained at a temperature of 220°C. The temperature profile of the airflow is given as fluid ()()expsV TyTTTy The airflow at 1 atm has a free stream velocity and temperature of 0.08 m/s and 20°C, respectively. Determine the heat flux on the plate surface and the convection heat transfer coefficient of the airflow.
Answers
Consider airflow over a plate surface maintained at a temperature of 220°C. The temperature profile of the airflow is given as T(y) = T∞ − (T∞−Ts)exp((−V/α_fluid)y).
The airflow at 1 atm has a free stream velocity and temperature of 0.08 m/s and 20°C, respectively. Determine the heat flux on the plate surface and the convection heat transfer coefficient of the airflow
Answer:
A) heat flux on the plate is;q_o = 11737.34 W/m²
B) convection heat transfer coefficient of the airflow is;h = 58.67 W/m².k
Explanation:
The temperature profile of the airflow is given as;
T(y) = T∞ − (T∞−Ts)exp((−V/α_fluid)y)
Let's differentiate with respect to y;
dT/dy = [[(T∞−Ts)V]/α](e^(-vy/α)
Where;
T∞ = 20°C
Ts = 220°C
V = 0.08 m/s
α is thermal diffusivity of air and from the table i attached at a temperature of 220°C, by interpolation it has a value of;
α = 5.33 x 10^(-5) m²/s
Thus, at y =0;
dT/dy = [[(20 − 220)0.08]/(5.33 x 10^(-5))](e^(0))
dT/dy = -300187.62 °C/m
A) Now, heat flux at y = 0 would be given by;
q_o = -k(dT/dy)
Where k is thermal conductivity
from the table attached at 220°C and by interpolation, the thermal conductivity k = 0.0391 W/m.k
Thus,
q_o = -0.0391(-300187.62)
q_o = 11737.34 W/m²
B) the convection heat transfer coefficient of the airflow is gotten from;
q_o = h(Ts - T∞).
Where h is the convection heat transfer coefficient of the airflow
Thus making h the formula, we have;
h = q_o/(Ts - T∞)
h = 11737.34/(220 - 20)
h = 58.67 W/m².k
The y and z keys swapping position is messing with your touch typing. You decide to write out your email as if the keys were in the correct position and then use Python to swap all ys and zs. Your task here is to write a function called fix_yz. This function takes a single argument which will be a string. Your function needs to return this string with all of the ys and zs swapped, and all of the Ys and Zs swapped. Here are some example calls to your function:
s = fix_yz('What did zou saz?')print(s)What did you say?s = fix_yz('Zour tip about the yoo was a great one!')print(s)Your tip about the zoo was a great one!s = fix_yz('We onlz have one week left')print(s)We only have one week left :(HintThe auto-marker is expecting you to submit only your fix_yz function definition. You should not include any calls to your function.
Answers
Answer:
# the function fix_yz is defined
# it takes a string as parameter
def fix_yz(word):
# new_word is to hold the new corrected string
new_word = ""
# loop through the string
# and check for any instance of y or z.
# if any instance is found, it is replaced accordingly
for each_letter in word:
if each_letter == 'z':
new_word += 'y'
elif each_letter == 'Z':
new_word += 'Y'
elif each_letter == 'y':
new_word += 'z'
elif each_letter == 'Y':
new_word += 'Z'
else:
new_word += each_letter
# the value of new string is returned
return new_word
Explanation:
The function is written in Python 3 and it is well commented. An image is attached showing the output of the given example.
The function take a string as input. It then loop through the string and check for any instance of 'y' or 'z'; if any instance is found it is swapped accordingly and then append to the new_word.
The value of bew_word is returned after the loop.
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. The differential height between the water columns connected to the two outlets of the probe is 0.126 m.Take the density of water to be 1000 kg/m3. The gas constant of air is R = 0.287 kPa-m3/kg-K.The air temperature and pressure in the duct are 352 K and 98 kPa, respectively.
Answers
Answer:
Flow velocity
50.48m/s
Pressure change at probe tip
1236.06Pa
Explanation:
Question is incomplete
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively
solution
In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe
please check attachment for complete solution and step by step explanation
You are provided the following information about a municipal wastewater treatment plant. This plant uses the traditional activated-sludge process.
Assume the microorganisms are 55 percent efficient at converting food to biomass, the organisms have a first-order death rate constant of 0.05/day, and the microbes reach half of their maximum growth rate when the BOD5 concentration is 10 mg/L. There are 150,000 people in the community (their wastewater production is 225 L/day-capita, 0.1 kg BOD5/capita-day). The effluent standard is BOD5 = 20 mg/L and TSS = 20 mg/L.
Suspended solids were measured as 4,300 mg/L in a wastewater sample obtained from the biological reactor, 15,000 mg/L in the secondary sludge, 200 mg/L in the plant influent, and 100 mg/L in the primary clarifier effluent. SRT is equal to 4 days.
(a) What is the design volume of the aeration basin (m3)?
(b) What is the plant
Answers
Answer:
Explanation:
Attached is the solution
A powerplant is emitting 80 g/s NO and has an effective stack height of 100 m. The windspeed is measured to be 4 m/s at a height of 10 m. It is a clear summer day and the sun is located directly overhead. Find: a. Ground-level NO concentration 2 km directly downwind of the powerplant. b. Maximum NO concentration at the ground-level. c. Ground-level NO concentration 2 km downwind and 0.1 km off the downwind axis.
Answers
Answer:
Explanation:
The step by step solution is in the attached file.
Utilizing the charge neutrality equation and the mass-action law derive an expression for the concentration of electrons and holes present inside a p-type compensated semiconductor material. How can the resulting expression be modified for practical cases where (????????AA − ????????DD) ≫ 2nn�
Answers
Answer:
The expression for the concentration of electrons is P = NA - ND
Explanation:
Please look at the solution in the attached Word file