Answer:
r = 1.922 mm
Explanation:
We are given;
Yield stress; σ = 250 MPa = 250 N/mm²
Force; F = 29 KN = 29000 N
Now, formula for yield stress is;
σ = F/A
A = F/σ
Where A is area = πr²
Thus;
r² = 2900/250π
r² = 3.6924
r = √3.6924
r = 1.922 mm
what's nested piezometer?
Explanation:
Nested piezometers indicate an upward flow if the elevation of the top of the water in the piezometer tube that penetrates the aquifer to the deeper point is greater than the elevation of the water in the shallower tube
Air at 25 m/s and 15°C is used to cool a square hot molded plastic plate 0.5 m to a side having a surface temperature of 140°C. To increase the throughput of the production process, it is proposed to cool the plate using an array of slotted nozzles with width and pitch of 4 mm and 56 mm, respectively, and a nozzle-to-plate separation of 40 mm. The air exits the nozzle at a temperature of 15°C and a velocity of 10 m/s.
Required:
a. Determine the improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate.
b. Would the heat rates for both arrangements change significantly if the air velocities were increased by a factor of 2?
c. What is the air mass rate requirement for the slotted nozzle arrangement?
Answer:
a. 2.30
b. decreases with increasing velocity.
c. 0.179 kg/s.
Explanation:
Without mincing let's dive straight into the solution to the question above.
[a].
The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:
The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.
While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.
Array of slot noozle = [10 × (2 × 0.004)]/ 20.92 × 10^-6] = 3824.
where A = 4/56 =0.714.
And Ar = [ 60 + 4 (40/2 × 4) - 2 ]^2 ]-1/2 = 0.1021.
N = 2/3 (0.1021)^3/4 [ 2 × 3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.
h = 24.3 × 0.030/0.004 = 91.1 W/m^2k.
Therefore; 659.6 × 0.030/0.5 = 39.0 W/m²k.
The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.
The slot noozle = 91.1 × 0.5 × 0.5 [ 140 -15] = 2846.87W.
The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.
[b].
2.3 [ (2^2/3)/ 2^4/5] = 2.1
Thus, it decreases with increasing velocity
[c].
The air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.
Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?
a bachelor’s degree then a master’s degree
vocational school certificate or master’s degree
on-the-job training or vocational school certificate
associate’s degree then a bachelor’s degree
Answer:
it is indeed C
Explanation:
Answer:
c
Explanation:
Automotive gas turbines have been under development for decades but have not been commonly used in automobiles. Yet helicopters routinely use gas turbines. Explore why different types of engines are used in these respective applications. Compare selection factors such as performance, power-to-weight ratio, space requirements, fuel availability, and environmental impact.
Required:
Summarize your findings in a report with at least three references.
Answer:
Gas turbines in Helicopters require lesser space.
Explanation:
[1] In terms of Space Requirements:
The gas used in helicopters requires lesser space as compared to Automotive gas turbines. The gas in automobile have higher thermal efficiency.
[2]. In terms of Environmental impact:
The occurrence of environmental solution is very slim when used in helicopters' engines.
[3]. In terms of power-to-weight ratio:
The vibrations in engines of helicopters make it to have lesser efficiency as compared to automobile.
[4]. In terms of Fuel availability:
Fuel is available. Automobile can make use of gas as fuel.
After earning a bachelor's degree, one must do which of the following before taking the PE examination to receive a Professional Engineering license?
i need some help with this
Java programing
def digit_sum(number):
sumOfDigits = 0
while number > 0:
sumOfDigits += number % 10
number = number // 10
return sumOfDigits
or you can use,
In [2]: digit_sum(10)
Out[2]: 1
In [3]: digit_sum(434)
Out[3]: 11
or you can use,
digits = "12345678901234567890"
digit_sum = sum(map(int, digits))
print("The equation is: ", '+'.join(digits))
print("The sum is:", digit_sum)
Have a great day <3
electrical engineering
Answer:
Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.
Explanation:
Derive the next state equations for each type (D, T, SR, and JK) of basic memory element. The next state equation is a symbolic equation describing the next state (Q ) as a function of the inputs (D,T,SR, or JK) and state (Q). In order to determine the next state equations for a a JK memory element, build a 3-variable Kmap with Q, J, and K as the inputs. The entries in the Kmap should be Q . Solving this Kmap will yield the next state equation. Show all work for full credit.
Answer:
Attached below is the derived next state equations
Explanation:
Attached below is the derived next state equations
used for the solution of the given problem.
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod is d = 25 mm and the outside diameter of the steel tube is D= 45 mm. The length of the composite column is L = 761 mm. A force P = 88 kN is applied at the top surface, distributed across both the rod and tube.
Required:
Determine the normal stress σ in the steel tube.
Answer:
Explanation:
From the information given:
[tex]E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN[/tex]
The total load is distributed across both the rod and tube:
[tex]P = P_1+P_2 --- (1)[/tex]
Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.
[tex]\delta_1=\delta_2[/tex]
[tex]\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}[/tex]
[tex]\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}[/tex]
[tex]P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})[/tex]
[tex]P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}[/tex]
[tex]P_2 = 6.6212 \ P_1[/tex]
Replace [tex]P_2[/tex] into equation (1)
[tex]P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN[/tex]
Finally, to determine the normal stress in aluminum rod:
[tex]\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}[/tex]
[tex]\sigma_1 = - 23.523 \ MPa}[/tex]
Thus, the normal stress = 23.523 MPa in compression.
