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.
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:
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:
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
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.
In a metal-oxide-semiconductor (MOS) device, a thin layer of SiO2 (density = 2.20 Mg/m3) is grown on a single crystal chip of silicon. How many Si atoms and how many O atoms are present per square millimeter of the oxide layer? Assume that the layer thickness is 160 nm.
Answer:
3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O
Explanation:
It is all about unit conversions.
The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.
160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm
By the way, now we can determine the volume of MOS, in order to work with density.
1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³
But density is mg/m³, so we convert mm³ to m³
1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³
Now, we apply density to determine the mass of MOS
Density = mass /volume → Density . volume = mass
1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg
To make more easier the calculate, we convert mg to g.
3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g
To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol
We need to know moles of Si and O₂ in the MOS
Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles
1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:
5.86×10⁻¹⁸ mol of SiO₂ may have:
(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si
(5.86×10⁻¹⁸ .2) /1 = 1.17×10⁻¹⁷ moles of O₂
Let's count the atoms (1 mol of anything contain NA particles)
5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si
1.17×10⁻¹⁷ mol of O₂ . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O
The number of Si and O atoms present per mm² of the oxide layer are respectively; 3.52 × 10⁶ atoms of Si and 7.05×10⁶ atoms of O
What is the number of atoms present?We are given;
Area of MOS device; A = 1 mm²
Thickness of layer; t = 160 nm = 1.6 × 10⁻⁴ mm
Formula for volume is;
V = Area * thickness
V = 1.6 × 10⁻⁴ mm × 1 mm² = 1.6 × 10⁻⁴ mm³
Converting volume to m³ gives;
V = 1.6 × 10⁻¹³ m³
Now, to get the mass of MOS, we will use the formula;
Mass = Density * volume
we are given density = 2.20mg/m³
Thus;
Mass = 1.6 × 10⁻¹³ m³ × 2.20mg/m³
Mass = 3.52 × 10⁻¹³ mg = 3.52×10⁻¹⁶ g
From periodic table, the molar mass of SiO₂ = 60.08 g/mol
Then;
Number of moles of MOS = (3.52 × 10⁻¹⁶ g)/60.08 g/mol
Number of moles of MOS = 5.86 × 10⁻¹⁸ moles
Now, 1 mol of SiO₂ is formed from 1 mol of Si and 2 mol of O. Thus;
5.86×10⁻¹⁸ mol of SiO₂ will have;
5.86×10⁻¹⁸ moles of Si
and (5.86×10⁻¹⁸ .2) = 11.72 × 10⁻¹⁸ moles of O₂
From Avogadro's number that 1 mol equals 6.02×10²³ atoms , we can say that;
5.86 × 10⁻¹⁸ mol of Si × 6.02 × 10²³ atoms/mol = 3.52 × 10⁶ atoms of Si
11.72 × 10⁻¹⁸ mol of O₂ × 6.02 × 10²³ atoms/mol = 7.05×10⁶ atoms of O
Read more about number of atoms at; https://brainly.com/question/3157958
