When framing a wall, temporary bracing is
used to support, plumb, and straighten the wall.
used to support, level, and straighten the wall.
used to square the wall before it is erected.
removed before the next level is constructed.
A long cylindrical black surface fuel rod of diameter 25 mm is shielded by a surface concentric to the rod. The shield has diameter of 50 mm, and its outer surface is exposed to surrounding air at 300 K with a convection heat transfer coefficient of 15 W/m2.K. Inner and outer surfaces of the shield have an emissivity of 0.05, and the gap between the fuel rod and the shield is a vacuum. If the shield maintains a uniform temperature of 335 K, determine the surface temperature of the fuel rod
Answer:
surface temp of fuel rod = 678.85 K
Explanation:
Given data :
D1 = 25 mm
D2 = 50 mm
T2 = 335 k
T∞ = 300 k
hconv = 0.15 w/m^2.k
ε2 = 0.05
ε1 = 1
Determine energy at Q23
Q23 = Qconv + Qrad
attached below is the detailed solution
Insert given values into equation 1 attached below to obtain the surface temperature of the fuel rod ( T1 )
A high-voltage discharge tube is often used to study atomic spectra. The tubes require a large voltage across their terminals to operate. To get the large voltage, a step-up transformer is connected to a line voltage (120 V rms) and is designed to provide 5000 V rms to the discharge tube and to dissipate 75.0 W. (a) What is the ratio of the number of turns in the secondary to the number of turns in the primary
Answer:
a. 41
b. i. 15 mA ii. 625 mA
c. 192 Ω
Explanation:
Here is the complete question
A high-voltage discharge tube is often used to study atomic spectra. The tubes require a large voltage across their terminals to operate. To get the large voltage, a step-up transformer is connected to a line voltage (120 V rms) and is designed to provide 5000 V (rms) to the discharge tube and to dissipate 75.0 W. (a) What is the ratio of the number of turns in the secondary to the number of turns in the primary? (b) What are the rms currents in the primary and secondary coils of the transformer? (c) What is the effective resistance that the 120-V source is subjected to?
Solution
(a) What is the ratio of the number of turns in the secondary to the number of turns in the primary?
For a transformer N₂/N₁ = V₂/V₁
where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V and V₂ = voltage of secondary coil = 5000 V
So, N₂/N₁ = V₂/V₁
N₂/N₁ = 5000 V/120 V = 41.6 ≅ 41 (rounded down because we cannot have a decimal number of turns)
(b) What are the rms currents in the primary and secondary coils of the transformer?
i. The rms current in the secondary
We need to find the current in the secondary from
P = IV where P = power dissipated in secondary coil = 75.0 W, I =rms current in secondary coil and V = rms voltage in secondary coil = 5000 V
P = IV
I = P/V = 75.0 W/5000 V = 15 × 10⁻³ A = 15 mA
ii. The rms current in the primary
Since N₂/N₁ = V₂/V₁ = I₁/I₂
where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V, V₂ = voltage of secondary coil = 5000 V, I₁ = current in primary coil and I₂ = current in secondary coil = 15 mA
So, V₂/V₁ = I₁/I₂
V₂I₂/V₁ = I₁
I₁ = V₂I₂/V₁
= P/V₁
= 75.0 W/120 V
= 0.625 A
= 625 mA
(c) What is the effective resistance that the 120-V source is subjected to?
Using V = IR where V = voltage = 120 V, I = current in primary = 0.625 A and R = resistance of primary coil
R = V/I
= 120 V/0.625 A
= 192 V/A
= 192 Ω
23 PM Sat Apr 10
42%
AA
A f2.app.edmentum.com
+
Previous
5
Next
Post Test: Manufacturing Busii
Submit Test
Reader Tools
0 Info
+ Save & Exit
Garth is a recruitment executive in a firm and knows the eight stages of recruitment. What activity or incident should Garth carry out or
expect to occur at each stage of the process?
place an advertisement in a job portal
vacancyWhat activity should Garth
Here are the eight stages of the recruitment process and what Garth might expect to occur or carry out at each stage in the explanation part.
What is recruitment?The process of identifying, attracting, and selecting qualified candidates for a job opening in an organisation is known as recruitment.
Here are the eight stages of the recruitment process, as well as what Garth might expect to happen or do at each stage:
Identifying the Need for the Position: Garth should review the company's staffing needs and determine if a position needs to be filled. Once a decision has been made, he should create a job description and identify the position's requirements.Garth should create a recruitment plan that includes a timeline for the recruitment process, a list of recruitment sources, and an advertising strategy.Garth should actively seek qualified candidates through various recruitment channels such as job boards, social media, referrals, and recruiting events.Screening Candidates: Garth should go over resumes, cover letters, and other application materials to see if candidates meet the job requirements. Garth should conduct interviews with the most qualified candidates to assess their skills, experience, and fit for the position. Garth should review all of the information gathered during the recruitment process and choose the best candidate for the position. Garth should ensure that the new hire has all of the necessary information and resources to succeed in their new role.Evaluating the Recruitment Process: Garth should go over the recruitment process to see where he can improve.Thus, these are the stages of recruitment.
For more details regarding recruitment, visit:
https://brainly.com/question/30086296
#SPJ7
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is 40 kh at 520 rev/min. The application factor is 1.4. The radial load is 2600 lbf. The reliability goal is 0.90.
Required:
Determine the C10 value in kN for this application and design factor.
Answer:
[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]
Explanation:
From the information given:
Life requirement = 40 kh = 40 [tex]40 \times 10^{3} \ h[/tex]
Speed (N) = 520 rev/min
Reliability goal [tex](R_D)[/tex] = 0.9
Radial load [tex](F_D)[/tex] = 2600 lbf
To find C10 value by using the formula:
[tex]C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}[/tex]
where;
[tex]x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}[/tex]
[tex]\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}[/tex]
The Weibull parameters include:
[tex]x_o = 0.02[/tex]
[tex](\theta - x_o) = 4.439[/tex]
[tex]b= 1.483[/tex]
∴
Using the above formula:
[tex]C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}[/tex]
[tex]C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}[/tex]
[tex]C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}[/tex]
[tex]C_{10} = 30962.449 \ lbf[/tex]
Recall that:
1 kN = 225 lbf
∴
[tex]C_{10} = \dfrac{30962.449}{225}[/tex]
[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]
A hypothetical metal alloy has a grain diameter of 1.7 102 mm. After a heat treatment at 450C for 250 min, the grain diameter has increased to 4.5 102 mm. Compute the time required for a specimen of this same material (i.e., d 0 1.7 102 mm) to achieve a grain diameter of 8.7 102 mm while being heated at 450C. Assume the n grain diameter exponent has a value of 2.1.
Answer:
the required time for the specimen is 1109.4 min
Explanation:
Given that;
diameter of metal alloy d₀ = 1.7 × 10² mm
Temperature of heat treatment T = 450°C = 450 + 273 = 723 K
Time period of heat treatment t = 250 min
Increased grain diameter 4.5 × 10² mm
grain diameter exponent n = 2.1
First we calculate the time independent constant K
dⁿ - d₀ⁿ = Kt
K = (dⁿ - d₀ⁿ) / t
we substitute
K = (( 4.5 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 250
K = (373032.163378 - 48299.511117) / 250
K = 1298.9306 mm²/min
Now, we calculate the time required for the specimen to achieve the given grain diameter ( 8.7 × 10² mm )
dⁿ - d₀ⁿ = Kt
t = (dⁿ - d₀ⁿ) / K
t = (( 8.7 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 1298.9306
t = ( 1489328.26061158 - 48299.511117) / 1298.9306
t = 1441028.74949458 / 1298.9306
t = 1109.4 min
Therefore, the required time for the specimen is 1109.4 min
a) The initial moisture content of a food product is 77% (wet basis), and the critical moisture content is 30% (wet basis). If the constant drying rate in a fluidized bed dryer is 0.1 kg water removed/m2-s, determine the time required for the product to begin the falling-rate drying period. The product has a cube shape with 5-cm sides; the initial product density is 950 kg/m3.
Answer:
≈ 53 seconds
Explanation:
calculate Time required for the product to begin the falling-rate drying period
Initial moisture content = 0.77 kg water /kg of product
= 3.35 kg water /kg solids
Critical moisture content = 0.3 kg water / kg product
= 0.43 kg water / kg solids
∴ amount of water to be removed = 3.35 - 0.43 = 2.95kg water /kg solids
next: calculate surface are a of product during drying
= (0.05 * 0.05 ) * 6
= 0.015 m^3
Drying rate = 0.1 kg water m^2.s^-1 * 0.015 m^3 = 1.5 * 10^-3 kg water s^-1
applying product density
initial product mass = 0.11875 * 0.23 = 0.0273kg solid
hence total amount of water to removed = 2.92 * 0.0273 = 0.07972 kg
therefore : Time required for the product to begin the falling-rate drying period
= 0.07972 / 1.5 * 10^-3
= 53 seconds