Answer:
[tex]T_{2}[/tex] = -9.24 °C
x = 0.1057
Explanation:
The tables used in this answer and explanation come from Fundamentals of Engineering Thermodynamics 9th Edition.
Using Table A-14: Properties of Saturated Ammonia (Liquid-Vapor): Pressure Table and the given [tex]P_{2}[/tex], [tex]T_{2}[/tex] can be determined by finding the temperature that corresponds with [tex]P_{2}[/tex] on the table. In this case, [tex]T_{2}[/tex] = -9.24 °C.
The quality of the refrigerant can be determined by using data from the same table and [tex]h_{2} =274.26[/tex] kJ/kg.
Necessary data (P=3bar):
[tex]h_{f}=137.42[/tex] kJ/kg
[tex]h_{g}=1431.47[/tex] kJ/kg
The formula to calculate quality is [tex]h_{2} =h_{f}+x(h_{g}-h_{f})[/tex].
Rearranging for x:
[tex]x=\frac{h_{2}-h_{f} }{h_{g}-h_{f} }= \frac{274.26-137.42}{1431.47-137.42}=0.1057[/tex]
Explain the process of energy conversion by describing how energy was converted from the windmill design brief. Discuss the different forms of energy and what technology was used to convert the energy from one form to another.
Answer:
Wind energy is converted to Mechanical energy which is then converted in to electrical energy
Explanation:
In a wind mill the following energy conversions take place
a) Wind energy is converted into Mechanical energy (rotation of rotor blades)
b) Mechanical energy is converted into electrical energy (by using electric motor)
This electrical energy is then used for transmission through electric lines.
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]
Two technicians are explaining what exhaust gas emissions tell you about engine operation. Technician A says that the higher the level of CO2 in the exhaust stream, the more efficiently the engine is operating. Technician B says that CO2 levels of 20 to 25 percent are considered acceptable. Who is correct?
A. Both Technicians A and B
B. Neither Technicians A and B
C. Technician A
D. Technician B
Technicians A is correct in the given scenario. The correct option is C.
What is exhaust gas?Exhaust gas is a byproduct of combustion that exits the tailpipe of an internal combustion engine.
It consists of a gas mixture that includes carbon dioxide (CO2), carbon monoxide (CO), nitrogen oxides (NOx), hydrocarbons (HC), and particulate matter (PM).
Technician B is mistaken. CO2 levels in the exhaust should be less than 15%, preferably between 13% and 14.5% for petrol engines and 11% to 13% for diesel engines.
High CO2 levels can actually indicate inefficient engine operation, as it means that not all of the fuel in the engine is being burned and is being wasted as exhaust.
Thus, C is the correct answer. A technician is correct.
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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 )
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Answer:
what?????
Explanation:
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.
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Which one of the following torque is produced by the spring in PMMC instrument?
O a. Damping
O b. Forcing
OC. Deflection
O d. Controlling
Answer:
A
Explanation:
Actually I don't know anything about American history, I chose it because South Africa is not in the least
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
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
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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.
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An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE
Answer:
true
Explanation:
true An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE
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 pump draws water from a reservoir through a 150 mm diameter suction pipe and delivers it to a 75 mm diameter discharge pipe.
a. The end of the pipe is 2 m below the free surface of the reservoir.
b. The pressure gage on the discharge pipe (located 2 m above the reservoir surface) reads 170 kPa.
c. The average velocity in the discharge pipe is 3 m/s.
Assume no head loss in the suction or discharge pipe. Determine the power added to the fluid by the pump.
Answer:
3.42 kW
Explanation:
calculate the mass flow rate of the water = density * velocity * area
= 13.3 kg/s
where Area of pipe = π/4 * d^2 = π/4 * 0.075^2 m = 0.0044 m^2
given that : 1 Kg = 1 N s^2 / m, 1 Watt = 1 N m /s
calculate : power produced by discharge pipe
Discharge Pressure x Volume flow =
Pressure * Area * velocity
P * 0.0044 * 3 = 2253 N m /s
calculate: power due the mass of water
mass of water * 2 = 258 N m/s
where: mass of water = density * volume
Calculate power produced due the velocity of exiting water
= m * V2/2 = 59.4 N m/s
hence The power added to fluid = 2253 watts
power added to the fluid by pump = 2253 / 0.75 = 3.42 kW