Answer:
11.94 V
Explanation:
Generally the regulated voltage drops as load increases. When the voltage has dropped by 0.5%, it will be 60 mV less than the nominal value:
12.0 V - 0.06 V = 11.94 V . . . . full load voltage
You have accumulated several parking tickets while at school, but you are graduating later in the year and plan to return to your home in another jurisdiction. A friend tells you that the authorities in your home jurisdiction will never find out about the tickets when you re-register your car and apply for a new license. What should you do?
Answer:
pay off the parking tickets
Explanation:
In the scenario being described, the best thing to do would be to pay off the parking tickets. The parking tickets stay under your name, and if they are not paid in time can cause problems down the road. For starters, if they are not paid in time the amount will increase largely which will be harder to pay. If that increased amount is also not paid, then the government will suspend your licence indefinitely which can later lead to higher insurance rates.
An air-conditioner which uses R-134a operates on the ideal vapor compression refrigeration cycle with a given compressor efficiency.
--Given Values--
Evaporator Temperature: T1 (C) = 9
Condenser Temperature: T3 (C) = 39
Mass flow rate of refrigerant: mdot (kg/s) = 0.027
Compressor Efficiency: nc (%) = 90
a) Determine the specific enthalpy (kJ/kg) at the compressor inlet.
Your Answer =
b) Determine the specific entropy (kJ/kg-K) at the compressor inlet
Your Answer =
c) Determine the specific enthalpy (kJ/kg) at the compressor exit
Your Answer =
d) Determine the specific enthalpy (kJ/kg) at the condenser exit.
Your Answer =
e) Determine the specific enthalpy (kJ/kg) at the evaporator inlet.
Your Answer =
f) Determine the coefficient of performance for the system.
Your Answer =
g) Determine the cooling capacity (kW) of the system.
Your Answer =
h) Determine the power input (kW)to the compressor.
Your Answer =
Answer:
A) 251.8 kj/kg
B) 0.9150 kj/kg-k
C) 155.4 kj/kg
F) 1.50
G) 3.95 kw
H) 2.6 kw
Explanation:
Given conditions :
air conditioner : R -134a
compressor efficiency (nc) = 90%.
T1 = 9⁰c, T3 = 39⁰c, mass flow rate = 0.027 kg/s
A) Specific enthalpy at the compressor inlet
at T = 9⁰c the saturated vapor (x) = 1
from the R-134a property table
h1 = 251.8 kj/kg
B ) specific entropy ( kj/kg-k) at the compressor inlet
at T = 9⁰c the saturated vapor (x) = 1
s = 0.9150 kj/kg-k ( from the R-134a property table )
C) specific enthalpy at the compressor exit
at T3 = 39⁰c , s2 = s1
has = 165.12 kj/kg
h2 = 155.4 kj/kg
attached below is the remaining solution to some of the problems
A two-lane, one-way ramp from an urban expressway with a design speed of 30 mi/h connects with a local road at a T-intersection. The turning roadway has a vertical curb on both sides. Determine the width of the turning roadway if the predominant turning vehicles are single unit trucks with some semi-trailers. Use 0.08 for super-elevation if applicable.
Answer:
30 feet
Explanation:
Given data :
design speed = 30 miles/h
super elevation = 0.08
determine the width of the turning roadway
calculate the value of R = V^2 / 15( e + p)
e = 0.08 , p = 0.2 , v = 30
R = (30)^2 / 15 ( 0.08 + 0.2 )
= 900 / 15 ( 0.28 )
≈ 215 ft
pavement width from the calculation above = 28 ft
width of the turning roadway = pavement width + 2 = 30 feet ( because there are two vertical widths joining up the main road at the T junction )
Write about traditional brick production in Pakistan
Answer:
Clay bricks are manufactured by mining and clay moulded blocks. There are 20,000 brick klins in Pakistan.
Explanation:
In Pakistan, the clay bricks are manufactured by mining and baking the clay moulded blocks in brick kilns. According to an estimate, the baking process emits about 1.4 pounds of carbon per brick made, but in Pakistan, because the systems are outdated, brick kilns are used, which is producing more than the average amount of the pollution.
There are about 20,000 brick kilns in Pakistan. The traditional brick production in Pakistan is consists of hand-made bricks which are first baked in Fixed Chimney Bull's Trench Kilns (FCBTK), this is the most widely used brick firing technology in South Asia.
There are some sections of the SDS that are not mandatory
True or false
Answer:
True
Explanation:
There are 16 in all and 4 are not mandatory
There are some sections of the SDS that are not mandatory. Thus, option A is correct.
What is SDS?A Safety Data Sheet, previously known as a Material Safety Data Sheet, is a thorough instructional report written by a dangerous product's producer or exporter. It gives a description of something like the product's chemical and physical characteristics.
Sections 1 through 11 and 16 must be used; sections 12 through 15 are optional but could be included. the section which will are not compulsory are:
Section 12: Environmental data
Section 13: Guidelines for Recycling
Section 14: Documentation regarding mobility
Section 15: Relevant data about regulations
In Safety Data Sheet, all the sections are not compulsory or mandatory. this statement is true. Therefore, option A is the correct option.
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Consider the following lines of Python:
>>> x = input()
>? 12
>>> x + “abc”
What is the output?
Answer:
I don't know
............................
The output will be an error message since the input() function in Python always takes input as a string, even if the entered value is an integer like 12.
Here,
It cannot directly concatenate a string with an integer.
In order to fix this, we need to convert the input string into an integer using the int() function,
```
x = int(input())
x = x + "abc" # this will give an error message
```If we fix the code, the correct output will be:
`>>> x = int(input())`
`12`
`>>> x + "abc"`
`TypeError: unsupported operand type(s) for +: 'int' and 'str'`
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In a series motor, the field electromagnet consists of A) a winding connected in parallel with the armature. B) a winding connected in parallel with the armature and a second winding connected in series with the armature. C) a winding connected in series with the armature. D) a winding connected in series with a separate dc power source.
Answer:
C) a winding connected in series with the armature.
Explanation:
In a series motor, an electromagnet is used as a stator to generate its magnetic field. The field coil of this stator are connected through a commutator in series with the rotor windings. This stator which is the armature windings will conduct AC even on a DC machine, due to the periodically reverses current direction (commutation) or due to electronic commutation (as in brushless DC motors).
Q1) Determine the force in each member of the
truss and state if the members are in tension or
compression.
Set P1 = 10 kN, P2=15 KN
Answer:
CD = DE = DF = 0BC = CE = 15 N tensionFA = 15 N compressionCF = 15√2 N compressionBF = 25 N tensionBG = 55/2 N tensionAB = (25√5)/2 N compressionExplanation:
The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.
At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.
Then link CE has 15 N tension.
The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.
In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.
At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.
At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.
The horizontal forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.
In summary, the link forces are ...
(25√5)/2 N compression in AB15 N tension in BC25 N tension in BF0 N in CD, DE, and EF15 N tension in CE15√2 compression in CF15 N compression in FA_____
Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.
9. A box contains (4) red balls, and (7) white balls ,we draw( two) balls with return , find 1. Show the sample space & n(s) ..... 2. Probability of all results that appeared in the sample space..
Answer:
The answers to your questions are given below.
Explanation:
The following data were obtained from the question:
Red (R) = 4
White (W) = 7
1. Determination of the sample space, S.
The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:
S = {R, R, R, R, W, W, W, W, W, W, W}
nS = 11
2. Determination of the probability of all results that appeared in the sample space.
From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:
i. Probability that the first draw is red and the second is also red.
P(R1) = nR/nS
Red (R) = 4
Space space (S) = 11
P(R1) = nR/nS
P(R1) = 4/11
P(R2) = nR/nS
P(R2) = 4/11
P(R1R2) = P(R1) x P(R2)
P(R1R2) = 4/11 x 4/11
P(R1R2) = 16/121
Therefore, the Probability that the first draw is red and the second is also red is 16/121.
ii. Probability that the first draw is red and the second is white.
Red (R) = 4
White (W) = 7
Space space (S) = 11
P(R) = nR/nS
P(R) = 4/11
P(W) = nW/nS
P(W) = 7/11
P(RW) = P(R) x P(W)
P(RW) = 4/11 x 7/11
P(RW) = 28/121
Therefore, the probability that the first draw is red and the second is white is 28/121.
iii. Probability that the first draw is white and the second is also white.
White (W) = 7
Space space (S) = 11
P(W1) = nW/nS
P(W1) = 7/11
P(W2) = nW/n/S
P(W2) = 7/11
P(W1W2) = P(W1) x P(W2)
P(W1W2) = 7/11 x 7/11
P(W1W2) = 49/121
Therefore, the probability that the first draw is white and the second is also white is 49/121.
iv. Probability that the first draw is white and the second is red.
Red (R) = 4
White (W) = 7
Space space (S) = 11
P(W) = nW/nS
P(W) = 7/11
P(R) = nR/nS
P(R) = 4/11
P(WR) = P(W) x P(R)
P(WR) = 7/11 x 4/11
P(WR) = 28/121
Therefore, the probability that the first draw is white and the second is red is 28/121.
Armature reaction in a dc machine A) is due to an increase of the armature voltage. B) occurs when the motor is connected to an ac power source. C) occurs when the motor is connected to a dc power source. D) is due to an increase of the armature current.
Answer:
D) is due to an increase of the armature current.
Explanation:
Option D is correct because on the DC motor, when the load increases, it leads to an increase in the armature current.
The armature current then sets up a magnetic flux which opposes the main field flux. The net field flux gets reduced. It is at this point, the armature reaction occurs.
Armature reaction is seen as the effect of magnetic flux which is usually set up by an armature current. This occurs when there is the distribution of flux under the main poles.
There are two effects the armature flux causes on the main field flux.
1. The main field flux is distorted by the armature reaction.
2. The magnitude of the main field flux is reduced by the armature flux.
Consider atmospheric air at 25 C and a velocity of 25 m/s flowing over both surfaces of a 1-m-long flat plate that is maintained at 125 C. Determine the rate of heat transfer per unit width from the plate for values of the critical Reynolds number corresponding to 105 , 5 105 , and 106 .
Answer:
Explanation:
Temperature of atmospheric air To = 25°C = 298 K
Free stream velocity of air Vo = 25 m/s
Length and width of plate = 1m
Temperature of plate Tp = 125°C = 398 K
We know for air, Prandtl number Pr = 1
And for air, thermal conductivity K = 24.1×10?³ W/mK
Here, charectorestic dimension D = 1m
Given value of Reynolds number Re = 105
For laminar boundary layer flow over flat plate
= 3.402
Therefore, hx = 0.08199 W/m²K
So, heat transfer rate q = hx×A×(Tp – To)
= 0.08199×1×(398 – 298)
A student lab group is brainstorming the design of an experiment that uses an ammeter (measures current) and different resistors to determine the effect of the resistance of a resistor upon the current in a simple circuit. Which Post-it note describes the most effective design?Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Replace the 10.0-ohm resistor with a 20.0-ohm resistor. Measure the new current. Continue replacing the resistor with a different resistor of known resistance. Measure the current for each resistor. Record all data. Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Move the ammeter to a different location in the circuit. Measure the current at this new location. Continue moving the ammeter to different locations within the circuit but be careful to keep the resistor in a fixed location. Measure and record all current values. Obtain a variety of batteries and build several circuits. Make sure that each circuit has at least one resistor and make sure that the resistance values are different in the different circuits. Place various ammeters in each circuit. Measure the number of batteries and the current for each of the circuits. Record the resistance values used in each of these circuits. Put a 10.0-ohm resistor in a circuit with a single D-cell. Measure the current in the circuit. Add a second D-cell and measure the current with two D-cells. Repeat trials for three, four, and five D-cells, being careful to get accurate current measurements for a fixed amount of resistance in each trial.
Answer:
Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Replace the 10.0-ohm resistor with a 20.0-ohm resistor. Measure the new current. Continue replacing the resistor with a different resistor of known resistance. Measure the current for each resistor. Record all data.
Explanation:
The only design that has resistance varying with everything else remaining the same is the first design. That would be what you'd want to do if you're exploring the effect of resistance on current.
Steam enters an adiabatic turbine at 800 psia and9008F and leaves at a pressure of 40 psia. Determine themaximum amount of work that can be delivered by thisturbine.
Answer:
[tex]w_{out}=319.1\frac{BTU}{lbm}[/tex]
Explanation:
Hello,
In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:
[tex]h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}[/tex]
Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:
[tex]s_2=s_1[/tex]
Thus, the liquid and liquid-vapor entropies are included to compute the quality:
[tex]x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965[/tex]
Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:
[tex]h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}[/tex]
Then, by using the first law of thermodynamics, the maximum specific work is computed via:
[tex]h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}[/tex]
Best regards.
When choosing building-construction materials, what kinds of materials would you choose, all other things being equal?
The building-construction materials are materials with a relatively large bulk modulus, would the choose of the constructor. Thus, option (e) is correct.
What is building?The term building refers to the large structure of the bricks. The building was the large structure made by the engineer and the architecture. There was the building are the people are the stay. The building was the used in the permanent basis. There are the different types of the building.
According to the building-construction materials are the engineers to the order of the bulk in the material. The bulk level of the material is the high discount rate of the order. The large amount of the material to the used in the building-construction, that are the order in the bulk of the production.
Therefore, option (e) is correct.
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Your question is incomplete, but most probably the full question was.
either materials with a large or a small bulk modulus.it doesn't matter as long as the building is not too tall.materials with a relatively small shear modulus.materials with a relatively small bulk modulus.materials with a relatively large bulk modulus.A fixed mass of saturated water vapor at 400 kpa is isothermally cooled until it is a saturated liquid. Calculate the amount of heat rejected during this process in kJ/kg.
The amount of heat rejected during the process in kJ/kg is 2133.4 kJ/kg.
What is heat?
Heat is defined as the process of kinetic energy being transferred from an energy source to a medium or an item. These three mechanisms radiation, conduction, and convection can all transmit energy. These three modes of heat transfer are typically combined, and it often happens on its own. The heat energy can be transformed into other types of energy, such as mechanical and kinetic energy. Thermal (heat) energy is a type of energy that is produced by the movement of particles and is a component of all matter.
The process rejects a certain amount of heat, which is
-q = w + Δu
q = - w - Δu
q = - 185.5 kJ/kg - 1948.9 kJ/kg
q = 2133.4 kJ/kg
Thus, the amount of heat rejected during the process in kJ/kg is 2133.4 kJ/kg.
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A rate of 0.42 minute per piece is set for a forging operation. The operator works on the job for a full eight-hour day and produces 1,500 pieces. Use a standard hour plan.
Required:
a. How many standard hours does the operator earn?
b. What is the operator's efficiency for the day?
c. If the base rate is 9.80 per hour, compute the earnings for the day.
d. What is the direct labor cost per piece at this efficiency?
e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?
Answer:
b. What is the operator's efficiency for the day?
AND
e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?
Explanation:
Conductivity is the reciprocal of what?
Cho biết tác dụng chung của các hệ giằng khung ngang nhà công nghiệp nhẹ 1 tầng 1 nhịp.
To prevent the bubble from popping, a second bubble is made with more total fluid. This makes the walls of the bubble thicker.
a. True
b. False
Summary of Possible Weather and Associated Aviation Impacts for Geographic/Topographic Categories Common in the Western United States.
Geographic/Topographic Descriptive Summary of Potential Aviation Impacts
Category of a Possible Weather That Could Impact Based on Weather
of Airport Location Aviation Operations
Along the US West coast,
with steep mountains to the east
(An example of this category is
Santa Barbara Airport, located
on the Southern California Coast,
at an elevation of 10 feet).
Within a valley in elevated terrain
surrounded by high mountains
(An example of this category is
Friedman Memorial Airport, located
in Central Idaho, at an elevation of 5300 feet).
In elevated terrain on the leeside of
high mountains
(An example of this category is Northern Colorado
Regional Airport, located in northern Colorado,
at an elevation of 5000 feet, on the leeside
of the Rocky mountains).
Answer: answer provided in the explanation section.
Explanation:
Weather phenomenons that would impart Aviation Operations in Santa Barbara -
1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.
2. The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.
3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.
4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.
This and more are some factors to look into when considering wheather conditions that would affect aviation operations.
I hope this was a bit helpful. cheers
Question 1 of 8.
What should you do to drive safely in the station?
Turn on your hazard lights
Roll your windows down and turn off the radio
Maintain idle speed (below 5mph)
All of the above
Mark for follow up
Answer: all of the above
Explanation:
To drive safely in the station, the correct option is D. All of the above.
It should be noted that when a person is driving, it's essential for the person to drive carefully in a station.
One should not smoke in a station. Also, it's important that the driver should turn on their hazard lights
Furthermore, it's important for one to roll their windows down, turn off the radio and also maintain idle speed.
In conclusion, the correct option is D.
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An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capacitor and a 301 Ω resistor.
(a) What is the impedance of the circuit?
(b) What is the rms current through the resistor?
(c) What is the average power dissipated in the circuit?
(d) What is the peak current through the resistor?
(e) What is the peak voltage across the inductor?
(f) What is the peak voltage across the capacitor?
The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?
Answer:
Explanation:
f = 50.0 Hz, L = 0.650 H, π = 3.14
C = 4.80 μF, R = 301 Ω resistor. V = 120volts
XL = wL = 2πfL
= 2×3.14×50* 0.650
= 204.1 Ohm
Xc= 1/wC
Xc = 1/2πfC
Xc = 1/2×3.14×50×4.80μF
= 1/0.0015072
= 663.48Ohms
1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2
√ 90601 + (459.38)^2
√ 90601+211029.98
√ 301630.9844
= 549.209
Z = 549.21Ohms
2. I=V/Z = 120/ 549.21Ohms =0.218Ampere
3. P=V×I = 120* 0.218 = 26.16Watt
Note that
I rms = Vrms/Xc
= 120/663.48Ohms
= 0.18086A
4. I(max) = I(rms) × √2
= 0.18086A × 1.4142
= 0.2557
= 0.256A
5. V=I(max) * XL
= 0.256A ×204.1
=52.2496
= 52.250volts
6. V=I(max) × Xc
= 0.256A × 663.48Ohms
= 169.85volts
7. Xc=XL
1/2πfC = 2πfL
1/2πfC = 2πf× 0.650
1/2×3.14×f×4.80μF = 2×3.14×f×0.650
1/6.28×f×4.8×10^-6 = 4.082f
1/0.000030144× f = 4.082×f
1 = 0.000030144×f×4.082×f
1 = 0.000123f^2
f^2 = 1/0.000123048
f^2 = 8126.922
f =√8126.922
f = 90.14 Hz
A resistor has code 104 printed on it .What is the resistive value of this resistor
Answer:
X = 1 (1st digit in the code)
Y = 0 (2nd digit)
Z = 4 (3rd multiplier digit)
104 → 10 × 10^4 Ω
→ 10 × 10000Ω
→ 100 kΩ
resistors are marked 104, 105, 205, 751, and 754. The resistor marked with 104 should be 100kΩ (10x10^4), 105 would be 1MΩ (10x10^5), and 205 is 2MΩ (20x10^5). 751 is 750Ω (75x10^1), and 754 is 750kΩ (75x10^4).
Here we need to understand how a code in a resistor gives us information on the resistor. Here we will see that the code means that the resistance is 100,000 Ω.
When we use numbers, let's assume that we have 3 single-digit numbers abc.
So if the code in our resistor is abc, this will mean that the resistance of the resistor is:
ab×10^c Ω
Using this general rule we can see that if the code is 104, then the resistance will be:
r = 10×10^4 Ω
= 100,000 Ω
Then we can conclude that the resistive value of this resistor is 100,000 Ω
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What improves the structured approach in design?
A team is adopting a structured approach in design which helps them to improve the ___ of the design.
Answer:
efficiency
Explanation:
Answer:
The correct answer is Efficiency.
Explanation:
I got it right on the plato test.
What's the resistance in a circuit that has a voltage of 60 V and a current of 2 A? A. 10 Ω B. 60 Ω C. 120 Ω D. 30 Ω
Answer:
D. Resistance = 30 ohms
Explanation:
Using Ohm's law
V = I times R
Given:
V = 60 V
I = 2 A
Resistance = V / I = 60 V / 20 A
Resistance = 30 ohms
if you are a mechanical engineer answer these questions:
1. Are communication skills (reading, writing, and speaking) necessary in this profession?
2. How are Communicative Competences integrated into this profession?
Answer:
1. Yes, they are all necessary.
2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.
Punctuate or edit the following sentences. Your punctuation and/or revisions should reflect best TW style and grammar writing practices.
1. The author an expert in cybersecurity will speak via Zoom this Wednesday.
2. Williams' book contains many illustrations, this makes it quick reading.
3. Based on the available evidence the university administrators have opted for a hybrid format for the fall quarter which begins September 20.
4. (Thesis statement) Free laptops should be offered to all students who need them.
I inferred you want literal editing of the text above.
Explanation:
Here's a correction of the sentences:
1. The author, an expert in cybersecurity will speak via Zoom on Wednesday.
In this sentence, punctuation mark ( , ) was added and the word 'this' was replaced with 'on'.
2. Williams' book contains many illustrations, which makes it easy to read.
Added punctuation and made a revision of the sentence.
3. Based on the available evidence, the university administrators have opted for a hybrid format for the fall, which begins September 20.
Mainly added punctuations to make the senstence clarer.
4. (Thesis statement) I believe Free laptops should be offered to all students who need them.
Made a few additions.
An 8-m long, uninsulated square duct of cross section 0.2m x 0.2m and relative roughness 10^-3 passes through the attic space of a house.. Hot air (80°C) enters an 8 m long un-insulated square duct (cross section 0.2 m x 0.2 m) that passes through the attic of a house at a rate of 0.15 m^3 /s. The duct is isothermal at a temperature of 60°C. Determine the rate of heat loss from the duct to the attic space and the pressure difference between the inlet and outlet sections of the duct.
Answer:
the rate of heat loss from the duct to the attic space = 1315.44 W
the pressure difference between the inlet and outlet sections of the duct = 7.0045 N/m²
Explanation:
We know that properties of air 80⁰C and 1atm (from appendix table) are;
density p = 0.9994 kg/m³, Specifice heat Cp = 1008 J/kg.⁰C
Thermal conductivity k = 0.02953 W/m.⁰C, Prandtl number Pr = 0.7154,
Kinematic viscosity v = 2.097 × 10⁻⁵ m²/s
haven gotten that, we calculate the hydraulic diameter of square duct
Dh = 4Ac / P { Ac = is cross sectional area of duct and P = perimeter}
now we substitute a² for Ac and 4a for P ( we know from the question that a = 0.2 m)
Dh = 4a² / 4a
Dh = 4(0.2)² / 4(0.2)
Dh = 0.2 m
Now we calculate the average velocity of air
Vₐ = Vˣ / Ac { vˣ = volume flow rate of air}
Vₐ = Vˣ / a² { Ac = a² }, we know that a = 0.2m₂, Vˣ = 0.15 m³
Vₐ = 0.15 / (0.2)²
Vₐ = 3.75 m/s
Next we calculate the Reynolds number
Re = Vₐ Dh / V
Re = (3.75 × 0.2) / 2.097× 10⁻⁵
Re = 35765.379
The Reynolds number IS GREATER than 10,000
so the flow is turbulent and entry length in this case is nearly 10 times the hydraulic diameter
Lh ≈ Lt ≈ 10D
= 10 × 0.2
= 2m
As this length is quite small when compared to the total of tube, we assume fully developed flow for the entire tube length.
Now we calculate the Nusselt number from this relation;
Nu = 0.023 Re⁰'⁸ Pr⁰'³
so we substitute for Re and Pr
Nu = 0.023(35765.379)⁰'⁸ (0.7154)⁰'³
Nu = 91.4
Now calculate the convective heat transfer coefficient
h = Nu × K/ Dh
we substitute
h = 91.4 × 0.02953 W/m.°C / 0.2 m
h = 13.5 W/m².°C
We calculate the surface area of the square duct
Aₓ = 4aL { L= length of duct}
we substitute
Aₓ = 4 × 0.2 × 8
Aₓ = 6.4 m²
Mass flow rate of air
m = pVˣ
we substitute again ( from our initials)
m = 0.9994 kg/m₃ × 0.15 m³/s
m= 0.150 kg/s
We calculate the exit temperature of the air from the duct
Te = Ts - (Ts -Ti) exp ( - hAₓ / mCp)
we know that
Ts = 60°C , Ti = 80°C, h = 13.5 W/m².°C , Aₓ = 6.4m², m = 0.150 kg/s , Cp = 1008 J/kg.°C
we substitute
Te = 60 - (60-80) exp(- ((13.5 × 6.4)/(0.15 × 1008))
Te = 71.3°
Now we calculate the rate of heat loss from the duct.
Q = mCp ( Ti -Te )
we substitute again
Q = 0.150 × 1008 × ( 80 - 71.3 )
Q = 1315.44 W
Next we calculate the estimated friction factors by using Haaland equation
1/√f = - 1.8log₁₀ [ 6.9/Re + (E/D)/3.7)¹'¹¹]
we know that E/D = relative roughness = 10⁻³
we substitute
so
1/√f = - 1.8log₁₀ [ (6.9/35765.379) + ( 10⁻³/3.7)¹'¹¹]
1/√f = - 1.8log₁₀ { 0.000192924 + 0.00010947}
1/√f = - 1.8log₁₀ 0.000302324
√f = 1/6.334
f = (1/6.334)²
f = 0.02492
We calculate the pressure difference between inlet and outlet sections of the duct
ΔPl = fLPVa² / Dh × 2
ΔPl = {0.02492 × 8 × 0.9994 × (3.75)²} / 0.2 × 2
ΔPl = 2.8018 / 0.4
ΔPl = 7.0045 N/m²
Therefore pressure deference is 7.0045 N/m²
A power screw is 30 mm in diameter and has a thread pitch of 5 mm. Find the thread depth, the thread width, the mean and root diameters, and the lead, provided that square threads are used. Assume single threads.
Answer:
thread depth = 2.5 mm
thread width = 2.5 mm
mean diameter = 27.5 mm
root diameter = 25 mm
lead of screw = 5 mm
Explanation:
given data
power screw diameter D = 30 mm
thread pitch P = 5 mm
solution
First, we get here thread depth fr square thread
thread depth = [tex]\frac{P}{2}[/tex] ......................1
thread depth = [tex]\frac{5}{2}[/tex]
thread depth = 2.5 mm
and
thread width for square thread
thread width = [tex]\frac{P}{2}[/tex] ......................2
thread width = [tex]\frac{5}{2}[/tex]
thread width = 2.5 mm
and
mean diameter is
mean diameter = D - [tex]\frac{P}{2}[/tex] ................3
mean diameter = 30 - [tex]\frac{5}{2}[/tex]
mean diameter = 27.5 mm
and
root diameter is
root diameter = D - P ....................4
root diameter = 30 - 5
root diameter = 25 mm
and
lead of screw for single thread so n = 1
so lead of screw = 1 × 5
lead of screw = 5 mm
Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic jump. The flow depth and velocity before the jump are 0.8m and 7m/s, respectively. Determine (a) the flow depth and the Froude number after the jump (b) the head loss (c) the dissipation ratio.
Answer:
A) Flow depth = 2.46 m, Froude number after jump = 0.464
B) head loss = 0.572 m
C) dissipation ratio = 0.173
Explanation:
Given data :
Velocity before jump ( v1 ) = 7 m/s
flow depth before jump ( y1 ) = 0.8 m
g = 9.81 m/s
Esi = 3.3 m ( calculated )
attached below is a detailed solution of the problem