/ Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a
mean velocity of 7 m/s, and is cooled by the water outside. If the average heat
transfer coefficient is 85 W/m2
°C and the tube temperature is nearly equal to the
water temperature of 5°C, determine the exit temperature of air and the rate of heat
transfer.
Answers
Answer:
A) EXIT TEMPERATURE = 14⁰C
b) rate of heat transfer of air = - 13475.78 = - 13.5 kw
Explanation:
Given data :
diameter of duct = 20-cm = 0.2 m
length of duct = 12-m
temperature of air at inlet= 50⁰c
pressure = 1 atm
mean velocity = 7 m/s
average heat transfer coefficient = 85 w/m^2⁰c
water temperature = 5⁰c
surface temperature ( Ts) = 5⁰c
properties of air at 50⁰c and at 1 atm
= 1.092 kg/m^3
Cp = 1007 j/kg⁰c
k = 0.02735 W/m⁰c
Pr = 0.7228
v = 1.798 * 10^-5 m^2/s
determine the exit temperature of air and the rate of heat transfer
attached below is the detailed solution
Calculate the mass flow rate
= p*Ac*Vmean
= 1.092 * 0.0314 * 7 = 0.24 kg/s
Related Questions
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.
Answers
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.
Answers
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
As the asteroid falls closer to the Earth's surface its _______ energy decreases and its _______ energy increases.
Answers
Answer:
As the asteroid falls closer to the Earth's surface its Gravitational Potential energy decreases and its Kinetic energy increases.
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.
Answers
Answer:
efficiency
Explanation:
Answer:
The correct answer is Efficiency.
Explanation:
I got it right on the plato test.
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).
Answers
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
Air at 1 atm, 15°C, and 60 percent relative humidity is first heated to 20 °C in a heating section and then humidified by introducing water vapor. The air leaves the humidifying section at 25°C and 65 percent relative humidity. Determine:
a. the amount of steam added to the air.
b. the amount of heal transfer to the air in the heating section.
Answers
Answer: a = change in w =0.00656
b = q = 5.1kj/kg
Explanation:
Find explanation in the attached file
The amount of steam added to the air a = change in w =0.00656 b = q = 5.1kj/kg
What is steam?The digital game retail and distribution service Steam is provided by Valve. In order to allow Valve to automatically update its games, it was first released as a software client in September 2003. In late 2005, it was expanded to include the distribution and sale of games from other publishers.
a) We can use the absolute humidity we and wg to determine the amount
of moisture added Aw.
Aww3-W2
Aw= 0.01291 -0.00635
Aw= 0.00656
b) To determine the heat transfer q we will need the enthalpies h and h2.
kJ
kg 9 = 36.2
kJ kg
31.1
q=5.1
kJ
kg
RESULT
Do = 0.00656
kJ
kg
9 = 5.1
Therefore, The amount of steam added to the air a = change in w =0.00656 b = q = 5.1kj/kg
Learn more about steam here:
https://brainly.com/question/15447025
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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.
Answers
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.
After a capacitor is fully chargerd, a small amount of current will flow though it. what is this current called?
Answers
Answer:
leakage
Explanation:
That current is "leakage current."
what scale model proves the initial concept?
Answers
Answer: A prototype
Explanation:
The scale model that proves the initial concept is called a domain model.
What is a scale model?A copy or depiction of something where all parts have the same dimensions as the original. A scale model is an image or copy of an object that is either larger or smaller than the object being represented's actual size.
A domain model is a type of conceptual model that is used to depict the structural elements and conceptual constraints within a domain of interest.
A domain model will include all of the entities, their attributes, and relationships, as well as the constraints that govern the conceptual integrity of the structural model elements that comprise that problem domain.
Therefore, a domain model is the scale model that proves the initial concept.
To learn more about the scale model, refer to the below link:
https://brainly.com/question/14341149
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Q1: You have to select an idea developing an application like web/mobile or industrial, it should be based on innovative idea, not just a simple CRUD application. After selecting the idea do the following: 1) How your project will be helpful and what problem this project addresses. (10-Marks) 2) Write down the requirements. (10Marks) 3) List the functional and non-functional requirements of your project. (10marks) 4) Which process model you will follow for this project and why? (10marks) 5) Draw the Level 0, and level 1 DFD of your application. (20marks)
Answers
Answer:
Creating an app is both an expression of our self and a reflection of what we see is missing in the world. We find ourselves digging deep into who we are, what we would enjoy working on, and what needs still need to be fulfilled. Generating an app idea for the first time can be extremely daunting. Especially with an endless amount of possibilities such as building a church app.
The uncertainty has always spawned a certain fear inside creators. The fear of creating something no one will enjoy. Spending hundreds of dollars and hours building something which might not bring back any real tangible results. The fear of losing our investment to a poor concept is daunting but not random. But simple app ideas are actually pretty easy to come by.
Great app idea generation is not a gift given to a selected few, instead, it is a process by which any of us are able to carefully explore step by step methods to find our own solution to any problem. Whether you are a seasoned creator or a novice, we have provided a few recommendations to challenge and aid you as you create your next masterpiece.
if I am right then make me brainliest
1. (16 points) True or False, one point each, Write down F (false) or T (true). ___ (01) In a mechanical design, it is recommended to use standard size/dimension to overcome uncertainties in stress or material strength
Answers
Answer:
True
Explanation:
I looked it up
Answer:
true
Explanation:
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?
Answers
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:
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 =
Answers
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 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observed to vibrate with a natural frequency of 9.3 HZ. What is the moment of inertia of the beam's cross section about its neutral axis?
Answers
Answer:
moment of inertia = 4.662 * 10^6 [tex]mm^4[/tex]
Explanation:
Given data :
Mass of machine = 400 kg = 400 * 9.81 = 3924 N
length of span = 3.2 m
E = 200 * 10^9 N/m^2
frequency = 9.3 Hz
Wm ( angular frequency ) = 2 [tex]\pi f[/tex] = 58.434 rad/secs
also Wm = [tex]\sqrt{\frac{g}{t} }[/tex] ------- EQUATION 1
g = 9.81
deflection of simply supported beam
t = [tex]\frac{wl^3}{48EI}[/tex]
insert the value of t into equation 1
W[tex]m^2[/tex] = [tex]\frac{g*48*E*I}{WL^3}[/tex] make I the subject of the equation
I ( Moment of inertia about the neutral axis ) = [tex]\frac{WL^3* Wn^2}{48*g*E}[/tex]
I = [tex]\frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}[/tex] = 4.662 * 10^6 [tex]mm^4[/tex]
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.
Answers
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.
The solid homogeneous cylinder is released from rest on the ramp. If θ= 40° , µs= 0.30 and µk= 0.20. Determine the magnitudes of the acceleration of the mass (W= 8lb) center G and the friction force exerted by the ramp on the cylinder.
Answers
Answer:
A) 13.80 ft/s^2
B) 1.714 Ib
Explanation:
Magnitude of acceleration center G
mass = W / g = 8 / 32.2 = 0.2484 Ib.s^2/ft
calculate the acceleration along x direction
A = ra
r = radius
a = angular acceleration
A = 6 in [tex]\frac{1 ft}{12 in}[/tex] * a
a= 2A
equation of the plane along the x-direction
w sin∅ - F = ma
8* sin40 - F = 0.2484 * a
hence F = 5.1423 - 0.2484 a
next find the moment of inertia along the z axis
I = 1/2 mr^2
= 1/2 * 0.2484 * (6/12)^2 = 0.03105 Ib.ft.s^2
Applying moment balance equation
F * r = inertia * a
(5.1423 - 0.2484 a)*0.5 = 0.03105 * 2A
2.57115 = 0.1863 A hence
A = 13.80 ft/s^2 ( acceleration of the cylinder )
B) Calculate the friction force exerted by the ramp on the cylinder
F = 5.1423 - 0.2484 A
= 5.1423 - 0.2484 ( 13.80 )
= 1.714 Ib
The magnitudes of the acceleration and the friction force are;
Acceleration = 13.8 ft/s²
Friction Force = 1.714 lb
The image of the solid homogeneous cylinder is missing and so i have attached it.
From the image we see that;
Weight; W = 8 lbRadius; r = 6 in = 0.5 ftWe are given;
Angle of incline; θ = 40°Coefficient of static friction; µ_s = 0.30 coefficient of kinetic friction; µ_k = 0.20We know that formula for weight is; W = mg
Thus; m = W/g
where g is acceleration due to gravity = 32.2 ft/s²
m = 8/32.2
mass; m = 0.2484 lb.s²/ft
Now, to get the acceleration along the x-axis, we will use the formula;
a = rα
where α is angular acceleration. Thus;
a = 0.5α
α = 2a ----- (eq 1)
Now, resolving forces along the x-direction gives;
W*sinθ - F = ma
Plugging in the relevant values;
8*sin 40 - F = 0.2484a
F = 8*sin 40 - 0.2484a -----(eq 2)
Now, moment of inertia of the cylinder along the z-axis is gotten from;
I = ¹/₂mr²
I = ¹/₂ × 0.2484 × 0.5²
I = 0.03105 lb.ft/s²
Taking equilibrium of moments we have;
F*r = I*α
Thus;
(8*sin 40 - 0.2484a)0.5 = 0.03105α
⇒ 2.57115 - 0.1242a = 0.03105α
⇒ 0.03105α + 0.1242a = 2.57115
From eq 1, α = 2a. Thus;
0.03105(2a) + 0.1242a = 2.57115
0.1863a = 2.57115
a = 2.57115/0.1863
a = 13.8 ft/s²
Formula for the friction force exerted by the ramp on the cylinder is;F = 8*sin 40 - 0.2484a
F = 5.1423 - 0.2484(13.8)
F = 1.714 lb
Read more about cylinder moment of inertia at; https://brainly.com/question/7020147
Consider an ideal gas undergoing a constant pressure process from state 1 to state
2 in a closed system. The specific heat capacities for this material depend on temperature in
the following way, cv = aT^b , cp = cT^d , where the constants a, b, c and d are known. Calculate
the specific entropy change, (s2 − s1), from state 1 to state 2.
Answers
Answer:
[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]
Explanation:
Hello,
In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy at constant pressure:
[tex]ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}[/tex]
Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:
[tex]\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,[/tex]
We obtain the expression to compute the specific entropy change:
[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]
Best regards.
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?
Answers
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.
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..
Answers
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.
A bona fide established commercial marketing agency is a business which is specifically devoted to public relations, advertising and promoting the services of a client. True or False
Answers
Answer:
True
Explanation:
Bona Fide is a Latin term which means in good faith or without any intention to deceive. The business established on a bona fide basis means that there is an absence of fraud. The marketing agency has devoted its services to public relations, advertising and promoting the services of clients. There is no intention of fraud in the business.
Conductivity is the reciprocal of what?
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Assume that the heat is transferred from the cold reservoir to the hot reservoir contrary to the Clausis statement of the second law. Prove that this violates the increase of entropy principle—as it should according to Clausius.
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Answer: hello attached below is the diagram which is part of your question
Total entropy change = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k it violates Clausius increase of entropy which is Sgen > 0
Explanation:
Clausius statement states that it is impossible to transfer heat energy from a cooler body to a hotter body in a cycle or region without any other external factors affecting it .
applying the increase in entropy principle to prove this
temp of cold reservoir (t hot)= 600 k
temp of hot reservoir(t cold) = 1220 k
energy (q) = 100 kj
total entropy change = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k
entropy change in cold reservoir = Q/t cold = 100 / 600 = -0.166 kj/k
entropy change in hot reservoir = Q / t hot = 100 / 1220 = 0.083 kj/k
hence it violates Clausius inequality of increase of entropy principle which is states that generated entropy has to be > 0
(2x+y)dx+(x-2y)dy=0 solve the differential equation
Answers
Answer: y' = - x'
Explanation:
Let f(x) = 2x + y
then f'(x) = 2 + y'
Let f(y) = x - 2y
then f'(y) = x' - 2
Given: f'(x) + f'(y) = 0
2 + y' + x' - 2 = 0
y' + x'= 0
y' = -x'
This can also be written as: [tex]\dfrac{dy}{dx}=-\dfrac{d}{dx}[/tex]
Define centrifugal pump. Give the construction and working of centrifugal pump.
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An AC voltage is represented by the relation v= 12. Determine the: (a) peak-to-peak voltage; (b) frequency; (c) root-mean-square voltage; (d) Period of the signal.
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Answer:
The answer is below
Explanation:
An AC voltage is represented by the relation v= 12 sin(500πt). Determine the:
The equation of an AC voltage is given as:
[tex]V=V_msin(2\pi ft)[/tex]
Where Vm is the maximum value of voltage and f is the frequency
From V= 12 sin(500πt), Vm = 12, 2πft = 500πt
(a) The peak to peak voltage is total amplitude (both the negative and positive amplitude) of the voltage, it is the difference between the positive amplitude and the negative amplitude. The peak to peak voltage ([tex]V{p-p}[/tex]) is given as:
[tex]V_{p-p}=2V_m=2*12=24\ V[/tex]
b) The frequency is the number of oscillation per second. The frequency (f) is gotten from:
2πft = 500πt
2f = 500
f = 500/2
f = 250 Hz
c) The root mean square voltage is the dc value of the voltage. It is given by:
[tex]V_{rms}=\frac{V_m}{\sqrt{2} }=\frac{12}{\sqrt{2} }=8.5\ V[/tex]
d) The period (T) is the time taken to complete one oscillation, it is given by:
[tex]T=\frac{1}{f}\\ \\T=\frac{1}{250} =0.004\ s[/tex]
Water at 20oC, with a free-stream velocity of 1.5 m/s, flows over a circular pipe with diameter of 2.0 cm and surface temperature of 80oC. Calculate the average heat transfer coefficient and the heat transfer rate per meter length of pipe.\
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Answer:
Average heat transfer coefficient = 31 kw/m^2 k
Heat transfer rate per meter length of pipe = 116.808 KW
Explanation:
water temperature = 20⁰c,
free-stream velocity = 1.5 m/s
circular pipe diameter = 2.0 cm = 0.02 m
surface temperature = 80⁰c
A) calculate average heat transfer coefficient
we apply the formula below :
m = αAv
A (area) = [tex]\pi /4 (d)^2[/tex]
m = 10^3 * [tex]\pi / 4 ( 0.02)^2[/tex] * 1.5
= 10^3 * 0.7857( 0.0004) * 1.5
= 0.4714 kg/s
Average heat transfer coefficient
h = [tex]\frac{m(cp)}{A}[/tex] , A = [tex]\pi DL[/tex]
L = 1 m , m = 0.4714 kgs , cp = 4.18
back to equation
h = [tex]\frac{0.4714*4.18}{\pi * 0.02 }[/tex] = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k
B) Heat transfer rate per meter length of pipe
Q = ha( ΔT ), a = [tex]\pi DL[/tex]
= 31 * 0.0628 * ( 80 - 20 )
= 31 * 0.0628 * 60 = 116.808 KW
A permanent-magnet dc motor has the following parameters: Ra = 0.3 Ω and kE = kT = 0.5 in MKS units. For a torque of up to 10 Nm, plot its steady state torque-speed characteristics for the following values of Va: 100 V, 75 V, and 50 V.
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Answer:
load speeds:
For V = 100 v speed = 188 rad/sec
For V = 75 v speed = 138 rad/sec
For V = 50 v speed = 88 rad/sec
Explanation:
Given data
Ra = 0.3 Ω
Ke = Kt = 0.5
torque = 10 Nm
using a constant torque = 10 Nm we can calculate the various load speed for the given values of 100 v , 75 v, 50 v
attached below is the detailed solutions and plot
A four-cylinder four-stroke engine is modelled using the air standard Otto cycle (two engine revolutions per cycle). Given the conditions at state 1, total volume (V1) of each cylinder, compression ratio (r), rate of heat addition (Q), and engine speed in RPM, determine the efficiency and other values listed below. The gas constant for air is R =0.287 kJ/kg-K.
T1 = 300 K
P1 = 100 kPa
V1 = 500 cm^3
r = 10
Q = 60 kW
Speed = 5600 RPM
Required:
a. Determine the total mass (kg) of air in the engine.
b. Determine the specific internal energy (kJ/kg) at state 1.
c. Determine the specific volume (m^3/kg) at state 1.
d. Determine the relative specific volume at state 1.
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Answer:
a) Mt = 0.0023229
b) = U1 = 214.07
c) = V₁ = 0.861 m³/kg
d) = Vr1 = 621.2
Explanation:
Given that
R = 0.287 KJ/kg.K, T1 = 300 K , P1 = 100 kPa , V1 = 500 cm³, r = 10 , Q = 60 kW , Speed N = 5600 RPM, Number of cylinders K = 4
specific heat at constant volume Cv = 0.7174 kJ/kg.K
Specific heat at constant pressure is 1.0045 Kj/kg.K
a) To determine the total mass (kg) of air in the engine.
we say
P1V1 = mRT1
we the figures substitute
(100 x 10³) ( 500 x 10⁻⁶) = m ( 0.287 x 10³) ( 300 )
50 = m x 86100
m = 0.00005 / 86100 = 0.0005807 ( mass of one cylinder)
Total mass of 4 cylinder
Mt = m x k
Mt = 0.0005807 x 4
Mt = 0.0023229
b) To determine the specific internal energy (kJ/kg) at state 1
i.e at T1 = 300
we obtain the value of specific internal energy U1 at 300 K ( state 1) from the table ideal gas properties of air.
U1 = 214.07
c) To determine the specific volume (m³/kg) at state 1.
we say
V₁ = V1/m
V₁ = (500 x 10⁻⁶) / 0.0005807
V₁ = 0.861 m³/kg
d) To determine the relative specific volume at state 1.
To obtain the value of relative specific volume at 300 K ( i.e state 1) from the table ideal gas properties of air.
At T1 = 300 k
Vr1 = 621.2
In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approximated as
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Answer:
Explanation:
In a particular application involving airflow over a heated surface, the boundary layer temperature distribution, T(y), may be approximated as:
[ T(y) - Ts / T∞ - Ts ] = 1 - e^( -Pr (U∞y / v) )
where y is the distance normal to the surface and the Prandtl number, Pr = Cpu/k = 0.7, is a dimensionless fluid property. a.) If T∞ = 380 K, Ts = 320 K, and U∞/v = 3600 m-1, what is the surface heat flux? Is this into or out of the wall? (~-5000 W/m2 , ?). b.) Plot the temperature distribution for y = 0 to y = 0.002 m. Set the axes ranges from 380 to 320 for temperature and from 0 to 0.002 m for y. Be sure to evaluate properties at the film temperature.
An ideal turbojet engine is analyzed using the cold air standard method. Given specific operating conditions determine the temperature, pressure, and enthalpy at each state, and the exit velocity.
--Given Values--
T1 (K) = 249
P1 (kPa) = 61
V1 (m/s) = 209
rp = 10.7
rc = 1.8
Required:
a. Determine the temperature (K) at state 2.
b. Determine the pressure (kPa) at state 2.
c. Determine the specific enthalpy (kJ/kg) at state 2.
d. Determine the temperature (K) at state 3.
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Answer:
a. the temperature (K) at state 2 is [tex]\mathbf{T_2 =270.76 \ K}}[/tex]
b. the pressure (kPa) at state 2 is [tex]\mathtt{ \mathbf{ p_2 = 81.79 \ kPa }}[/tex]
c. the specific enthalpy (kJ/kg) at state 2 is [tex]\mathbf{h_2 = 271.84 \ kJ/kg}}[/tex]
d. the temperature (K) at state 3 is [tex]\mathbf{ T_3 = 532.959 \ K}[/tex]
Explanation:
From the given information:
T1 (K) = 249
P1 (kPa) = 61
V1 (m/s) = 209
rp = 10.7
rc = 1.8
The objective is to determine the following:
a. Determine the temperature (K) at state 2.
b. Determine the pressure (kPa) at state 2.
c. Determine the specific enthalpy (kJ/kg) at state 2.
d. Determine the temperature (K) at state 3.
To start with the specific enthalpy (kJ/kg) at state 2.
By the relation of steady -flow energy balance equation for diffuser (isentropic)
[tex]h_1 + \dfrac{V_1^2}{2}=h_2+\dfrac{V^2_2}{2}[/tex]
[tex]h_1 + \dfrac{V_1^2}{2}=h_2+0[/tex]
[tex]h_2=h_1 + \dfrac{V_1^2}{2}[/tex]
For ideal gas;enthalpy is only a function of temperature, hence [tex]c_p[/tex]T = h
where;
[tex]h_1[/tex] is the specific enthalpy at inlet = [tex]c_pT_1[/tex]
[tex]h_2[/tex] is the specific enthalpy at outlet = [tex]c_pT_2[/tex]
[tex]c_p[/tex] = 1.004 kJ/kg.K or 1004 J/kg.K
Given that:
[tex]T_1[/tex] (K) = 249
[tex]V_1[/tex] (m/s) = 209
∴
[tex]h_2=C_pT_1+ \dfrac{V_1^2}{2}[/tex]
[tex]h_2=1004 \times 249+ \dfrac{209^2}{2}[/tex]
[tex]h_2 = 249996+21840.5[/tex]
[tex]\mathbf{\mathtt{h_2 = 271836.5 \ J/kg}}[/tex]
[tex]\mathbf{h_2 = 271.84 \ kJ/kg}}[/tex]
Determine the temperature (K) at state 2.
SInce; [tex]\mathtt{h_2 = c_pT_2 = 271.84 \ kJ/kg}[/tex]
[tex]\mathtt{ c_pT_2 = 271.84 \ kJ/kg}[/tex]
[tex]\mathtt{T_2 = \dfrac{271.84 \ kJ/kg}{ c_p}}[/tex]
[tex]\mathtt{T_2 = \dfrac{271.84 \ kJ/kg}{1.004 \ kJ/kg.K}}[/tex]
[tex]\mathbf{T_2 =270.76 \ K}}[/tex]
Determine the pressure (kPa) at state 2.
For isentropic condition,
[tex]\mathtt{ \dfrac{T_2}{T_1}= \begin {pmatrix} \dfrac{p_2}{p_1} \end {pmatrix} ^\dfrac{k-1}{k}}[/tex]
where ;
k = specific heat ratio = 1.4
[tex]\mathtt{ \dfrac{270.76}{249}= \begin {pmatrix} \dfrac{p_2}{61} \end {pmatrix} ^\dfrac{1.4-1}{1.4}}[/tex]
[tex]\mathtt{ 1.087389558= \begin {pmatrix} \dfrac{p_2}{61} \end {pmatrix} ^\dfrac{0.4}{1.4}}[/tex]
[tex]\mathtt{ 1.087389558 \times 61 ^ {^ \dfrac{0.4}{1.4} }}=p_2} ^\dfrac{0.4}{1.4}}[/tex]
[tex]\mathtt{ 3.519487255=p_2} ^\dfrac{0.4}{1.4}}[/tex]
[tex]\mathtt{ \mathbf{ p_2 = \sqrt[0.4]{3.519487255^{1.4}} }}[/tex]
[tex]\mathtt{ \mathbf{ p_2 = 81.79 \ kPa }}[/tex]
d. Determine the temperature (K) at state 3.
For the isentropic process
[tex]\mathtt{\dfrac{T_3}{T_2} = \begin {pmatrix} \dfrac{p_3}{p_2} \end {pmatrix}^{\dfrac{k-1}{k}}}[/tex]
where;
[tex]\mathtt{\dfrac{p_3}{p_2} }[/tex] is the compressor ratio [tex]\mathtt{r_p}[/tex]
Given that ; the compressor ratio [tex]\mathtt{r_p}[/tex] = 10.7
[tex]\mathtt{\dfrac{T_3}{T_2} = \begin {pmatrix} r_p \end {pmatrix}^{\dfrac{k-1}{k}}}[/tex]
[tex]\mathtt{\dfrac{T_3}{270.76} = \begin {pmatrix} 10.7 \end {pmatrix}^{\dfrac{1.4-1}{1.4}}}[/tex]
[tex]\mathtt{\dfrac{T_3}{270.76} = \begin {pmatrix} 10.7 \end {pmatrix}^{^ \dfrac{0.4}{1.4}}}[/tex]
[tex]\mathtt{{T_3}{} =270.76 \times\begin {pmatrix} 10.7 \end {pmatrix}^{^ \dfrac{0.4}{1.4}}}[/tex]
[tex]\mathbf{ T_3 = 532.959 \ K}[/tex]
