Answer:
"Process capability" is the correct answer.
Explanation:
The Process Capability seems to be a method of measuring of how and why the framework performs concerning something like the successful objectives. This same capacity is characterized as that of the client's voice over procedure speech.Through using functionality indicators it analyses the performance with an in-control process with the permissible range.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
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.
The magnitude of the normal acceleration is
A) proportional to radius of curvature.
B) inversely proportional to radius of curvature.
C) sometimes negative.
D) zero when velocity is constant.
Answer:
b. inversely proportional to radius of curvature
Explanation:
In curvilinear motions, the normal acceleration which is also called the centripetal acceleration is always directed towards the center of the circular path of motion. This acceleration has a magnitude that is directly proportional to the square of the speed of the body undergoing the motion and inversely proportional to the radius of the curvature of the motion path. The centripetal or normal acceleration a, can be given by;
a = [tex]\frac{v^2}{r}[/tex]
Where;
v = speed of the body
r = radius of curvature.
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.
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]
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
Define centrifugal pump. Give the construction and working of centrifugal pump.
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 )
In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approximated as
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.
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
Answer:
True
Explanation:
I looked it up
Answer:
true
Explanation:
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?
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]
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.
Conductivity is the reciprocal of what?
After a capacitor is fully chargerd, a small amount of current will flow though it. what is this current called?
Answer:
leakage
Explanation:
That current is "leakage current."
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.
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
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.\
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 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:
what scale model proves the initial concept?
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
#SPJ2
The radial component of acceleration of a particle moving in a circular path is always:________ a. negative. b. directed towards the center of the path. c. perpendicular to the transverse component of acceleration d. all of the above
Answer:
d. all of the above
Explanation:
There are two components of acceleration for a particle moving in a circular path, radial and tangential acceleration.
The radial acceleration is given by;
[tex]a_r = \frac{V^2}{R}[/tex]
Where;
V is the velocity of the particle
R is the radius of the circular path
This radial acceleration is always directed towards the center of the path, perpendicular to the tangential acceleration and negative.
Therefore, from the given options in the question, all the options are correct.
d. all of the above
There are different type of acceleration. The radial component of acceleration of a particle moving in a circular path is always negative, directed towards the center of the path and perpendicular to the transverse component of acceleration.
Radial acceleration is simply known as the rate of change of angular velocity where the direction is towards the center about whose circumference, the body tend to shift.
This is due to because of the centripetal force. So centripetal force is the reason for a radial acceleration.
Learn more from
https://brainly.com/question/13308560
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.
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
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]
A cylinder is to be cast out of aluminum. The diameter of the disk is 500 mm and its thickness is 20 mm. The mold constant 2.0 sec/mm2 in Chvorinov's formula to calculate the solidification time.
Required:
a. Calculate the minimum time (minutes) for the casting to solidify.
b. Discuss if the result in part (a) is the same when casting grey cast iron.
Answer:
a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min
b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)
Explanation:
Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²
first we find the volume and Area;
Volume V = πD²t / 4
Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³
Area A = 2πD²/ 4 + πDt
Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}
Area A = 392,699.08 + 31,415.93
Area A = 424,115 mm²
a)
Chvorinov’s rule
T(aluminium) = Cm (V/A) ²
T(aluminium) = 2.0 × (3,926,991 / 424,115) ²
T(aluminium) = 171.5 s = 2.86 min
∴ the minimum time (minutes) for the casting to solidify is 2.86 min
b)
For cast iron
Cm (mold constant = 1.488 sec/mm²)
Chvorinov’s rule
T(iron) = Cm (V/A) ²
T(iron) = 1.488 × (3,926,991 / 424,115) ²
T(iron) = 127.5720s = 2.13 min
Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)
A single-threaded power screw is 35 mm in diameter with a pitch of 5 mm. A vertical load on the screw reaches a maximum of 5 kN. The coefficients of friction are 006 for the collar and 009 for the threads, while the frictional diameter of the collar is 45 mm. Find the overall efficiency and the torque to raise and lower the load for
Answer:
the torque required to RAISE the load is Tr = 18.09 Nm
the torque required to LOWER the load is Tl = 10.069 ≈ 10.07 Nm
the Overall Efficiency e = 0.2199 ≈ 0.22
Explanation:
Given that; F = 5 kN, p = 5mm, d = 35mm
Dm = d - p/2
Dm = 35 - ( 5/2) = 35 - 2.5
DM = 32.5mm
So the torque required to RAISE the load is
Tr = ( 5 × 32.5)/2 [(5 + (π × 0.09 × 32.5)) / ( (π × 32.5) - ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]
Tr = 81.25 × (14.1892 / 101.6518) + 6.75
Tr = 11.3414 + 6.75
Tr = 18.09 Nm
the torque required to LOWER the load is
Tl = ( 5 × 32.5)/2 [(π × 0.09 × 32.5) - 5) / ( (π × 32.5) + ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]
Tl = 81.25 × 4.1892 / 102.5518 + 6.75
Tl = 3.3190 + 6.75
Tl = 10.069 ≈ 10.07 Nm
So since torque required to LOWER the load is positive
that is, the thread is self locking
Therefore the efficiency is
e = ( 5 × 5 ) / ( 2π × 18.09 )
e = 25 / 113.6628
e = 0.2199 ≈ 0.22
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.
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.
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
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.
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
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.
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]
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 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
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.
The structure of a house is such that it loses heat at a rate of 5400 kJ/h per degree Cdifference between the indoors and outdoors. A heat pump that requires a power input of 6 kW isused to maintain this house at 21 C. Determine the lowest outdoor temperature for which the heatpump can meet the heating requirements of this house
Answer: Tl = - 13.3°C
the lowest outdoor temperature is - 13.3°C
Explanation:
Given that;
Temperature of Th = 21°C = 21 + 273 = 294 K
the rate at which heat lost is Qh = 5400 kJ/h°C
the power input to heat pump Wnet = 6 kw
The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;
COPhp = Th/(Th - Tl)
COPhp = Qh/Wnet
Qh/Wnet = Th/(Th -Tl)
the amount of heat loss is expressed as
Qh = 5400/3600(294 - Tl)
the temperature of sink
( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)
now solving the equation
Tl = 259.7 - 273
Tl = - 13.3°C
so the lowest outdoor temperature is - 13.3°C
