Answer:
minimum current level required = 8975.95 amperes
Explanation:
Given data:
diameter = 5.5 mm
length = 5.0 mm
T = 0.3
unit melting energy = 9.5 j/mm^3
electrical resistance = 140 micro ohms
thickness of each of the two sheets = 3.5mm
Determine the minimum current level required
first we calculate the volume of the weld nugget
v = [tex]\frac{\pi }{4} * D^2 * l[/tex] = [tex]\frac{\pi }{4} * 5.5^2 * 5[/tex] = 118.73 mm^3
next calculate the required melting energy
= volume of weld nugget * unit melting energy
= 118.73 * 9.5 = 1127.94 joules
next find the actual required electric energy
= required melting energy / efficiency
= 1127 .94 / ( 1/3 ) = 3383.84 J
TO DETERMINE THE CURRENT LEVEL REQUIRED use the relation below
electrical energy = I^2 * R * T
3383.84 / R*T = I^2
3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2
therefore 8975.95 = I ( current )
Define centrifugal pump. Give the construction and working of centrifugal pump.
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:
Based on the hardness values determined in Part 1, what is the tensile strength (in MPa) for each of the alloys?
(a) 0.25 wt%C with spheroidite,
(b) 0.25 wt%C with coarse pearlite,
(c) 0.60 wt%C with fine pearlite, and
(d) 0.60 wt%C with coarse pearlite.
The tolerance to all answers is +/-10 %.
Answer:
a. 115
b. 135
c. 220
d. 185
Explanation:
Spheriodite is microscopic constituents in some steels which is composed of spherically shaped cementide particle. It is most ductile and softest type of steel. Pearlite is two phased lamellar compose of alternating layer of ferrite and cementite. It is hard and strong but not tough. It is applied on cutting tools like chopper, blades and knives.
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]
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
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)
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
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.
Air at 30 C, 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate of 3 kg/min and mixes with a saturated moist air stream entering at 5 C, 1 bar with a mass flow rate of 5 kg/min. A single mixed stream exits at 1 bar. Determine (a) the relative humidity and temperature, in C, of the exiting stream. (b) the rate of exergy destruction, in kW, for T0
Answer:
A) The relative humidity : 0.818 (81.8%), Temperature at C = 14.4⁰c
B) The rate of energy destruction = 0.0477 kw
Explanation:
Given data :
at point 1 : m1 = 3 kg/min , T1 = 30⁰c, p1 = 1 bar, ∅ = 0.50 ( 50%)
at point 2 : T2 = 5⁰c, P2 = 1 bar, m2 = 5 kg/min
at point 3 : p3 = 1 bar
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.
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
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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
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 after jump = 2.46 m, Froude number after jump = 0.464
B) head loss = 0.572 m
C) dissipation ratio = 0.173
Explanation:
Given data :
width of channel = 10-m
velocity of before jump (V1) = 7 m/ s
flow depth before jump (y1) = 0.8 m
A) determining the flow depth and the Froude's number after the jump
attached below is the solution
B) head loss
HL = Y1 -Y2 + [tex]\frac{V_{1} ^2 - V_{2} ^2}{2g}[/tex] = 0.8 - 2.46 + [tex]\frac{49 - 5.1984}{19.62}[/tex] = 0.572
C) dissipation ratio
HL / Es1 = 0.572 / 3.3 = 0.173
Es1 = 0.8 + [tex]\frac{7^2}{2*9.81}[/tex] = 0.173
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
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.
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.
Answer:
The answer is below
Explanation:
1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:
[tex]n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm[/tex]
2) The speed of the rotor is the motor speed. The slip is given by:
[tex]Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm[/tex]
3) The frequency of the rotor is given as:
[tex]f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz[/tex]
4) At standstill, the speed of the motor is 0, therefore the slip is 1.
The frequency of the rotor is given as:
[tex]f_r=slip*f_s\\f_r=1*60=60\ Hz[/tex]
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.
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.
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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
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.
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.
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.
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.
The fins attached to a heat exchanger-surface are determined to have an effectiveness of 0.9. Do you think the rate of heat transfer from the surface has increased or decreased as a result of the addition of these fins?
Answer:
The rate of heat transfer has increased.
Explanation:
Heat transfer rate is the rate at which heat energy is dissipated to the ambient from a hot body. The rate of heat transfer is proportional to the available surface area for heat exchange. This means that the greater the exposed surface area for heat exchange, the greater the rate at which heat is lost to the ambient. In introducing the fins to the heat exchange system (fins have a large surface area to volume ratio for maximum exposure to the ambient), one maximizes the available surface area for heat exchange between the material and the ambient, increasing the rate of heat transfer.
Input resistance of a FET is very high due to A) forward-biased junctions have high impedance B) gate-source junction is reverse-biased C) drain-source junction is reverse-biased D) none of the above
Answer:
B) gate-source junction is reverse-biased
Explanation:
FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow" in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".
In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".
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
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.
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
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.
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
Suppose that the resistors in the various circuit diagrams represented the resistances of lightbulbs. When a lightbulb "burns out," the circuit is open through that particular component, that is R is infinite. Would the remaining bulbs continue to burn?
Answer:
The Remaining Bulbs will either burn out( draw more current ) or Not burn out depending on the arrangement of the bulbs in the circuit
Explanation:
The Remaining Bulbs will either burn more brightly ( draw more current ) or Burn less brightly ( draw less current).and this depends on the arrangement of the light bulbs in the various circuit.
If the light bulbs are connected in series the remaining bulbs will burn out as soon as any light bulb burns out and this is because bulbs connected in series receive the same amount of current ,
If the light bulbs are connected in parallel the remaining bulbs will not burn out because each bulb receives current based on its resistance.
As the asteroid falls closer to the Earth's surface its _______ energy decreases and its _______ energy increases.
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.
Answer:
efficiency
Explanation:
Answer:
The correct answer is Efficiency.
Explanation:
I got it right on the plato test.
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.
