To have a 99.9% chance that the component does not fail before 1,000,000 cycles, the maximum stress level (σmax) should be set to 17.15 ksi or lower.
Given:
- Material: 2024-T4 aluminum
- Ultimate tensile strength: 70 ksi
- Minimum stress level: 0 ksi
- Desired reliability: 99.9% for 1,000,000 cycles
Step 1: Determine the endurance limit (σe) for the material.
For aluminum alloys, the endurance limit is typically taken as the stress level at which the S-N curve (stress vs. number of cycles to failure) becomes horizontal.
A common approximation for the endurance limit of aluminum alloys is:
σe ≈ 0.35 × Ultimate tensile strength
σe ≈ 0.35 × 70 ksi = 24.5 ksi
Step 2: Adjust the endurance limit for the desired reliability.
The endurance limit is typically determined for a reliability of 50%. To achieve a higher reliability, we need to apply a correction factor.
For a reliability of 99.9%, the correction factor is approximately 0.7.
Adjusted endurance limit = σe × 0.7 = 24.5 ksi × 0.7 = 17.15 ksi
Step 3: Determine the maximum stress level (σmax) based on the adjusted endurance limit.
For a fully reversed stress cycle (minimum stress = 0), the maximum stress level should not exceed the adjusted endurance limit.
σmax ≤ 17.15 ksi
Therefore, to have a 99.9% chance that the component does not fail before 1,000,000 cycles, the maximum stress level (σmax) should be set to 17.15 ksi or lower.
The high electrical conductivity of copper is an important design factor that helps improve the energy efficiency of electric motors. This is important because motors and motor-driven systems are significant consumers of electricity, accounting for 43% - 46% of all global electricity consumption and 69% of all electricity used by industry. Inefficient motors waste electrical energy and are indirect contributors to greenhouse gas emissions. ElectroSpark, Inc. has been developing a new copper die-cast rotor technology specifically for premium efficiency motors, replacing the standard aluminum rotor. There are multiple reasons for doing so, including the possibility that the motor will consume less energy. They designed an experiment to test their idea in a common ¾ Horse power (HP) motor that is normally manufactured with an aluminum rotor. They designed a copper rotor that fit in their ¾ HP motor housing and ran a production line for a day producing the motors. They randomly selected 20 copper-rotor motors from that output and 20 aluminum-rotor motors produced from the same line the day before. These 40 motors were all run for 8 hours a day for 30 days and the energy consumed was measured in total Kilowatt Hours (example data below, using alpha=.05):
Copper: 560.145 539.673 556.834 559.873
Aluminium: 564.674 573.912 553.385 574.078
What is the correct hypothesis to test the problem described in this scenario?
A. H0: μD (copper-aluminum) ≥ 0; H1: μD (copper-aluminum) < 0
B. H0: μ_copper – μ_aluminum ≥ 0; H1: μ_copper – μ_aluminum < 0
C. H0: μD (copper-aluminum) ≥ 0; H1: μD (copper-aluminum) > 0
D. H0: μ_copper – μ_Aluminum ≤ 0; H1: μ_copper – μ_aluminum > 0
Answer:
B
Explanation:
This is a two sample t-test and not a matched pair t-test
null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors
alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.
So, option D is rejected
The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.
Option A and C are also rejected
You are asked by your college crew to estimate the skin friction drag in their eight-seat racing shell. The hull of the shell may be approximated as half a circular cylinder with 450 mm diameter and 7.32 m length. The speed of the shell through the water is 6.71 m/s. Estimate the location of the transition from laminar to turbulent flow in the boundary layer on the hull of the shell. Calculate the thickness of the turbulent boundary layer at the rear of the hull. Determine the total skin friction drag on the hull under the given conditions
Answer:
The total skin friction drag on the hull under these conditions is 276N
Explanation:
In this question, we are asked to determine the total skin friction drag on the hull under the given conditions.
Please check attachment for complete solution and step by step explanation
Suppose we have a database for an investment firm, consisting of the following attributes: B (broker), O (office of a broker), I (investor), S (stock), Q (quantity of stock owned by an investor), and D (dividend paid by a stock), with the following functional dependencies: SD, I B.IS a Find a key for the relation scheme R(B,O,S,Q,I,D). b. Find a decomposition of R into third normal form, having a lossless join and preserving dependencies c.Find a lossless join decomposition of R into Boyce-Codd normal form.
Answer:
Given, FDs are:
S -> D
I -> B
IS -> Q
B -> O
a)
"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.
The only candidate key is: IS
IS -> ISBDQO
b)
Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)
c)
Decomposition of R into BCNF:
Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).
R1 is in BCNF
Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).
R21is in BCNF
Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).
R221 is in BCNF.
R222 is in BCNF.
The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)
We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)
Explanation:
The answer above is rendered in a very explanatory way.
Software, such as a word processor, search engine, or mobile interface, typically includes plug-in support specific to a language to aid with spelling. In this assignment, you will implement a class that provides general language support; such a class could presumably be (re)used in these broader software applications. For the purpose of spell checking, a simple language model is a set of valid words. By convention, a language specification may include both capitalized and uncapitalized words. A word that is is entirely lowercased in the language specification can be used in either capitalized or uncapitalized from (e.g., if 'dog'is in the language specification, then both 'dog' and 'Dog' are legitimate usages). However, any word that includes one or more uppercased letters in the original language reflects a form that cannot be modified (e.g., 'Missouri' is acceptable but 'missouri' is not; 'NATO' is acceptable, but neither 'Nato', 'nato', nor 'nAto' would be acceptable). The goals of the new class will be to answer the following types of queries: • Is a given string a legitimate word in the language? (based on the above conventions regarding capitalization) • Given a string, which may or may not be in the language, produce a list of suggestions that are valid words in the language and reasonably "close" to the given string in terms of spelling. (We will say more below, about the notion of distance between words.) Formally, you are to provide a file named language_tools.py that defines aLanguageHelper class with the following three methods. _init__(self, words) The words parameter can be any iterable sequence of strings that define the words in the language. For example, the parameter may be a list of strings, or a file object that has one word per line. All you should assume about this parameter is that you are able to do a loop, for w in words: to access its entries. The class is responsible for recording all words from the language into an internal data representation, and stripping any extraneous whitespace from each entry (such as newline characters that will appear in a file). For the sake of efficiency, we recommend that you store the language words in a Python set instance. (We discuss sets in a later section.) _contains_(self, query) The query parameter is a string. This method should determine whether the string is considered a legitimate word, returning True if the word is contained in the language and False otherwise. This method should adhere to the aforementioned conventions regarding capitalized and uncapitalized words. For example, dog, Dog and Missouri are contained in the English language, yet missouri and Missourri are not. The _contains_ special method is used by Python to support the in operator. It allows the standard syntax "Missouri' in language which is implicitly translated by Python to the internal call language. _contains_('Missouri') presuming that language is an instance of our LanguageHelper class. getSuggestions (self, query) Given a query string, this method should return an alphabetical list of "nearby" words in the language. Doing a good job at offering suggestions is the most difficult part of writing a good language helper. We discuss this aspect of the project in a later section. S = set() create a new set instance (which is initially an empty set). s.add(value) adds the given value to the set (value will be a string in our application). value in s returns True if the given value is currently in the set, and False otherwise.
Answer:
Check the explanation
Explanation:
class LanguageHelper:
language=set()
#Constructor
def __init__(self, words):
for w in words:
self.language.add(w)
def __contains__(self,query):
return query in self.language
def getSuggestionns(self,query):
matches = []
for string in self.language:
if string.lower().startswith(query) or query.lower().startswith(string) or query.lower() in string.lower():
matches.append(string)
return matches
lh = LanguageHelper(["how","Hi","What","Hisa"])
print('how' in lh)
print(lh.getSuggestionns('hi'))
===========================================
OUTPUT:-
==================
True
['Hisa', 'Hi']
====
A commercial enclosed gear drive consists of a 20o spur pinion having 16 teeth driving a 48-tooth gear. The pinion speed 350 rev/min, the face width 2 in, and the diametral pitch 6 teeth/in. The gears are grade 1 steel, through-hardened at 240 Brinell, made to No. 6 quality standards, uncrowned, and are to be accurately and rigidly mounted. Assume a pinion life of 108 cycles and a reliability of 0.90. Determine the AGMA bending and contact stresses and the corresponding factors of safety if 5 hp is to be transmitted.
Answer:
Check the explanation
Explanation:
Generally, when there is a pair of gears meshing, the minor gear is referred to as the pinion gear. in addition, it involves the meshing of cylindrical gear with a rack in a rack-and-pinion system which transforms to linear motion from a rotational motion.
Rack and pinion gears are mostly utilized in converting rotation into linear motion.
Kindly check the attached images below for the full step by step explanation to the question above.
The mathematical form of the stream-function for 2-D SSSF satisfies the Laplace Equation inside a 2-D region in space (or inside a control volume) with specific boundary conditions on the entire control surface enclosing the region, The values for the velocity components at any given point in the flow field are determined from:
A. The velocity components at arty point are found by solving the clinical Armour, equation
B. The velocity components at any pant are found by integrating the stream function around the boundary (control surface)
C. The velocity components at any pant are found by integrating the stream functran over the entire flow field region (control volume)
D The velocity components at any point are found from the first partial derivatives of the stream function at the given point
Answer:
D. The velocity components at any point are found from the first partial derivatives of the stream function at the given point
Explanation:
The stream function can be used to plot the streamlines of the flow and find the velocity. For two-dimensional flow the velocity components can be calculated in Cartesian coordinates by u = −∂ψ/∂y and v = ∂ψ/∂x,
where u and v are the velocity vectors (components) in the x and y directions, respectively, ψ is the stream function.
From the above, the velocity components at any point are found from the first partial derivatives of the stream function at the given point.
The fins are 90% efficient. Determine the minimum free-stream velocity the fan needs to supply to avoid overheating. Assume the flow is laminar over the entire finned surface of the transformer and check your assumption at the end. Assume the same convection heat transfer coefficient for the finned and unfinned area. Use 40 [C] for the film temperature.
Answer:
The minimum required free stream velocity required to dissipate 12W via fins is V∞ = 0.0020378 m/s = 2.0378 mm/s.
Explanation:
Given:-
- The dimension of transformer surface ( L , w , H ) = ( 10 cm long , 6.2 cm wide, 5 cm high )
- The dimensions of the fin : ( l , h , t ) = (10 cm long , 5 mm high, 2 mm thick )
- The total number of fins, n = 7
- The convection heat transfer coefficient of the finned and unfinned area = h.
- The efficiency of fins, ε = 0.9 ( 90% )
- The transformer fin base temperature, Tb = 60°C
- The free temperature of air, T∞ = 25°C
- The free stream velocity of air = U∞
Find:-
Determine the minimum free-stream velocity the fan needs to supply to avoid overheating. ( U∞ )
Solution:-
- Since the convection heat transfer coefficient of the finned and unfinned area i.e the fins and the transformer base are at the same temperature (Tb).
- The theoretical heat transfer ( Q_th ) rate from the fins can be calculated from the following convection cooling relation.
Q_th = h*As*[ Tb - T∞ ]
Where,
As : The total available surface area available for heat transfer.
Surface area of the fins (As1)As1 = n * { 2* [ ( l * h ) + ( t * h ) ] + ( l * t ) }
As1 = 7* { 2* [ ( 0.5 * 10 ) + ( 0.2 * 0.5 ) + ( 10 * 0.5 ) }
As1 = 106.4 cm^2 .... 0.01064 m^2
Surface area of the unfinned part of base (As2)As2 = Total base area - Finned top plane area
As2 = ( L * w ) - n* ( l * t ) = ( 10 * 6.2 ) - 7* ( 10 * 0.5 )
As2 = 27 cm^2 .... 0.0027 m^2
- Therefore, the total available surface area (As) is:
As = As1 + As2
As = 0.01064 + 0.0027
As = 0.01334 m^2
- The heat transfer coefficient (h) using convection heat transfer relation:
Q* ε = h*As*[ Tb - T∞ ]
h = Q* ε / [As*[ Tb - T∞ ] ]
h = (12*0.9) / [ 0.01334*( 60 - 25 ) ]
h = 23.13129 W/m^2K
- The air properties at film temperature:
T = 40 C
Viscosity ν = 1.6982 m^2 / s
Thermal conductivity, k = 0.027076 W/mK
Prandlt Number Pr = 0.71207
- The Nusselt number for the convection heat transfer for the transformer along the fins (Assumed flat plate):
Nu = h*L / k
Nu = 23.13129*0.1 / 0.027076
Nu = 85.43
- The correlation for Nusselt number between flow conditions and viscosity effects of the flow (Re & Pr) for a isothermal flat plate - Laminar Flow is given:
[tex]Nu = 0.664*Re^\frac{1}{2} *Pr^\frac{1}{3} \\\\Re^\frac{1}{2} = \frac{Nu}{0.664*Pr^\frac{1}{3}} \\\\\\Re = \sqrt{\frac{Nu}{0.664*Pr^\frac{1}{3}}} \\\\\\Re = \sqrt{\frac{85.43063}{0.664*0.71207^\frac{1}{3}}}\\\\Re = 12.00[/tex]
- The reynold number denotes the characteristic of the flow by the following relation:
Re = V∞*L / ν
V∞ = Re*ν / L
V∞ = 12*1.6982*10^-5 / 0.1
V∞ = 0.0020378 m/s .... = 2.0378 mm/s
You’re engineering an energy-efficient house that will require an average of 6.85 kW to heat on cold winter days. You’ve designed a photovoltaic system for electric power, which will supply on average 2.32 kW. You propose to heat the house with an electri-cally operated groundwater-based heat pump. What should you specify as the minimum acceptable COP for the pump if the pho-tovoltaic system supplies its energy?
Answer:
2.95 approximately 3
Explanation:
For a heat pump,
COP = Q/W
Where Q = power needed for heating process
W = power input into heat pump.
Power for heating Q = 6.85 kW
Proposed power input to heat pump W = 2.32 kW
Minimum COP = 6.85/2.32 = 2.95
Approximately 3
"A northbound freeway segment is on a 4% upgrade from station 430+20 to 450+00 and has two 11-ft wide lanes, a 5-ft right shoulder, and has a ramp density of 1 per mile in the 3 miles before and after station 440+10. The peak-hour factor is 0.9. Northbound traffic during the peak hour is 2550 cars, 300 STs, and 300 TTs. Determine the density and LOS of the freeway segment."
Answer: D = 23.09 pc/mi/h
LOS = C
Explanation:
we will begin by solving this with a step by step process for easy understanding;
given that the freeway has two lanes = 11 ft wide
and has a width of 5 ft right shoulder.
ramp density = 1 per mile in 3m
substituting the free flow speed value gives us;
Free flow speed = 75 - 1.9 - 0.6 - (3.22 × 1∧0.84) = 69.28 mph
we have that the length of the road = (1980 ft) × (1 mile / 5280 ft)
L = 0.375 mile
The next thing we will do is to calculate the proportion of bus and the truck
Pt = 300 + 300 / (2400 + 300 + 300)
Pt = 600/3000 = 20%
following up, we will consider the length and percentage of the buses and the trucks
Et = 2.0, Pr = 0, Er = 0
to calculate the percentage of truck
Ft = 1 / 1 + Pt (Et -1 ) + Pr (Er -1)
Ft = 1 / 1+0.2 (2-1) + 0 = 0.833
Truck percentage = 0.833
To the determine the traffic volume,
Vt = V / PHF × N × Ft × Fp
Vt = 2400/ (0.9×2×0.833×1) = 1600 pc/lane/hour
But the Density of the freeway is given thus;
D = Vp / FFS .............(1)
but to get the FFS, we will consider the graph of flow rate vs speed and show the level of service
FFS = 69.28 mi/h
From the above expression in (1) we have that
D = 1600/69.28 = 23.09 pc/mi/h
D = 23.09 pc/mi/h
now we have that the the density of the freeway segment is 23.09 pc/mi/h, we can thus safely say that the level of service (los) = C
cheers i hope this helps
5/25 At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing at a constant rate of 0.2 m/s. If vA = 0.6 m/s and θ = 35°, determine the velocity of slider B.
The image of the question is missing, so i have attached it
Answer:
Velocity of slider B; = - 0.176 m/s
Explanation:
We are given;
Length of (AB) = 0.75 m
Rate of increase of length; (AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
We want to find vB;
Looking at the image attached, we can use the trigonometric ratio to find OA
Thus;
Sin θ = (OA)/(AB)
So, Sin 35° = (OA)/(AB)
(OA) = (AB)Sin 35°
(OA) = 0.75•Sin 35°
(OA) = 0.75•0.5736
(OA) = 0.43 m
Also, we can use the same system to find (OB)
Thus;
Cos θ = (OB)/(AB)
Cos 35° = (OB)/(AB)
(OB) = (AB)Cos 35°
(OB) = 0.75•Cos 35°
(OB) = 0.75•0.8192
(OB) = 0.6144 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation;
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
Divide through by 2 to give;
(AB)*(AB)' = (OA)*vA + (OB)*vB
vB = ((AB)*(AB)' - (OA)*vA) / (OB)
We now have ;
vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s)/(0.614 m)
vB = - 0.176 m/s
Implement this C program by defining a structure for each payment. The structure should have at least three members for the interest, principle and balance separately. And store all the payments in a structure array (the max size of which could be 100). Name this C program as loanCalcStruct.c
Answer:
Explanation:
check the attached files for the solution and output result.
Which of these actions can be taken to minimize number of victims or prevent injury? (check all that apply) A. Hire good lawyers to defend those responsible for the disater. B. Regular inspections of man-made structures by licensed inspectors. C. Alarm systems and plans for evacuation and shelter of all communities neighboring a site of a potential disaster. D. Include fail-safe mechanisms in the design of life-critical systems
Answer:
The options that apply are:
B, C and D.
Explanation:
There have been a number of accidents all over the world resulting from Acts of God, professional negligence amongst other things.
These may not be avoided completely but the actions above speak to how they can be mitigated or reduced.
Cheers!
Water at 20◦C is pumped through 1000 ft of 0.425 ft diameter pipe at a volumetric flowrate of 1 ft3/s through a cast iron pipe that connects to connects two reservoirs. The elevation difference between the two reserviors is 120 ft. Find the pumping power delivered to the water. The minor losses only include a wide-open globe valve with KL = 10. Use a density of 1.94 slug/ft3 and a dynamic viscosity of 2.32 × 10−5 lbf·s/ft2 .
Answer:
7582.9 ft.Ibf/s
Explanation:
Given
L=1000ft,d=0.425ft,Q=1ft^3/s,z2-z1=120ft,Kl=10,d=1.94slug/ft^3, vicosity u= 2.32*10-5ibf.s/ft2
Reynold Re= Density*diameter*velocity/ viscosity
But Q=AV
V= 4/3.142*0.425=2.99ft/s
Re= 1.94*0.425*2.99/2.32*10-5)=106455.3
Friction factor=1/√f=-1.8log[((e/d)/3.7)^1.11+6.9/Re] is very neglible hence equals 0
Pump head Hp= z2-z1+v^2/2g[FL/f+KL]
Hp=120+2.99^2/2*32.2(0+10)=121.4ft
Pump power = density*g*Q*hp
1.94*32.2*121.4=7582.9 ft.Ibf/s
An incandescent lightbulb is an inexpensive but highly inefficient device that converts electrical energy into light. It converts about 10 percent of the electrical energy it consumes into light while converting the remaining 90 percent into heat. The glass bulb of the lamp heats up very quickly as a result of absorbing all that heat and dissipating it to the surroundings by convection and radiation.
Consider an 8-cm-diameter 60-W lightbulb in a room at 258C. The emissivity of the glass is 0.9. Assuming that 10 percent of the energy passes through the glass bulb as light with negligible absorption and the rest of the energy is absorbed and dissipated by the bulb itself by natural convection and radiation, determine the equilibrium temperature of the glass bulb. Assume the interior surfaces of the room to be at room temperature.
An incandescent light bulb's equilibrium temperature is determined by considering energy conversion, dissipation processes, and emissivity of the glass.
Explanation:An incandescent light bulb converts electrical energy into light and heat. In the given scenario, the glass bulb absorbs and dissipates heat through convection and radiation. To determine the equilibrium temperature of the glass bulb, we need to consider energy conversion and emissivity.
We know that 10% of the energy passes through the glass bulb as light while the remaining 90% is absorbed and dissipated. By calculating the energy balance and accounting for the emissivity of the glass, the equilibrium temperature of the glass bulb can be found.
Factors like the wattage of the bulb, its size, and the room temperature play a role in determining the final equilibrium temperature of the glass bulb based on the energy conversion and dissipation processes involved.
Water flows with an average speed of 6.5 ft/s in a rectangular channel having a width of 5 ft The depth of the water is 2 ft.
Part A
Determine the specific energy.
Express your answer to three significant figures and include the appropriate units.
E =
SubmitRequest Answer
Part B
Determine the alternate depth that provides the same specific energy for the same volumetric flow.
Express your answer to three significant figures and include the appropriate units.
Answer:
specific energy = 2.65 ft
y2 = 1.48 ft
Explanation:
given data
average speed v = 6.5 ft/s
width = 5 ft
depth of the water y = 2 ft
solution
we get here specific energy that is express as
specific energy = y + [tex]\frac{v^2}{2g}[/tex] ...............1
put here value and we get
specific energy = [tex]2 + \frac{6.5^2}{2\times 9.8\times 3.281}[/tex]
specific energy = 2.65 ft
and
alternate depth is
y2 = [tex]\frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})[/tex]
and
here Fr² = [tex]\frac{v1}{\sqrt{gy}} = \frac{6.5}{\sqrt{32.8\times 2}}[/tex]
Fr² = 0.8025
put here value and we get
y2 = [tex]\frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})[/tex]
y2 = 1.48 ft
For the following transfer function, derive expressions for the real and imaginary part for s = jω in terms of the frequency variable ω. Then write a MATLAB script to plot the imaginary part versus the real part (and its reflection about the real axis) for a frequency ω range of 10−2 to 102 radians per second. Verify that your plots match the output of the nyquist function in MATLAB. • G(s) = 1 /(s+0.5)(s+1)(s+2) Suppose G represents an open-loop plant transfer function. Use your plot to determine the Gain Margin for the closed-loop system, i.e., determine how much the loop gain could be increased before the closed-loop becomes unstable
Answer:
See all solutions attached as picture.
Explanation:
It is well explanatory
A short-circuit experiment is conducted on the high-voltage side of a 500 kVA, 2500 V/250 V, single-phase transformer in its nominal frequency. The short-circuit voltage is found as 100 V and the short-circuit current and power are 110 A and 3200 W, respectively. Find the series impedance of the transformer referred to its low voltage side.
Given Information:
Primary secondary voltage ratio = 2500/250 V
Short circuit voltage = Vsc = 100 V
Short circuit current = Isc = 110 A
Short circuit power = Psc = 3200 W
Required Information:
Series impedance = Zeq = ?
Answer:
Series impedance = 0.00264 + j0.00869 Ω
Step-by-step explanation:
Short Circuit Test:
A short circuit is performed on a transformer to find out the series parameters (Z = Req and jXeq) which in turn are used to find out the copper losses of the transformer.
The series impedance in polar form is given by
Zeq = Vsc/Isc < θ
Where θ is given by
θ = cos⁻¹(Psc/Vsc*Isc)
θ = cos⁻¹(3200/100*110)
θ = 73.08°
Therefore, series impedance in polar form is
Zeq = 100/110 < 73.08°
Zeq = 0.909 < 73.08° Ω
or in rectangular form
Zeq = 0.264 + j0.869 Ω
Where Req is the real part of Zeq and Xeq is the imaginary part of Zeq
Req = 0.264 Ω
Xeq = j0.869 Ω
To refer the impedance of transformer to its low voltage side first find the turn ratio of the transformer.
Turn ratio = a = Vp/Vs = 2500/250 = 10
Zeq2 = Zeq/a²
Zeq2 = (0.264 + j0.869)/10²
Zeq2 = (0.264 + j0.869)/100
Zeq2 = 0.00264 + j0.00869 Ω
Therefore, Zeq2 = 0.00264 + j0.00869 Ω is the series impedance of the transformer referred to its low voltage side.
An insulated rigid tank is divided into two compartments by a partition. One compartment contains 7 kg of oxygen gas at 40°C and 100 kPa, and the other compartment contains 4 kg of nitrogen gas at 20°C and 150 kPa. Now the partition is removed, and the two gases are allowed to mix. Determine
(a) the mixture temperature and
(b) the mixture pressure after equilibrium has been established.
Answer:
(a) The mixture temperature, T₃ is 305.31 K
(b) The mixture pressure, P₃ after establishing equilibrium is 114.5 kPa
Explanation:
Here we have the initial conditions as
Oxygen compartment
Mass of oxygen = 7 kg
Molar mass of oxygen = 32.00 g/mol
Pressure in compartment, P₁ = 100 kPa
Temperature of oxygen, T₁ = 40 °C = 313.15 K
Number of moles of oxygen, n₁ is given by
[tex]Number \ of\ moles \ of \ oxygen, n_1 = \frac{Mass \ of \ oxygen}{Molar \ mass \ of \ oxygen} = \frac{7000}{32} = 218.75 \ moles[/tex]
From the universal gas equation, we have;
P·V = n·R·T
[tex]V_1 = \frac{n_1RT_1}{P_1} =\frac{218.75 \times 8.3145 \times 313.15 }{100000 } = 5.696 m^3[/tex]
For the Nitrogen compartment, we have
Mass of nitrogen = 4 kg
Molar mass of oxygen = 28.0134 g/mol
Pressure in compartment, P₂ = 150 kPa
Temperature of oxygen, T₂ = 20 °C = 293.15 K
Number of moles of nitrogen, n₂ is given by
[tex]Number \ of\ moles \ of \ nitrogen, n_2 = \frac{Mass \ of \ nitrogen}{Molar \ mass \ of \ nitrogen} = \frac{4000}{28.0134} = 142.79 \ moles[/tex]
From the universal gas equation, we have;
P·V = n·R·T
[tex]V_2 = \frac{n_2RT_2}{P_2} =\frac{142.79\times 8.3145 \times 293.15 }{150000 } = 2.32 m^3[/tex]
Therefore
We have for nitrogen
[tex]\frac{c_p}{c_v} = 1.4[/tex]
[tex]c_p - c_v = 296.8 J/KgK[/tex]
Therefore;
[tex]Nitrogen, \ c_p = 1036 \ J/(kg \cdot K)[/tex]
[tex]Nitrogen, \ c_v = 740\ J/(kg \cdot K)[/tex]
The molar heat capacities of Nitrogen are therefore as follows;
[tex]Nitrogen, \ \tilde{c_p} = 29.134 \ kJ/(kmol \cdot K)[/tex]
[tex]Nitrogen, \ \tilde{c_v} = 20.819 \ kJ/(kmol \cdot K)[/tex]
For oxygen we have
[tex]Oxygen, \ \tilde{c_p} = 29.382 \ kJ/(kmol \cdot K)[/tex]
[tex]Nitrogen, \ \tilde{c_v} = 21.068 \ kJ/(kmol \cdot K)[/tex]
The final volume, V₃ then becomes
V₃ = V₁ + V₂ = 5.696 m³ + 2.32 m³ = 8.016 m³
(a) For adiabatic mixing of gases the final temperature of the mixture is then found as follows
Therefore before mixing
U₁ = [tex]\sum \left (n_i\tilde{c_v}_i T_i \right )[/tex] = 0.21875 × 21.068 × 313.15 + 0.14279×20.819×293.15 = 2,314.65 kJ
After mixing, we have
U₂ = [tex]T_3 \sum \left (n_i\tilde{c_v}_i \right )[/tex] = T (0.21875 × 21.068 + 0.14279×20.819) = T×7.58137001
Therefore the final temperature, T is then
[tex]T_3 = \frac{\sum \left (n_i\tilde{c_v}_i T_i \right )}{\sum \left (n_i\tilde{c_v}_i \right )} =\frac{2,314.65 }{7.58137001} =305.30761550 \ K[/tex]
The mixture temperature, T₃ = 305.31 K
(b) The mixture pressure, P₃ after equilibrium has been established is given as
[tex]P_3 = \frac{n_3 \tilde{R}T_3}{V_3}[/tex]
Where:
n₃ = n₁ + n₂ = 0.21875 + 0.14279 = 0.36154 kmol = 361.54 moles
[tex]\tilde{R}[/tex] = 8.3145 J/(gmol·K)
Therefore ,
[tex]P_3 = \frac{361.54 \times 8.3145 \times 305.31 }{8.016 } = 114,492.1766706961 Pa[/tex]
P₃ ≈ 114.5 kPa.
1. A spur gear made of bronze drives a mid steel pinion with angular velocity ratio of 13 /2 : 1. Thepressure angle is 14 1/2° . It transmits 5 kW at 1800 r.p.m. of pinion. Considering only strength, design the smallest diameter gears and find also necessary face width. The number of teeth should not be less than 15 teeth on either gear. The elastic strength of bronze may be taken as 84 MPa and of steel as 105 MPa. Lewis factor for 14 1/2° pressure angle may be taken as y =0.6840.124 –No. of teeth.
Answer:
Given data
w1/w2=6.5/1
Power=5 KW
wp=1800 rpm
angle=14 degrees
Based on above values,the minimum diameter=30 mm
Diesel-fueled generators are frequently used as backup electrical power sources for homes and hospitals. Consider a Diesel powered generator with an efficiency of 39 percent for an engine speed of idle to about 1,800 rpm (revolutions per minute). If Diesel fuel has a chemical formula of C12H23.
a. Determine the chemical reaction for 1 kmol of Diesel fuel burning with the stoichiometric amount of air.
b. For each kg of Diesel fuel burned, how much CO2 is generated, in kg?
c. Find the higher heating value (HHV) for C12H23 at 25°C, 1 atm.
d. Calculate the amount of Diesel fuel, in kg and in gallons, required to produce a power output of 18 kW to a home for a period of 8 h.
e. Comment on your results.
Answer:
a. C12H23 + 84.5 moles of air —-> 12CO2(g)+ 11.5H2O(g)
b. 3.2kg of CO2 per 1kg of C12H23
c. HVV of C12H23 is -1724.5 KCal/mol
d. Total weight required is 30.742kg
e. The amount of CO2 produced per kg of C12H23 is too much. CO2 is harmful to the environment and should be produced in weights as low as possible
Explanation:
Please check attachment for complete solution and step by step explanation
Evaluate the performance of the proposed heat pump for three locations Using R134a. Discuss the effect of outdoor temperature on the performance of the heat pump. What happens to the COP if the heat exchangers are only 80% effective, Philadelphia only. Discuss improvements to the design for Philadelphia that will increase the COP closer to the theoretical max (you must show supporting numbers). Your report should include tables that list the information at each state, the evaluated work, heat transfer, and calculated COP values. Sample calculations of your work should be included in an appendix.
Answer:Table 2.2: Differences in runstitching times (standard − ergonomic).
1.03 -.04 .26 .30 -.97 .04 -.57 1.75 .01 .42
.45 -.80 .39 .25 .18 .95 -.18 .71 .42 .43
-.48 -1.08 -.57 1.10 .27 -.45 .62 .21 -.21 .82
A paired t-test is the standard procedure for testing this null hypothesis.
We use a paired t-test because each worker was measured twice, once for Paired t-test for
each workplace, so the observations on the two workplaces are dependent. paired data
Fast workers are probably fast for both workplaces, and slow workers are
slow for both. Thus what we do is compute the difference (standard − er-
gonomic) for each worker, and test the null hypothesis that the average of
these differences is zero using a one sample t-test on the differences.
Table 2.2 gives the differences between standard and ergonomic times.
Recall the setup for a one sample t-test. Let d1, d2, . . ., dn be the n differ-
ences in the sample. We assume that these differences are independent sam-
ples from a normal distribution with mean µ and variance σ
2
, both unknown.
Our null hypothesis is that the mean µ equals prespecified value µ0 = 0
(H0 : µ = µ0 = 0), and our alternative is H1 : µ > 0 because we expect the
workers to be faster in the ergonomic workplace.
The formula for a one sample t-test is
t =
¯d − µ0
s/√
n
,
where ¯d is the mean of the data (here the differences d1, d2, . . ., dn), n is the The paired t-test
sample size, and s is the sample standard deviation (of the differences)
s =
vuut
1
n − 1
Xn
i=1
(di − ¯d )
2 .
If our null hypothesis is correct and our assumptions are true, then the t-
statistic follows a t-distribution with n − 1 degrees of freedom.
The p-value for a test is the probability, assuming that the null hypothesis
is true, of observing a test statistic as extreme or more extreme than the one The p-value
we did observe. “Extreme” means away from the the null hypothesis towards
the alternative hypothesis. Our alternative here is that the true average is
larger than the null hypothesis value, so larger values of the test statistic are
extreme. Thus the p-value is the area under the t-curve with n − 1 degrees of
freedom from the observed t-value to the right. (If the alternative had been
µ < µ0, then the p-value is the area under the curve to the left of our test
Explanation: The curve represents the sum total of the evaluation
The rainfall rate in a certain city is 20 inches per year over an infiltration area that covers 33000 acres. Twenty percent of the rainfall percolates into the groundwater, with the remaining 80% running off into the river. The city uses 83000 acre-ft per year, some of which comes from the river and the remainder from groundwater. The groundwater volume of fresh water is currently at 1.1 × 105 acre-ft and is expected to last for 30 years before being completely depleted of fresh water (assume uniform withdrawal each). Assuming the groundwater lasts exactly as expected, determine the rate at which water is being withdrawn from the river by the city.
Answer:
The rate at which water is being withdrawn from the river by the city is 57353 acre-ft/y
Explanation:
Please look at the solution in the attached Word file
Water flows steadily through a fire hose and nozzle. The hose is 75 mm inside diameter, and the nozzle tip is 25 mm inside diameter; water gage pressure in the hose is 510 kPa, and the stream leaving the nozzle is uniform. The exit speed 32 m/s and pressure is atmospheric. Determine the force transmitted by the coupling between the nozzle and hose. (25 points)
Answer:
R = 1804 N
Explanation:
Given:-
- The density of water, ρ = 997 kg/m^3
- The inside diameter of the hose, dh = 75 mm
- The gauge pressure of water in the hose, P1 = 510 KPa
- The exit speed of the water, V2 = 32 m/s
- The inside diameter of the nozzle tip, dn = 25 mm
- The atmospheric pressure (gauge), P2 = 0 KPa ... P = 1 atm (Absolute).
Find:-
Determine the force transmitted by the coupling between the nozzle and hose.
Solution:-
- We will first develop a control surface at the hose-nozzle interface.
- Assuming steady and one dimensional flow - (x-direction).
- Since there are no fictitious unbalanced forces acting on the fluid flow due to roughness of hose any any losses of energy from the fluid are negligible.
- The use of conservation of momentum of fluid flow is valid for an isolated system, where the flow of fluid into the control volume is denoted by (-) and the flow of fluid going out of the control volume is denoted by (+):
- The principle of conservation of momentum, the pair of equal force (Newton's third law) act on the control volume at (nozzle-hose) interface:
R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)
Where, Q: Flow rate
V1: The velocity of fluid in hose
A1: Cross sectional area of the hose
A2: Cross sectional area of the nozzle exit
- We see that the reaction force (R) that acts on nozzle-hose interface is due to changes in dynamic and hydrostatic pressures.
- Compute the required quantities Q, A1 and A2 and V1 using the given data:
- The flow rate Q for any flow in the hose can be given, where the cross sectional area of hose (A1)is:
[tex]A1 =\pi\frac{d_h^2}{4} = \pi\frac{0.075^2}{4} \\\\A1 = 0.00441 m^2\\\\\\[/tex]
- The cross sectional area of the nozzle tip with diameter dn = 25 mm is:
[tex]A2 =\pi\frac{d_n^2}{4} = \pi\frac{0.025^2}{4} \\\\A2 = 0.00049 m^2\\\\\\[/tex]
- The flow rate (Q) can now be calculated:
[tex]Q = A2*V2\\\\Q = (0.00049)*(32)\\\\Q = 0.01570 \frac{m^3}{s}[/tex]
- Since, the density of the water does not vary along the direction of flow, the flow rate (Q) remains constant throughout. So from continuity equation we have:
[tex]Q = A2*V2 = A1*V1\\\\V1 = \frac{Q}{A1} = \frac{0.0157}{0.00441} \\\\V1 = 3.56189 \frac{m}{s}[/tex]
- Now use the calculated quantities and compute the pair of reaction force at the nozzle-hose interface:
R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)
R = (997)*(0.01570)*(32-3.56189) + (0 - 510*0.00441)*1000
R = 445.13889 - 2,249.1
R = - 1803.961 ≈ -1,804 N
- Here the negative sign denotes the direction of in which the force (R) is exerted. Since, (-) denotes into the control volume it acts opposite to the flow of water.
The coupling between the nozzle and hose is -1.81N
This question relates to flow rate of a liquid
Data given:
The density of water = 997kg/m^3
The inside diameter of the hose = 75mm = 0.0075m
The gauge pressure of water in the hose = 510kPa
The exit speed of the water = 32m/s
The inside diameter of the nozzle tip = 25mm = 0.0025m
The atmospheric pressure = 0kPa or 1atm
Let's calculate the inlet velocity
[tex]v_1=v_2=A_2/A_1\\v_1=V_2(\frac{d_2}{d_1})^2\\v_1=32(\frac{25}{75})^2\\v_1=3.50m/s[/tex]
Calculating the force transmitted by coupling between the nozzle and hose
[tex]R_x+p_1gA_1=v_1[-|pv_1A_1|]+v_2[|pv_2A_2|]\\[/tex]
μ[tex]_1[/tex]=[tex]v_1[/tex] and μ[tex]_2[/tex] =[tex]v_2[/tex]
[tex]R_x=-p_1gA_1-v_1pv_1A_1+v_2pv_2A_2\\R_x=-p_1gA+pv_2A_2(v_2-v_1)\\R_x=-510*10^3N/m^3*\frac{\pi }{4}(0.075m)^2+997kg/m^3*32m/s*\frac{\pi }{4} (0.025m)^2(32-3.50)=-1805=-1.81kN[/tex]
The force between the nozzle and hose is -1.81
Learn more about flow rate;
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Each of the two drums and connected hubs of 13-in. radius weighs 210 lb and has a radius of gyration about its center of 30 in. Calculate the magnitude of the angular acceleration of each drum. Friction in each bearing is negligible.
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The angular acceleration for first drum [tex]\alpha = 0.792 rad/s^2[/tex]
The angular acceleration for the second drum is [tex]\alpha =1.262[/tex]
Explanation:
From the question we are told that
Their radius of the drum is [tex]r = 13 in = \frac{13}{12} ft = 1.083ft[/tex] each
The weight is [tex]W = 210 lb[/tex]
The mass is [tex]M = \frac{210 lb}{32.2 ft /s^2} = 6.563\ lb s^2 ft^{-1}[/tex]
Their radius of gyration is [tex]z=30 in= \frac{30 }{12} = 2.5 ft[/tex]
The free body diagram of a drum and its hub and 30lb and in the case the weight is connect to the hub separately is shown on the second uploaded image
The T in the diagram is the tension of the string
Now taking moment about the center of the the drum P we have
[tex]\sum M_p = I_p \alpha[/tex]
=> [tex]T * r = Mz^2 * \alpha[/tex]
Where r is the radius ,z is the radius of gyration about the center O , M is the mass of the drum including the hub, and [tex]\alpha[/tex] is the angular acceleration
Inputting
[tex]T * 1.083 = 6.563 * 2.5^2 \alpha[/tex]
=> [tex]T = 37.87\alpha[/tex]
Considering the force equilibrium in the vertical direction (Looking at the second free body diagram now )
The first on is
[tex]\sum F_y = ma[/tex]
=> [tex]30lb - T = m(r \alpha )[/tex]
Where m is the mass of the hanging block which has a value of
[tex]m = \frac{30lb}{32.2 ft/s^2} = 0.9317 \ lb ft^{-1} s^2[/tex]
a is the acceleration of the hanging block
inputting values we have
[tex]30- 37.87 \alpha = 0.9317* 1.083 \alpha[/tex]
[tex]30 = 37.87\alpha + \alpha[/tex]
[tex]\alpha = \frac{30}{38.87 }[/tex]
[tex]\alpha = 0.792 rad/s^2[/tex]
So the angular acceleration for first drum [tex]\alpha = 0.792 rad/s^2[/tex]
The free body diagram of a drum and its hub when the only on the string is 30lb is shown on the third uploaded image
So here we would take the moment about O
[tex]\sum M_o = I_O \alpha[/tex]
So [tex]\sum M_o = 30* 1.083[/tex]
and [tex]I = M z^2[/tex]
Therefore we will have
[tex]30 * 1.083 = (Mz^2 )\alpha[/tex]
inputting values
[tex]30 * 1.083 = 6.563 * 2.5^2 \alpha[/tex]
[tex]32.49=41.0\alpha[/tex]
[tex]\alpha =\frac{41}{32.49}[/tex]
[tex]\alpha =1.262[/tex]
So the angular acceleration for the second drum is [tex]\alpha =1.262[/tex]
5. Steel balls 50 mm in diameter are annealed by heating to 1200 K and then slowly cooling to 450 K in an air environment for which the ambient temperature is 300 K and h = 20 W/m2 ·K. Assuming the properties of the steel to be k = 40 W/m·K, rho = 7800 kg/m3 , and c = 600 J/kg·K. Estimate the time required for this cooling process.
Answer:
time required for cooling process = 0.233 hours
Explanation:
In Transient heat conduction of a Sphere, the formula for Biot number is;
Bi = hL_c/k
Where L_c = radius/3
We are given;
Diameter = 12mm = 0.012m
Radius = 0.006m
h = 20 W/m²
k = 40 W/m·K
So L_c = 0.006m/3 = 0.002m
So,Bi = 20 x 0.002/40
Bi = 0.001
The formula for time required is given as;
t = (ρVc/hA)•In[(T_i - T_(∞))/(T - T_(∞))]
Where;
A is Area = πD²
V is volume = πD³/6
So,
t = (ρ(πD³/6)c/h(πD²))•In[(T_i - T_(∞))/(T - T_(∞))]
t = (ρDc/6h)•In[(T_i - T_(∞))/(T - T_(∞))]
We are given;
T_i = 1200K
T_(∞) = 300K
T = 450K
ρ = 7800 kg/m³
c = 600 J/kg·K
Thus, plugging in relevant values;
t = (7800 x 0.012 x 600/(6 x20) )•In[(1200 - 300)/(450 - 300)]
t = 468•In6
t = 838.54 seconds
Converting to hours,
t = 838.54/3600
t = 0.233 hours
A piston-cylinder assembly contains 2 lb of air at a temperature of 540 °R and a pressure of 1 atm. The air compressed to a state where the temperature is 840 °R and the pressure is 6 atm. During the compression, there is a heat transfer from the air to the surroundings equal to 20 Btu.
1. Using the ideal gas model for air, determine the work during the process in Btu.
Answer:
123.9 Btu
Explanation:
The energy balance on the air is:
∆E = E2 − E1 = ∆KE + ∆PE + ∆U = Q + W
ignore ∆KE and ∆PE,
W = ∆U − Q = m(u2 − u1) − Q; (u2 − u1 = 51.94 Btu/lb)
ideal gas properties is attached
W = (2 lb)(143.98 − 92.04) Btu/lb − (− 20 Btu) = 123.9 Btu
u2 − u1 ≈ cv(T2 − T1) = (0.173 Btu/lb°R)(840 − 540) °R = 51.9 Btu/lb
The net work done in compressing the air as given is; W = -123.8 Btu
What is the Energy Balance equation?The equation for Energy Balance in thermodynamics is;
Q - W = ΔU
where;
Q is
ΔU is change in the internal energy of the system
Q is the net heat transfer
W is Net work done
Now, ΔU can also be written as;
ΔU = mC_v(T₂ - T₁)
C_v for air is 0.173 Btu/bm.R
We are given;
m = 2 lb
T₁ = 540 °R
T₂ = 840 °R
Q = -20 Btu (negative because heat is transferred to the surrounding)
Thus;
ΔU = 2 * 0.173 * (840 - 540)
ΔU = 103.8 Btu
Work done during the process is;
W = Q - ΔU
W = -20 - 103.8
W = -123.8 Btu
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A shaft is loaded in bending and torsion such that Ma = 70 N ? m, Ta = 45 N ? m, Mm = 55 N ? m, and Tm = 35 N ? m. For the shaft, Su = 700 MPa and Sy = 560 MPa, and a fully corrected endurance limit of Se = 210 MPa is assumed. Let Kf = 2.2 and Kfs = 1.8. With a design factor of 2.0 determine the minimum acceptable diameter of the shaft using
a. ) DE-Gerber criterion.
b) DE-ASME Elliptic criterion.
c) DE-Soderberg criterion.
d) DE-Goodman criterion.
Answer:
Please see the attached picture for the compete answer.
Explanation:
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm, and its flight angle = 18°. Head pressure at the die end of the barrel = 10.0(10^6) Pa. Viscosity of the polymer melt = 100 Pa- s
(a) Find the volume flow rate of plastic at the die end of the barrel.
________ x 10^-6 m^3/s
Answer:
Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 × [tex]10^6[/tex] Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85 [tex]\times 10^{-3}[/tex] )²× 9 [tex]\times 10^{-3}[/tex] × sin18 × cos18
Qd = 94.305 × [tex]10^{-6}[/tex] m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 × [tex]10^{6}[/tex] × π × 85 [tex]\times 10^{-3}[/tex] × ( 9 [tex]\times 10^{-3}[/tex] )³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 × [tex]10^{-6}[/tex] m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 × [tex]10^{-6}[/tex] - 85.2 × [tex]10^{-6}[/tex]
Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s
The density of a liquid is to be determined by an old 1-cm-diameter cylindrical hydrometer whose division marks are completely wiped out. The hydrometer is first dropped in water, and the water level is marked. The hydrometer is then dropped into the other liquid, and it is observed that the mark for water has risen 1.4 cm (hw) above the liquid–air interface. If the height of the original water mark is 23 cm (hl + hw), determine the density of the liquid.
Answer:
1064.8 kg/m³
Explanation:
Weight of the hydrometer = ρghA where ρ is the density, g is acceleration due to gravity, h is the submerged height and A is the cross sectional area.
W in water = ρwghwA
W in liquid = (ρliq)g hliq A where the cross sectional area is constant
W in water = W in liquid
(ρw)ghwA = (ρliq)g hliq A where ρw is density of water, ρliq is the density of liquid and hw and hliq are the heights of the liquid and that water. g acceleration due to gravity cancel on both sides as well as the constant A
pliq = [tex]\frac{hw}{hliq}[/tex] × 1000 kg /m³ ( density of water) =( [tex]\frac{23}{23-1.4}[/tex]) × 1000 = 1064.8 kg/m³
A completely reversible heat pump produces heat at a rate of 100 kW to warm a house maintained at 25℃. The exterior air, which is at 0℃, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.
Answer:
1) 0. 03 kW/K
2) the value is grater than zero so it satisfies the second law of thermodynamic (states that rate of entropy change must be equal to or greater than zero) .
Explanation:
Rate of entropy change S = dQ/dT
= Q(1/T1 - 1/T2)
T2 = 25°C = 298 K
T1 = 0°C = 273 K
Q = 100 kW
S = 100( 1/273 - 1/298)
S = 100(0.0003) = 0. 03 kW/K