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]
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.
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]
All hermetic compressors require a crankshaft seal.
True
False
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)
An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capacitor and a 301 Ω resistor.
(a) What is the impedance of the circuit?
(b) What is the rms current through the resistor?
(c) What is the average power dissipated in the circuit?
(d) What is the peak current through the resistor?
(e) What is the peak voltage across the inductor?
(f) What is the peak voltage across the capacitor?
The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?
Answer:
Explanation:
f = 50.0 Hz, L = 0.650 H, π = 3.14
C = 4.80 μF, R = 301 Ω resistor. V = 120volts
XL = wL = 2πfL
= 2×3.14×50* 0.650
= 204.1 Ohm
Xc= 1/wC
Xc = 1/2πfC
Xc = 1/2×3.14×50×4.80μF
= 1/0.0015072
= 663.48Ohms
1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2
√ 90601 + (459.38)^2
√ 90601+211029.98
√ 301630.9844
= 549.209
Z = 549.21Ohms
2. I=V/Z = 120/ 549.21Ohms =0.218Ampere
3. P=V×I = 120* 0.218 = 26.16Watt
Note that
I rms = Vrms/Xc
= 120/663.48Ohms
= 0.18086A
4. I(max) = I(rms) × √2
= 0.18086A × 1.4142
= 0.2557
= 0.256A
5. V=I(max) * XL
= 0.256A ×204.1
=52.2496
= 52.250volts
6. V=I(max) × Xc
= 0.256A × 663.48Ohms
= 169.85volts
7. Xc=XL
1/2πfC = 2πfL
1/2πfC = 2πf× 0.650
1/2×3.14×f×4.80μF = 2×3.14×f×0.650
1/6.28×f×4.8×10^-6 = 4.082f
1/0.000030144× f = 4.082×f
1 = 0.000030144×f×4.082×f
1 = 0.000123f^2
f^2 = 1/0.000123048
f^2 = 8126.922
f =√8126.922
f = 90.14 Hz
When replacing a timing belt, many experts and vehicle manufacturers recommend that all of the following should be replaced except the
A. water pump
B. camshaft oil seal(s).
C. camshalt sprocket
D. tensioner assembly
Answer:
Correct Answer:
A. water pump
Explanation:
Timing belt in a vehicle helps to ensure that crankshaft, pistons and valves operate together in proper sequence. Timing belts are lighter, quieter and more efficient than chains that was previously used in vehicles.
Most car manufacturers recommended that, when replacing timing belt, tension assembly, water pump, camshaft oil seal should also be replaced with it at same time.
A charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 36 N/C. The electric field 4 cm from the wire is:
Answer:
New electric field = 18 N/C
Explanation:
Given:
Length (E1) = 2 cm
New length (E2) = 4 cm
Electric field = 36 N/C
Find:
New electric field
Computation:
New electric field = 36 [2 / 4]
New electric field = 36 [1/2]
New electric field = 18 N/C
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 )
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
(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]
"The transistor base-emitter voltage (VBE) a. increases with an increase in temperature. b. is not affected by temperature change. c. decreases with an increase in temperature. d. has no effect on collector current."
Answer:
C) Decreases with an increase in temperature
Explanation:
As the temperature of a transistor increases, the thermal runaway property of the transistor becomes more significant and the transistors, conducting more freely as a result of the rise in temperature, causes an increase in the collector current or leakage current. The transistor base-emitter voltage decreases as a result.
With increased heating due to heavy current flow, the transistor is damaged.
A plate is supported by a ball-and-socket joint at A, a roller joint at B, and a cable at C. How many unknown support reactions are there in this problem?
Answer:
There are five (5) unknown support reactions in this problem.
Explanation:
A roller joint rotates and translates along the surface on which the roller rests. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. This allows the roller to move in a single plane along the surface where it rests.
A cable support provides support in one direction, parallel, and in opposite direction to the load on it. There exists a single reaction from the cable pointed upwards.
A ball-and-socket joint have reaction forces in all 3 cardinal directions. This allows it to move in the x-y-z plane.
The total unknown reactions on the member are five in number.
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 phase sequence of a 3-phase system for which VAN = 120 /90o V and VBN = 120 /210o V is:_______
a) bca
b) abc
c) cab
d) acb
Answer:
b. abc
Explanation:
The phase rotation and its sequence is order in which voltage is reached to their respective sources. The wave forms of a polyphase flow through AC source to the panel. The three phase system has two possible sequence which can be a-b-c or c-b-a. The correct answer is therefore b.
Where are revolved sections placed in a print? A) in between break lines B) cutting planes are used to identify their locations C) in between section lines D) stand alone
Answer:
B. Cutting planes are used to identify their locations.
Explanation:
Revolved view is a cross section view of revolved 90 degrees around a cutting plane projections. The revolved view of print will differ from a cross sectional view. It includes a line nothing the axis of revolution for the view. The correct answer is B. The revolved section in the prints has cutting planes that will be used to identify their location.
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.
Strain gage is a device that senses the strain of the structure. The property of the strain gage that is used to correlate with the strain to be measured is
Answer:
resistance
Explanation:
A strain gauge changes resistance with applied strain.
The plate is made of steel having a density of 7910 kg/m3 .If the thickness of the plate is 9 mm , determine the horizontal and vertical components of reaction at the pin A and the tension in cable BC.
Answer:
Horizontal component ( Ax) = 0
vertical component ( Ay ) = 1794.87 N
Explanation:
Attached below is the detailed solution and the free body diagram of the question
given data:
density = 7910 kg/m3
thickness of plate = 9 mm = 0.009
determine the horizontal and vertical components of the reaction at the pin A and the tension in the cable BC
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."
Technician A says that when the malfunction indicator light or service engine light is on you should retrieve the diagnostic trouble code and follow the manufacturers recommended procedure. Technician B says that all obd-ll monitors must have the enabled criteria achieved before a test is performed. Who is correct?
Answer:both
Explanation:
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
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:
Conductivity is the reciprocal of what?
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.
B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget that is 5.5 mm in diameter and 5.0 mm thick after 0.3 sec welding time. The unit melting energy for a certain sheet metal is 9.5 J/mm3 . The electrical resistance between the surfaces is 140 micro ohms, and only one third of the electrical energy generated will be used to form the weld nugget (the rest being dissipated), determine the minimum current level required.
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 )
After clamping a buret to a ring stand, you notice that the set-up is tippy and unstable. What should you do to stabilize the set-up
Answer:
Move the buret clamp to a ring stand with a larger base.
Explanation:
A right stand is used for titration experiments in the laboratory. It holds the burette firmly during experiments so that accurate readings can be taken.
The right stand is made up of support base, vertical stainless steel, clamp with adjustable screw that holds on to the vertical rod.
The clamp is used to hold the burette in place.
If after clamping a buret to a ring stand, you notice that the set-up is tippy and unstable, the best action will be to move the buret clamp to a ring stand with a larger base.
The larger base provides a better center of gravity and stabilises the setup
The natural variation of a process relative to the variation allowed by the design specifications is known as
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.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
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
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
