Answer:
Can you make friend with me ?
how to solve circuit theory using mesh analysis
Explanation:
Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.
This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.
__
Example
Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...
mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)
mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs
mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0
You will note that the matrix of equation coefficients is symmetric.
__
In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.
By using order of magnitude analysis, the continuity and Navier-Stokes equations can be simplified to the Prandtl boundary-layer equations. For steady, incompressible, and two-dimensional flow, neglecting gravity, the result is delta u/ delta x + delta v/ delta y= 0; u delta u/ delta x +v delta u/ delta y= -1/p(delta u/ delta x)+ v delta^2 u/ delta y^2 Use L and V0 as characteristic length and velocity, respectively. Non-dimensionalize these equations and identify the similarity parameters that result.
Answer: Attached below is the well written question and solution
answer:
i) Attached below
ii) similar parameter = [tex]\frac{V}{VoL } = 1 / Re[/tex]
Explanation:
Using ; L as characteristic length and Vo as reference velocity
i) Nondimensionalize the equations
ii) Identifying similarity parameters
the similar parameters are = [tex]\frac{V}{VoL } = 1 / Re[/tex]
Attached below is the detailed solution
Calculate the pressure of dry O2 if the total pressure of O2 generated over water is measured to be 698 Torr and the temperature is 30.1 oC. P(H2O) = 19.8 torr.
If the volume of the O2 sample in the question above was 56.3 ml, what volume would the dry O2 occupy at 755 torr (assume the temp was unchanged).
Answer:
[tex]V_2=46mL[/tex]
Explanation:
From the question we are told that:
Pressure over Water [tex]P=698 Torr[/tex]
Temperature [tex]T= 30.1 \textdegree C[/tex]
Pressure of Water [tex]P(H2O) = 19.8 torr.[/tex]
Volume of O2 [tex]O_2=56.3[/tex]
Pressure of Dry O2 [tex]P_(0)=755torr[/tex]
Generally the equation for Total Pressure is mathematically given by
[tex]P_t = P_O + P_H[/tex]
Therefore
[tex]P_O=P_t-P_H[/tex]
[tex]P_O=638-19.8[/tex]
[tex]P_O=618.2torr[/tex]
Generally the equation for Ideal gas is mathematically given by
[tex]P_1*V_1 = P_2*V_2[/tex]
[tex]V_2=\frac{P_1*V_1}{P_2}[/tex]
Therefore
[tex]V_2=\frac{ 618.2*56.3}{755}[/tex]
[tex]V_2=46mL[/tex]
Hence,The volume would the dry O2 occupy at 755 torr
[tex]V_2=46mL[/tex]
A recessed luminaire bears no marking indicating that it is ""Identified for Through- Wiring."" Is it permitted to run branch-circuit conductors other than the conductors that supply the luminaire through the integral junction box on the luminaire?
Answer:
No it is not permitted
Explanation:
It is not permitted because as per NEC 410.21 policy no other conductor is allowed to be passed through integral junction box luminaries unless such conductor supply recessed luminaries.
The marking will show that the Luminaries is of the right construction or right installation to ensure that the the conductors ( in the outer boxes ) will not be exposed to temperatures greater than the conductor rating, hence the lack of marking makes it not to be permitted.
Maggie discovered that a pipe in her basement has sprung a leak. She calls a plumber but in the meantime she grabs a roll of duct tape and wraps it around the pipe to stop the water from leaking. The duct tape in this situation is similar to _____ in that it/they _____.
Answer:
The answer is "Option a".
Explanation:
Myelination was the myelinization mechanism of a neuron axon. The endothelium is enveloped all around the axon and isolates the axon that inhibits the neuronal message from leaking with the other neuronal axons. Inside this example, therefore, its tubes tape worked similarly to those of myelin sheath, which stops brain transmission.
The Myelination was the myelinization mechanism of the neuron axon. Thus the option A is correct.
What is a Myelination?It is the process by which the brain's oligodendrocytes make layers of myelin and is wound around the neural axons and is seen as a layer of insulation for the transmission electrical potential down to the neuronal axon.
Find out more information about the sentence.
brainly.com/question/5636726
Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domestic wastewater from an influent BOD concentration of 250 mg/L to an effluent concentration of 10 mg/L. X (MLVSS) = 3000 mg/L, and the kinetics are first order and not variable order. The first order equation you must use to calculate the specific substrate utilization rate is q = K S where S is the effluent BOD concentration and K is the first order BOD degradation rate constant. The value of K is 0.04 L/(day*mg). What is the required reactor volume in MG (millions of gallons)? All the choices below are in units of MG.
0.4
1.0
0.2
4.8
Answer:
1.0MG
Explanation:
to solve this problem we use this formula
S₀-S/t = ksx --- (1)
the values have been given as
concentration = S₀ = 250mg
effluent concentration = S= 10mg
value of K = 0.04L/day
x = 3000 mg
when we put these values into this equation,
250-10/t = 0.04x10x3000
240/t = 1200
we cross multiply from this stage
240 = 1200t
t = 240/1200
t = 0.2
remember the question says that 5MGD is required to be treated
so the volume would be
v = 0.2x5
= 1.0 MG
. Bơm kiểu piston tác dụng đơn có áp suất p=0,64 Mpa và lưu lượng Q=3,5 l/s. Xác định tốc độ quay của trục bơm và công suất của bơm nếu biết đường kính piston D=150 mm; bán kính tay quay R=60 mm; hiệu suất thể tích của bơm là 0=0,94; hiệu suất chung của bơm b=0,80.
Answer:
not understand language
2 Air enters the compressor of a cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rates of exergy destruction in the compressor and turbine, respectively, each in kW, for T0 5 300 K.
State three types of maintenance.
Answer:
Tradicionalmente, se han distinguido 5 tipos de mantenimiento, que se diferencian entre sí por el carácter de las tareas que incluyen:
Explanation:
Mantenimiento Correctivo: Es el conjunto de tareas destinadas a corregir los defectos que se van presentando en los distintos equipos y que son comunicados al departamento de mantenimiento por los usuarios de los mismos.
Mantenimiento Preventivo: Es el mantenimiento que tiene por misión mantener un nivel de servicio determinado en los equipos, programando las intervencions de sus puntos vulnerables en el momento más oportuno. Suele tener un carácter sistemático, es decir, se interviene aunque el equipo no haya dado ningún síntoma de tener un problema.
Mantenimiento Predictivo: Es el que persigue conocer e informar permanentemente del estado y operatividad de las instalaciones mediante el conocimiento de los valores de determinadas variables, representativas de tal estado y operatividad. Para aplicar este mantenimiento, es necesario identificar variables físicas (temperatura, vibración, consumo de energía, etc.) cuya variación sea indicativa de problemas que puedan estar apareciendo en el equipo. Es el tipo de mantenimiento más tecnológico, pues requiere de medios técnicos avanzados, y en ocasiones, de fuertes conocimientos matemáticos, físicos y/o técnicos.
Mantenimiento Cero Horas (Overhaul): Es el conjunto de tareas cuyo objetivo es revisar los equipos a intervalos programados bien antes de que aparezca ningún fallo, bien cuando la fiabilidad del equipo ha disminuido apreciablemente de manera que resulta arriesgado hacer previsiones sobre su capacidad productiva. Dicha revisión consiste en dejar el equipo a Cero horas de funcionamiento, es decir, como si el equipo fuera nuevo. En estas revisiones se sustituyen o se reparan todos los elementos sometidos a desgaste. Se pretende asegurar, con gran probabilidad un tiempo de buen funcionamiento fijado de antemano.
Mantenimiento En Uso: es el mantenimiento básico de un equipo realizado por los usuarios del mismo. Consiste en una serie de tareas elementales (tomas de datos, inspecciones visuales, limpieza, lubricación, reapriete de tornillos) para las que no es necesario una gran formación, sino tal solo un entrenamiento breve. Este tipo de mantenimiento es la base del TPM (Total Productive Maintenance, Mantenimiento Productivo Total).
A steel component with ultimate tensile strength of 800 MPa and plane strain fracture toughness of 20 MPam is known to contain a tunnel (internal) crack of length 1.4 mm. This alloy is being considered for use in a cyclic loading application for which the design stresses vary from 0 to 410 MPa. Would you recommend this alloy for this application
Complete question:
A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?
a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.
c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.
d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.
Answer:
A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
Explanation:
we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.
thank you!
Consider CO at 500 K and 1000 kPa at an initial state that expands to a final pressure of 200 kPa in an isentropic manner. Report the final temperature in units of K and using three significant digits.
Answer:
[tex]T_2=315.69k[/tex]
Explanation:
Initial Temperature [tex]T_1=500K[/tex]
Initial Pressure [tex]P_1=1000kPa[/tex]
Final Pressure [tex]P_2=200kPa[/tex]
Generally the gas equation is mathematically given by
[tex]\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}[/tex]
Where
n for [tex]CO=1.4[/tex]
Therefore
[tex]\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}[/tex]
[tex]T_2=315.69k[/tex]
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the ____________ of the pipe.
Answer:
Inlet
Explanation:
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the INLET of the pipe.
This is because the friction factor is experienced at the highest level when a laminar forced flow is at the tube inlet where the thickness of the boundary layer is zero. Also, the friction factor decreases step by step at a lower rate to the fully augmented value.
Side milling cutter is an example of ______ milling cutter.
Answer:
special type
Explanation:
As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.
lists at least 6 units of measuring atmospheric pressure
Answer:
On my console displays for the ISS visiting vehicles, three units are used. The Americans use pounds per square inch (psi). The Russians use kilopascals (kPa). The Japanese use Torr - millimeters of mercury (mmHg). A fourth unit is simply the atmosphere, or multiples of it. So, for example, sea level air pressure (which is what we use onboard ISS) is defined as 1 atmosphere. That is equivalent to 14.7 psi, 101.3 kPa, or 760 mmHg.Here N represents newton which is SI unit of Force which is same as Kg.m/s2." role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">2.2.
m represent metre which is SI unit of length.
Kg represent Kilogram which is SI unit of Mass.
m2SI" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">2SI2SI unit of Area.
Hope it helps.
Thanks.
Answer:
Pounds per square inch (psi)
Kilopascals (kPa)
Millimeters of mercury (mmHg)
Pascal (Pa)
Megapascal (MPa)
Atmospheric pressure (atm)
Hope this helps!
The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Sieve analysis Sieve Size No. 4 (4.75 mm) No. 10 (2.00 mm) No. 40 (0.425 mm) No. 200 (0.075 mm) Percent passing by weight 80 60 30 10 Atterberg limits Liquid limit (LL) Plastic limit (PL 31 25
(a) Classify this soil according to USCS system, providing the group symbol for it. Show how you arrive at the final classification.
(b) According to USCS system, what is a group name for this soil?
(c) Is this a clean sand? If not, explain why.
Answer: hello the complete question is attached below
answer:
A) Group symbol = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Explanation:
A) Classifying the soil according to USCS system
( using 2nd image attached below )
description of sand :
The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also
The soil is a sand given that more than 50% particles passed from No 4 sieve
The soil can be a clean sand given that fines ≤ 12%
The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time
Group symbol as per the 2nd image attached below = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 41:1. The primary coil is plugged into a standard 120-V outlet. The current in the secondary coil is 1.2 x 10-3 A. Find the power consumed by the air filter.
Answer:
5.9 watts
Explanation:
The secondary voltage is the primary voltage multiplied by the turns ratio:
(120 V)(41) = 4920 V
The power is the product of voltage and current:
(4920 V)(1.2·10^-3 A) = (4.92)(1.2) W = 5.904 W
The power consumed is about 5.9 watts.
what are the main subsystem of GSM
network
Explanation:
Network and Switching Subsystem (NSS)
Base-Station Subsystem (BSS)
Mobile station (MS)
Operation and Support Subsystem (OSS)
A one electron species, Xm, where m is the charge of the one electron species and X is the element symbol, loses its one electron from its ground state when it absorbs 7.84×10−17 J of energy. Using the prior information, the charge of the one electron species is?
Answer:
c +5
Explanation:
we have difference in energy =
2.18x10⁻¹⁸ x z² / n²
now n = 1
amount of energy absorbed Δdelta = 7.84×10−17 J
7.84×10⁻¹⁷ = 2.18x10⁻¹⁸ x z²
we divide through by 2.18x10⁻¹⁸
z² = 7.84×10⁻¹⁷ / 2.18x10⁻¹⁸
z² = 35.9633
z = √35.9633
z = 5.9969
≈ 6
charge = atomic number 6 - number of electrons available in the element 1
= 6-1 = 5
from the calculations above, the charge of the one electron specie would be c +5
A security engineer deploys a certificate from a commercial CA to the RADIUS server for use with the EAP-TLS wireless network. Authentication is failing, so the engineer examines the certificate's properties:
Issuer: (A commercial CA)
Valid from: (yesterday’s date)
Valid to: (one year from yesterday’s date)
Subject: CN=smithco.com
Public key: RSA (2040 bits)
Enhanced key usage: Client authentication (1.3.6.1.5.7.3.2)
Key usage: Digital signature, key encipherment (a0)
Which of the following is the MOST likely cause of the failure?
A. The certificate is missing the proper OID.
B. The certificate is missing wireless authentication in key usage.
C. The certificate is self-signed.
D. The certificate has expired.
Answer:
The MOST likely cause of the failure is:
A. The certificate is missing the proper OID.
Explanation:
OID (Object Identifier) is a globally unique group of characters, alphanumeric or numeric identifier, registered under the ISO registration standard to reference a specific object or object class (an entity). In computing, OID allows a server or end-user to retrieve an object without identifying the specific physical data location. Since OID is system-generated, it is immutable and can only be assigned to one object. It cannot be shared. Its presence in the certificate would have enabled the security engineer to authenticate the certificate.
Dalton needs to prepare a close-out report for his project. Which part of the close-out report would describe
how he would plan and manage projects in the future?
Select an answer:
project highlights
major changes and risks
summary of schedule and cost performance
summary of project management effectiveness
Answer:
Dalton
The part of the close-out report that would describe how he would plan and manage projects in the future is:
summary of project management effectiveness
Explanation:
The Project Close-out Report is a project management document, which identifies the variances from the baseline plans. These variances are specified in terms of project performance, project cost, and schedule. The project close-out report records the completion of the project and the subsequent handover of project deliverables to others. The project management effectiveness summary details the project's objectives and the achievements recorded, including the lessons learned.
1025 steel wire is stretched with a stress of 70 MPa at room temperature 20 C. If th length is held constant, to what temperature in 'C and 'F must the wire be heated to reduce the stres to 17 MPa?
Check attached pictureCheck attached pictureCheck attached pictureCheck attached picture
You are responsible for notifying the DMV within 5 days of the sale using a Notice of Release of Liability form if you sell or transfer a vehicle to someone else.
Answer:
True
Explanation:
After you have sold or have made the transfer of ownership of a motor vehicle to another person, you are required to fill a notice of transfer and release of liability. You do this as a notification to let DMV be aware of a change of ownership of this vehicle. And it also serves to protect the previous owner from liabilities such as parking violations or traffic violations. This notification should be done within 5 Days of the transfer of ownership.
Thank you.
A structure is designed using 4 circular columns. Due to a quirky design, the four columns will all carry different loads of 1800 N, 2100 N, 2275 N, and 2200 N. A factor of safety of 5 is used to design the columns. The diameter of each of the columns is supposed to be 50 cm, at most. Determine the maximum height of the structure (i.e. the column height) so that the structure will not fail. Assume that all columns may be modeled as Euler columns for your analysis. Assume a pinned-pinned boundary condition for your analysis, and assume the elastic modulus of the column material is 10 MPa.
Answer:
5.16 M
Explanation:
Loads ; 1800N, 2100N, 2275N, 2200N
safety factor = 5
diameter of each column = 50 cm = 0.5 m
Elastic modulus = 10 MPa
Calculate the max height of structure
moment of inertia for a circular section ( I ) = πd^4 / 64
lets represent the required maximum height of the column as L
Applying Euler column theory
The bucker load of the column = ( attached below )
attached below is the remaining solution
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Answer:
[tex]D=0.41m[/tex]
Explanation:
From the question we are told that:
Discharge rate [tex]V_r=0.35 m3/s[/tex]
Distance [tex]d=4km[/tex]
Elevation of the pumping station [tex]h_p= 140 m[/tex]
Elevation of the Exit point [tex]h_e= 150 m[/tex]
Generally the Steady Flow Energy Equation SFEE is mathematically given by
[tex]h_p=h_e+h[/tex]
With
[tex]P_1-P_2[/tex]
And
[tex]V_1=V-2[/tex]
Therefore
[tex]h=140-150[/tex]
[tex]h=10[/tex]
Generally h is give as
[tex]h=\frac{0.5LV^2}{2gD}[/tex]
[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
Therefore
[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]
[tex]D=0.41m[/tex]
Question 1: Determine the maximum load P the steel bracket can withstand if the steel bracket has a circular cross section with a diameter of 1.2 in, and has an allowable normal stress of allow
Complete Question
Complete Question is attached below
Answer:
[tex]P=1124.2ibf[/tex]
Explanation:
From the question we are told that:
Diameter [tex]d=1.2in[/tex]
Allowable Normal stress [tex]\sigma=27.5ksi=27.5 * 10^3 psi[/tex]
Generally the equation for Bending Stress is mathematically given by
[tex]\phi= \frac{32M}{ \pi d^3}[/tex]
[tex]\phi= {32 * 4 P}{\pi * 1.2^3}[/tex]
[tex]\phi=23.58 psi[/tex]
Generally the equation for Direct Normal Stress is mathematically given by
[tex]\sigma'=\frac{4P}{ \phi * 1.2^2}[/tex]
[tex]\sigma'= 0.88P psi[/tex]
Therefore
Total Normal stress
[tex]\sigma_T=23.58 + 0.88[/tex]
[tex]\sigma_T=24.46P[/tex]
Generally the equation for Allowable Stress is mathematically given by
[tex]\sigma=\sigma_T P[/tex]
[tex]P=\frac{\sigma}{\sigma_T}[/tex]
[tex]P=\frac{27.5 * 10^3}{24.46P}[/tex]
[tex]P=1124.2ibf[/tex]
A Rankine power generation cycle operates with steam as the working fluid. The turbine produces 100 MW of power using 89 kg/s of steam entering at 700C and 5MPa. The steam exits the turbine at 0.10135 MPa. Saturated liquid water exits the condenser and is pumped back to 5Mpa before it is fed to an isobaric boiler. a. Draw a schematic of the cycle. Number the streams and label any constraints across the units. b. The turbine operates adiabatically but not reversibly. What is the temperature of the steam exiting the turbine
Your organization recently purchased 20 Android tablets for use by the organization's management team. To increase the security of these devices, you want to ensure that only specific apps can be installed. Which of the following would you implement?
A. Credential Manager.
B. App whitelisting.
C. App blacklisting.
D. Application Control.
thì nghiệm nén xác định cường độ của bê tông trên ba mẫu thí nghiệm hình trụ HxD=300x150(mm). kết quả thu được lực phá hoại P1=45200daN, P2=46800daN, P3=46000daN. hãy xác định cường độ chịu nén của bê tông theo TCNV 3118:1993
spanish
Explanation:
the above question is written in spanish
A technician wants to implement a dual factor authentication system that will enable the organization to authorize access to sensitive systems on a need-to-know basis. What should be implemented during the authorization stage?
Answer: Biometrics
Explanation:
Dual factor authentication refers to an electronic authentication method whereby a user will only be granted an access to an application or a website after the user has successfully been able to present two pieces of evidence which then grants access to the application or website.
Since the technician wants to implement a dual factor authentication system, the biometrics should be implemented during the authorization stage.
Biometrics refers to the body measurements and the calculations that are related to the characteristics of humans. Biometric authentication is used as a form of identification.
At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops power at a rate of (a) 1000 kW, (b) 750 kW, (c) 0 kW. For each case, apply Eq. 5.13 on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
Answer:
a. impossible
b. possible and reversible
c. possible and irreversible
Explanation:
a. 1000kw
Qh - Wnet
we have
QH = 1500
wnet = 1000
1500 - 1000
= 500kw
σcycle = [tex]-[\frac{QH}{TH} -\frac{QC}{TC} ][/tex]
Qh = 1500
Th = 1000
Tc = 500
Qc = 500
[tex]-[\frac{1500}{1000} -\frac{500}{500} ][/tex]
solving this using LCM
= -0.5
the cycle is impossible since -0.5<0
b. 750Kw
Qc = 1500 - 750
=750Kw
Qh = 1500
Th = 1000
Tc = 500
Qc = 750
σ-cycle
[tex]-[\frac{1500}{1000} -\frac{750}{500} ]\\= 1.5 -1.5\\= 0[/tex]
This cycle is possible and it is also reversible
c. 0 kw
Qc = 1500-0
= 1500
Qh = 1500
Th = 1000
Tc = 500
Qc = 1500
σ- cycle
[tex]-[\frac{1500}{1000} -\frac{1500}{500} ]\\-(1.5-3)\\-(-1.5)\\= 1.5[/tex]
1.5>0
so this cycle is possible and irreversible
