Answer:
a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.
Explanation:
Maximum theoretical efficiency for a power cycle ([tex]\eta_{r}[/tex]), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:
[tex]\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\%[/tex] (1)
Where:
[tex]T_{C}[/tex] - Temperature of the cold reservoir, in Kelvin.
[tex]T_{H}[/tex] - Temperature of the hot reservoir, in Kelvin.
The maximum theoretical efficiency associated with this power cycle is: ([tex]T_{C} = 400\,K[/tex], [tex]T_{H} = 1200\,K[/tex])
[tex]\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{r} = 66.667\,\%[/tex]
In exchange, real efficiency for a power cycle ([tex]\eta[/tex]), no unit, is defined by this expression:
[tex]\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\%[/tex] (2)
Where:
[tex]Q_{C}[/tex] - Heat released to cold reservoir, in kilojoules.
[tex]Q_{H}[/tex] - Heat gained from hot reservoir, in kilojoules.
[tex]W_{C}[/tex] - Power generated within power cycle, in kilojoules.
A power cycle operates irreversibly for [tex]\eta < \eta_{r}[/tex], reversibily for [tex]\eta = \eta_{r}[/tex] and it is impossible for [tex]\eta > \eta_{r}[/tex].
Now we proceed to solve for each case:
a) [tex]Q_{H} = 900\,kJ[/tex], [tex]W_{C} = 450\,kJ[/tex]
[tex]\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 50\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
b) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{C} = 300\,kJ[/tex]
[tex]\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 66.667\,\%[/tex]
Since [tex]\eta = \eta_{r}[/tex], the power cycle operates reversibly.
c) [tex]W_{C} = 600\,kJ[/tex], [tex]Q_{C} = 400\,kJ[/tex]
[tex]\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 60\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
d) Since [tex]\eta > \eta_{r}[/tex], the power cycle is impossible.
P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diameter is 3 cm. Estimate (a) the mass flow rate through the nozzle and (b) the Mach number at the throat.
Answer:
a) [tex]m=0.17kg/s[/tex]
b) [tex]Ma=0.89[/tex]
Explanation:
From the question we are told that:
Pressure [tex]P=60kPa[/tex]
Diameter [tex]d=3cm[/tex]
Generally at sea level
[tex]T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4[/tex]
Generally the Power series equation for Mach number is mathematically given by
[tex]\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}[/tex]
[tex]\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]Ma=0.89[/tex]
Therefore
Mass flow rate
[tex]\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]\rho=0.848kg/m^3[/tex]
Generally the equation for Velocity at throat is mathematically given by
[tex]V=Ma(r*T_0\sqrt{T_e}[/tex])
Where
[tex]T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}[/tex]
[tex]T_e=248[/tex]
Therefore
[tex]V=0.89(1.4*288\sqrt{248})\\\\V=284[/tex]
Generally the equation for Mass flow rate is mathematically given by
[tex]m=\rho*A*V[/tex]
[tex]m=0.84*\frac{\pi}{4}*3*10^{-2}*284[/tex]
[tex]m=0.17kg/s[/tex]
The input sin(20) is sampled at 20 ms intervals by using impulse train sampling: i. Construct the input and sampled signal spectra.
Solution :
Let [tex]$x(t) = \frac{\sin (20 \pi t)}{\pi t}$[/tex]
[tex]$T_s = 20$[/tex] ms, so [tex]$f_s=\frac{1}{T_s}[/tex]
[tex]$=\frac{1}{20}$[/tex]
= 0.05 kHz
[tex]$f_s=50 $[/tex] Hz , ws = [tex]$2 \pi f_s = 100 \pi$[/tex] rad/s
We know that,
FT → [tex]$\frac{\sin (20 \pi \omega)}{\pi \omega}$[/tex]
The sampled signal is :
[tex]$XS(\omega) = \frac{1}{T_s} \sum_{k=- \infty}^{\infty}X (\omega-k\omega S)[/tex]
So, [tex]$XS(\omega) = \frac{1}{20 \times 10^{-3}} \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
[tex]$XS(\omega) = 50 \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
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
A 2-stage dcv that has an internal pilot does not work well (if at all) on
Answer:
i really font onow why tbh eot you
Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the altitude of Indianapolis, find the Mach number of these cars and estimate whether compressibility might affect their aerodynamics.
Answer:
- the Mach number is 0.24.
- Compressibility becomes effective when Mach number is greater than 0.3, the Mach number of the race cars is less than 0.3, hence, compressibility will not affect their aerodynamics.
Explanation:
Given the data in the question;
Average speed V = 185 miles per hour = ( 185 /2.237 ) m/s = 82.7 m/s
From Almanac, we can find that Indianapolis is at 220 m altitude.
So from table, at that altitude, the standard speed of sound will be 339.4 m/s .
Mach number of the race car will be;
Mach Number = Velocity / sound speed
we substitute
Mach Number = ( 82.7 m/s ) / ( 339.4 m/s )
Mach Number = 0.24
Therefore the Mach number is 0.24.
We know that, compressibility becomes effective when the Mach number is greater than 0.3.
Since the Mach number of the race cars is less than 0.3, compressibility will not affect their aerodynamics.
8- Concentration polarization occurs on the surface of the.......
a- cathode.
b- anode.
C- both
d-ption 4
Explanation:
Concentration overpotential, ηc,
I hope it helps you
determine if the fluid is satisfied
Which statement describes the relay between minerals and rocks ?
Answer:
•○●□■hey hi!■□●○•Explanation:
Minerals and rocks are the same. Aggregates of minerals form rocks. Minerals determine the texture of a rock. Most rocks are made of a single mineral type.
☆♡hope this helps♡☆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.
If a corporation is socially responsible, it will develop and implement a sustainability plan and communicate it to stakeholders.
True
False
Answer:
True
Explanation:
All big companies are pretty much required in today's day and ages to complete these reports whether they truly believe it.
I want to explain what 2000 feet looks like to young children so that they can imagine it in class
Answer:
maybe take a really common toy kids play with or often see, find the average height for the toy and do the math to see how many of those toys stacked ontop of eachother would make up 2000 feet. For example (this isn't accurate btw just an idea of what it would sound like but) "Have you ever seen a barbie doll? well if you stack 400 barbie dolls ontop of their head it would be equal to 2000 feet."
Explanation:
sometimes taking common or beloved objects children have into your examples makes them have a better image of how small or how big something is.
State the factor that influence the frequency of the induced emf of an alternating quantity
Explanation:
conductor, flux, and movement of conductor in magnetic field are some of the factors that induce emf.
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
deep invasion lead to???
Answer:
2
Explanation:
usjshshshjeoeieieiiekeiekeiekekekekekkske
Answer:
2
Explanation:
High separation between deep...........
The impeller shaft of a fluid agitator transmits 20 kW at 430 rpm. If the allowable shear stress in the impeller shaft must be limited to 65 MPa, determine (a) the minimum diameter required for a solid impeller shaft. (b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 36 mm. (c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)
Given :
Power, P = 20 kW
Speed, N = 430 rpm
Allowable shear stress, τ = 65 MPa
Torque in the shaft is given by :
[tex]$P=\frac{2 \pi NT}{60}$[/tex]
[tex]$T=\frac{60 \times 20 \times 10^3}{2 \pi \times 430}$[/tex]
T = 444.37 N.m
Diameter of the solid shaft is
[tex]$d=\sqrt[3]{\frac{16 T}{\pi \tau}}[/tex]
[tex]$d=\sqrt[3]{\frac{16 \times 444.37}{3.14 \times 65}}[/tex]
[tex]$d=\sqrt[3]{34.83} $[/tex]
d = 3.265 m
d = 326.5 mm
Internal diameter of the hollow shaft is :
[tex]$\frac{T}{\frac{\pi}{32} \left( d_0^4 - d_i^4 \right)}=\frac{\tau}{d_0/2}$[/tex]
[tex]$\frac{444.37}{\frac{3.14}{32} \left( 0.036^4 - d_i^4 \right)}=\frac{65 \times 10^6}{0.036/2}$[/tex]
[tex]$\frac{444.37}{0.09 \left( 1.6 \times 10^{-6} - d_i^4 \right)}=\frac{65 \times 10^6}{0.018}$[/tex]
[tex]$\frac{7.99}{ \left( 1.6 \times 10^{-6} - d_i^4 \right)}=5850000$[/tex]
[tex]$1.3\times 10^{-6} = 1.6 \times 10^{-6} - d_i^4 \right)}$[/tex]
[tex]$d_i^4=300000$[/tex]
[tex]$d_i = 23.40$[/tex] mm
Percentage savings in the weight is given by :
Percentage saving = [tex]$\frac{W_{solid}-W_{hollow}}{W_{solid}}\times100$[/tex]
[tex]$=\frac{V_{solid}-V_{hollow}}{V_{solid}}\times100$[/tex]
[tex]$=\frac{d^2 - (d_0^2 - d_i^2)}{d^2} \times 100$[/tex]
[tex]$=\frac{(326.5)^2 - (0.036^2 - (32.40)^2)}{(326.5)^2} \times 100$[/tex]
[tex]$=\frac{106602 - \left(1.29 \times 10^{-3} - 1049.76 \right)}{106602} \times 100$[/tex]
[tex]$=\frac{106602 - 1049 }{106602} \times 100$[/tex]
[tex]$=\frac{105553 }{106602} \times 100$[/tex]
= 99.01 %
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.
What is protection scheme?
Answer:
The objective of a protection scheme is to keep the power system stable by isolating only the components that are under fault, whilst leaving as much of the network as possible still in operation.
Explanation:
The devices that are used to protect the power systems from faults are called protection devices.
can you guys please introduce yourself
Answer: why?
Explanation:
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.
Troy must keep track of the amount of refrigerant he uses from a 50-pound cylinder to ensure that accurate
records are kept. He used 13 pounds on a systein for Ms. Jones and 9 pounds on a system for a commercial
client. How many pounds should he have left in the cylinder?
tof
Troy should have
pounds of refrigerant left in the cylinder.
baon naid Thamar basic
محمود احمد مجد
12
اهداء ما در
Answer:
Amount of gas still in cylinder = 28 pound
Explanation:
Given:
Amount of gas in cylinder = 50 pound
Amount of gas used in Ms. Jones system = 13 pound
Amount of gas used in client system = 9 pound
Find:
Amount of gas still in cylinder
Computation:
Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system
Amount of gas still in cylinder = 50 - 13 - 9
Amount of gas still in cylinder = 28 pound
A binary system of species 1 and 2 consists of vapor and liquid phases in equilibrium
at temperature T. The overall mole fraction of species 1 in the system is z1 = 0.65. At
temperature T, lnγ1 = 0.67 x2
2; lnγ2 = 0.67 x1
2; P1
sat = 32.27 kPa; and P2
sat = 73.14 kPa.
Assuming the validity of Eq. (13.19),
Final PDF to printer
13.10. Problems 511
smi96529_ch13_450-523.indd 511 01/06/17 03:27 PM
(a) Over what range of pressures can this system exist as two phases at the given T and z1?
(b) For a liquid-phase mole fraction x1 = 0.75, what is the pressure P and what molar
fraction of the system is vapor?
(c) Show whether or not the system exhibits an azeotrope
Write the code using the do-while loop to force the user to enter a number in the range [20,50]
Answer:
Mark as brainlist pls hello
17- The cathodic polarization is ..... *
O
a- activation.
O
b- concentration
O c- both.
A 40kg steel casting (Cp=0.5kJkg-1K-1) at a temperature of 4500C is quenched in 150kg of oil (Cp=2.5kJkg-1K-1) at 250C. If there are no heat losses, what is the change in entropy of?
(i) The casing.
(ii) The oil.
(iii) Both considered.
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]
an atom that gained an electron is called
Answer:
Hey mate......
Explanation:
This is ur answer.....
When an atom gains/loses an electron, the atom becomes charged, and is called an ion.Hope it helps!
Brainliest pls!
Follow me :)
Hãy trình bày các bộ phận chính trong một bộ điều khiển điện tử (ECU) dùng trên ô tô. Cho biết công dụng của từng thành phần.
Answer:
sorry but I can't understand this Language.
Explanation:
unable to answer sorry
You toss two coins. If you get heads with the first coin, you stop. If you get tails, you toss it again. The second coin is tossed regardless. What is the ratio of heads to tails?
Answer:
1:1
Explanation:
A coin has either a head or tail.
Thus, probability of head = ½
Probability of tail = ½
The first coin is tossed twice and probability of head or tail on both tosses is still ½. Thus, they will have a ratio 1:1.
The second coin is still tossed regardless and thus continuously and so should have same ratio of 1:1.
Therefore, the ratio of the both of them must, also be 1:1.
The number-average molecular weight of a poly (styrene-butadiene) alternating copolymer is 1,350,000 g/mol. What is the average number of styrene and butadiene repeat units per molecule.
a) 6,806
b) 6,944
c) 4,801
d) 8,544
Increasing following distance to
when encountering other motorists who follow too closely
is an example of appropriate implementation of the IPDE defensive driving strategy for the
maintenance of an appropriate Safety Cushion.
Two-seconds
Enree-seconds
Four-seconds
Twenty-seconds
Answer:
Increasing following distance to Four-seconds when encountering other motorists who follow too closely is an example of appropriate implementation of the IPDE defensive driving strategy for the maintenance of an appropriate Safety Cushion.
Explanation:
Maintaining the required safety cushion by utilizing the IPDE defensive driving strategy to manage the nine to fifteen space driving zones involves continuous scanning. Therefore, motorists should be able to identify objects and hazards in the driving scene, line of sight, and path of travel. They should predict points of driving conflicts. They should determine appropriate and safe driving actions to take, when, and where. Finally, action is required to ensure that conflicts are avoided.
