The arrival rate at a parking lot is 6 veh.min. Vehicles start arriving at 6:00PM and when the queue reaches 36 vehicles, service begins. If company policy is that total vehicle delay should be equal to 500 veh-min, what is the departure rate?

Answers

Answer 1

Answer:

Departure rate = 7.65 vehicle/min

Explanation:

See the attached file for the calculation.

The Arrival Rate At A Parking Lot Is 6 Veh.min. Vehicles Start Arriving At 6:00PM And When The Queue
The Arrival Rate At A Parking Lot Is 6 Veh.min. Vehicles Start Arriving At 6:00PM And When The Queue
The Arrival Rate At A Parking Lot Is 6 Veh.min. Vehicles Start Arriving At 6:00PM And When The Queue

Related Questions

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 .

Answers

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

Utilizing the charge neutrality equation and the mass-action law derive an expression for the concentration of electrons and holes present inside a p-type compensated semiconductor material. How can the resulting expression be modified for practical cases where (????????AA − ????????DD) ≫ 2nn�

Answers

Answer:

The expression for the concentration of electrons is P = NA - ND

Explanation:

Please look at the solution in the attached Word file

Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a cold reservoir at 250 K and rejects 950 KJ to a hot reservoir at 300 K while receiving 70 kJ of work to operate. Draw a schematic of the cycle clearly indicating the hot and cold reservoir and the direction of heat and work transfers. (10 pts.)

Answers

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

[tex]COP_{i} = \frac{250\,K}{300\,K-250\,K}[/tex]

[tex]COP_{i} = 5[/tex]

The real Coefficient of Performance is:

[tex]COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}[/tex]

[tex]COP_{r} = 12.571[/tex]

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

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.

Answers

Answer:

Please see the attached picture for the compete answer.

Explanation:

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.

Answers

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.

Jasper and Gemma are going to play on a teeter totter. Gemma gets on first. When Jasper gets on, Gemma moves into the air, but she does not move to the top. Which statement could correctly explain the forces acting on the teeter totter? assume that Jasper and Gemma are the same distance from one another.

Answers

Answer:

A) the forces are balanced because Jasper weighs the same as Gemma

Explanation:

Answer:

A. The forces are balanced because Jasper weighs the same as Gemma.

Explanation:

Took the test

In your new role at Wayne Industries, you have been given the freedom to propose and develop your own project ideas. You have an idea for a frictionless piston-cylinder assembly. Your design requires the expansion of 0.25 m3 of air at 3.3 MPa and 280 oC to 180 kPa in a slow, isothermal process. In order to be approved for further development, you need to submit calculations for the values of the heat transferred in the process.

Answers

Answer:

461.65 KJ/Kg

Explanation:

In this question, we are asked to calculate the values of heat transferred in the process.

Please check attachment for complete solution and step by step explanation

A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa√m. If it is known that the largest surface crack is 0.5 mm long and that the value of Y is 1.1, which of the following can be said about this plate when a tensile stress of 1000 MPa is applied?

A. It is not possible to determine whether or not the plate will fracture.
B. The plate will definitely not fracture
C. The plate will definitely fracture.

Answers

Answer:

option B is correct. Fracture will definitely not occur

Explanation:

The formula for fracture toughness is given by;

K_ic = σY√πa

Where,

σ is the applied stress

Y is the dimensionless parameter

a is the crack length.

Let's make σ the subject

So,

σ = [K_ic/Y√πa]

Plugging in the relevant values;

σ = [50/(1.1√π*(0.5 x 10^(-3))]

σ = 1147 MPa

Thus, the material can withstand a stress of 1147 MPa

So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.

The ice on the rear window of an automobile is defrosted by attaching a thin, transparent, film type heating element to its inner surface. By electrically heating this element, a uniform heat flux may be established at the inner surface. The inside air temperature and convection heat transfer coefficient are T,i = 25°C and hi = 25 W/(m2 -K), while the outside air temperature is T,o = -10°C. For 4-mm thick window glass, determine (a) the electric power required per unit window area to maintain an inner surface temperature of 15°C and (b) the thermal conductivity of the windshield.

Answers

Answer:

A)Q = 1208.33 W/m²

B)K = 0.138 W/m.K

Explanation:

We are given;

inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K

outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K

Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K

Thickness, L = 4mm = 0.004m

convection heat transfer coefficient ; hi = 25 W/(m².K)

A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;

Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]

Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;

Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]

Q = 583.33 + 625

Q = 1208.33 W/m²

B) The formula for thermal conductivity is;

K = (QL)/(AΔT)

Where;

K is the thermal conductivity in W/m.K

Q is the amount of heat transferred through the material

L is the distance between the two isothermal planes

A is the area of the surface in square meters

ΔT is the difference in temperature in Kelvin

ΔT = 298K - 263K = 35K

Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;

k = (Q/A) x (L/ΔT)

K = 1208.33 x (0.004/35)

K = 0.138 W/m.K

2.) For a 20‐mm‐diameter tube with either water or unused engine oil flowing through it, find: a.)The mean velocity, hydrodynamic entry length, and thermal entry length for each of the fluids with a temperature of 300 K if the mass flow rate is 0.01 kg/s. b.)The mass flow rate, hydrodynamic entry length, and thermal entry length for each fluid at 400 K and a mean velocity of 0.02 m/s.

Answers

Answer:

a.) The mean velocity = 0.0318 m/s

    The  hydrodynamic entry length = 0.636 m

     The  thermal entry length = 0.004 m

(b) The mass flow rate = 0.0051 kg/s

    The hydrodynamic entry length = 0.028 m

     The  thermal entry length = 1.419 m

Explanation:

See the attached files for the calculation.

Neglecting the presence of friction, air drag, and other inefficiencies, how much gasoline is consumed when a 1300 kg automobile accelerates from rest to 80 km/h. Assume the density and enthalpy of gasoline are 680 kg/m3 and 45 MJ/kg respectively. Express your answer in the units of mL. Show all work and use SI units for full credit. Box your final answer(s)

Answers

Answer:

Explanation:

Given that, .

Mass of car is

M = 1300kg

Velocity of car

V = 80km/h = 80 × 1000/3600

V = 22.22m/s

Calculate the kinetic energy of the vehicle as follows:

K.E = ½ MV²

K.E = ½ × 1300 × 22.22²

K.E = 320,987.65 J

Given that,

Enthalpy is 45MJ / kg

h = 45MJ / kg

Then, enthalpy is given as.

Enthalpy = Energy / mass

h = E / m

45 × 10^6 = 320,987.65 / m

m = 320,987.65 / 45 × 10^6

m = 7.133 × 10^-3 kg

m = 7.133 mg

Also, given that, density is 680kg/m³

Density is given as

Density = mass / Volume

ρ = m / v

Then, v = m / ρ

v = 7.133 × 10^-3 / 680

v = 1.049 × 10^-5 m³

We know that

1mL = 10^-6 m³

Therefore,

v = 1.049 × 10^-5 m³ × 1mL / 10^-6m³

v = 10.49 mL

The function below takes a single string parameter called sentence. Your function should return True if the sentence contains at least one copy of each of the following vowels: a, e, i, o, and u. Otherwise, return False. The vowels can be either upper case or lower case.

student.py
Bef contains_all_values (sentence): 1 w

Answers

Final answer:

The question asks for a function to check if a sentence contains all five vowels at least once, regardless of case sensitivity. A solution involves creating a set of vowels and comparing it to a set of found vowels in the sentence, returning true if all vowels are present.

Explanation:

The question relates to determining whether a given sentence contains all five vowels (a, e, i, o, u) at least once, ignoring case sensitivity. This problem is typically solved using a function that iterates through each character in the sentence, checks if it is a vowel, and then keeps track of whether all vowels have been encountered. The essential steps involve converting the sentence to lowercase (to ignore case sensitivity), then checking for the presence of each vowel. A simple approach is to use a set to keep track of the vowels found, and once the set contains all five vowels, the function can return True. Otherwise, it returns False after checking the entire sentence.

An example implementation could be:

def contains_all_vowels(sentence):
   vowels = set('aeiou')
   found = set(c.lower() for c in sentence if c.lower() in vowels)
   return found == vowels

This code creates a set of vowels and then iterates over the sentence, adding each encountered vowel to another set. If, by the end of the sentence, the second set is equal to the set of all vowels, the function returns True; otherwise, it returns False.

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.

Answers

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']

====

Consider airflow over a plate surface maintained at a temperature of 220°C. The temperature profile of the airflow is given as fluid ()()expsV TyTTTy     The airflow at 1 atm has a free stream velocity and temperature of 0.08 m/s and 20°C, respectively. Determine the heat flux on the plate surface and the convection heat transfer coefficient of the airflow.

Answers

Consider airflow over a plate surface maintained at a temperature of 220°C. The temperature profile of the airflow is given as T(y) = T∞ − (T∞−Ts)exp((−V/α_fluid)y).

The airflow at 1 atm has a free stream velocity and temperature of 0.08 m/s and 20°C, respectively. Determine the heat flux on the plate surface and the convection heat transfer coefficient of the airflow

Answer:

A) heat flux on the plate is;q_o = 11737.34 W/m²

B) convection heat transfer coefficient of the airflow is;h = 58.67 W/m².k

Explanation:

The temperature profile of the airflow is given as;

T(y) = T∞ − (T∞−Ts)exp((−V/α_fluid)y)

Let's differentiate with respect to y;

dT/dy = [[(T∞−Ts)V]/α](e^(-vy/α)

Where;

T∞ = 20°C

Ts = 220°C

V = 0.08 m/s

α is thermal diffusivity of air and from the table i attached at a temperature of 220°C, by interpolation it has a value of;

α = 5.33 x 10^(-5) m²/s

Thus, at y =0;

dT/dy = [[(20 − 220)0.08]/(5.33 x 10^(-5))](e^(0))

dT/dy = -300187.62 °C/m

A) Now, heat flux at y = 0 would be given by;

q_o = -k(dT/dy)

Where k is thermal conductivity

from the table attached at 220°C and by interpolation, the thermal conductivity k = 0.0391 W/m.k

Thus,

q_o = -0.0391(-300187.62)

q_o = 11737.34 W/m²

B) the convection heat transfer coefficient of the airflow is gotten from;

q_o = h(Ts - T∞).

Where h is the convection heat transfer coefficient of the airflow

Thus making h the formula, we have;

h = q_o/(Ts - T∞)

h = 11737.34/(220 - 20)

h = 58.67 W/m².k

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.

Answers

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

The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter dd. Express your answer in terms of some or all of the variables rrr, RRR, ddd, and TTT. nn

Answers

Answer:

n = 2r³/Rd²

Explanation:

See the attached file for the derivation.

Ethanol is produced in a one-liter batch fermenter by Saccharomyces cerevisiae (yeast). To begin the batch process, glucose and ammonia are added to the reactor, which has been inoculated by the yeast. The reaction produces glycerol and cell mass, in addition to ethanol, carbon dioxide, and water according to the (unbalanced) reaction below. Note that that the mass ratio of glycerol produced to glucose consumed is 0.2556. Also note that the mass ratio of H.O produced to NH, consumed is 1.058 Glucose + Ammonia → Glycerol + Ethanol + Yeast Cell Mass + Carbon Dioxide + Water a CH (8) + b NH(g)}p CH 0 (1) + (CHO) +ỊCH K04 Nụ (8) + $ 0 g) + t H (1) Balance the equation and fill out the table below. (1 point cach, 6 total) a- 1 b = 115 g of NH, are added to the reactor. At the end of the batch, 640 g of ethanol has been produced. How much glucose was added (assume complete conversion of the limiting reactant). mlar B 16 points) BONUS QUESTION USING CHAPTER 4 CONCEPTS: The standard heat of combustion (AHC) for the yeast is-22.1 kJ/mol. In order to maintain a constant temperature, how much heat must be added or removed from the reactor? Both the sign and magnitude of your answer will be graded J 3 points)

Answers

Answer:

Check the explanation

Explanation:

The process of Alcoholic fermentation involves the converting a single mole of glucose into two moles of carbon dioxide and two moles of ethanol, and in the process producing two moles of ATP. The total chemical formula for alcoholic fermentation is: C6H12O6 → 2 C2H5OH + 2 CO. Sucrose is a dimer of fructose and glucose molecules.

Kindly check the attached image below to see the full step by step explanation to the question above.

You are provided the following information about a municipal wastewater treatment plant. This plant uses the traditional activated-sludge process.

Assume the microorganisms are 55 percent efficient at converting food to biomass, the organisms have a first-order death rate constant of 0.05/day, and the microbes reach half of their maximum growth rate when the BOD5 concentration is 10 mg/L. There are 150,000 people in the community (their wastewater production is 225 L/day-capita, 0.1 kg BOD5/capita-day). The effluent standard is BOD5 = 20 mg/L and TSS = 20 mg/L.

Suspended solids were measured as 4,300 mg/L in a wastewater sample obtained from the biological reactor, 15,000 mg/L in the secondary sludge, 200 mg/L in the plant influent, and 100 mg/L in the primary clarifier effluent. SRT is equal to 4 days.

(a) What is the design volume of the aeration basin (m3)?

(b) What is the plant

Answers

Answer:

Explanation:

Attached is the solution

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.

Answers

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

Read more about Energy Balance at; https://brainly.com/question/25329636

A minor road intersects a major 4-lane divided road with a design speed of 55mph and a median width of 8 feet. The intersection is controlled with a stop sign on the minor road. If the design vehicle is a passenger car, determine the minimum sight distance required on the major road that will allow a stopped vehicle on the minor road to safely turn left if the approach grade on the minor road is 3%.

Answers

Answer:

stopping sight diatnce is =158 meters

Explanation:

The Design Speed =55 MPH(Miles per Hour)

1 miles per hour =0.447 meter pe second

then 55MPH=55*0.447=24.58 meter per second

Consider recation time =2.5 second

Consider Co efficinet friction is 0.35

As per the Stopping sight distance based on the Breaking distance+ Lag distance

SSD=Vt+(V^2/(2gf))

SSD=24.58+(24.58^2/(2*9.81*0.35))

The total stopping sight distance is =150 meters

if we consider the Approaching garde on the minor road is 3%

SSD=Vt+(V^2/(2g(f-n/100))

SSD=24.58+(24.58^2/(2*9.81*(0.35-3/100))

then stopping sight diatnce is =158 meters

See attachment for workings

A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of 24 kpsi. After some calculation, the designer has selected a standard 1/2 in diameter (d) bar. The factor of safety (n) is ____. Round the answers to three significant digits.

Answers

Answer:

[tex]n = 2.36[/tex]

Explanation:

The stress experimented by the circular bar is:

[tex]\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)[/tex]

[tex]\sigma = 10.186\,kpsi[/tex]

The safety factor is:

[tex]n = \frac{24\,kpsi}{10.186\,kpsi}[/tex]

[tex]n = 2.36[/tex]

A small vehicle is powered by a pulsejet. The available net thrust is 6000 N and the traveling speed is 200 km/hr. The gases leave the engine with an average velocity (Ve) of 360 m/s Assume pressure equilibrium exists at the outlet plane and the fuel to air ratio is 0.06.

a. Compute the mass flow rate required
b. Calculate the inlet area (assume To is 16 °C and Po is one atmosphere)
C. Calculate the thrust power
d. Calculate the propulsive efficiency

Answers

Answer:

a) The mass flow rate is 19.71 kg/s

b) The inlet area is 0.41 m²

c) The thrust power is 333.31 kW

d) The propulsive efficiency is 26.7%

Explanation:

Please look at the solution in the attached Word file.

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.

Answers

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.

Boron fibers (of total mass 5kg, rho B = 2.3g/cc, σy B = 55MPa) are uniaxially introduced into 8kg of an Al matrix (rho Al = 2.7g/cc, σy Al = 10MPa). There is no porosity before and after composite fabrication. Calculate the density of the composite, yield strengths parallel and perpendicular to fiber orientations.

Answers

Answer:

Explanation:

Check attachment for step by step solution

Five hundred gallons of 89-octane gasoline is obtained by mixing 87-octane gasoline with 92-octane gasoline. (a) Write a system of equations in which one equation represents the total amount of final mixture required and the other represents the amounts of 87- and 92-octane gasoline in the final mixture. Let x and y represent the numbers of gallons of 87- and 92-octane gasoline, respectively. (b) Use a graphing utility to graph the two equations in part (a) in the same viewing window. As the amount of 87-octane gasoline increases, how does the amount of 92-octane gasoline change? (c) How much of each type of gasoline is required to obtain the 500 gallons of 89-octane gasoline?

Answers

Explanation:

a) The total volume equals the sum of the volumes.

500 = x + y

The total octane amount equals the sum of the octane amounts.

89(500) = 87x + 92y

44500 = 87x + 92y

b) desmos.com/calculator/ekegkzllqx

As x increases, y decreases.

c) Use substitution or elimination to solve the system of equations.

44500 = 87x + 92(500−x)

44500 = 87x + 46000 − 92x

5x = 1500

x = 300

y = 200

The required volumes are 300 gallons of 87 gasoline and 200 gallons of 92 gasoline.

Design a Mealy machine for a 20 cent candy dispensing machine which accepts nickel (n) , dime (d) and quarter(q) . It gives candy as well change in the form of nickels only. To help you, here are the some of the elements of the machine: States: 0, 5, 10 and 15 represent the amount of money already inserted in the machine. n, d and q representing coins inserted in the machine Output: c0 (candy and no change), c1 (one nickel as change), c2(two nickels as change ), c3 (three nickels as change), and c4 (four nickels as change)

Answers

Find the answer in the attachment

The brakes on a vehicle work OK for a while, then the vehicle slows because the brakes self- applied. Technician A says that an overfilled master cylinder could be the cause. Technician B says that a blocked vent port (compensating port) could be the cause. Which technician is correct?


A. Technician A only

B. Technician B only

C. Both technician A and B

D/ Neither technician A nor B

Answers

Answer:

C. Both technician A and B

Explanation:

If the master cylinder is overfilled it will not allow enough room for the brake fluid to expand due to heat expansion. This blocks the vent port.  If a vent port is not open, brake fluid pressure will increase as brakes heat up.  This will cause the brakes to self apply, cause more heat in the brake fluid and the vehicle will slow down.

There, we can conclude that Both technician A and B are correct.

Answer:

C. Both technician A and B

Explanation:

The event that made both cylinders to be over filled especially the master cylinder and the blocking of the vent port, this will cause the vehicle brake to apply itself after just a little motion of the vehicle.

Therefore both technicians are correct from the information given above.

Hence, we can boldly say the correct answer is C. ie Both technician A and B

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

Answers

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!

The entire population of a given community is examined, and all who are judged to be free from bowel cancer are questioned extensively about their diets. These people then are followed for several years to see whether or not their eating habits will predict their risk of developing bowel cancer.

a. Cross-sectional study
b. Case-control study
c. Prospective cohort study
d. Historical prospective cohort study
e. Clinical trial
f. Community trial

Answers

Answer:

type of study design is Prospective cohort study

Explanation:

This study follows overtime in a group of similar people who differ from some of the factors in the study to determine how factors may affect the outcome rate. Tests may vary depending on the hypothesis given by Cretin patients who are cigarette patients, who are most likely to be smokers, then most likely to be over 20 years of age, with a high rate of lung cancer. The effective cause of the disease is determined by the method of screening of the individuals below.

thermodynamics A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor to develop net cycle power of 3 MW. Steam exits the reactor core at 100 bar, 620°C and expands through the turbine to the condenser pressure of 1 bar. Saturated liquid exits the condenser and is pumped to the reactor pressure of 100 bar. Isentropic efficiencies of the turbine and pump are 87% and 78%, respectively. Cooling water enters the condenser at 15°C with a mass flow rate of 114.79 kg/s. Determine: (a) the percent thermal efficiency. (b) the temperature of the cooling water exiting the condenser, in °C.

Answers

Answer:

(a) the percent thermal efficiency is 27.94%

(b) the temperature of the cooling water exiting the condenser is 31.118°C

Explanation:

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