Hydrogen peroxide, H2O2, enters a gas generator at 25 Celsius, 500 kPa, at the rate of 0.1 kg/s and is decomposed to steam and oxygen exiting at 800 K, 500 kPa. The resulting mixture is expanded through a turbine to atmospheric pressure, 100 kPa. Determine the power output of the turbine and the heat transfer rate in the gas generator. The enthalpy of formation of liquid H2O2 is −187 583 kJ/kmol.

Answer:

[tex]\phi = 155.57[/tex]

Explanation:

from figure

taking summation of force in x direction be zero

[tex]\sum x = 0 [/tex]

[tex]F_D = Tsin \theta[/tex]  .....1

[tex]\frac{c_d \rho v^2 A}{2} =Tsin \theta[/tex]

taking summation of force in Y direction be zero

[tex]F_B - W-  Tcos \theta[/tex]

[tex]T = \frac{F_B -W}{cos \theta}[/tex] .........2

putting T value in equation 1

[tex]F_D - \frac{F_B -W}{cos \theta} sin\theta[/tex]

[tex]F_D = \rho g V ( 1 -Sg) tan \theta [/tex].........3

[tex]F_D = \rho g [\frac{\pi d^3}{6}] ( 1 -Sg) tan \theta [/tex]

[tex]tan \theta = \frac{6 c_D \rho v&2 A}{ 2 \rho g V \pi D^3 (1- Sg)}[/tex]

Water at 10 degree C  has kinetic viscosity v = 1.3 \times 10^{-6} m^2/s

Reynold number

[tex]Re = \frac{ VD}{\nu} = \frac{10\times 0.5}{1.3 \times 10^{-6}} = 3.84 \times 10^6[/tex]

so for Re =[tex] 3.84 \times 10^6 [/tex]  cd is 0.072

[tex]tan \theta = \frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}[/tex]

[tex]\theta = tan^{-1} [\frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}][/tex]

[tex]\theta = - 65.57 degree[/tex]

[tex]\phi = 90 - (-65.57) = 1557.57[/tex] degree

Answer:

104,576 cycles

Explanation:

Step 1: identify given parameters

Ultimate strength of steel ([tex]S_{ut}[/tex])= 120 Kpsi

stress amplitude ([tex]\alpha_{a}[/tex])= 70 kpsi

life of the specimen (N) = ?

[tex]N = (\frac{\alpha_{a}}{a})^\frac{1}{b}[/tex]

where a and b are coefficient of fatigue cycle

Step 2: calculate the the endurance limit of specimen

[tex]S_{e} = 0.5*S_{ut}[/tex]

[tex]S_{e}[/tex] = 0.5*120 = 60 kpsi

Step 3: calculate coefficient 'a'

[tex]a=\frac {(0.8XS_{ut})^2}{S_{e}}[/tex]

[tex]a=\frac {(0.8X120)^2}{60}[/tex]

[tex]a= 153.6 kpsi

Step 4: calculate the coefficient 'b'

[tex]b =-\frac{1}{3}log(\frac{f*S_{ut} }{S_{e}})[/tex]

[tex]b =-\frac{1}{3}log(\frac{0.8*120}{60})[/tex]

[tex]b =-0.0680

Step 5: calculate the life of the specimen

[tex]N=(\frac{\alpha_{a}}{a})^\frac{1}{b}[/tex]

[tex]N=(\frac{70}{153.6})^\frac{1}{-0.068}[/tex]

[tex]N=104,576 cycles [/tex]

∴ the life (N) of the steel specimen is 104,576 cycles

Answer:

[tex]P_m_i_n= 356.9 KPa[/tex]

Explanation:

Coefficient of performance (COP) of refrigeration cycle is given by:

[tex]COP=\frac{1}{\frac{T_H}{T_L}-1 }[/tex]

We are given:

[tex]T_H=1.2T_L[/tex]

[tex]COP=\frac{1}{1.2-1 }[/tex]

COP= 5

We can also write Coefficient of performance (COP) of refrigeration cycle  as:

[tex]COP_R=\frac{Q_L}{W_i_n}[/tex]

Amount of heat absorbed by low temperature reservoir can be found as:

[tex]Q_L=COP_R * W_i_n[/tex]

[tex]Q_L=5 * 22 KJ[/tex]

[tex]Q_L=110 KJ[/tex]

According to first law of thermodynamics amount of heat rejected by hot reservoir is given by:

[tex]Q_H=Q_L + W_i_n[/tex]

[tex]Q_H=110 KJ + 22 KJ[/tex]

[tex]Q_H=132 KJ[/tex]

We are given the mass of 0.96 kg. So,

[tex]q_H=\frac{Q_H}{m}[/tex]

[tex]q_H=\frac{132 KJ}{0.96Kg}[/tex]

[tex]q_H=137.5 KJ/Kg[/tex]

Since it is a saturated liquid-vapour mixture [tex]q_H=h_f_g[/tex].

[tex]q_H=h_f_g=137.5 KJ/Kg[/tex]

From Refrigerant 134-a tables [tex]T_H[/tex] at [tex]h_f_g=137.5 KJ/Kg[/tex] is 61.3 C. (We calculated this by interpolation)

Converting [tex]T_H[/tex] from Celsius to Kelvin:

[tex]61.3^{o} C+273 = 334.3^{o} K[/tex]

[tex]T_H= 334.3^{o} K[/tex]

We are given:

[tex]T_H=1.2T_L[/tex]

[tex]T_L=\frac{T_H}{1.2}[/tex]

[tex]T_L=\frac{334.3}{1.2}[/tex]

[tex]T_L=278.58^{o} K[/tex]

Converting [tex]T_L[/tex] from Kelvin to Celsius:

[tex]278.58^{o} K-273 = 5.58^{o} C [/tex]

[tex]T_L= 5.58^{o} C [/tex]

From Refrigerant 134-a tables [tex]P_m_i_n[/tex] at [tex]T_L=5.58^{o} C[/tex] is 356.9 KPa. (We calculated this by interpolation).

[tex]P_m_i_n= 356.9 KPa[/tex]

The minimum pressure in the Carnot refrigeration cycle is 356.9 kPa.

Using,

Coefficient of performance (COP) of the refrigeration cycle. The COP is a measure of how efficient the refrigeration cycle is.

It is given by the formula:

COP = 1 / ([tex]T_H[/tex] /[tex]T_L[/tex] - 1),

where

[tex]T_H[/tex] is the maximum absolute temperature and [tex]T_L[/tex] is the minimum absolute temperature.

We are told that [tex]T_H[/tex] is 1.2 times [tex]T_L[/tex].

COP = 1 / (1.2 - 1).

Simplifying

COP = 5.

We can also express the COP as the ratio of the heat absorbed by the low-temperature reservoir to the net work input.

Using this relationship,

The heat absorbed ([tex]Q_L[/tex]) is equal to 5 times the net work input, which is 5 * 22 kJ

= 110 kJ.

According to the first law of thermodynamics, the heat rejected by the hot reservoir ([tex]Q_H[/tex]) is equal to the sum of the heat absorbed ([tex]Q_L[/tex]) and the net work input (22 kJ).

So,

[tex]Q_H[/tex] = 110 kJ + 22 kJ

= 132 kJ.

Since we are given the mass of the refrigerant (0.96 kg), we can find the heat transfer per unit mass ([tex]q_H[/tex]) using QH/m.

[tex]q_{H}[/tex]= 132 kJ / 0.96 kg

= 137.5 kJ/kg.

Since the refrigerant is in a saturated liquid-vapor mixture state, [tex]q_H[/tex] represents the enthalpy change from liquid to vapor.

By looking up the tables for refrigerant-134a, we can find the corresponding temperature ([tex]T_H[/tex]) at [tex]q_H[/tex] = 137.5 kJ/kg.

After conversion to Kelvin, [tex]T_H[/tex] is found to be 334.3 K.

Using the relationship [tex]T_H[/tex] = 1.2[tex]T_L[/tex],

Dividing [tex]T_H[/tex] by 1.2,  

[tex]T_L[/tex] = 334.3 K / 1.2 =

278.58 K.

Converting this back to Celsius, we get [tex]T_L[/tex] = 5.58 °C.

Finally, by looking up the tables for refrigerant-134a, we can find the minimum pressure ([tex]P_{min}[/tex]) at [tex]T_L[/tex] = 5.58 °C.

The value is determined to be 356.9 kPa.

So, the minimum pressure in the cycle is 356.9 kPa.

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Answer: -1543371.65837 W

= - 1543.372 kW.

Explanation:

Using the equation;

Q= EσA -------------------------------------------------------------------------------------------(1).

Where Q= net rate of radiation heat transfer between the floor and the ceiling of the furnace, σ = Boltzmann's constant, A= area of the cube, E = emissitivity.

Recall that the emissitivity of a black body is equals to one(1).

From the question, the parameters given are; The view factor from the ceiling to the floor of the furnace,F12 = 0.2, σ = 5.67 × 10-8., A= (4.45×4.45) m.

Slotting in the parameters into the equation;

Q= EσA[T(2)^4 - (T(1)^4] ---------------------------------------------------------------------(2).

Therefore, Q= (1)× (5.67×10^-8) × (4.45×4.45) m × [(547)^4 - (1100)^4]

= 0.0000011228 × (89526025681 - 1.4641×10^12).

= 0.0000011228×(-1.374574×10^12)

= -1543371.65837 W

= -1543.372 kW.

Answer:

False

Explanation:

The Engineer of Record for all projects shall be a registered Professional Engineer (P.E.)

Some Detailed Responsibilities of Engineer of Record

The Engineer of record shall attend

All these responsibilties require his physical presence which cannot be achieved through electronic communication.

Answer:

Options B and E

Explanation:

To give sustainable environmental design when considering time of convergence in layer 3, an engineer must consider the loss of a valid forwarding path and the table updates since these will determine whether the design becomes fit or not.

Answer:

Increasing the amount of fiber in your diet can aid in achieving and maintaining a healthy weight. This is correct if you are consuming less than 25-30gms of fiber per day. exceding this limit won't be beneficial.

Consuming a high-fiber diet most likely promotes the health of the digestive system. This is correct. Fibers are important for the digestive system´s health, especially for intestines and colon.

Fiber and other carbohydrates like starch and sugar are digested and absorbed in the same manner. This is Incorrect. Fiber is absorbed and digested at a much slower rate than sugar or starch.

Consuming a diet high in dietary fiber increases LDL "the bad" cholesterol. This is incorrect. Consuming a diet high in dietary fiber would decrease the LDL.

Most American women consume more than 20 g of fiber per day, and most American men consume more than 30 g per day. This is incorrect. The data obtained by the University of California San Francisco said that currently the amount of fiber intake by Americans adults is about 15g a day, which is half the recommended amount.

Answer:

Consuming a high-fiber diet most likely promotes the health of the digestive system.  

Increasing the amount of fiber in your diet can aid in achieving and maintaining a healthy weight.

Explanation:

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:

Explanation:

effective delay = delay when no traffic x [tex]\frac{100}{100- network\r usage}[/tex]

effective delay = [tex]84.1 \times \frac{100}{100-39.3}=138.55024711697ms[/tex]

Answer:

[tex]\left | Z_{in}} \right | = 524.72 \approx  525[/tex]ohm

Explanation:

Given data:

capacitor = 0.5 micro farad

resistor = 500 ohm

Frequency = 2000 Hz

Impedance of any circuit is calculated by using following equation

[tex]Z_{in} = Resistor - \frac{j}{\omega_c}[/tex]

         [tex] = 500 - \frac{j}{2\pi*2000*0.5 \mu}[/tex]

           =500 - j 159.155

[tex]\left | Z_{in}} \right | = \sqrt{(500^2 + 159.155^2)}[/tex]

[tex]\left | Z_{in}} \right | = 524.72 \approx  525[/tex]ohm

Answer:

Resultant force = 639 kN and it acts at 0.99m from the bottom of the conduit

Explanation:

The pressure is given as 200 KPa and the specific gravity of the liquid is 1.6.

The resultant force acting on the vertical plate, Ft, is equivalent to the sum of the resultant force as a result of pressurized air and resultant force due to oil, which will be taken as F1 and F2 respectively.

Therefore,

Ft = F1 + F2

According to Pascal's law which states that a change in pressure at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere, the air pressure will act on the whole cap surface.

To get F1,

F1 = p x A

= p x (πr²)

Substituting values,

F1 = 200 x π x 1²

F1 = 628.32 kN

This resultant force acts at the center of the plate.

To get F2,

F2 = Π x hc x A

F2 = Π x (4r/3π) x (πr²/2)

Π - weight density of oil,

A - area on which oil pressure is acting,

hc - the distance between the axis of the conduit and the centroid of the semicircular area

Π = Specific gravity x 9.81 x 1000

Therefore

F2 = 1.6 x 9.81 x 1000 x (4(1)/3π) x (π(1)²/2)

F2 = 10.464 kN

Ft = F1 + F2

Ft = 628.32 + 10.464

Ft = 638.784 kN

The resultant force on the surface is 639 kN

Taking moments of the forces F1 and F2 about the centre,

Mo = Ft x y

Ft x y = (F1 x r) + F2(1 - 4r/3π)

Making y the subject,

y = (628.32 + 10.464(1 - 4/3π)/ 638.784

y = 0.993m

An Independent consultant working on the design of an electrical distribution system for a corporation is not exempt from licensure. Applicants with foreign degrees must prove ABET equivalency, and the FBPE verifies experience through multiple documentation and procedures.

The activity that is not exempt from licensure pursuant to Chapter 471, F.S. is 'An Independent consultant working on the design of an electrical distribution system project for Progress Energy Corp.' This scenario indicates an individual offering engineering services to the public, which requires a license. On the other hand, individuals practicing engineering on their own properties, full-time employees of a utility company, and civil engineers employed by the U.S. Army Corps of Engineers typically fall under exemptions to licensure requirements in many states, including Florida.

For applicants with degrees from foreign institutions to document "substantial equivalency to ABET criteria to the FBPE," they can achieve this by providing a transcript from their institution, providing a notarized certification of completed credit hours as per the specified rules, getting the evaluation from a service provider approved by the FBPE, or passing the Principles & Practice examination. To verify an applicant's experience, the FBPE requires various forms of documentation and follows several procedures, including adherence to specific rules, consideration of personal references, and requiring evidence of employment in the engineering profession – essentially, they consider all of the above methods.

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Copper cables are used to communicate binary data by altering the voltage between two ranges.

A pair of Twisted copper cables send data through a network by transmitting pulses of electricity that represent binary data.

On this note, to make sure cables are transmitting information in a way that can be understood by the recipient, they follow the Ethernet standards. This twisted pair cables are commonly known as Ethernet cables.

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Answer / Explanation:

On the basis of the test result, Ultimate strength which is mostly known as the ultimate tensile strength is the strength attached to the ability or capacity of a structural element or material used in the test to withstand elongation forces or pull force applied to it.  

WHILE,

True stress at fracture can be classified as stress or load associated to the point where yielding or fracture occurred divided by the cross-sectional area at the yield point.

Answer:

a)75.8%

b)2.517KW/K

Explanation:

Hello!

To solve this problem follow the steps below

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

1. use thermodynamic tables to find the following variables.

a.enthalpy and entropy at the turbine entrance

h1=Enthalpy(Water;T=500;P=6000)

=3422KJ/kg

s1=Entropy(Water;T=500;P=6000)

=6.881KJ/kgK

b. enthalpy and ideal entropy at the turbine outlet

h2i=Enthalpy(Water;s=6.881;P=20)

=2267KJ/kg

s2i=s1=6.881KJ/kgK

2. uses the output power and the first law of thermodynamics to find the real enthalpy at the turbine's output

W=m(h1-h2)

h2=h1-W/m

h2r=3422-2626/3=2546.6KJ/kg

3.

find efficiency with the following equation

[tex]eficiency=\frac{h1-h2r}{h1-h2i}[/tex]

[tex]\frac{h1-h2r}{h1-h2i}=\frac{3422-2546.6}{3422-2267} =0.758=75.8%[/tex]

4.

find the real entropy at the turbine exit

s2=Entropy(Water;h=2546,6;P=20)=7.72KJ/kgK

5.Finally find the entropy generated, using the following equation

ΔS=m(s2-s1)=(3kg/s)(7.72  KJ/kgK-6.881 KJ/kgK)=2.517KW/K

Answer:

A) The charge accumulated on upper plate for t>0 is

[tex]q(t)=50[1-e^{-2500t}]\mu C[/tex]

B) The total charge that accumulates at the upper terminal is 50μC

C)  If the current is stopped at t = 0.5 ms then total charge stored on upper terminal is 35.67μC

Explanation:

Given that:

[tex]I(t)=0 \quad \quad \quad \quad \quad t<0\\\\I(t)= 125e^{-2500t} \quad t\geq 0[/tex]

A) The charge that accumulates at the upper terminal for t > 0:

As we know

[tex]q(t)=\int {I(t)} \, dt[/tex]

for t > 0

[tex]q(t)=\int\limits^t_0 {I(t)} \, dt\\q(t)=\int\limits^t_0 {125e^{-2500t} mA} \, dt\\q(t)=(125\times 10^{-3})[\frac{e^{-2500t}}{-2500}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}+1]\\\\[/tex]

The charge accumulated on upper plate for t>0 is

[tex]q(t)=50[1-e^{-2500t}]\mu C---(1)[/tex]

B)  The total charge that accumulates at the upper terminal can be found by substituting t → ∞ in equation (1)

[tex]q(t)=( 50\times 10^{-6})[1-e^{-2500(\infty)}]\\q(t)=(50\times 10^{-6})[1-0]\\q(t) =50\mu C[/tex]

C)  If the current is stopped at t = 0.5 ms then

[tex]q(t)=( 50\times 10^{-6})[1-e^{-2500(0.5\times10^{-3})}]\\q(0.5ms)=35.67\mu C[/tex]

Answer:e. support reinforcing bars in beams and slabs, prior to concrete placement.

Explanation: Chairs are used for construction of foundations,large steel supports,deck constructions and for underground works. Chairs can be for rebar support ( rebar support chairs),post tension chairs.

Bolsters are usually long made of metallic materials mainly used as support for construction of different infrastructures like roads it helps to ensure the concretes and other construction materials stay firmly connected. Image 1 is a metal be chair

Image 2 is a metal bolster

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s       and ρ= 850 kg/m³  

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ =  ρv

   =850 x 0.00062

   = 0.527 kg/m·s  

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

[tex]Q = \dfrac{\Delta P \pi D^4}{128 \mu L}[/tex]

[tex]Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}[/tex]

Q = 5.06 x 10⁻⁸ m³/s

Answer:

The right answer is d.

Explanation:

The amount of play that a ball joint can have (if any) depends on the application of the joint itself and the suspension system. For each vehicle, the manufacturer must warn wich is the maximum tolerance of play. For some load-carrying ball joints, the amount of play should be minimum or null (like in some heavy-duty vehicles). Therefore technician A is wrong. In most cases in urban vehicles, it is acceptable some level of play for follower ball joints (these joints are normally unloaded). Therefore the technician B is also wrong. That means neither technicians A nor B statements are correct.

There are different functions of Technician. The two Technician are wrong so Neither A nor B is the right answer.

In Ball Joint Inspection , using the old rule of thumb which state that ball joints if more than .050 inches of play are worn and it does will not hold true for all vehicles.

There are some ball joints which does not have visible play while others can hold up to . 250 inch or more of play and still be work well.

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Answer:

a) 0.26

b) 1077 MPa

Explanation:

a) The following equation can be used to determine the volume fraction:

[tex]\frac{F_f}{F_m} =\frac{E_fV_f}{E_m(1-V_f)}[/tex]

[tex]\frac{0.97}{1-0.97} =\frac{260V_f}{2.8(1-V_f)}[/tex]

[tex]32.3 = \frac{260V_f}{2.8-2.8V_f}[/tex]

[tex]V_f = 0.26[/tex]

b) Tensile strength can be found by using the following equation:

[tex]\sigma_{cl} = \sigma_m(1-V_f)+\sigma_fV_f = 50*(1-0.26)+4000*0.26 = 1077[/tex] MPa

Answer:

The Hilbert transform we use in mathematics it usually apply on a function consisting of a real variable and produces another function of a real variable.

Explanation:

The heat transfer for the process is -20.9 kJ, and the entropy production is 1 kJ/K.

The heat transfer for the given ammonia process is -20.9 kJ. To calculate the entropy production, we can use the equation: ΔS = Q/T, where temperature T is in Kelvin. Given Q = 40 kJ, and the average temperature is 40°C, the entropy production is 1 kJ/K.

Answer:

Explanation:

Organization structure:

refers to the idea of how people are supposed to work and coordinate in an organization to maintain a healthy and effective work environment.

There are various types of organizational structures which depends on several factors. There is no single best organization structure. Each structure has its own advantages and disadvantages. In order to select a structure the organization's vision, mission, culture values, goals are to be identified first.

Answer:

So, we have to analyze the whole circuit system provided, then we will get to know that its an RL-Circuit along with that we can easily pick the best option from the given, which is mentioned below:

Explanation:

Option (1) 3- 4e-2t+e-8t is the best option to chose from the given options.

To determine the net power developed, the rate of heat addition, and the thermal efficiency for an air-standard Brayton cycle with specified inlet conditions, compressor pressure ratio, maximum cycle temperature, and efficiencies, detailed thermodynamic calculations considering the given parameters are required.

The Brayton cycle is a thermodynamic cycle that describes the workings of a constant-pressure heat engine, commonly used in jet engines and gas turbine engines. The cycle involves air entering a compressor, being compressed, then heated in a combustor before expanding through a turbine, and finally being released to the environment. This question asks to calculate the net power developed, the rate of heat addition in the combustor, and the thermal efficiency of the cycle given specific conditions and efficiencies of the compressor and turbine.

Unfortunately, without the necessary thermodynamic equations and properties of air provided, along with the specific steps and intermediate states of the cycle, it is impossible to perform the detailed calculations needed to answer this question precisely. Typically, solving this problem would involve using the thermodynamic relations for an ideal gas, the isentropic process equations for the compressor and turbine, and the definitions of isentropic efficiency. The conservation of energy principle would be applied to find the net work output and heat added, and subsequently, the thermal efficiency would be calculated.

As such, this question requires a detailed understanding of thermodynamics, specifically the principles governing the Brayton cycle and the equations for calculating work done, heat transfer, and efficiency in thermodynamic cycles.

Answer:

240.83 ft

Explanation:

The distances AE and BF will be equal = 182.58 ft

The area of the lot will be the product of the depth and the average of the two frontages

= ( 350 * (220 + 260)/2) = 84000 ft

Half of the area becomes 42000 ft

<A = arctan (20/350) = 3. 3°

Hence 42000=h/2*(220 + 220 + 2h*tan3.3)

Solving, we obtain h= 182.28ft

EF = 220 + (2*182.28*tan 3.3)

= 240.83ft

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

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Answer:

Fy = -11267.294 lbf

Explanation:

given data

nozzle flow =  30 degrees

discharges the water =  20 degrees C

volume of water =  100 lb

Area of flange = 1.0 ft²

Area of nozzle = 0.50 ft²

Volume of area flange = 1.8 ft³

Vertical height flange to nozzle = 2 ft

solution

we will apply here continuity equation that is

A1 × V1 = A2 × V2    .............1

put here value and we get volume V1 that is

V1 = [tex]\frac{0.5\times 125}{1}[/tex]

V1 = 62.5 ft/s

and

now we will apply here Bernoulli equation that is

[tex]\frac{p1}{\gamma 1} + \frac{V1^2}{2g} + z1 = \frac{p2}{\gamma 2} + \frac{V2^2}{2g} + z2[/tex]      .............................2

put here value and we will get

p1 = 0 + [tex]\frac{62.4}{2\times 32.2}(125^2 - 62.5^2) + 62.4 (2)[/tex]

p1 = 11479.614 psf

so here moment in y will be

∑ Fy = m [  (Vo)y - (Vi)x ]

so here we get

p1 ×A1 + Fy - Wn - Ww = [tex]\rho[/tex] Q  [ V2 × sin30 - V1 ]

put here value and we get Fy

1147.614 × 1 + Fy - 100 - (62.4 × 1.8) = (1.94) × (0.5 ×130) × (125sin30 - 62.5)

solve it we get

Fy = -11267.294 lbf

Answer:

the volume of the oxygen tank is

V ox = 0.844 m³

the volume of the nitrogen tank is

V ni = 2.359 m³

the final pressure is

P = 255. 534 kPa

Explanation:

taking into account that

n=  m/M

where

n= number of moles , m = mass, M = molecular weight

then

n oxigen = m ox / M ox = 6.0 kg/(32 gr/mol) *1000gr/kg = 187.5 moles

n nitrogen = m ni / M ni = 4.0 kg/(28 gr/mol) *1000gr/kg = 142.857 moles

from the ideal gas law

P*V=n*R*T

where P= absolute pressure, V= volume occupied by the gas , R= ideal gas constant= 8.314 J/(mol K) , T= absolute temperature

V=n*R*T/P

replacing values

for the oxygen tank , T ox= 25°C= 298 K , P = 550 kPa= 550000 Pa ,

V ox =n*R*T/P = 187.5 mol* 8.314 J/(mol K)* 298 K/ 550000 Pa = 0.844 m³

V ox = 0.844 m³

for the nitrogen tank, T ni=  298 K ,  P = 150000 Pa

V ni =n*R*T/P = 142.857 mol* 8.314 J/(mol K)* 298 K/ 150000 Pa = 2.359 m³

V ni = 2.359 m³

when the gases mix , they occupy a volume of

V = V ox + V ni = 0.844 m³ + 2.359 m³ = 3.203 m³

and total number of moles of gas of the mixture is

n = n oxigen + n nitrogen = 187.5 moles + 142.857 moles = 330.357 moles

therefore

P*V=n*R*T

P = n*R*T/V

replacing values

P = n*R*T/V = 330.357 mol*8.314 J/(mol K)* 298 K/ 3.203 m³ *1 kPa/1000Pa = 255. 534 kPa

P = 255. 534 kPa

Answer:

a) the molar fraction of neon at the exit is

xₙ= 0.735

and carbon monoxide

xₓ = 0.265

b) the final temperature is

T =  410.55 K

c) the rate of entropy production is

ΔS = 1.83 KW/K

Explanation:

denoting n for neon and x for carbon monoxide:

a) from a mass balance, the molar fraction of neon at the exit is:

outflow mass neon=inflow mass neon

xₙ = outflow mass neon/ (total outflow of mass) = inflow mass neon/ (total outflow of mass) = (1 kg/seg / 20.18 kg/kmol) / (1 kg/seg / 20.18 kg/kmol + 0.5 kg/seg / 28.01 kg/kmol) = 0.735

and the one of carbon monoxide is

xₓ = 1-xₙ = 1-0.735 = 0.265

b) from the first law of thermodynamics applied to an open system, then

Q - Wo = ΔH + ΔK +  ΔV

where

Q= heat flow to the chamber = 0 ( insulated)

Wo= external work to the chamber = 0 ( there is no propeller to mix)

ΔH = variation of enthalpy

ΔK = variation of kinetic energy ≈ 0 ( the changes are small with respect to the one of enthalpy)

ΔV = variation of kinetic energy ≈ 0 ( the changes are small with respect to the one of enthalpy)

therefore

ΔH = 0 → H₂ - H₁ = 0 → H₂=H₁

if we assume ideal has behaviour of neon and carbon monoxide, then

H₁ = H ₙ₁ + H ₓ₁ = mₙ₁*cpₙ*Tn + mₓ₁*cpₓ*Tc

H₂ = (m ₙ+mₓ)*cp*T  

for an ideal gas mixture

cp = ∑ cpi xi

therefore

mₙ*cpₙ*Tₙ + mₓ*cpₓ*Tₙ₁ = (m ₙ+mₓ)*∑ cpi xi*T

mₙ/(m ₙ+mₓ)*cpₙ*Tₙ + mₓ/(m ₙ+mₓ)*cpₓ*Tₓ = T ∑ cpi xi

xₙ* cpₙ*Tₙ +xₓ*cpₓ*Tₓ = T*( xₙ* cpₙ+xₓ*cpₓ)

T= [xₙ* cpₙ/( xₙ* cpₙ+xₓ*cpₓ)]*Tₙ₁ +[xₓ*cpₓ/( xₙ* cpₙ+xₓ*cpₓ)]*Tₓ₁

denoting

rₙ = xₙ* cpₙ/( xₙ* cpₙ+xₓ*cpₓ)

and

rₓ= xₓ*cpₓ/( xₙ* cpₙ+xₓ*cpₓ)

T= rₙ *Tₙ +rₓ*Tₓ

for an neon , we can approximate its cv through the cv for an monoatomic ideal gas  

cvₙ= 3/2 R , R= ideal gas constant=8.314 J/mol K=

since also for an ideal gas: cpₙ - cvₙ = R → cpₙ = 5/2 R

for the carbon monoxide ,  we can approximate its cv through the cv for an diatomic ideal gas

cvₓ= 7/2 R → cpₓ = 9/2 R

replacing values

rₙ = xₙ* cpₙ/( xₙ* cpₙ+xₓ*cpₓ)  = xₙ₁*5/2 R/ ( xₙ₁*5/2 R+xₓ*9/2 R) =

xₙ₁*5/(xₙ₁*5 + xₓ*9) = 5xₙ₁/(5 + 4*xₓ) = 5*0.735/(5+ 4*0.265) =0.598

since

rₙ + rₓ =1  → rₓ = 1-rₙ = 1- 0.598 = 0.402

then

T =  rₙ *Tₙ +rₓ*Tₓ  = 0.598 * 300 K + 0.402 * 575 K = 410.55 K

c) since there is no entropy changes due to heat transfer , the only change in entropy is due to the mixing process

since for a pure gas mixing process

ΔS = n*Cp* ln T₂/T₁ -n*R ln (P₂/P₁)

but P₂=P₁ (P=pressure)

ΔS = n*Cp* ln T₂/T₁ = n*Cp*ln T₂ - n*Cp*ln T₁ = S₂-S₁

for a gas mixture as end product

ΔS = (nₓ+nₙ)*Cp*ln T - (nₓ*Cpₓ*ln Tₓ + nₙ*Cpₙ*ln Tₙ)

ΔS = nₙ*Cpₙ ( (nₓ+nₙ)*Cp/[nₙ*Cpₙ]* ln T - ( nₓ*Cpₓ/ (nₙ*Cpₙ) *ln Tₓ + ln Tₙ)

ΔS = nₙ*Cpₙ [ 1/rₙ * ln T -( (rₓ/rₙ)*ln Tₓ + ln Tₙ)]

replacing values ,

ΔS = nₙ*Cpₙ [ 1/rₙ * ln T -( (rₓ/rₙ)*ln Tₓ + ln Tₙ)] = (1 kg/s/ 20.18*10 kg/kmol)* 5/2* 8.314 kJ/kmol K *[ 1/0.598 * ln 410.55 K-( 0.402/0.598  *ln 575 K + ln 300K)]

= 1.83 KW/K