Answer:
Fluid flow in a pipe, the pressure decrease due to throttle valve from 10 bar to 100 kpa if the specific volume of fluid increase from 0.3 m3 /kg to 1.8 m3/kg. Find the change in internal energy during the process
Explanation:
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An example of a transient analysis involving the 1st law of thermodynamics and conservation of mass is the filling of a compressed air tank. Assume that an air tank is being filled using a compressor to a pressure of 5 atm, and that it is being fed with air at a temperature of 25°C and 1 atm pressure. The compression process is adiabatic. Will the temperature of the air in the tank when it is done being filled i.e. once the pressure in the tank reaches 5 atm), be greater than, equal to, or less that the temperature of the 25°C air feeding the compressor?
A. Greater than 25°C
B. Unable to determine
C. Same as 25°C
D. Less than 25°C
Answer:
The temperature will be greater than 25°C
Explanation:
In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.
mathematically
Change in the internal energy of a system ΔU = ΔQ + ΔW
in an adiabatic process, ΔQ = 0
therefore
ΔU = ΔW
where ΔQ is the change in heat into the system
ΔW is the work done by or done on the system
when work is done on the system, it is conventionally negative, and vice versa.
also W = pΔv
where p is the pressure, and
Δv = change in volume of the system.
In this case, work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C
Determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 225 and at the same time have a ductility of at least 12%EL. Justify your decision
Answer:
First we determine the tensile strength using the equation;
Tₓ (MPa) = 3.45 × HB
{ Tₓ is tensile strength, HB is Brinell hardness = 225 }
therefore
Tₓ = 3.45 × 225
Tₓ = 775 Mpa
From Conclusions, It is stated that in order to achieve a tensile strength of 775 MPa for a steel, the percentage of the cold work should be 10
When the percentage of cold work for steel is up to 10,the ductility is 16% EL.
And 16% EL is greater than 12% EL
Therefore, it is possible to cold work steel to a given minimum Brinell hardness of 225 and at the same time a ductility of at least 12% EL
A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 mL of the base. Assuming complete neutralization of the acid,
1) What was the normality of the acid solution?
2) What was the molarity of the acid solution?
Answer:
a. 0.4544 N
b. [tex]5.112 \times 10^{-5 M}[/tex]
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
[tex]H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O[/tex]
[tex]NaOH\ Mass = Normality \times equivalent\ weight \times\ volume[/tex]
[tex]= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL[/tex]
= 0.27264 g
[tex]NaOH\ mass = \frac{mass}{molecular\ weight}[/tex]
[tex]= \frac{0.27264\ g}{40g/mol}[/tex]
= 0.006816 mol
Now
Moles of [tex]H_2SO_4[/tex] needed is
[tex]= \frac{0.006816}{2}[/tex]
= 0.003408 mol
[tex]Mass\ of\ H_2SO_4 = moles \times molecular\ weight[/tex]
[tex]= 0.003408\ mol \times 98g/mol[/tex]
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
[tex]= \frac{acid\ mass}{equivalent\ weight \times volume}[/tex]
[tex]= \frac{0.333984 g}{49 \times 0.015}[/tex]
= 0.4544 N
b. And, the acid solution molarity is
[tex]= \frac{moles}{Volume}[/tex]
[tex]= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}[/tex]
= 0.00005112
=[tex]5.112 \times 10^{-5 M}[/tex]
We simply applied the above formulas
The volume of the 0.3200 M, NaOH required to neutralize the H₂SO₄, is
21.30 mL, which gives the following acid solution approximate values;
1) Normality of the acid solution is 0.4544 N
2) The molarity of the acid is 0.2272
How can the normality, molarity of the solution be found?Molarity of the NaOH = 0.3200 M
Volume of NaOH required = 21.30 mL
1) The normality of the acid solution is found as follows;
The chemical reaction is presented as follows;
H₂SO₄(aq) + 2NaOH (aq) → Na₂SO₄ (aq) + H₂O
Number of moles of NaOH in the reaction is found as follows;
[tex]n = \dfrac{21.30}{1,000} \times 0.3200 \, M = \mathbf{0.006816 \, M}[/tex]
Therefore;
The number of moles of H₂SO₄ = 0.006816 M ÷ 2 = 0.003408 M
[tex]Normality = \mathbf{ \dfrac{Mass \ of \, Acid \ in \ reaction}{Equivalent \ mass \times Volume \ of \ soltute}}[/tex]
Which gives;
[tex]Normality = \dfrac{ 98 \times 0.003408 }{49 \times 0.015} = \mathbf{0.4544}[/tex]
The normality of the acid solution, H₂SO₄(aq), N ≈ 0.45442) The molarity is found as follows;
[tex]Molarity = \dfrac{0.003408 \, moles}{0.015 \, L} = \mathbf{0.2272 \, M}[/tex]
The molarity of the acid solution is 0.2272 MLearn more about the normality and the molarity of a solution here:
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For each of the following stacking sequences found in FCC metals, cite the type of planar defect that exists:
a. . . . A B C A B C B A C B A . . .
b. . . . A B C A B C B C A B C . . .
Copy the stacking sequences and indicate the position(s) of planar defect(s) with a vertical dashed line.
Answer:
a) The planar defect that exists is twin boundary defect.
b) The planar defect that exists is the stacking fault.
Explanation:
I am using bold and underline instead of a vertical line.
a. A B C A B C B A C B A
In this stacking sequence, the planar defect that occurs is twin boundary defect because the stacking sequence at one side of the bold and underlined part of the sequence is the mirror image or reflection of the stacking sequence on the other side. This shows twinning. Hence it is the twin boundary inter facial defect.
b. A B C A B C B C A B C
In this stacking sequence the planar defect that occurs is which occurs is stacking fault defect. This underlined region is HCP like sequence. Here BC is the extra plane hence resulting in the stacking fault defect. The fcc stacking sequence with no defects should be A B C A B C A B C A B C. So in the above stacking sequence we can see that A is missing in the sequence. Instead BC is the defect or extra plane. So this disordering of the sequence results in stacking fault defect.
The closed feedwater heater of a regenerative Rankine cycle is to heat 7000 kPa feedwater from 2608C to a saturated liquid. The turbine supplies bleed steam at 6000 kPa and 3258C to this unit. This steam is condensed to a saturated liquid before entering the pump. Calculate the amount of bleed steam required to heat 1 kg of feedwater in this unit.
Answer:
the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s
Explanation:
Given that:
Pressure of the feed water = 7000 kPa
Temperature of the closed feedwater heater = 260 ° C
Pressure of of the turbine = 6000 kPa
Temperature of the turbine = 325 ° C
The objective is to calculate the amount of bleed steam required to heat 1 kg of feedwater in this unit.
From the table A-4 of saturated water temperature table at temperature 260° C at state 1 ;
Enthalpies:
[tex]h_1 = h_f = 1134.8 \ kJ/kg[/tex]
From table A-6 superheated water at state 3 ; the value of the enthalpy relating to the pressure of the turbine at 6000 kPa and temperature of 325° C is obtained by the interpolating the temperature between 300 ° C and 350 ° C
At 300° C; enthalpy = 2885.6 kJ/kg
At 325° C. enthalpy = 3043.9 kJ/kg
Thus;
[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-{h_{300^0}}}{{h_{350^0}}- {h_{300^0}}}[/tex]
[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]
[tex]\dfrac{25}{50}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]
[tex]h_{325^0} = 2885.6 + \dfrac{25}{50}({3043.9-2885.6 )[/tex]
[tex]h_{325^0} = 2885.6 + 0.5({3043.9-2885.6 )[/tex]
[tex]h_{325^0} =2964.75 \ kJ/kg[/tex]
At pressure of 7000 kPa at state 6; we obtain the enthalpies corresponding to the pressure at table A-5 of the saturated water pressure tables.
[tex]h_6 = h_f = 1267.5 \ kJ/kg[/tex]
From state 4 ;we obtain the specific volume corresponding to the pressure of 6000 kPa at table A-5 of the saturated water pressure tables.
[tex]v_4 = v_f = 0.001319\ m^3 /kg[/tex]
However; the specific work pump can be determined by using the formula;
[tex]W_p = v_4 (P_5-P_4)[/tex]
where;
[tex]P_4[/tex] = pressure at state 4
[tex]P_5[/tex] = pressure at state 5
[tex]W_p = 0.001319 (7000-6000)[/tex]
[tex]W_p = 0.001319 (1000)[/tex]
[tex]W_p =1.319 \ kJ/kg[/tex]
Using the energy balance equation of the closed feedwater heater to calculate the amount of bleed steam required to heat 1 kg of feed water ; we have:
[tex]E_{in} = E_{out} \\ \\ m_1h_1 +m_3h_3 + m_3W_p = (m_1+m_3)h_6[/tex]
where;
[tex]m_1 = 1 \ kg[/tex]
Replacing our other value as derived above into the energy balance equation ; we have:
[tex]1 \times 1134.8 +m_3 \times 2964.75 + m_3 \times 1.319 = (1+m_3)\times 1267.5[/tex]
[tex]1134.8 + 2966.069 \ m_3 = 1267.5 + 1267.5m_3[/tex]
Collect like terms
[tex]2966.069 \ m_3- 1267.5m_3 = 1267.5-1134.8[/tex]
[tex]1698.569 \ m_3 =132.7[/tex]
[tex]\ m_3 = \dfrac{132.7}{1698.569}[/tex]
[tex]\mathbf{ m_3 = 0.078 \ kg/s}[/tex]
Hence; the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s
Air enters the compressor of an ideal 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. 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.
Answer:
(a) 48.2 %
(b) 0.4137
(c) 2385.9 kW
Explanation:
The given values are:
Initial pressure,
p₁ = 100 kPa
Initial temperature,
T₁ = 300 K
Mass,
M = 6 kg/s
Pressure ration,
r = 10
Inlent temperature,
T₃ = 1400 K
Specific heat ratio,
k = 1.4
At T₁ and p₁,
⇒ [tex]c_{p}=1.005 \ KJ/Kg.K[/tex]
Process 1-2 in isentropic compression, we get
⇒ [tex]\frac{T_{2}}{T_{1}}=(\frac{p_{2}}{p_{1}})^{\frac{k-1}{k}}[/tex]
[tex]T_{2}=(\frac{p_{2}}{p_{1}})^{\frac{k-1}{k}}. T_{1}[/tex]
On putting the estimated values, we get
[tex]=(10)^{\frac{1.4-1}{1.4}}(300)[/tex]
[tex]=579.2 \ K[/tex]
Process 3-4,
⇒ [tex]\frac{T_{4}}{T_{3}}=(\frac{p_{4}}{p_{3}})^{\frac{k-1}{k}}[/tex]
[tex]T_{4}=(\frac{1}{10})^{\frac{1.4-1}{1.4}}(1400)[/tex]
[tex]=725.13 \ K[/tex]
(a)...
The thermal efficiency will be:
⇒ [tex]\eta =\frac{\dot{W_{t}}-\dot{W_{e}}}{\dot{Q_{in}}}[/tex]
[tex]\eta=1-\frac{\dot{Q_{out}}}{\dot{Q_{in}}}[/tex]
⇒ [tex]\dot{Q_{in}}=\dot{m}(h_{1}-h_{2})[/tex]
[tex]=\dot{mc_{p}}(T_{3}-T_{2})[/tex]
[tex]=6\times 1005\times (1400-579.2)[/tex]
[tex]=4949.4 \ kJ/s[/tex]
⇒ [tex]\dot{Q_{out}}=\dot{m}(h_{4}-h_{1})[/tex]
[tex]=6\times 1.005\times (725.13-300)[/tex]
[tex]=2563.5 \ KJ/S[/tex]
As we know,
⇒ [tex]\eta=1-\frac{\dot{Q_{out}}}{\dot{Q_{in}}}[/tex]
On putting the values, we get
[tex]=1-\frac{2563.5}{4949.4}[/tex]
[tex]=0.482 \ i.e., \ 48.2 \ Percent[/tex]
(b)...
Back work ratio will be:
⇒ [tex]bwr=\frac{\dot{W_{e}}}{\dot{W_{t}}}[/tex]
Now,
⇒ [tex]\dot{W_{e}}=\dot{mc_{p}}(T_{2}-T_{1})[/tex]
On putting values, we get
[tex]=6\times 1.005\times (579.2-300)[/tex]
[tex]=1683.6 \ kJ/s[/tex]
⇒ [tex]\dot{W_{t}}=\dot{mc_{p}}(T_{3}-T_{4})[/tex]
[tex]=6\times 1.005\times (1400-725.13)[/tex]
[tex]=4069.5 \ kJ/s[/tex]
So that,
⇒ [tex]bwr=\frac{1683.6}{4069.5}=0.4137[/tex]
(c)...
Net power is equivalent to,
⇒ [tex]\dot{W}_{eyele}=\dot{W_{t}}-\dot{W_{e}}[/tex]
On substituting the values, we get
[tex]= 4069.5-1683.6[/tex]
[tex]=2385.9 \ kW[/tex]
Following are the solution to the given points:
Given :
Initial pressure [tex]p_1 = 100\ kPa \\\\[/tex]
Initial temperature [tex]T_1 = 300\ K \\\\[/tex]
Mass flow rate of air [tex]m= 6\ \frac{kg}{s}\\\\[/tex]
Compressor pressure ratio [tex]r =10\\\\[/tex]
Turbine inlet temperature [tex]T_3 = 1400\ K\\\\[/tex]
Specific heat ratio [tex]k=1.4\\\\[/tex]
Temperature [tex]\ T_1 = 300\ K[/tex]
pressure [tex]p_1 = 100\ kPa\\\\[/tex]
[tex]\to c_p=1.005\ \frac{kJ}{kg\cdot K}\\\\[/tex]
Process 1-2 is isen tropic compression
[tex]\to \frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}} \\\\[/tex]
[tex]\to T_2=(\frac{P_2}{P_1})^{\frac{k-1}{k}} \ T_1 \\\\[/tex]
[tex]=(10)^{\frac{1.4-1}{1.4}} (300)\\\\ =(10)^{\frac{0.4}{1.4}} (300) \\\\[/tex]
[tex]\to T_2 = 579.2\ K \\\\[/tex]
Process 3-4 is isen tropic expansion
[tex]\to \frac{T_4}{T_3}=(\frac{P_4}{P_3})^{\frac{k-1}{k}}\\\\ \to T_4=(\frac{1}{10})^{\frac{1.4-1}{1.4}} (1400)\\\\\to T_4= 725.13\ K \\\\[/tex]
For point a:
The thermal efficiency of the cycle:
[tex]\to \eta = \frac{W_i-W_e}{Q_{in}} \\\\\to \eta = \frac{Q_{in}- Q_{out}}{Q_{in}}\\\\\to \eta =1 - \frac{Q_{out}}{Q_{in}} \\\\\to Q_{in}= m(h_3-h_1) = mc_p (T_4-T_1) =(6)(1.005)(725.13-300) = 2563 \ \frac{kJ}{S}\\\\\to \eta =1- \frac{Q_{out}}{Q_{in}}\\\\[/tex]
[tex]=1-\frac{2563.5}{4949.4}\\\\ = 0.482\\\\[/tex]
[tex]\eta = 48.2\%\\\\[/tex]
For point b:
The back work ratio
[tex]\to bwr =\frac{W_e}{W_t}[/tex]
Now
[tex]\to W_e =mc_p (T_2 -T_1)[/tex]
[tex]=(6) (1.005)(579.2 -300)\\\\ =1683.6 \ \frac{kJ}{S}\\\\[/tex]
[tex]\to W_t=mc_p(T_3-T_4)[/tex]
[tex]=(6)(1.005)(1400 - 725.13)\\\\ = 4069.5 \frac{KJ}{s}[/tex]
[tex]\to bwr =\frac{W_s}{W_t}= \frac{1683.6}{4069.5}=0.4137[/tex]
For point c:
The net power developed is equal to
[tex]\to W_{cycle} = W_t-W_e \\\\[/tex]
[tex]= ( 4069.5-1683.6)\\\\ = 2385.9 \ kW\\[/tex]
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1. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle.
i. True
ii. False
2. The cosine rule is used when we are given either a) three sides or b) two sides and the included angle.
i. True
ii. False
Answer:
A. Yes
B. Yes
Explanation:
We want to evaluate the validity of the given assertions.
1. The first statement is true
The sine rule stipulates that the ratio of a side and the sine of the angle facing the side is a constant for all sides of the triangle.
Hence, to use it, it’s either we have two sides and an angle and we are tasked with calculating the value of the non given side
Or
We have two angles and a side and we want to calculate the value of the side provided we have the angle facing this side in question.
For notation purposes;
We can express the it for a triangle having three sides a, b, c and angles A,B, C with each lower case letter being the side that faces its corresponding big letter angles
a/Sin A = b/Sin B = c/Sin C
2. The cosine rule looks like the Pythagoras’s theorem in notation but has a subtraction extension that multiplies two times the product of the other two sides and the cosine of the angle facing the side we want to calculate
So let’s say we want to calculate the side a in a triangle of sides a, b , c and we have the angle facing the side A
That would be;
a^2 = b^2 + c^2 -2bcCosA
So yes, the cosine rule can be used for the scenario above
Aggregate blend composed of 65% coarse aggregate (SG 2.701), 35% fine aggregate (SG 2.625)
Compacted specimen weight in air = 1257.9 g, submerged weight = 740.0 g, SSD weight = 1258.7 g
Compacted specimen contains 5.0% asphalt by total weight of the mix with Gb = 1.030
Theoretical maximum specific gravity = 2.511
Bulk specific gravity of the aggregate __________
Bulk specific gravity of the compacted specimen__________
Percent stone __________
Effective specific gravity of the stone__________
Percent voids in total mix__________
Percent voids in mineral aggregate__________
Percent voids filled with asphalt__________
Answer:
2.6742.42891.695%2.5923.305%11.786%78.1%Explanation:
coarse aggregate (ca) = 65%, SG = 2.701
Fine aggregate = 35%, SG = 2.625
A) Bulk specific gravity of aggregate
= [tex]\frac{65*2.701 + 35*2.625}{100} = 2.674[/tex]
B) Wm = 1257.9 g { weight in air }
Ww = 740 g { submerged weight }
therefore Bulk specific gravity of compacted specimen
= [tex]\frac{Wm}{Wm-Ww}[/tex] = [tex]\frac{1257.9}{1257.9 - 740 }[/tex] = 2.428
Theoretical specific gravity = 2.511
Percent stone
= 100 - asphalt content - Vv
= 100 - 5 - 3.305 = 91.695%
c) percent of void
= [tex]\frac{9.511-2.428}{2.511} * 100[/tex] Vv = 3.305%
d) let effective specific gravity in stone
= [tex]\frac{91.695*unstone+ 5 *1.030 }{96.695} = 2.511[/tex]
= Instone = 2.592 effective specific gravity of stone
e) Vv = 3.305%
f ) volume filled with asphalt (Vb) = [tex]\frac{\frac{Wb}{lnb} }{\frac{Wm}{Inm} } * 100[/tex]
Vb = [tex]\frac{5 * 2.428}{1.030 * 100} * 100[/tex]
Vb = 11.786 %
Volume of mineral aggregate = Vb + Vv
VMA = 11.786 + 3.305 = 15.091
g) percent void filled with alphalt
= Vb / VMA * 100
VMA = 11.786 + 3.305 = 15.091
percent void filled with alphalt
= Vb / VMA * 100 = (11.786 / 15.091) * 100 = 78.1%
The first choice for how to reduce or eliminate a hazard is: a) Engineering controls b) Workplace controls c) Personal protective equipment d) Administrative controls
Answer:
The correct answer would be a) Engineering Controls.
Explanation:
If the controls are handled correctly, you can reduce and eliminate hazards so no one gets hurt. Engineering controls are absolutely necessary to prevent hazards.
Hope this helped! :)
Personal protective equipment (PPE) is appropriate for controlling hazards
PPE are used for exposure to hazards when safe work practices and other forms of administrative controls cannot provide sufficient additional protection, a supplementary method of control is the use of protective clothing or equipment. PPE may also be appropriate for controlling hazards while engineering and work practice controls are being installed.
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Which of the following reduces friction in an engine A)wear B)drag C)motor oil D)defractionation
It is motor oil, as oil is used to reduce friction
Anytime scaffolds are assembled or __________, a competent person must oversee the operation.
a. Drawn
b. Disassembled
c. Thought
d. Made
When scaffolds are now being construct or deconstruct, a competent person must supervise the work and train everybody who'll be assisting, and the further discussion can be defined as follows:
The competent person is also responsible for proposing whether fall protection is required for each scaffold erected. In constructing a scaffold, there are specific criteria for the ground the scaffold is constructed. On the products and components used to build the scaffold, its height in relation to the foundation. It's platform's design, and whether or not high efficiency is needed to supervise the installation.Therefore, the final answer is "Option B".
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Solid solution strengthening is achieved byGroup of answer choicesstrain hardening restricting the dislocation motion increasing the dislocation motion increasing the grain boundary g
Answer:
B. restricting the dislocation motion
Explanation:
Solid solution strengthening is a type of alloying that is carried out by the addition of the atoms of the element used for the alloying to the crystallized lattice structure of the base metal, which the metal that would be strengthened. The purpose of this act is to increase the strength of metals. It actually works by impeding or restricting the motion in the crystal lattice structure of metals thus making them more difficult to deform.
The solute atoms used for strengthening could be interstitial or substitutional. The interstitial solute atoms work by moving in between the space in the atoms of the base metal while the substitutional solute atoms make a replacement with the solvent atoms in the base metal.
Consider a double-pipe counter-flow heat exchanger. In order to enhance its heat transfer, the length of the heat exchanger is doubled. Will the effectiveness of the exchanger double?
Answer:
effectiveness of the heat exchanger will not be double when the length of the heat exchanger is doubled.
Because effectiveness depends on NTU and not necessarily the length of the heat exchanger
. The job of applications engineer for which Maria was applying requires (a) excellent technical skills with respect to mechanical engineering, (b) a commitment to working in the area of pollution control, (c) the ability to deal well and confidently with customers who have engineering problems, (d) a willingness to travel worldwide, and (e) a very intelligent and well-balanced personality. List 10 questions you would ask when interviewing applicants for the job.
Answer:
Tell us about your self Are your confident that you are the right candidate for this positionwhy should i hire youDo you like working under supervisionHow do you like to work ( in a group or individually )What is your ultimate workplace goalwhat are your future plansWhat do you expect from the Organization when given the jobDo you like taking on critical problemsHow long can you work in this positionExplanation:
For a job of applications engineer which require excellent technical skills, commitment to working , ability to deal well and confidently with customers a willingness to travel and very intelligent and well-balanced personality.
The ten questions you should ask Maria to determine if she is qualified for the job are :
Tell us about your self ( functions you have )Are your confident that you are the right candidate for this positionwhy should i hire youDo you like working under supervisionHow do you like to work ( in a group or individually )What is your ultimate workplace goalwhat are your future plansWhat do you expect from the Organization when given the jobDo you like taking on critical problemsHow long can you work in this positionA ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lane is provided on the local road to allow vehicles from the ramp to turn right onto the local road without stopping. The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle. Determine the width of the turning roadway if the design vehicle is a single-unit truck. Use 0.08 for superelevation.
Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
[tex]R = \dfrac{u^2}{15(e+f_s)}[/tex]
where;
R= radius
[tex]f_s[/tex] = coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
[tex]f_s[/tex] = 0.20
So;
[tex]R = \dfrac{30^2}{15(0.08+0.20)}[/tex]
[tex]R = \dfrac{900}{15(0.28)}[/tex]
[tex]R = \dfrac{900}{4.2}[/tex]
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
why is the peak value of the rectified output less than the peak value of the ac input and by how much g
Answer:
The Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Explanation:
This is what we called HALF WAVE RECTIFIER in which the Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Therefore this is the formula for Half wave rectifier
Vrms = Vm/2 and Vdc
= Vm/π:
Where,
Vrms = rms value of input
Vdc = Average value of input
Vm = peak value of output
Hence, half wave rectifier is a rectifier which allows one half-cycle of an AC voltage waveform to pass which inturn block the other half-cycle which is why this type of rectifiers are often been used to help convert AC voltage to a DC voltage, because they only require a single diode to inorder to construct.
A long corridor has a single light bulb and two doors with light switch at each door. design logic circuit for the light; assume that the light is off when both switches are in the same position.
Answer and Explanation:
Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:
A B C (output)
0 0 0
0 1 1
1 0 1
1 1 0
The logic circuit is shown below
C = A'B + AB'
If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.
Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answer:
The volume percentage of graphite is 10.197 per cent.
Explanation:
The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:
[tex]\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%[/tex]
Where:
[tex]V_{gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
[tex]V_{fe}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
The expression is expanded by using the definition of density and subsequently simplified:
[tex]\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%[/tex]
Where:
[tex]m_{fe}[/tex], [tex]m_{gr}[/tex] - Masses of the ferrite and graphite phases, measured in grams.
[tex]\rho_{fe}, \rho_{gr}[/tex] - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.
[tex]\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%[/tex]
[tex]\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%[/tex]
If [tex]\rho_{gr} = 2.3\,\frac{g}{cm^{3}}[/tex], [tex]\rho_{fe} = 7.9\,\frac{g}{cm^{3}}[/tex], [tex]m_{gr} = 3.2\,g[/tex] and [tex]m_{fe} = 96.8\,g[/tex], the volume percentage of graphite is:
[tex]\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%[/tex]
[tex]\%V_{gr} = 10.197\,\%V[/tex]
The volume percentage of graphite is 10.197 per cent.
Following are the solution to the given points:
[tex]\to C_{Gr} = 100\\\\ \to C_{\alpha}= 0[/tex]From [tex]Fe-F_{\frac{e}{3}} c[/tex] diagram.
[tex]\to W_{\alpha} =\frac{C_{Gr}-C_{o}}{C_{Gr}-C_{\alpha}}[/tex]
[tex]= \frac{100-3.6}{100-0} \\\\= \frac{100-3.6}{100} \\\\= \frac{96.4}{100} \\\\=0.964[/tex]
Calculating the weight fraction of graphite:
[tex]\to W_{Gr}=\frac{C_0 - c_d}{C_{Gr} -c_d}[/tex]
[tex]= \frac{3.6-0}{100-0} \\\\ = \frac{3.6}{100} \\\\= 0.036[/tex]
Calculating the volume percent of graphite:
[tex]\to V_{Gr}=\frac{\frac{W_{Gr}}{P_{Gr}}}{\frac{w_{\alpha}}{P_{\alpha}}+ \frac{W_{Gr}}{P_{Gr}}}[/tex]
[tex]=\frac{\frac{0.036}{2.3}}{\frac{0.964}{7.9}+\frac{0.036}{2.3}}\\\\=0.11368 \times 100\%\\\\=11.368\%[/tex]
Therefore, the final answer is "0.964, 0.036, and 11.368%"
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The value of an SMT capacitor is signified by a
Answer:
Working volttage
Explanation:
SMT electrolytic capacitors are marked with working voltage. The value of these capacitors is measured in micro farads. It is a surface mount capacitor which is used for high volume manufacturers. They are small lead less and are widely used. They are placed on modern circuit boards.
Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 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 each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.
Answer:
a. [tex]\eta _{th}[/tex] = 77.65%
b. bwr = 6.5%
c. 3538.986 kW
d. -163.169 kJ
Explanation:
a. The given property are;
P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa
p₄/p₁ = 10
P₂/P₁ = p₄/p₃ = √10
p₂ = 100·√10
[tex]T_{2s}[/tex] = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K
T₂ = T₁ + ([tex]T_{2s}[/tex] - T₁)/[tex]\eta _c[/tex] = 300 + (416.85 - 300)/0.8 = 446.0625 K
p₄ = 10×p₁ = 10×100 = 1000 kPa
p₄/p₃ = √10 =
p₃ = 100·√10
T₃ = 300 K
T₃/[tex]T_{4s}[/tex] = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)
[tex]T_{4s}[/tex] = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K
T₄ = T₃ + ([tex]T_{4s}[/tex] - T₃)/[tex]\eta _c[/tex] = 300 + (215.905- 300)/0.8 = 194.881 K
The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28
T₄ = 446.0625 K
T₆ = 1400 K
[tex]T_{7s}[/tex]/T₆ = (1/√10)^(0.4/1.4)
[tex]T_{7s}[/tex] = 1400×(1/√10)^(0.4/1.4) = 1007.6 K
T₇ = T₆ - [tex]\eta _t[/tex](T₆ - [tex]T_{7s}[/tex]) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K
T₈ = 1400 K
T₉ = 1086.08 K
T₅ = T₄ + [tex]\epsilon _{regen}[/tex](T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K
[tex]\eta _{th}[/tex] =(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))
(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765
[tex]\eta _{th}[/tex] = 77.65%
b. Back work ratio, bwr = [tex]bwr = \dfrac{w_{c,in}}{w_{t,out}}[/tex]
((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))
40.9435/627.84 = 6.5%
c. [tex]w_{net, out} = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)][/tex]
Power developed is given by the relation;
[tex]\dot m \cdot w_{net, out}[/tex]
[tex]\dot m \cdot w_{net, out}[/tex]= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW
d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)
-163.169 kJ
Identify the advantages of using 6 tube passes instead of just 2 of the same diameter in shell-and-tube heat exchanger.What are the advantages and disadvantages of using 6 tube passes instead of just 2 of the same diameter?
Answer:
Please check explanation for answer
Explanation:
Here, we are concerned with stating the advantages and disadvantages of using a 6 tube passes instead of a 2 tube passes of the same diameter:
Advantages
* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface
* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too
Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.
Disadvantages
* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.
* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of manufacturing costs
is used to determine the shear stress at point P over the section supporting a downward shear force in the -y direction. What is Q
Answer:
Transverse shear stress formula
Explanation:
Transverse shear stress also known as the beam shear, is the shear stress due to bending of a beam.
Generally, when a beam is made to undergo a non-uniform bending, both bending moment (I) and a shear force (V) acts on its cross section or width (t).
Transverse shear stress formula is used to determine the shear stress at point P over the section supporting a downward shear force in the -y direction.
Mathematically, the transverse shear stress is given by the formula below;
[tex]T' = \frac{VQ}{It}[/tex]
Also note, T' is pronounced as tau.
Where;
V is the total shear force with the unit, Newton (N).
I is the Moment of Inertia of the entire cross sectional area with the unit, meters square (m²).
t is the thickness or width of cross sectional area of the material perpendicular to the shear with the unit centimeters (cm).
Q is the statical moment of area.
Mathematically, Q is given by the formula;
[tex]Q = y'P^{*} = ∑y'P^{*}[/tex]
Where [tex]P^{*}[/tex] is the section supporting a downward shear force in the y' direction.
For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in. An external load of 80 kips is applied to the entire joint. Assume the load is equally distributed to all the bolts. It has been determined to use 1/2 in- 13 UNC grade 8 bolts with rolled threads. Assume the bolts are preloaded to 75% of the proof load. Clearly state any assumptions.
(a) Determine the yielding factor of safety,
(b) Determine the overload factor of safety,
(c) Determine the factor of safety baserd on joint seperation.
Answer:
nP ≈ 4.9 nL = 1.50Explanation:
GIVEN DATA
external load applied (p) = 85 kips
bolt stiffness ( Kb ) = 3(10^6) Ibf / in
Member stiffness (Km) = 12(10^6) Ibf / in
Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8
Number of bolts = 6
assumptions
for unified screw threads UNC and UNF
tensile stress area ( A ) = 0.1419 in^2
SAE specifications for steel bolts for grade 8
we have
Minimum proff strength ( Sp) = 120 kpsi
Minimum tensile strength (St) = 150 Kpsi
Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips
Given the following values
Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip
Preload stress
αi = 0.75Sp = 0.75 * 120 = 90 kpsi
stiffness constant
C = [tex]\frac{Kb}{Kb + Km}[/tex] = [tex]\frac{3}{3+2}[/tex] = 0.2
A) yielding factor of safety
nP = [tex]\frac{sPAt}{Cp + Fi}[/tex] = [tex]\frac{120* 0.1419}{0.2*14.17 + 12.771}[/tex]
nP = 77.028 / 15.605 = 4.94 ≈ 4.9
B) Determine the overload factor safety
[tex]nL = \frac{SpAt - Fi}{CP}[/tex] = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17
= 17.028 - 12.771 / 2.834
= 1.50
Calculate the camacitance-to-neutral in F/m and the admittance-to-neutral in S/km for the three-phase line in problem Neglect the effect of the earth plane.
Answer:
The answer is given below
Explanation:
A 60 Hz three-phase, three-wire overhead line has solid cylindrical conductors arranged in the form of an equilateral triangle with 4 ft conductor spacing. Conductor diameter is 0.5 in.
Given that:
The spacing between the conductors (D) = 4 ft
1 ft = 0.3048 m
D = 4 ft = 4 × 0.3048 m = 1.2192 m
The conductor diameter = 0.5 in
Radius of conductor (r) = 0.5/2 = 0.25 in = 0.00635 m
Frequency (f) = 60 Hz
The capacitance-to-neutral is given by:
[tex]C_n=\frac{2\pi \epsilon_0}{ln(\frac{D}{r} )} =\frac{2\pi *8.854*10^{-12}}{ln(1.2192/0.00635)}=1.058*10^{-11}\ F/m[/tex]
The admittance-to-neutral is given by:
[tex]Y_n=j2\pi fC_n=j*2\pi *60*1.058*10^{-11}*\frac{1000\ m}{1\ km}=j3.989*10^{-6}\ S/km[/tex]
The effectiveness of a heat exchanger is defined as the ratio of the maximum possible heat transfer rate to the actual heat transfer rate.
a. True
b. False
Answer:
False
Explanation:
Because
The effectiveness (ϵ) of a heat exchanger is defined as the ratio of the actual heat transfer to the maximum possible heat transfer.
For a fluid flowing through a pipe assuming that pressure drop per unit length of pipe (P/L) depends on the diameter of the pipe , the velocity of fluid, the density of fluid and the viscosity of the fluid. Show that = ∅ ൬ ൰
Answer:
Explanation:
La vaca
El pato
which solution causes cells to shrink
It is to be noted that a hypertonic solution have the capacity to make cells to shrink.
What happens in a hypertonic solution?In a hypertonic solution, the concentration of solutes (e.g., salts, sugars) outside the cell is higher than inside the cell.
As a result, water moves out of the cell through osmosis, trying to equalize the concentration, causing the cell to lose water and shrink.
This process is commonly observed in biology when examining the effect of different solutions on cells, such as in red blood cells or plant cells.
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For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _______ and using the Maximum Shear Stress Theory the material will __________
a. fail / not fail
b. fail /fail
c. not fail/fail
d. not fail/not fail
Answer:
Option A - fail/ not fail
Explanation:
For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______
Commutation is the process of converting the ac voltages and currents in the rotor of a dc machine to dc voltages and currents at its terminals. True False
Answer:
false
Explanation:
the changing of a prisoner sentence or another penalty to another less severe
If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially, the block is at rest
Answer:
115.2 W
Explanation:
The computation is shown below:
As we know that
Power = F . v
[tex]F_H = F cos \theta[/tex]
[tex]F_H = 30 \frac{4}{5}[/tex]
[tex]F_H = 24N[/tex]
Now we solve for V
[tex]V = V_0 + at[/tex] a = 24N ÷ 20Kg
But V_0 = 0 a = 1.2 m/s^2
F_H = ma V = 0 + (1.2) (4)
a = F_H ÷ m V = 4.8 m/s
Therefore
Power = F_Hv
= (24) (4.8)
= 115.2 W
By applying the above formuals we can get the power