Answer:
a) [tex]h_c = 0.1599 W/m^2-K[/tex]
b) [tex]H_{loss} = 5.02 W[/tex]
c) [tex]T_s = 302 K[/tex]
d) [tex]\dot{Q} = 25.125 W[/tex]
Explanation:
Non horizontal pipe diameter, d = 25 cm = 0.25 m
Radius, r = 0.25/2 = 0.125 m
Entry temperature, T₁ = 304 + 273 = 577 K
Exit temperature, T₂ = 284 + 273 = 557 K
Ambient temperature, [tex]T_a = 25^0 C = 298 K[/tex]
Pipe length, L = 10 m
Area, A = 2πrL
A = 2π * 0.125 * 10
A = 7.855 m²
Mass flow rate,
[tex]\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600} = 1.25 kg/sec[/tex]
Rate of heat transfer,
[tex]\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W[/tex]
a) To calculate the convection coefficient relationship for heat transfer by convection:
[tex]\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K[/tex]
Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.
c) The surface temperature of the pipe:
Smear coefficient of the pipe, [tex]k_c = 0.8[/tex]
[tex]\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K[/tex]
b) Heat loss from the pipe to the environment:
[tex]H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W[/tex]
d) The required fan control power is 25.125 W as calculated earlier above
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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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
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16% of the cylinder volume at bottom dead center and the crankshaft rotates at 2400 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.5 lbf/in. 2 and a temperature of 60 8 F at the beginning of compression. The maximum temperature in the cycle is 5200 8 R.
Based on this model,
1- Write possible Assumptions no less than three assumptions
2- Draw clear schematic for this problem
3- Determine possible Assumptions no less than three assumptions
4- Draw clear schematic for this problem.
5- calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.
Answer:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The net work per cycle is 845.88 kJ/kg
The power developed in horsepower ≈ 45374 hP
Explanation:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m
Stroke length = 3.4 in. = 0.08636 m.
The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³
The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³
The clearance volume, v₂ = 9.59 × 10⁻⁵ m³
p₁ = 14.5 lbf/in.² = 99973.981 Pa
T₁ = 60 F = 288.706 K
[tex]\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}[/tex]
Otto cycle T-S diagram
T₂ = 288.706*[tex]6.25^{0.393}[/tex] = 592.984 K
The maximum temperature = T₃ = 5200 R = 2888.89 K
[tex]\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}[/tex]
T₄ = 2888.89 / [tex]6.25^{0.393}[/tex] = 1406.5 K
Work done, W = [tex]c_v[/tex]×(T₃ - T₂) - [tex]c_v[/tex]×(T₄ - T₁)
0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg
The power developed in an Otto cycle = W×Cycle per second
= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.
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
Which of the following are the main psychological domains?
Answer:
Domain 1: Biological (includes neuroscience, consciousness, and sensation) Domain 2: Cognitive (includes the study of perception, cognition, memory, and intelligence) Domain 3: Development (includes learning and conditioning, lifespan development, and language) i hope this helps you.
An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectively, and the respective atomic weights are 22.7 and 91.4 g/mol, what is the density (in g/cm3) of this material?
A. 0.438g/cm3
B. 0. 571g/cm3
C. 1.75g/cm3
D. 3.50g/cm3
Answer:
c) 1.75 g/cm³
Explanation:
Given that
Radii of the A ion, r(c) = 0.137 nm
Radii of the X ion, r(a) = 0.241 nm
Atomic weight of the A ion, A(c) = 22.7 g/mol
Atomic weight of the X ion, A(a) = 91.4 g/mol
Avogadro's number, N = 6.02*10^23 per mol
Solution is attached below
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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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
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
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.
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.
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
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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When the value of the output cannot be determined even if the value of the controllable input is known, the model is:_________
a. analog.
b. digital.
c. stochastic.
d. deterministic.
Answer:
c. stochastic.
Explanation:
A stochastic model is a tool in statistics, used to estimate the probability distributions of intended outcomes by the allowance of random variation in any number of the inputs over time. For a stochastic model, Inputs to a quantitative model are uncertain, and the value of the output from a stochastic model cannot be easily determined, even if the value of the input that can be determined is known. The distributions of the resulting outcomes of a stochastic model is usually due to the large number of simulations involved, and it is widely used as a statistical tool in the life sciences.
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
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:
The power of force F is 115.2 W
Explanation:
Use following formula
Power = F x V
[tex]F_{H}[/tex] = F cos0
[tex]F_{H}[/tex] = (30) x 4/5
[tex]F_{H}[/tex] = 24N
Now Calculate V using following formula
V = [tex]V_{0}[/tex] + at
[tex]V_{0}[/tex] = 0
a = [tex]F_{H}[/tex] / m
a = 24N / 20 kg
a = 1.2m / [tex]S^{2}[/tex]
no place value in the formula of V
V = 0 + (1.2)(4)
V = 4.8 m/s
So,
Power = [tex]F_{H}[/tex] x V
Power = 24 x 4.8
Power = 115.2 W
A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 in. wide by 12 in. deep. The beam carries a 9-kip concentrated load at midspan. Which point has the largest Q value at section a–a?
Answer:
point B where [tex]Q_B = 101.25 \ in^3[/tex] has the largest Q value at section a–a
Explanation:
The missing diagram that is suppose to be attached to this question can be found in the attached file below.
So from the given information ;we are to determine the point that has the largest Q value at section a–a
In order to do that; we will work hand in hand with the image attached below.
From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.
We also have block partitioned into different point segments . i,e A,B,C, D
For point A ;
Let Q be the moment of the Area A;
SO ; [tex]Q_A = Area \times y_1[/tex]
where ;
[tex]y_1 = (6 - \dfrac{1.5}{2})[/tex]
[tex]y_1 = (6- 0.75)[/tex]
[tex]y_1 = 5.25 \ in[/tex]
[tex]Q_A =(L \times B) \times y_1[/tex]
[tex]Q_A =(6 \times 1.5) \times 5.25[/tex]
[tex]Q_A =47.25 \ in^3[/tex]
For point B ;
Let Q be the moment of the Area B;
SO ; [tex]Q_B = Area \times y_2[/tex]
where ;
[tex]y_2 = (6 - \dfrac{1.5 \times 3}{2})[/tex]
[tex]y_2= (6 - \dfrac{4.5}{2}})[/tex]
[tex]y_2 = (6 -2.25})[/tex]
[tex]y_2 = 3.75 \ in[/tex]
[tex]Q_B =(L \times B) \times y_1[/tex]
[tex]Q_B=(6 \times 4.5) \times 3.75[/tex]
[tex]Q_B = 101.25 \ in^3[/tex]
For point C ;
Let Q be the moment of the Area C;
SO ; [tex]Q_C = Area \times y_3[/tex]
where ;
[tex]y_3 = (6 - \dfrac{1.5 \times 2}{2})[/tex]
[tex]y_3 = (6 - 1.5})[/tex]
[tex]y_3= 4.5 \ in[/tex]
[tex]Q_C =(L \times B) \times y_1[/tex]
[tex]Q_C =(6 \times 3) \times 4.5[/tex]
[tex]Q_C=81 \ in^3[/tex]
For point D ;
Let Q be the moment of the Area D;
SO ; [tex]Q_D = Area \times y_4[/tex]
since there is no area about point D
Area = 0
[tex]Q_D =0 \times y_4[/tex]
[tex]Q_D = 0[/tex]
Thus; from the foregoing ; point B where [tex]Q_B = 101.25 \ in^3[/tex] has the largest Q value at section a–a
. 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 positionIdentify 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
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]
help mhee why are you u an enigner
Answer:
help me why are you an enginer
Explanation:
because lives
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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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.
The temperature of water is 45 what does the measurement represent
Answer:
degree of hotness of coldness of a substance
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_______
Find the heat flow from the composite wall as shown in figure. Assume one dimensional flow KA=150 W/m°C , KB=25 W/m°C, KC=60 W/m°C , KD=60 W/m°C
Answer:
The heat flow from the composite wall is 1283.263 watts.
Explanation:
The conductive heat flow through a material, measured in watts, is represented by the following expression:
[tex]\dot Q = \frac{\Delta T}{R_{T}}[/tex]
Where:
[tex]R_{T}[/tex] - Equivalent thermal resistance, measured in Celsius degrees per watt.
[tex]\Delta T[/tex] - Temperature gradient, measured in Celsius degress.
First, the equivalent thermal resistance needs to be determined after considering the characteristics described below:
1) B and C are configurated in parallel and in series with A and D. (Section II)
2) A and D are configurated in series. (Sections I and III)
Section II
[tex]\frac{1}{R_{II}} = \frac{1}{R_{B}} + \frac{1}{R_{C}}[/tex]
[tex]\frac{1}{R_{II}} = \frac{R_{B}+R_{C}}{R_{B}\cdot R_{C}}[/tex]
[tex]R_{II} = \frac{R_{B}\cdot R_{C}}{R_{B}+R_{C}}[/tex]
Section I
[tex]R_{I} = R_{A}[/tex]
Section III
[tex]R_{III} = R_{D}[/tex]
The equivalent thermal resistance is:
[tex]R_{T} = R_{I} + R_{II}+R_{III}[/tex]
The thermal of each component is modelled by this:
[tex]R = \frac{L}{k\cdot A}[/tex]
Where:
[tex]L[/tex] - Thickness of the brick, measured in meters.
[tex]A[/tex] - Cross-section area, measured in square meters.
[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Celsius degree.
If [tex]L_{A} = 0.03\,m[/tex], [tex]L_{B} = 0.08\,m[/tex], [tex]L_{C} = 0.08\,m[/tex], [tex]L_{D} = 0.05\,m[/tex], [tex]A_{A} = 0.01\,m^{2}[/tex], [tex]A_{B} = 3\times 10^{-3}\,m^{2}[/tex], [tex]A_{C} = 7\times 10^{-3}\,m^{2}[/tex], [tex]A_{D} = 0.01\,m^{2}[/tex], [tex]k_{A} = 150\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]k_{B} = 25\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]k_{C} = 60\,\frac{W}{m\cdot ^{\circ}C}[/tex] and [tex]k_{D} = 60\,\frac{W}{m\cdot ^{\circ}C}[/tex], then:
[tex]R_{A} = \frac{0.03\,m}{\left(150\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}[/tex]
[tex]R_{A} = \frac{1}{50}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{B} = \frac{0.08\,m}{\left(25\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (3\times 10^{-3}\,m^{2})}[/tex]
[tex]R_{B} = \frac{16}{15}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{C} = \frac{0.08\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (7\times 10^{-3}\,m^{2})}[/tex]
[tex]R_{C} = \frac{4}{21}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{D} = \frac{0.05\,m}{\left(60\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (0.01\,m^{2})}[/tex]
[tex]R_{D} = \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{I} = \frac{1}{50} \,\frac{^{\circ}C}{W}[/tex]
[tex]R_{III} = \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{II} = \frac{\left(\frac{16}{15}\,\frac{^{\circ}C}{W} \right)\cdot \left(\frac{4}{21}\,\frac{^{\circ}C}{W}\right)}{\frac{16}{15}\,\frac{^{\circ}C}{W} + \frac{4}{21}\,\frac{^{\circ}C}{W}}[/tex]
[tex]R_{II} = \frac{16}{99}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{T} = \frac{1}{50}\,\frac{^{\circ}C}{W} + \frac{16}{99}\,\frac{^{\circ}C}{W} + \frac{1}{12}\,\frac{^{\circ}C}{W}[/tex]
[tex]R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}[/tex]
Now, if [tex]\Delta T = 400\,^{\circ}C - 60\,^{\circ}C = 340\,^{\circ}C[/tex] and [tex]R_{T} = \frac{2623}{9900}\,\frac{^{\circ}C}{W}[/tex], the heat flow is:
[tex]\dot Q = \frac{340\,^{\circ}C}{\frac{2623}{9900}\,\frac{^{\circ}C}{W} }[/tex]
[tex]\dot Q = 1283.263\,W[/tex]
The heat flow from the composite wall is 1283.263 watts.
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%
If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I send
Answer:
32000 bits/seconds
Explanation:
Given that :
there are 16 signal combinations (states) = 2⁴
bits n = 4
and a baud rate (number of signals/second) = 8000/second
Therefore; the number of bits per seconds can be calculated as follows:
Number of bits per seconds = bits n × number of signal per seconds
Number of bits per seconds = 4 × 8000/second
Number of bits per seconds = 32000 bits/seconds
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
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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