Answer:
25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
Explanation:
Given data
steel tube : outer diameter = 50-mm
power transmitted = 100 KW
frequency(f) = 34 Hz
shearing stress ≤ 60 MPa
Determine tube thickness
firstly we calculate the ; power, angular velocity and torque of the tube
power = T(torque) * w (angular velocity)
angular velocity ( w ) = 2[tex]\pi[/tex]f = 2 * [tex]\pi[/tex] * 34 = 213.71
Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s
next we calculate the inner diameter using the relation
[tex]\frac{J}{c_{2} } = \frac{T}{t_{max} }[/tex] = 467.92 / (60 * 10^6) = 7.8 * 10^-6 m^3
also
c2 = (50/2) = 25 mm
[tex]\frac{J}{c_{2} }[/tex] = [tex]\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )[/tex] = [tex]\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ][/tex]
therefore; 0.025^4 - [tex]c^{4} _{1}[/tex] = 0.050 / [tex]\pi[/tex] (7.8 *10^-6)
[tex]c^{4} _{1}[/tex] = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)
39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8
[tex]c_{1} = \sqrt[4]{26.66 * 10^{-8} }[/tex] =
THE TUBE THICKNESS
[tex]c_{2} - c_{1}[/tex] = 25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
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.
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
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.
Select True/False for each of the following statements regarding aluminum / aluminum alloys: (a) Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor. (b) Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum. (c) Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic. (d) Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation. (e) The relatively low melting point of aluminum is often considered a significant limitation for high-temperature structural applications.
Explanation:
(a) Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor.
False, aluminium is not susceptible to any corrosion by the presence of water vapor.
(b) Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their micro structures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.
True.
(c) Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.
False, aluminium is stable at high temperatures and does not oxidizes.
(d) Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.
False,pure aluminium is not resistant to the creep deformation.
(e) The relatively low melting point of aluminum is often considered a significant limitation for high-temperature structural applications.
False.
In this exercise, we have to analyze the statements that deal with aluminum and its properties, thus classifying it as true or false:
A) False
B) True
C) False
D) False
E) True
Analyzing the statements we can classify them as:
(a) For this statement we can say that it is False, aluminium is not susceptible to any corrosion by the presence of water vapor.
(b) For this statement we can say that it is True.
(c) For this statement we can say that it is False, aluminium is stable at high temperatures and does not oxidizes.
(d) For this statement we can say that it is False, pure aluminium is not resistant to the creep deformation.
(e) For this statement we can say that it is True.
See more about aluminum properties at brainly.com/question/12867973
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
Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the inlet temperature for each turbine stage is 1400 K. The pressure ratios across each turbine stage are equal. The turbine stages and compressor 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.
Answer:
a. 47.48%
b. 35.58%
c. 2957.715 KW
Explanation:
[tex]T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}[/tex]
T₁ = 300 K
[tex]\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }[/tex]
[tex]T_{2s}[/tex] = 579.21 K
T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K
T₃ = T₂ + [tex]\epsilon _{regen}[/tex](T₅ - T₂)
T₄ = 1400 K
Given that the pressure ratios across each turbine stage are equal, we have;
[tex]\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{5s}[/tex] = 1400×[tex]\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }[/tex] = 1007.6 K
T₅ = T₄ + ([tex]T_{5s}[/tex] - T₄)/[tex]\eta _t[/tex] = 1400 + (1007.6- 1400)/0.8 = 909.5 K
T₃ = T₂ + [tex]\epsilon _{regen}[/tex](T₅ - T₂)
T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K
T₆ = 1400 K
[tex]\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{7s}[/tex] = 1400×[tex]\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }[/tex] = 1007.6 K
T₇ = T₆ + ([tex]T_{7s}[/tex] - T₆)/[tex]\eta _t[/tex] = 1400 + (1007.6 - 1400)/0.8 = 909.5 K
a. [tex]W_{net \ out}[/tex] = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg
Heat supplied is given by the relation
cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg
Thermal efficiency of the cycle = (Net work output)/(Heat supplied)
Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%
b. [tex]bwr = \dfrac{W_{c,in}}{W_{t,out}}[/tex]
bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%
c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW
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
A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-vapor mixture with a quality of 0.94 at a pressure of 325 kPa and a velocity of 15 m/s. In addition, there is heat transfer from the turbine to the surroundings for 560 kW. Find the power produced by the turbine and express it in kW?
Answer:
The power produced by the turbine is 23309.1856 kW
Explanation:
h₁ = 3755.39
s₁ = 7.0955
s₂ = sf + x₂sfg =
Interpolating fot the pressure at 3.25 bar gives;
570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175
2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345
h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg
Power output of the turbine formula =
[tex]Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ][/tex]
Which gives;
[tex]560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ][/tex]
= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856
[tex]- \dot{W }[/tex] = -22749.1856 - 560 = -23309.1856 kJ
[tex]\dot{W }[/tex] = 23309.1856 kJ
Power produced by the turbine = Work done per second = 23309.1856 kW.
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
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
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacuum. The two sides are separated by a piston that is initially held in place by the pins. The pins are removed and the gas suddenly expands until it hits the stops. What happens to the internal energy of the gas?
a. internal energy goes up
b. internal energy goes down
c. internal energy stays the same
d. we need to know the volumes to make the calculation
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.