The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation. For A992 steel, G = 11 × 103 ksi. (1) Determine the angle of twist of B with respect to D.(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places
Answers
Answer:
the angle of twist of B with respect to D is -1.15°
the angle of twist of C with respect to D is 1.15°
Explanation:
The missing diagram that is supposed to be added to this image is attached in the file below.
From the given information:
The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.
For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²
The objective are :
1) To determine the angle of twist of B with respect to D
Considering the Polar moment of Inertia at the shaft [tex]J\tau[/tex]
shaft [tex]J\tau[/tex] = [tex]\dfrac{\pi}{2}r^4[/tex]
where ;
r = 1 in /2
r = 0.5 in
shaft [tex]J \tau[/tex] = [tex]\dfrac{\pi}{2} \times 0.5^4[/tex]
shaft [tex]J\tau[/tex] = 0.098218
Now; the angle of twist at B with respect to D is calculated by using the expression
[tex]\phi_{B/D} = \sum \dfrac{TL}{JG}[/tex]
[tex]\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}[/tex]
where;
[tex]T_{CD} \ \ and \ \ L_{CD}[/tex] are the torques at segments CD and length at segments CD
[tex]{T_{BC} \ \ and \ \ L_{BC}}[/tex] are the torques at segments BC and length at segments BC
Also ; from the diagram; the following values where obtained:
[tex]L_{BC}}[/tex] = 2.5 in
[tex]J\tau[/tex] = 0.098218
G = 11 × 10⁶ lb/in²
[tex]T_{BC[/tex] = -60 lb.ft
[tex]T_{CD[/tex] = 0 lb.ft
[tex]L_{CD[/tex] = 5.5 in
[tex]\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}[/tex]
[tex]\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}[/tex]
[tex]\phi_{B/D} = \dfrac{-21600}{1079980}[/tex]
[tex]\phi_{B/D} =[/tex] − 0.02 rad
To degree; we have
[tex]\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}[/tex]
[tex]\mathbf{\phi_{B/D} = -1.15^0}[/tex]
Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction
Thus; the angle of twist of B with respect to D is 1.15°
(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places
For the angle of twist of C with respect to D; we have:
[tex]\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}[/tex]
[tex]\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}[/tex]
[tex]\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}[/tex]
[tex]\phi_{C/D} = \dfrac{21600}{1079980}[/tex]
[tex]\phi_{C/D} =[/tex] 0.02 rad
To degree; we have
[tex]\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}[/tex]
[tex]\mathbf{\phi_{C/D} = 1.15^0}[/tex]
Related Questions
Many HVACR industry publications are published by
Answers
Answer:
HVACR Industry Trade Groups
Explanation:
In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?
Answers
Answer:
No, the velocity profile does not change in the flow direction.
Explanation:
In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.
how does a TV'S screen work
Answers
Answer:
A TVS screen works when the pixels are switched on electronically using liquid crystals to rotate polarized light.
Explanation:
Air flows along a horizontal, curved streamline with a 20 foot radius with a speed of 100 ft/s. Determine the pressure gradient normal to the streamline.
Answers
Answer:
- 1.19 lb/ft^3
Explanation:
You are given the following information;
Radius r = 20 ft
Speed V = 100 ft/s
You should use Bernoulli equation pertaining to streamline. That is, normal to streamline.
The pressure gradient = dp/dn
Where air density rho = 0.00238 slugs per cubic foot.
Please find the attached files for the solution and diagram.
How old are you? answer this question plz lol I will mark someone as brainliest
Answers
Answer:
100000000000000000000000
A gold vault has 3 locks with a key for each lock. Key A is owned by the
manager whilst Key B and C are in the custody of the senior bank teller
and the trainee bank teller respectively. In order to open the vault door at
least two people must insert their keys into the assigned locks at the same
time. The trainee bank teller can only open the vault when the bank
manager is present in the opening.
i) Determine the truth table for such a digital locking system (4 marks)
ii) Derive and minimize the SOP expression for the digital locking system
Answers
Answer:
i) Truth Table:
A | B | C | O
0 | 0 | 0 | 0
0 | 0 | 1 | 0
0 | 1 | 0 | 0
0 | 1 | 1 | 0 (condition 2 not satisfied)
1 | 0 | 0 | 0
1 | 0 | 1 | 1 (both conditions satisfied)
1 | 1 | 0 | 1 (both conditions satisfied)
1 | 1 | 1 | 1 (both conditions satisfied)
ii) The minimized sum of products (SOP) expression is
O = AC + AB
Explanation:
We have three inputs A, B and C
Let O is the output.
We are given two conditions to open the vault door:
1. At least two people must insert their keys into the assigned locks at the same time.
2. The trainee bank teller (C) can only open the vault when the bank manager (A) is present in the opening.
i) Construct the Truth Table
A | B | C | O
0 | 0 | 0 | 0
0 | 0 | 1 | 0
0 | 1 | 0 | 0
0 | 1 | 1 | 0 (condition 2 not satisfied)
1 | 0 | 0 | 0
1 | 0 | 1 | 1 (both conditions satisfied)
1 | 1 | 0 | 1 (both conditions satisfied)
1 | 1 | 1 | 1 (both conditions satisfied)
ii) SOP Expression using Karnaugh-Map:
A 3 variable Karnaugh-map is attached.
The minimized sum of products (SOP) expression is
O = AC + AB
The orange pair corresponds to "AC" and the purple pair corresponds to "AB"
Bonus:
The above expression may be realized by using two AND gates and one OR gate.
Please refer to the attached logic circuit diagram.
A complex Brayton-cycle power plant using intercooling, reheat, and regeneration is analyzed using the cold air standard method. Air is compressed from State 1 to State 2 using a compressor with a pressure ratio of RP1. An intercooler is used to cool the air to State 3 before entering a second compressor with a pressure ratio of RP2. The compressed air exits at State 4 and is preheated in a regenerator that uses the exhaust air from the low pressure turbine. The preheated air enters the combustor at State 5 and is heated to State 6 where it enters the high pressure turbine. The air exits the turbine at State 7 and is heated in a reheat combustor to State 8. The air expands in a low pressure turbine to State 9 where it enters the counterflow regenerator with an effectiveness of RE. Given the specified operating conditions determine the efficiency and other values listed below. The specific heat ratio and gas constant for air are given as k
Answers
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Example 1: the two dimensional points P1(0,0) and P2(1,0) and the two tangents P', (1,1) and P2 (0,-1).find the equation of the curve P(u).
Answers
Answer: (0,0)+ (1,0)= 1 lines upwards( suggesting that this is a line graph not saying it is but as an example) an (1,1) and (0,-1) all make a small square ( as this is a 2 dimensional graph that it has a negative side too,(below the positive side)) i hope this helps and is what you are looking for
Explanation:
For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because (select all that are correct
Answers
Answer:
hello the answer options are missing here are the options
A)The thickness of the heated region near the plate is increasing
B)The velocities near the plates are increasing
C)The fluid temperature near the plate are increasing
ANSWER : all of the above
Explanation:
Laminar flow is the flow of a type of fluid across the surface of an object following regular paths and it is unlike a turbulent flow which flows in irregular paths (encountering fluctuations)
For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because :
The thickness of the heated region near the plate is increasingThe velocities near the plates are increasingThe fluid temperature near the plate are increasingwhat is the difference between erratic error and zero error
Answers
The negative mark is balanced by a positive mark on the set key scale while the jaws are closed.
It is common practice to shut the jaws or faces of the system before taking some reading to guarantee a zero reading. If not, please take care of the read. This read is referred to as "zero defect."
There are two forms of zero error:
zero-mistake positive; and
Non-null mistake.
----------------------------
Hope this helps!
Brainliest would be great!
----------------------------
With all care,
07x12!
Air enters a compressor operating at steady state at 176.4 lbf/in.^2, 260°F with a volumetric flow rate of 424 ft^3/min and exits at 15.4 lbf/in.^2, 80°F. Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in hp
Answers
Answer:
[tex]W_s =[/tex] 283.181 hp
Explanation:
Given that:
Air enters a compressor operating at steady state at a pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 and Temperature [tex]T_1[/tex] at 260°F
Volumetric flow rate V = 424 ft^3/min
Air exits at a pressure [tex]P_2[/tex] = 15.4 lbf/in.^2 and Temperature [tex]T_2[/tex] at 80°F.
Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:
[tex]Q_{cv}[/tex] = -6800 Btu/h = - 1.9924 kW
Using the steady state energy in the process;
[tex]h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
where;
[tex]g(z_2-z_1) =0[/tex] and [tex]\dfrac{1}{2}(v^2_2-v_1^2) = 0[/tex]
Then; we have :
[tex]h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
[tex]h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}[/tex]
[tex]{m}(h_2 - h_1) ={Q_{cv} - W_s}[/tex]
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex] ----- (1)
Using the relation of Ideal gas equation;
P₁V₁ = mRT₁
Pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 = ( 176.4 × 6894.76 ) N/m² = 1216235.664 N/m²
Volumetric flow rate V = 424 ft^3/min = (424 × 0.0004719) m³ /sec
= 0.2000856 m³ /sec
Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K
Gas constant R=287 J/kg K
Then;
1216235.664 N/m² × 0.2000856 m³ /sec = m × 287 J/kg K × 399.817 K
[tex]m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }[/tex]
m = 2.121 kg/sec
The change in enthalpy:
[tex]m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)[/tex]
[tex]= 2.121* 1.005* ( 399.817 -299.817)[/tex]
= 213.1605 kW
From (1)
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex]
[tex]W_s =[/tex] - 1.9924 kW + 213.1605 kW
[tex]W_s =[/tex] 211.1681 kW
[tex]W_s =[/tex] 283.181 hp
The power input is [tex]W_s =[/tex] 283.181 hp
The liquid-phase reaction A + B → C follows an elementary rate law and is carried out isothermally in a flow system. The concentrations of A and B feed streams are 2 M before mixing. The volumetric flow rate of each stream is 5 dm3 /min and the entering temperature is 300 K. The streams are mixed immediately before entering. Two reactors are available: One is a gray 200.0 dm3 CSTR that can be heated to 77°C or cooled to 0°C, and the other is a white 800.0 dm3 PFR operated at 300 K that cannot be heated or cooled but can be painted red or black. (Note: k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol.) How long would it take to achieve 90% conversion in a 200 dm3 batch reactor with CA ° = CB ° = 1 ???? after mixing at a temperature of 70°C?
Answers
Answer:
1.887 minutes
Explanation:
We are given k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol = 20000 cal/mol
To solve this, first of all let's calculate the rate constant(k);
For this question, The formula is;
K(t) = k(300K) × exp[(E/R)((1/300) - (1/T2))]
R is gas constant = 1.987 cal/mol.K
For temperature of 70°C which is = 70 + 273K = 343K, we have;
K(343) = 0.07 × exp[(20000/1.987)((1/300) - (1/343))]
K(343) = 4.7 dm³/mol.min
The design equation is;
dX/dt = -(rA/C_Ao) = K•(C_Ao)²•(1 - X)²/(C_Ao) = (KC_Ao)(1 - X)²
Since there is no change in volume by cause of the state at which the reaction is carried out, that is liquid. Thus, integrating and solving for time for a 90% conversion we obtain;
(0.9,0)∫dX/(1 - X)².dX = (KC_Ao)((t, 0)∫dt
So, we'll get;
0.9/(1 - 0.9) = 4.77 × 1 × t
t = 9/4.77
t = 1.887 minutes
An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process, air is at 95 kPa and 27 degree Celsius.
(a) Determine the temperature after the heat-addition process.
(b) Determine the thermal efficiency.
(c) Determine the mean effective pressure. Solve the problem in the constant heat supposition.
Answers
Answer:
a) T₃ = 1818.8 K
b) η = 0.614 = 61.4%
c) MEP = 660.4 kPa
Explanation:
a) According to Table A-2 of The ideal gas specific heat of gases, the properties of air are as following:
At 300K
The specific heat capacity at constant pressure = [tex]c_{p}[/tex] = 1.005 kJ/kg.K,
The specific heat capacity at constant volume = [tex]c_{v}[/tex] = 0.718 kJ/kg.K
Gas constant R for air = 0.2870 kJ/kg·K
Ratio of specific heat k = 1.4
Isentropic Compression :
[tex]T_{2}[/tex] = [tex]T_{1}[/tex] [tex](v1/v2)^{k-1}[/tex]
= 300K ([tex]16^{0.4}[/tex])
[tex]T_{2}[/tex] = 909.4K
P = Constant heat Addition:
[tex]P_{3}v_{3} / T_{3} = P_{2} v_{2} /T_{2}[/tex]
[tex]T_{3}=v_{3}/v_{2}T_{2}[/tex]
2[tex]T_{2}[/tex] = 2(909.4K)
= 1818.8 K
b) [tex]q_{in}[/tex] = [tex]h_{3}-h_{2}[/tex]
= [tex]c_{p}[/tex] ([tex]T_{3}[/tex] - [tex]T_{2}[/tex])
= (1.005 kJ/kg.K)(1818.8 - 909.4)K
= 913.9 kJ/kg
Isentropic Expansion:
[tex]T_{4}[/tex] = [tex]T_{3}[/tex] [tex](v3/v4)^{k-1}[/tex]
= [tex]T_{3}[/tex] [tex](2v_{2} /v_{4} )^{k-1}[/tex]
= 1818.8 K (2 / 16[tex])^{0.4}[/tex]
= 791.7K
v = Constant heat rejection
[tex]q_{out}[/tex] = μ₄ - μ₁
= [tex]c_{v} ( T_{4} - T_{1} )[/tex]
= 0.718 kJ/kg.K (791.7 - 300)K
= 353 kJ/kg
η[tex]_{th}[/tex] = 1 - [tex]q_{out}[/tex] / [tex]q_{in}[/tex]
= 1 - 353 kJ/kg / 913.9 kJ/kg
= 1 - 0.38625670
= 0.6137
= 0.614
= 61.4%
c) [tex]w_{net}._{out}[/tex] = [tex]q_{in}[/tex] - [tex]q_{out}[/tex]
= 913.9 kJ/kg - 353 kJ/kg
= 560.9 kJ/kg
[tex]v_{1} = RT_{1} /P_{1}[/tex]
= (0.287 kPa.m³/kg/K)*(300 K) / 95 kPa
= 86.1 / 95
= 0.9063 m³/kg = v[tex]_{max}[/tex]
[tex]v_{min} =v_{2} = v_{max} /r[/tex]
Mean Effective Pressure = MEP = [tex]w_{net,out}/v_{1} -v_{2}[/tex]
= [tex]w_{net,out}/v_{1}(1-1)/r[/tex]
= 560.9 kJ/kg / (0.9063 m³/kg)*(1-1)/16
= (560.9 kJ / 0.8493m³) (kPa.m³/kJ)
= 660.426 kPa
Mean Effective Pressure = MEP = 660.4 kPa
The temperature after the addition process is 1724.8k, the thermal efficiency of the engine is 56.3% and the mean effective pressure is 65.87kPa
Assumptions made:
The air standard assumptions are madeThe kinetic and potential energy changes are negligibleThe air in the system is an ideal gas with variable or different specific heat capacity.a) The temperature after the addition process:
Considering the process 1-2, Isentropic expansion
at
[tex]T_1=300k\\u_1=214.07kJ/kg\\v_o_1=621.3\\v_o_2=\frac{v_2}{v_1} *v_o_1[C.R=16]=v_2/v_1\\v_o_2=(v_2/v_1)v_o_1=1/16*621.2=38.825[/tex]
From using this value, v[tex]_o_2[/tex]=38.825, solve for state point 2;
[tex]T_2=862.4k\\h_2=890.9kJ/kg[/tex]
Considering the process 2-3 (state of constant heat addition)
[tex]\frac{p_3v_3}{t_3}=\frac{p_2v_2}{t_2} \\\\T_3=\frac{P_3V_3T_2}{V_2} \\T_3=(\frac{V_3}{V_2}) T_2\\\frac{v_3}{v_2}=2\\T_3=2(862.4)=1724.8k\\[/tex]
NB: p[tex]_3[/tex]≈p[tex]_2[/tex]
b) The thermal efficiency of the engine is
Q[tex]_i_n[/tex]=h[tex]_3-h_2[/tex] = 1910.6-890.9=1019.7kJ/kg
Considering process 3-4,
[tex]v_o_4=\frac{v_A}{v_2}\\ v_o_3 =\frac{V_a}{V_2}*\frac{v_2}{v_3}\\v_o_3=\frac{16}{2}*4.546\\v_o_3=36.37;v_4=659.7kJ/kg[/tex]
Q[tex]_o_u_t=v_4-u_1=659.7-214.07=445.3kJ/kg[/tex]
nth = [tex]1-\frac{Q_o_u_t}{Q_i_n}=1-\frac{445.63}{1019.7}=0.5629*100=56.3%[/tex]%
The thermal efficiency is 56.3%
W[tex]_n_e_t[/tex]=[tex]Q_i_n-Q_o_u_t=574.07kJ/kg[/tex]
[tex]v_1=\frac{RT_1}{p_1}=\frac{0.287*300}{95}=0.906m^3/kg\\v_2=v_1/16=0.05662m^3/kg\\[/tex]
Therefore, the mean effective pressure of the system engine is
[tex]\frac{W_n_e_t}{v_1-v_2}=675.87kPa[/tex]
The mean effective pressure is 65.87kPa as calculated above
Learn more about mean effective pressure
https://brainly.com/question/19309495
When checking the resistance of a dual voltage wye motor, there should be ____ resistance readings. 1) twelve 2) six 3) three
Answers
Answer:
1) twelve
Explanation:
The dual voltage motors are used in day to day operations. The wye is connected with 9 lead motors. Maximum resistance can be obtained if the resistance are connected in series. To check resistance of dual voltage wye motor there must be twelve resistance readings of 1 ohm each.
What is the final temperature after compression of a diesel cycle if the initial temperature is 32C and clearance is 8%
Answers
Answer:
863 K
Explanation:
See the attachment
Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabatically to an exit state of 1 bar, 160°C. Kinetic and potentialenergy effects are negligible. Determine for the turbine (a) the powerdeveloped, in kW, (b) the rate of entropy production, in kW/K, and (c)the isentropic turbine efficiency
Answers
Answer:
A) W' = 178.568 KW
B) ΔS = 2.6367 KW/k
C) η = 0.3
Explanation:
We are given;
Temperature at state 1;T1 = 360 °C
Temperature at state 2;T2 = 160 °C
Pressure at state 1;P1 = 10 bar
Pressure at State 2;P2 = 1 bar
Volumetric flow rate;V' = 0.8 m³/s
A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;
Specific volume;v1 = 0.287322 m³/kg
Mass flow rate of water vapour at turbine is defined by the formula;
m' = V'/v1
So; m' = 0.8/0.287322
m' = 2.784 kg/s
Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific enthalpy;h1 = 3179.46 KJ/kg
Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific enthalpy;h2 = 3115.32 KJ/kg
Now, since stray heat transfer is neglected at turbine, we have;
-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2 - h1)
Plugging in relevant values, the work of the turbine is;
W' = -2.784(3115.32 - 3179.46)
W' = 178.568 KW
B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific entropy: s1 = 7.3357 KJ/Kg.k
Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific entropy; s2 = 8.2828 KJ/kg.k
The amount of entropy produced is defined by;
ΔS = m'(s2 - s1)
ΔS = 2.784(8.2828 - 7.3357)
ΔS = 2.6367 KW/k
C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;
h2s = 2966.14 KJ/Kg
Energy equation for turbine at ideal process is defined as;
Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Again, Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2s - h1)
W' = -2.784(2966.14 - 3179.46)
W' = 593.88 KW
the isentropic turbine efficiency is defined as;
η = W_actual/W_ideal
η = 178.568/593.88 = 0.3
A two-dimensional flow field described by
V = (2x^2y + x)1 + (2xy^2 + y + 1 )j
where the velocity is in m/s when x and y are in meters. Determine the angular rotation of a fluid element located at x 0.5 m, y 1.0 m.
Answers
Answer:
the answer is
Explanation:
We now focus on purely two-dimensional flows, in which the velocity takes the form u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) With the velocity given by (2.1), the vorticity takes the form ω = ∇ × u = ∂v ∂x − ∂u ∂y k. (2.2) We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence ∂v ∂x − ∂u ∂y = 0. (2.3) We have already shown in Section 1 that this condition implies the existence of a velocity potential φ such that u ≡ ∇φ, that is u = ∂φ ∂x, v = ∂φ ∂y . (2.4) We also recall the definition of φ as φ(x, y, t) = φ0(t) + Z x 0 u · dx = φ0(t) + Z x 0 (u dx + v dy), (2.5) where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is even easier to establish when we restrict our attention to two dimensions. If we consider two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, Green’s Theorem implies thatIf a sky diver decides to jump off a jet in Arkansas
with the intention of floating through Tennessee to
North Carolina, then completing his journey in a
likely manner back to Arkansas by drifting North
from his last point. What state would be the third t
be drifted over and what is the estimated distance
between the zone and then drop point?
Answers
Answer:
The answer to this question can be defined as follows:
Explanation:
The sky driver began his sky journey from Arkansas, drove across the Tennessee River then landed in North Carolina. He returned to both the north in the very same direction. He began with NC, traveled through Tennessee, eventually lands in Arkansas. But North Carolina has been in the third state on which skydiver was traveling over, and It's also more than 700 miles from Arkansas to the NC.
Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at 20 ℃ throughout the year, while the walls of the room are nominally at 27 ℃ and 14 ℃ in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of 32 ℃ throughout the year and to have an emissivity of 0.90. The coefficient associated with heat transfer by natural convection between the person and the room air is approximately 2 W/m2∙K.
Answers
Answer:
radiative heat loss substantially increases as the wall temperature declines
Explanation:
The body's heat loss due to convection is ...
(2 W/m^2·K)((32 -20)K) = 24 W/m^2
__
The body's heat loss due to radiation in the summer is ...
[tex]\epsilon\sigma(T_b^4-T_w^4)\quad\text{where $T_b$ and $T_w$ are body and wall temperatures ($^\circ$K)}\\\\0.90\cdot 5.6703\cdot 10^{-8}(305.15^4-300.15^4)\,\text{W/m$^2$}\\\\\approx 28.3\,\text{W/m$^2$}[/tex]
The corresponding heat loss in the winter is ...
[tex]0.90\cdot 5.6703\cdot 10^{-8}(305.15^4-287.15^4)\,\text{W/m$^2$}\\\\\approx 95.5\,\text{W/m$^2$}[/tex]
Then the total of body heat losses to surroundings from convection and radiation are ...
summer: 24 +28.3 = 52.3 . . . W/m^2
winter: 24 +95.5 = 119.5 . . . W/m^2
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It is reasonable that a person would feel chilled in the winter due to the additional radiative loss to the walls in the winter time. Total heat loss is more than doubled as the wall temperature declines.
A piston cylinder device contains 5 kg of Refrigerant 134a at 600 kPa and 80 C. The refrigerant is now cooled at constant pressure until it reaches a liquid-vapor mixture state with a quality of 0.3. How much heat was extracted in the process?
Answers
Answer:
The answer is 920 kJ
Explanation:
Solution
Given that:
Mass = 5kg
Pressure = 600 kPa
Temperature = 80° C
Liquid vapor mixture state (quality) = 0.3
Now we find out the amount of heat extracted in the process
Thus
Properties of RI34a at:
P₁ = 600 kPa
T₁ = 80° C
h₁ = 320 kJ/kg
So,
P₁ = P₂ = 600 kPa
X₂ =0.3
h₂ = 136 kJ/kg
Now
The heat removed Q = m(h₁ -h₂)
Q = 5 (320 - 136)
Q= 5 (184)
Q = 920 kJ
Therefore the amount of heat extracted in the process is 920 kJ
A 1/150 scale model is to be usedin a towing tank to study the water motion near the bottom of a shallow channel as a large barge passes over. (See Video V7.16.) Assume that the model is operated in accordance with the Froude number criteria for dynamic similitude. The prototype barge moves at a typical speed of 15 knots. (a) At what speed (in ft/s) should the model be towed
Answers
Answer:
The speed will be "3.58 ft/s". The further explanation is given below.
Explanation:
Number of knots
= 15
For the similarity of Froude number:
⇒ [tex]\frac{V_{m}}{\sqrt{g_{m}l_{m}} }=\frac{V}{\sqrt{gl} }[/tex]
Here,
[tex]l = length[/tex]
[tex]g_{m}=g[/tex]
⇒ [tex]\frac{V_{m}}{V}=\sqrt{\frac{l_{m}}{l} }[/tex]
[tex]V_{m}=\sqrt{\frac{1}{50} }\times number \ of \ knots[/tex]
[tex]=\sqrt{\frac{1}{50}}\times 15[/tex]
[tex]=2.12 \ knots[/tex]
Now,
⇒ [tex]1 \ knots=0.514\times 3.281[/tex]
[tex]=1.69 \ ft/s[/tex]
So that,
⇒ [tex]V_{m}=2.12\times 1.69[/tex]
[tex]=3.58 \ ft/s[/tex]
Describe the components of a stream's load and how is each component is transported. Discuss at least two factors that affect the transportation of the load and evaluate the impact of each on components of a stream load.
Answers
Answer:
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Explanation:
In real world, sampling and quantization is performed in an analog to digital converter (ADC) and reconstruction is performed in a digital to analog converter (DAC). Which of the following statements hold true (fs denotes the sampling frequency)?
a. the reconstruction filter can be found in the DAQ
b. the antialiasing filter removes all frequencies of the continuous-time analog input signal that are above fs/2
c. the DAC needs to know the sampling frequency of the ADC to correctly reconstruct the signal.
d. the reconstructed continuous-time signal only contains frequencies up to fs/2
Answers
Answer:
b
Explanation:
a) ADC is located on DAQ filter but not the reconstruction filter
b) to remove aliasing, the sampling rate must be greater than or equal ot twice the highest frequency component in the input signal. In other words, all frequencies in input sgnal are less than fs/2. Therefore, frequencies greater than fs/2 are removed by anti-aliasing filter
c) the DAC can have different sampling rate from ADC
Suppose a student carrying a flu virus returns to an isolated college campus of 9000 students. Determine a differential equation governing the number of students x(t) who have contracted the flu if the rate at which the disease spreads is proportional to the number of interactions between students with the flu and students who have not yet contracted it. (Usek > 0for the constant of proportionality and x forx(t).)
Answers
Answer:
dx/dt = kx(9000-x) where k > 0
Explanation:
Number of students in the campus, n = 9000
Number of students who have contracted the flu = x(t) = x
Number of students who have bot yet contracted the flu = 9000 - x
Number of Interactions between those that have contracted the flu and those that are yet to contract it = x(9000 - x)
The rate of spread of the disease = dx/dt
Note: the rate at which the disease spread is proportional to the number of interactions between those that have contracted the flu and those that have not contracted it.
[tex]\frac{dx}{dt} \alpha [x(9000 -x)]\\[/tex]
Introducing a constant of proportionality, k:
dx/dt = kx(9000-x) where k > 0
Consider a 2-shell-passes and 8-tube-passes shell-and-tube heat exchanger. What is the primary reason for using many tube passes
Answers
Answer:
See explanation
Explanation:
Solution:-
- The shell and tube heat exchanger are designated by the order of tube and shell passes.
- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.
- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.
- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.
- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:
U ∝ v^( 0.8 ) .... ( turbulence )
- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.
Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).
Question 44
What should you do if you encounter a fishing boat while out in your vessel?
A
Make a large wake nearby.
B
Avoid making a large wake.
с
Pass on the side with the fishing lines.
D
Pass by close to the anglers.
Submit Answer
Answers
Answer:
The answer is B. Avoid making a large wake.
Explanation:
When passing a fishing boat it is important to maintain a minimal wake due to the dangers a large wake could pose to the fishing boat you are passing, it is part of maintaining safety on the water.
You can not pass on the sides with the fishing lines also, and you are supposed to communicate to the fishing boat before taking the appropriate action.
A rectangular steel bar 37.5 mm wide and 50 mm thick is pinned at each end and subjected to axial compression. The bar has a length of 1.75 m. The modulus of elasticity is 200 Gpa. What is the critical buckling load
Answers
Answer:
The critical buckling load is [tex]\mathbf{P_o = 141.61 \ kN}[/tex]
Explanation:
Given that:
the width of the rectangular steel = 37.5 mm = 0.0375 m
the thickness = 50 mm = 0.05 m
the length = 1.75 m
modulus of elasticity = 200 Gpa = 200 10⁹ × Mpa
We are to calculate the critical buckling load [tex]P_o[/tex]
Using the formula:
[tex]P_o = \dfrac{\pi ^2 E I}{L^2}[/tex]
where;
[tex]I = \dfrac{0.0375^3*0.05}{12}[/tex]
[tex]I = 2.197 * 10^{-7}[/tex]
[tex]P_o = \dfrac{\pi ^2 *200*10^9 * 2.197*10^{-7}}{1.75^2}[/tex]
[tex]P_o = 141606.66 \ N[/tex]
[tex]\mathbf{P_o = 141.61 \ kN}[/tex]
The critical buckling load is [tex]\mathbf{P_o = 141.61 \ kN}[/tex]
If the resistance reading on a DMM'S meter face is to 22.5 ohms in the range selector switch is set to R X 100 range, what is the actual measure resistance of the circuit?
Answers
Answer:
The answer is 2.25 kΩ
Explanation:
Solution
Given that:
The resistance reading on a DMM'S meter face = 22.5 ohms
The range selector switch = R * 100 range,
We now have to find the actual measure resistance of the circuit which is given below:
The actual measured resistance of the circuit is=R * 100
= 22.5 * 100
=2.25 kΩ
Hence the measured resistance of the circuit is 2.25 kΩ
A student proposes a complex design for a steam power plant with a high efficiency. The power plant has several turbines, pumps, and feedwater heaters. Steam enters the first turbine at T1 (the highest temperature of the cycle) and saturated liquid exits the condenser at T7 (the lowest temperature of the cycle). The rate of heat transfer to the boiler (the only energy input to the system)is Qb. Determine the maximum possible efficiency and power output for this complex steam power plant design.
Answers
Answer:
Hello your question lacks some values here are the values
T1 = 500⁰c, T7 = 70⁰c, Qb = 240000 kj/s
answer : A) 56%
B) 134400 kw ≈ 134.4 Mw
Explanation:
Given values
T1 (tmax) = 500⁰c = 773 k
T7(tmin) = 70⁰c = 343 k
Qb = 240000 kj/s
A) Determine the maximum possible efficiency
[tex]n_{max}[/tex] = 1 - [tex]\frac{tmin}{tmax}[/tex] * 100
= 1 - ( 343 / 773 )
= 1 - 0.44 = 0.5562 * 100 ≈ 56%
B) Determine the power output for this complex steam power plant design
[tex]p_{out}[/tex] = Qb * max efficiency
= 240000 kj/s * 56%
= 240000 * 0.56 = 134400 kw ≈ 134.4 Mw
What is the criteria for a guard having to be used on a machine?
Answers
The criteria for a guard having to be used on a machine is;
As a safety measure If the operation exposes you to an injury.
When operating a machine, there are possibilities that the operator could be injured or exposed to injury.
Due to the possible safety issues when operating a machine, the Occupational Safety and Health Administration (OSHA) in their 29 code mandated that a safeguard must be put at each machine to ensure that there is adequate safety that prevents or minimizes the risk of getting injured.
Read more on Occupational Safety and Health Administration (OSHA) rules at; https://brainly.com/question/17069021