Answer:
D. Overrule any other laws and traffic control devices.
Explanation:
Laws and traffic control devices are undoubtedly compulsory to be followed at every point in time to control traffic and other related situations. However, there are cases when certain instructions overrule these laws and traffic control devices. For example, when a traffic police is giving instructions, and though the traffic control devices too (such as traffic lights) are displaying their own preset lights to control some traffic, the instructions from the traffic police take more priority. This is because at that point in time, the instructions from the traffic control devices might not be just applicable or sufficient.
Also, in the case of instructions given by construction flaggers, these instructions have priority over those from controlling devices. This is because during construction traffic controls are redirected from the norms. Therefore, the flaggers such be given more importance.
Answer:
D. Overrule any other laws and traffic control devices.
Explanation:
Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa and 800 K, respectively. Assuming an ideal gas behaviour, find the entropy change of the carbon dioxide by assuming that the specific heats are constant. For the gas, take Cp = 0.846 kJ/kg.K and R = 0.1889 kJ/kg.K
Answer:
[tex]$ S_2 - S_1 = -0.1104 \: \: kJ/kg.K$[/tex]
The entropy change of the carbon dioxide is -0.1104 kJ/kg.K
Explanation:
We are given that carbon dioxide undergoes a process in a closed system.
We are asked to find the entropy change of the carbon dioxide with the assumption that the specific heats are constant.
The entropy change of the carbon dioxide is given by
[tex]$ S_2 - S_1 = C_p \ln (\frac{T_2}{T_1}) - R\ln (\frac{P_2}{P_1}) $[/tex]
Where Cp is the specific heat constant
Cp = 0.846 kJ/kg.K
R is the universal gas constant
R = 0.1889 kJ/kg.K
T₁ and T₂ is the initial and final temperature of carbon dioxide.
P₁ and P₂ is the initial and final pressure of carbon dioxide.
[tex]$ S_2 - S_1 = 0.846 \ln (\frac{800}{400}) - 0.1889\ln (\frac{2000}{50}) $[/tex]
[tex]$ S_2 - S_1 = 0.846(0.69315) - 0.1889(3.6888) $[/tex]
[tex]$ S_2 - S_1 = 0.5864 - 0.6968 $[/tex]
[tex]$ S_2 - S_1 = -0.1104 \: \: kJ/kg.K$[/tex]
Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K
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
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]
When checking the resistance of a dual voltage wye motor, there should be ____ resistance readings. 1) twelve 2) six 3) three
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.
A dual cycle which operates on air with a compression ratio of 16 has alow pressure of 200 kpa and a low temperature of 200 c . If tha cutoff ratio is 2 and the pressure ratio is 1.3 calculate tha thermal efficiency the heat input amd tha work output
Answer:
1. The thermal efficiency of the cycle is 57.913%
2. The heat input is 32165.68 kJ/kg
3. The work output is -18.628.11 kJ/kg
Explanation:
1. T₁ = 200C = 473.15 K
The thermal efficiency of the cycle is given by the relation;
[tex]\eta =\dfrac{k \beta ^{\gamma }-1}{\left [ \left ( k-1 \right ) \right + \gamma k(\beta -1)]r_{v}^{\gamma -1}}[/tex]
Where:
k = Pressure ratio = p₃/p₂ = 1.3
γ = Specific heat ratio = 1.4
β = v₄/v₃
[tex]r_v[/tex] = v₁/v₂
Which gives;
η = (1.3*(16/2)^1.4 - 1)/(((1.3 - 1) + 1.4*1.3*((16/2)-1))*16^0.4) = 0.57913
η = 57.913%
The thermal efficiency of the cycle = 57.913%
2. The heat input is given by the following relation;
Q = cv(T₂ - T₁) + cp(T₄ - T₃)
T₂/T₁ = (v₁/v₂)^(γ - 1) = 16^0.4 = 3.03
Therefore, T₂ = T₁×3.03= 473.15*3.03= 1434.323 K
T₃/T₂ = p₃/p₂ = 1.3
T₃ = T₂ ×1.3 = 1434.323 *1.3 = 1864.62 K
v₄/v₃ = T₄/T₃ = 16/2 = 8
T₄ = T₃×8 = 1864.62 × 8 = 14916.959 K
At 473.15 K, cv = 0.7386 cp = 1.026 kJ/(kg*K)
At 1434.323 K, cv = 0.9158 cp = 1.203
At 1864.62 K, cv = 0.9535 cp = 1.239
At 1864.62 K, cv = 0.874 cp = 1.24
Taking average values, we have
(0.7386 + 0.9158)/2= 0.8272 kJ/(kg*K)
cp = 1.24
The heat input is then found as follows;
Q = 0.874*(1434.323 - 473.15) + 2.4*(14916.96 - 1864.62) = 32165.68 kJ/kg
The heat input = 32165.68 kJ/kg
3. The work output is found as follows;
Given that η = -W/Q, we have;
-W = Q × η = 0.57913*32165.68 = -18.628.11 kJ/kg.
The work output = -18.628.11 kJ/kg.
Based on the calculations, the thermal efficiency for this dual cycle is 62.2%.
Given the following data:
Compression ratio = 16.Pressure = 200 kPa.Temperature = 200°C.Cutoff ratio = 2.Pressure ratio = 1.3.How to calculate the thermal efficiency.Mathematically, the thermal efficiency for a dual cycle is given by this equation:
[tex]\eta= 1-\frac{1}{r^{k-1}} [\frac{r_pr_c^k -1}{kr_p(r_c-1)+r_p-1} ]\\\\\eta = 1-\frac{1}{16^{1.4-1}} [\frac{1.3(2)^{1.4} -1}{1.4(1.3)(2-1)+1.3-1}]\\\\\eta = 1-\frac{1}{16^{0.4}} [\frac{3.43 -1}{2.12}]\\\\\eta = 1-\frac{1}{3.03}\times [\frac{2.43}{2.12}]\\\\\eta =0.622[/tex]
Thermal efficiency = 62.2%.
How to calculate the heat input.First of all, we would determine the temperatures as follows:
[tex]T_2 = T_1(\frac{V_1}{V_2} )^{k-1}\\\\T_2 = 473 \times (16 )^{1.4-1}\\\\T_2 = 473 \times (16 )^{0.4}\\\\T_2 = 1434\;K[/tex]
[tex]T_3 = T_2(\frac{P_3}{P_2} )\\\\T_3 = 1434(1.3)\\\\T_3 = 1864\;K[/tex]
[tex]T_4 = T_3(\frac{V_4}{V_3} )\\\\T_4 = 1864(2.0)\\\\T_4 = 3728\;K[/tex]
Now, we can determine the heat input:
[tex]q_{in}=c_v(T_3-T_2)+c_p(T_4-T_3)\\\\q_{in}=0.717(1864-1434)+1.00(3728-1864)\\\\q_{in}=2172 \;kJ/kg.[/tex]
How to calculate the work output.[tex]W_{out}=nq_{in}\\\\W_{out}= 0.622 \times 2172\\\\W_{out}=1350\;kJ/kg.[/tex]
Read more on thermal efficiency here: https://brainly.com/question/13577244
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.
Answer:
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Explanation:
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).)
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
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
[tex]
<html>
<head>
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Time Picker</title>
</head>
<body>
<!--24 Hours format-->
<input type="time" placeholder="Enter Time" />
<input type="date">
</body>
</html>
[/tex]
what is the difference between erratic error and zero error
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!
For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because (select all that are correct
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 increasingExample 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).
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:
Which greenhouse gas is produced by commercial refrigeration and air conditioning systems?
carbon dioxide
Ofluorinated gas
O nitrous oxide
O methane
Answer:
B- Fluorinated gas
Explanation:
Answer:
B.) fluorinated gas
Explanation:
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occurs at an average outer surface temperature of 315 K at the rate of 30 kJ per kg of air flowing. Kinetic and potential energy effects are negligible. Assuming the air is modeled as an ideal gas with variations in specific heat, determine (a) the rate power is developed, in kJ per kg of air flowing, and (b) the rate of entropy production within the turbine, in kJ/K per kg of air flowing.
Answer:
A)W'/m = 311 KJ/kg
B)σ'_gen/m = 0.9113 KJ/kg.k
Explanation:
a).The energy rate balance equation in the control volume is given by the formula;
Q' - W' + m(h1 - h2) = 0
Dividing through by m, we have;
(Q'/m) - (W'/m) + (h1 - h2) = 0
Rearranging, we have;
W'/m = (Q'/m) + (h1 - h2)
Normally, this transforms to another equation;
W'/m = (Q'/m) + c_p(T1 - T2)
Where;
W'/m is the rate at which power is developed
Q'/m is the rate at which heat is flowing
c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k
T1 is initial temperature
T2 is exit temperature
We are given;
Q'/m = -30 kj/kg (negative because it leaves the turbine)
T1 = 980 k
T2 = 670 k
Plugging in the relevant values;
W'/m = -30 + 1.1(980 - 670)
W'/m = 311 KJ/kg
B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;
(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0
Dividing through by m gives;
((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0
Rearranging, we have;
σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)
Under the conditions given in the question, this transforms normally to;
σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)
σ'_gen/m is the rate of entropy production in kj/kg
We are given;
p2 = 100 kpa
p1 = 400 kpa
T_boundary = 315 K
For an ideal gas, R = 0.287 KJ/kg.K
Plugging in the relevant values including the ones initially written in answer a above, we have;
σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))
σ'_gen/m = 0.0952 + 0.4183 + 0.3979
σ'_gen/m = 0.9113 KJ/kg.k
An 1,840 W toaster, a 1,420 W electric frying pan, and a 70 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket).
A) What current is drawn by each device?
B) Will this combination blow the 15-A fuse?
Answer:
A)
Current drawn by toaster = 15.33 A
Current drawn by electric frying pan = 11.83 A
Current drawn by lamp = 0.58 A
B)
The fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.
Explanation:
There are three devices plugged into the same outlet.
Toaster = 1840 W
Electric frying pan = 1420 W
Lamp = 70 W
Since the three devices are connected in parallel therefore, the voltage across them will be same but the current drawn by each will be different.
A) What current is drawn by each device?
The current flowing through the device is given by
I = P/V
Where P is the power and V is the voltage.
Current drawn by toaster:
I = 1840/120
I = 15.33 A
Current drawn by electric frying pan:
I = 1420/120
I = 11.83 A
Current drawn by lamp:
I = 70/120
I = 0.58 A
B) Will this combination blow the 15-A fuse?
The total current drawn by all three devices is
Total current = 15.33 + 11.83 + 0.58
Total current = 27.74 A
Therefore, the fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.
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.
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
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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
Answer:
see the attached truth tableOpens = AB + ACExplanation:
i) In the attached truth table, TRUE means the respective key owner is present and their keys are inserted at the same time. The "Opens" column is TRUE when two owners are present, not including the case where the only two owners present are B and C.
__
ii) The second attachment is a Karnaugh map of the truth table. The circled terms are ...
Opens = AB +AC
Consider a 2-shell-passes and 8-tube-passes shell-and-tube heat exchanger. What is the primary reason for using many tube passes
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 ).
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.
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 thatDetermine the combined moment about O due to the weight of the mailbox and the cross member AB. The mailbox weighs 3.2 lb and the uniform cross member weighs 10.3 lb. Both weights act at the geometric centers of the respective items. The moment will be positive if counterclockwise, negative if clockwise.
Answer:
Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib
Answer : moment of inertia = 186.7 Ib - in
Explanation:
Given data
weight of the mailbox = 3.2 Ib
weight of the uniform cross member = 10.3 Ib
The origin is of mailbox and cross member is 0
The perpendicular distance from Y axis of centroid of the mailbox
= 4 + (25/2) = 16.5"
The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"
therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)
= 52.8 + 133.9 = 186.7 Ib-in
The combined moment about O due to the weight of the mailbox and the cross member AB is; M_o = 122.4 lb.in (ccw)
We are given;
Weight of mailbox; W_m = 3.2 lb
Weight of uniform cross member; W_c = 10.3 lb
Now, from the attached diagram, let us calculate the geometric location of the mailbox and uniform cross section from point O.
Geometric location of mailbox from point O; g_m = 3 + (19/2) = 12.5 in
Geometric location of cross member from point O;
g_c = (¹/₂(1 + 19 + 3 + 7)) - 7
g_c = 8 in
Thus. combined moment about point O is;
M_o = (W_m × g_m) + (W_c × g_c)
M_o = (3.2 × 12.5) + (10.3 × 8)
M_o = 122.4 lb.in
Since positive then it is counterclockwise. Thus;
M_o = 122.4 lb.in (ccw)
The image of this question is missing and so i have attached it.
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how does a TV'S screen work
Answer:
A TVS screen works when the pixels are switched on electronically using liquid crystals to rotate polarized light.
Explanation:
Consider fully developed laminar flow in a circular pipe. If the viscosity of the fluid is reduced by half by heating while the flow rate is held constant, how will the pressure drop change
Answer:
The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.
Explanation:
For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.
Q = π(ΔPR⁴/8μL)
where Q = volumetric flowrate
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe
ΔP = μ(8QL/πR⁴)
ΔP = Kμ
K = (8QL/πR⁴) = constant (for this question)
ΔP = Kμ
K = (ΔP/μ)
So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).
μ₁ = (μ/2)
The new pressure drop (ΔP₁) is then
ΔP₁ = Kμ₁ = K(μ/2)
Recall,
K = (ΔP/μ)
ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)
Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.
Hope this Helps!!!
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.
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.
Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.20 MPa and 258C at a rate of 0.07 kg/s, and it leaves at 1.2 MPa and 708C. The refrigerant is cooled in the condenser to 448C and 1.15 MPa, and it is throttled to 0.21 MPa. Disregarding any heat transfer and pressure drops in the connecting lines between the components, show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the COP of the refrigerator.
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?
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 manometer connected to a Pitot-static tube has a difference in the height of the two columns of water of 10 cm when the Pitot-static tube is placed in a flow of air at standard sea level conditions. What is the velocity of the airflow
Answer:
v = 40 m/s
Explanation:
Difference in the height of the two columns:
[tex]h_m = \triangle h = 10 cm \\h_m = 0.1 m[/tex]
Air density, [tex]\rho_{air} = 1.225 kg/m^3[/tex]
Water density, [tex]\rho_{w} = 1000 kg/m^3[/tex]
The velocity of air flow is given by the equation:
[tex]v = \sqrt{2gh_m (\frac{p_w}{p_{air}} - 1) } \\\\v = \sqrt{2*9.81*0.1 (\frac{1000}{1.225} - 1) }\\\\v = 40 m/s[/tex]
What is the criteria for a guard having to be used on a machine?
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.
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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
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]
What is the final temperature after compression of a diesel cycle if the initial temperature is 32C and clearance is 8%
Answer:
863 K
Explanation:
See the attachment
Tech A says that a gear set that has a drive gear with 9 teeth and a driven gear with 27 teeth has a gear ratio of 3:1. Tech B says that the drive gear is also called the output gear. Who is correct?
Answer:
Tech A is correct.
Explanation:
Gears are toothed wheels that can be used to transmit power. When two or more gears are in tandem, a gear train is formed.
Gear ratio = [tex]\frac{number of teeth of the driven gear}{Number of teeth of the driving gear}[/tex]
= [tex]\frac{27}{9}[/tex]
= [tex]\frac{3}{1}[/tex]
Gear ratio = 3:1
The driver gear is called the input gear since it transfers its power to the driven gear. While the driven gear is called the output gear because it produces an effect due to both gears.
Tech A is correct.
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.
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
__
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.
If 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?
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.