Answer:
[tex]Q_2 = 32[/tex] mL/s
Explanation:
Given :
The flow is incompressible viscous flow.
The initial flow rate, [tex]Q_1[/tex] = 1 mL/s
Initial diameter, [tex]D_1= D_0[/tex]
Initial length, [tex]L_1=L_0[/tex]
The initial pressure difference to maintain the flow, [tex]P_1=P_0[/tex]
We know for a viscous flow,
[tex]$\Delta P = \frac{32 \mu V L}{D^2}$[/tex]
[tex]$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$[/tex]
[tex]$Q \propto \Delta P \times D^4$[/tex]
[tex]$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$[/tex]
[tex]$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$[/tex]
[tex]$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$[/tex]
[tex]$\frac{1}{Q_2}= \frac{1}{32}$[/tex]
∴ [tex]Q_2 = 32[/tex] mL/s
The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s
We are given;
Initial flow rate; Q1 = 1 mL/s
Initial uniform diameter; D0
Initial Length; L0
Initial Pressure difference; P0
Relationship between pressure, flow rate and diameter for vicious flow is given by;
Q1/Q2 = (P1/P2) × (D1/D2)⁴
Where;
Q1 is initial flow rate
Q2 is final flow rate
P1 is initial pressure difference
P2 is final pressure difference
D1 is initial diameter
D2 is final diameter
We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;
P2 = 2P0
D2 = 2D0
Thus;
1/Q2 = (P0/2P0) × (D0/2D0)⁴
>> 1/Q2 = ½ × (½)⁴
1/Q2 = 1/32
Q2 = 32 mL/s
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The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Sieve analysis Sieve Size No. 4 (4.75 mm) No. 10 (2.00 mm) No. 40 (0.425 mm) No. 200 (0.075 mm) Percent passing by weight 80 60 30 10 Atterberg limits Liquid limit (LL) Plastic limit (PL 31 25
(a) Classify this soil according to USCS system, providing the group symbol for it. Show how you arrive at the final classification.
(b) According to USCS system, what is a group name for this soil?
(c) Is this a clean sand? If not, explain why.
Answer: hello the complete question is attached below
answer:
A) Group symbol = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Explanation:
A) Classifying the soil according to USCS system
( using 2nd image attached below )
description of sand :
The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also
The soil is a sand given that more than 50% particles passed from No 4 sieve
The soil can be a clean sand given that fines ≤ 12%
The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time
Group symbol as per the 2nd image attached below = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
A steam turbine generator unit is used to produce electricity, where steam enters the turbine with a velocity of 30 m/s and enthalpy (internal energy) of 3348 kJ/kg. The 1 steam leaves the turbine as a mixture of vapor and liquid having a velocity of 60 m/s and an enthalpy of 2550 kJ/kg. The flow through the turbine is adiabatic, and changes in elevation are negligible. Determine the work output from the turbine, if the mass flow rate is 5kg/s.
Answer:
3.983 MW
Explanation:
Given that:
At the inlet:
Velocity (v₁) = 30 m/s
Enthalpy (h₁) = 3348 kJ/kg
At the outlet:
Velocity (v₂) = 60 m/s
Enthalpy (h₂) = 2550 kJ/kg
Mass flow rate (m) = m₁ = m₂ = 5kg/s
According to the steady flow energy equation:
[tex]Q+ m_1 (h_1 + \dfrac{v_1^2}{2000}+ \dfrac{gz_1}{1000} )= m_2(h_2+\dfrac{v_2^2}{2000}+\dfrac{gz_2}{1000})+W_{shaft}[/tex]
Since the elevation (z) is negligible and flow via the turbine is adiabatic:
Then,
Q = 0 and z₁ = z₂
∴
[tex]W_{shaft} = (mh_1-mh_2) + (\dfrac{mv_1^2-mv_2^2}{2000})[/tex]
[tex]W_{shaft} = ((5*3348) -(5*2550)) + (\dfrac{(5*(30)^2)-(5*(60)^2)}{2000})[/tex]
[tex]W_{shaft} = (16740-12750) + (\dfrac{4500-18000}{2000})[/tex]
[tex]W_{shaft} = (16740-12750) + (-6.75)[/tex]
[tex]W_{shaft} = 3983.25 \ kW[/tex]
[tex]\mathbf{W_{shaft} = 3.983 \ MW}[/tex]
hỗ trợ mình với được không các bạn
Answer:
Explanation:
Be bop
You are given a pot of hot tea (350 mL) at a temperature of 85°C and a quantity of ice at -12°C. Determine the absolute minimum amount of ice you can add to the hot tea so that at equilibrium you have iced tea (ie. a very, very small amount of ice and some water). You can assume tea has the same density (1.00 g/mL) and specific heat (4190 J/kgK) as liquid water. The heat of fusion of H2O is 3.33x10^5 J/kg. The specific heat of ice is 2090 J/kgK.
Answer:
348 g
Explanation:
The heat gained by the ice equals the heat lost by the hot tea.
Also, since we require a very, very small amount of ice and some water, to still have some ice, the temperature of the hot tea has to decrease to the melting point of ice which is 0°C. So, the final temperature of the mixture is 0 °C.
So, mc(T - T') + mL = -MC(T - T") where m = mass of ice, c = specific heat of ice = 2090 J/kgK., T = final temperature of mixture = 0 °C, T' = initial temperature of ice = -12 °C, L = heat of fusion of H2O = 3.33 × 10⁵ J/kg, M = mass of hot tea = ρV where ρ = density of tea = 1.00 g/mL and V = volume of hot tea = 350 mL. So, M = ρV = 1.00 g/mL × 350 mL = 350 g = 0.350 kg, C = specific heat of tea = 4190 J/kgK and T" = initial temperature of tea = 85 °C.
Making m subject of the formula, we have
mc(T - T') + mL = -MC(T - T")
m[c(T - T') + L] = -MC(T - T")
m = -MC(T - T")/[c(T - T') + L]
substituting the values of the variables into the equation, we have
m = -MC(T - T")/[c(T - T') + L]
m = -0.350 kg × 4190 J/kgK(0 °C - 85 °C.)/[2090 J/kgK(0 °C. - (-12 °C)) + 3.33 × 10⁵ J/kg]
m = -1466.5 J/kK(- 85 °C)/[2090 J/kgK(0 °C + 12 °C)) + 3.33 × 10⁵ J/kg]
m = 124652.5 J/[2090 J/kgK(12 °C) + 3.33 × 10⁵ J/kg]
m = 124652.5 J/[25080 J/kg + 3.33 × 10⁵ J/kg]
m = 124652.5 J/358080 J/kg
m = 0.3481 kg
m = 348.1 g
m ≅ 348 g
So, the minimum amount of ice to be added is 348 g.
beacuse thye want them to hav egoood and thye wn thme tto
Answer:
I don't understand the question
Tech A says that when performing a cylinder leakage test, the piston should be positioned at bottom dead center. Tech B says that a blown head gasket would generally leak past the exhaust valve. Who is correct
The causes of low compression in an engine cylinder and leakage past the exhaust can be identified by performing mechanical tests on the engine
The correct technician is Tech B; Blown head gasket would generally leak past the exhaust valve
The reason for arriving at the above selection is as follows:
A cylinder leakage test is generally performed after there is an indication of low compression from one or more cylinders. The test allows the identification of the part of the cylinder that is leaking.
The cylinder that shows sign of low compression is often the location that the cylinder leakage test is performed, although, the test can be performed on every cylinder
A cylinder leakage test is valid only when
The temperature of engine is about the temperature range of the engine in operationThe cylinder under test is at the Top Dead Center (TDC) position, such that the position of the piston in the cylinder is at the top or highest point of its strokeTherefore, the positioning of the piston in the bottom dead center during a leakage test as mentioned by Tech A is not correct
The route of coolant in the engine are sealed by the cylinder head gasket, where there is a blown head gasket, the coolant will be allowed to leak into the cylinders and into the exhaust, causing white smoke and sweet odor of antifreeze
Therefore, Tech B is correct, coolant leakage into the exhaust is an indication of a blown head gasket
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Cho biết tác dụng chung của các hệ giằng khung ngang nhà công nghiệp nhẹ 1 tầng 1 nhịp.
Which of the following conditions were present in over 80% of paddling fatalities from 1995-2000?
Answer:
80% of the people that were killed weren't wearing a safety flotation device ( in correct terminology Personal Flotation Device, or PFD )
Explanation:
Hence they drowned due to the lack of safety.
The condition present in over 80% of paddling fatalities is that the victims were said to be not wearing a personal flotation device (PFD).
What is paddling in the sea?The act of paddling is known to be a sport or something one can do for fun or for games.
Conclusively, a lot of paddling fatalities 1995-2000 was due to the fact that the people who were engage in the sport or act were not putting on a personal flotation device (PFD) and this can lead to death or drowning.
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A 2-stage dcv that has an internal pilot does not work well (if at all) on
Answer:
i really font onow why tbh eot you
what is tracer lathe machine
Answer: The tracer lathe is a roughing operation for the output shaft on rear wheel drive transmissions.
Explanation:
A steel component with ultimate tensile strength of 800 MPa and plane strain fracture toughness of 20 MPam is known to contain a tunnel (internal) crack of length 1.4 mm. This alloy is being considered for use in a cyclic loading application for which the design stresses vary from 0 to 410 MPa. Would you recommend this alloy for this application
Complete question:
A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?
a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.
c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.
d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.
Answer:
A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.
Explanation:
we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.
thank you!
Why is logging done during drilling?
Answer:
Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible
which type of clectrical circuit is represented by this diagram?
Answer:
parallel
Explanation:
All components in this circuit are tied in parallel. Each component experiences the same voltage from one terminal to the other. It is a parallel circuit.
Work to be performed can come from the work package level of the work breakdown structure as well as other sources. Which of these is NOT a source of authorized work to be performed?
a. Scope creep
b. Defect repairs
c. Preventive actions
d. Corrective actions
A project is meant to achieve clearly defined goals within present constraints
The correct option for the source of work in a project which is NOT a source of authorized work to be performed is option (a)
a. Scope creep
The reason for the selecting the above option is as follows;
In project management, alongside work planned in a work package, authorized work to be performed can come from;
The repairing of defects found in completed work and which is a requirement for the project to be approvePreventive actions, that are meant to mitigate factors that affect the project timeline and costCorrective actions which include actions meant to rectify parts of the project that produce an error outcomeA scope creep, however, is generally is due to the inclusion of the end
users of the project at a time that is behind schedule, insufficient analysis of
the requirements, and project complexity underestimation. and consists of
changes made after a project has commenced
A scope creep therefore, changes the amount of work to be done which
therefore alters the project timeline, and due to project, budgetary, and
end user constraints, CANNOT be a source of authorized work to be
performed
The correct option for the source that is NOT a authorized work to be performed is option a. Scope creep
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A resistor has code 104 printed on it .What is the resistive value of this resistor
Answer:
X = 1 (1st digit in the code)
Y = 0 (2nd digit)
Z = 4 (3rd multiplier digit)
104 → 10 × 10^4 Ω
→ 10 × 10000Ω
→ 100 kΩ
resistors are marked 104, 105, 205, 751, and 754. The resistor marked with 104 should be 100kΩ (10x10^4), 105 would be 1MΩ (10x10^5), and 205 is 2MΩ (20x10^5). 751 is 750Ω (75x10^1), and 754 is 750kΩ (75x10^4).
Here we need to understand how a code in a resistor gives us information on the resistor. Here we will see that the code means that the resistance is 100,000 Ω.
When we use numbers, let's assume that we have 3 single-digit numbers abc.
So if the code in our resistor is abc, this will mean that the resistance of the resistor is:
ab×10^c Ω
Using this general rule we can see that if the code is 104, then the resistance will be:
r = 10×10^4 Ω
= 100,000 Ω
Then we can conclude that the resistive value of this resistor is 100,000 Ω
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Rafel knows that lessons learned is a valuable aid to future projects. When should he and his team address
lessons learned on a project?
Select an answer:
Add lessons learned as a topic in status meetings
Review past lessons learned so a new one does not have to be created,
Create lessons learned at the end of the project.
Brainstorm lessons learned at the beginning of a project
Answer: Create lessons learned at the end of the project.
Explanation:
Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.
The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.
anxiety: a. is never normal. b. is common of many psychological disorders c. is identical to fear d. is a modern development, unlikely to have roots in human history
Answer:
B
Explanation:
Anxiety is very common especially nowadays but it's especially common in psychological disorders
A mechanic claims to have developed a car engine that runs on water instead of gasoline. What is your response to this claim?
The claim will be impractical. The reason behind this is provided below throughout the explanation segment.
Inside the same combustion chamber, fresh water can't be pressurized even more than fuel as well as air. Consequently, the cylinder does not burn due to a problem with the compressor.To accomplish the development processes, a machine requires any sort of combustible which might cause combustion as well as create a tremendous quantity of power.
Thus, the idea of using water instead of petroleum appears unworkable throughout traditional engines.
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an atom that gained an electron is called
Answer:
Hey mate......
Explanation:
This is ur answer.....
When an atom gains/loses an electron, the atom becomes charged, and is called an ion.Hope it helps!
Brainliest pls!
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. Bơm kiểu piston tác dụng đơn có áp suất p=0,64 Mpa và lưu lượng Q=3,5 l/s. Xác định tốc độ quay của trục bơm và công suất của bơm nếu biết đường kính piston D=150 mm; bán kính tay quay R=60 mm; hiệu suất thể tích của bơm là 0=0,94; hiệu suất chung của bơm b=0,80.
Answer:
not understand language
what does mean setbacks with MRT station?
A O.1m³ rigid tank contains steam initially as 500k pa and 200°C. The steam is now allowed to cool until the temperature drops to 50°C. Determine the amount of heat transfer during the process and the final pressure in the tank
Answer:
Check it out here
Explanation:
Functional and nonfunctional requirements documents are used to _____.
define the financial budget of a system
define the purpose of a system
facilitate communication between the users and a system
help exercise control over the inner workings of the firm.
Answer:
Non-functional requirements when defined and executed well will help to make the system easy to use and enhance the performance
Explanation:
True or false: You can create a network with two computers.
Answer
True
if a person is injured at the hospital during a natural disaster a correct action to take is
The impeller shaft of a fluid agitator transmits 20 kW at 430 rpm. If the allowable shear stress in the impeller shaft must be limited to 65 MPa, determine (a) the minimum diameter required for a solid impeller shaft. (b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 36 mm. (c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)
Given :
Power, P = 20 kW
Speed, N = 430 rpm
Allowable shear stress, τ = 65 MPa
Torque in the shaft is given by :
[tex]$P=\frac{2 \pi NT}{60}$[/tex]
[tex]$T=\frac{60 \times 20 \times 10^3}{2 \pi \times 430}$[/tex]
T = 444.37 N.m
Diameter of the solid shaft is
[tex]$d=\sqrt[3]{\frac{16 T}{\pi \tau}}[/tex]
[tex]$d=\sqrt[3]{\frac{16 \times 444.37}{3.14 \times 65}}[/tex]
[tex]$d=\sqrt[3]{34.83} $[/tex]
d = 3.265 m
d = 326.5 mm
Internal diameter of the hollow shaft is :
[tex]$\frac{T}{\frac{\pi}{32} \left( d_0^4 - d_i^4 \right)}=\frac{\tau}{d_0/2}$[/tex]
[tex]$\frac{444.37}{\frac{3.14}{32} \left( 0.036^4 - d_i^4 \right)}=\frac{65 \times 10^6}{0.036/2}$[/tex]
[tex]$\frac{444.37}{0.09 \left( 1.6 \times 10^{-6} - d_i^4 \right)}=\frac{65 \times 10^6}{0.018}$[/tex]
[tex]$\frac{7.99}{ \left( 1.6 \times 10^{-6} - d_i^4 \right)}=5850000$[/tex]
[tex]$1.3\times 10^{-6} = 1.6 \times 10^{-6} - d_i^4 \right)}$[/tex]
[tex]$d_i^4=300000$[/tex]
[tex]$d_i = 23.40$[/tex] mm
Percentage savings in the weight is given by :
Percentage saving = [tex]$\frac{W_{solid}-W_{hollow}}{W_{solid}}\times100$[/tex]
[tex]$=\frac{V_{solid}-V_{hollow}}{V_{solid}}\times100$[/tex]
[tex]$=\frac{d^2 - (d_0^2 - d_i^2)}{d^2} \times 100$[/tex]
[tex]$=\frac{(326.5)^2 - (0.036^2 - (32.40)^2)}{(326.5)^2} \times 100$[/tex]
[tex]$=\frac{106602 - \left(1.29 \times 10^{-3} - 1049.76 \right)}{106602} \times 100$[/tex]
[tex]$=\frac{106602 - 1049 }{106602} \times 100$[/tex]
[tex]$=\frac{105553 }{106602} \times 100$[/tex]
= 99.01 %
2 Air enters the compressor of a cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rates of exergy destruction in the compressor and turbine, respectively, each in kW, for T0 5 300 K.
Explain the 11 sections that a typical bill of quantity is divided into
Answer:
The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.
Magnetic particle testing can be done only on metals that can be magnetized.
True or false
what is the best book for mechanic of materials?
Answer:
As per my experience Egor Popov is one the best book for mechanics of materials.
Explanation:
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