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Anna71 [15]
3 years ago
8

A displacement transducer has the following specifications: Linearity error ± 0.25% reading Drift ± 0.05%/○C reading Sensitivity

error ± 0.25% reading Excitation 10 – 25 V dc Output 0 – 5 V dc Range 0 – 5 cm The transducer output is to be indicated on a voltmeter having a stated accuracy of ± 0.1% reading with a resolution of 10 μV. The system is to be used at room temperature, which can vary by ±10○C. Estimate an uncertainty in a nominal displacement of 2 cm at the design stage. Assume 95% confidence. Assume sensitivity is K = 1 V/cm.
Engineering
1 answer:
White raven [17]3 years ago
8 0

Answer:

The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

Explanation:

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

assume from specifications that k = 5v/5cm

                                                         = 1v/cm

u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2

      = 0.01225v

v = 2v * 0.001

  = 0.002v

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

= ((0.01225)^2 + (0.002)^2)^(1/2)

= 0.0124 cm

Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

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Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u
Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × 10^{-2} N/m

so average radius = \frac{diameter}{2} =  100 µm = 100 ×10^{-6} m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r²    .....................1

here r is radius and  Δp pressure difference and σ surface tension

Δp = \frac{2 \sigma }{r}    

put here value

Δp = \frac{2*2.69*10^{-2}}{100*10^{-6}}  

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0
3 years ago
Explain why Chloe's design needs to be redone in the following scenario, and recommend the techniques she needs to include in he
Alenkinab [10]

Answer:

She believes he's weak and won't do what needs to be done to become a king.

Explanation:

She says ('I fear thy nature') and calls him 'too full o' th' milk of human kindness' which reflects that she feels his kindness makes him weak and may prevent him from proceeding the plan. Thus, she manipulates him to keep his kindness aside and do what she wishes him to do. She rather belittles him to get her purpose solved and

7 0
3 years ago
Read 2 more answers
A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri
babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are  73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

T_{BE}-W=0 hence

T_{BE}=W=20*9.81=196.2 N

Therefore, tension in the cable, T_{BE}=196.2 N

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0

24.525-39.24+0.2D_x=0

D_x=73.575 N

Similarly,

A_x-D_y=0

A_x=73.575 N

Therefore, both reactions at A and D are 73.575 N

7 0
3 years ago
The maximum stress that a bar will withstand before failing is called • Rapture Strength • Yield Strength • Tensile Strength • B
konstantin123 [22]

Answer: Rupture strength

Explanation: Rupture strength is the strength of a material that is bearable till the point before the breakage by the tensile strength applied on it. This term is mentioned when there is a sort of deformation in the material due to tension.So, rupture will occur before whenever there are chances of failing and the material is still able to bear stresses before failing.  

7 0
3 years ago
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
Bad White [126]

Answer:

the net work per cycle \mathbf{W_{net} = 0.777593696}  Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume V_2 = 0.16 V_1

the crankshaft N rotates at a speed of  2400 RPM.

At the beginning of the compression , temperature T_1 = 60 F = 519.67 R    

and;

Otto cycle with a pressure =  14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

V_1-V_2 = \dfrac{\pi}{4}D^2L

V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)

V_1-0.16V_1= 36.55714291

0.84 V_1 =36.55714291

V_1 =\dfrac{36.55714291}{0.84 }

V_1 =43.52040823 \ in^3 \\ \\  V_1 = 43.52 \ in^3

V_1 = 0.02518 \ ft^3

the mass in air ( lb) can be determined by using the formula:

m = \dfrac{P_1V_1}{RT}

where;

R = 53.3533 ft.lbf/lb.R°

m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R  \times 519 .67 ^0 R}

m = 0.0018962 lb

From the tables  of ideal gas properties at Temperature 519.67 R

v_{r1} =158.58

u_1 = 88.62 Btu/lb

At state of volume 2; the relative volume can be determined as:

v_{r2} = v_{r1}  \times \dfrac{V_2}{V_1}

v_{r2} = 158.58 \times 0.16

v_{r2} = 25.3728

The specific energy u_2 at v_{r2} = 25.3728 is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

v_{r3} = 0.1828

u_3 = 1098 \ Btu/lb

To determine the relative volume at state 4; we have:

v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}

v_{r4} =0.1828 \times \dfrac{1}{0.16}

v_{r4} =1.1425

The specific energy u_4 at v_{r4} =1.1425 is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

W_{net} = Heat  \ supplied - Heat  \ rejected

W_{net} = m(u_3-u_2)-m(u_4 - u_1)

W_{net} = m(u_3-u_2- u_4 + u_1)

W_{net} = m(1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (410.08)

\mathbf{W_{net} = 0.777593696}  Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the  four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

W = 4 \times N'  \times W_{net

where ;

N' = \dfrac{2400}{2}

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec ×  0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0
3 years ago
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