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3 years ago

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The following displacement (cm), output (V) data have been recorded during a calibration of an LVDT. Displacement (cm), output (

V): 0.00 cm, 0.20 V; 0.125 cm, 0.60 V; 0.500 cm, 1.32 V; 0.769 cm, 1.95V; 1.019 cm; 2.33 V. Calculate the sensitivity of the LVDT expressed in (V/mm). (Please express your answer in V/mm and round up the answer to two (2) significant decimal digits.)

1 answer:

nadya68 [22]3 years ago

4 0

Answer:

2.08V/cm

Explanation:

Plot the points on a graph. Draw the line of best fit. Calculate the gradietn of line of best fit.

Attached is the graph plotted on excel.

The equation of the line is

Votlage= 2.08× distance + 0.276

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What is the difference between the pressure head at the end of a 150m long pipe of diameter 1m coming from the bottom of a reser

uysha [10]

Answer:

$\frac {p_2- p_1}{\rho g} = 31.06 m$

Explanation:

from bernoulli's theorem we have

$\frac{p_1}{\rho g} + \frac{v_1^{2}}{2g} +z_1 = \frac{p_2}{\rho g} + \frac{v_2^{2}}{2g} +z_2 + h_f$

we need to find pressure head difference i.e.

$\frac {p_2- p_1}{\rho g} = (z_1 - z_2) - h_f$

where h_f id head loss

$h_f = \frac{flv^{2}}{D 2g}$

velocity v = $\frac{1}{n} * R^{2/3} S^{2/3}$

$S = \frac{\delta h}{L} = \frac{40}{150} = 0.267$

hydraulic mean radius R = $\frac{A}{P} = \frac{hw}{2h+w}$

$R = \frac{40*1}{2*40+1} = 0.493 m$

so velocity is = $\frac{1}{0.013} * 0.493^{2/3} 0.267^{1/2}$

v = 24.80 m/s

head loss

$h_f = \frac{0.0019*150*24.80^{2}}{1* 2*9.81}$

$h_f =8.93 m$

pressure difference is

$\frac {p_2- p_1}{\rho g} = 40 - 8.93 = 31.06 m$

$\frac {p_2- p_1}{\rho g} = 31.06 m$

4 0

3 years ago

You have designed a treatment system for contaminant Z. The treatment system consists of a pipe that feeds into a CSTR. The pipe

IRISSAK [1]

Answer:

0.667 per day.

Explanation:

Our values here are

$Q=10m^3/dV_p=15m^3\\V_{cstr}=60m^3\\c_p=2500mg/L\\c_{cstr}=500mg/L$

Degradation constant=k and is unknown.

We calculate the concentration through the formula,

$cc_{cstr} =\frac{c_{in}}{1+K(V/Q)} \\cc_{cstr}=\frac{c_p}{1+K*\frac{V_{csrt}}{Q}}$

Replacing values we have

$1+k(\frac{60}{10})=\frac{2500}{500}\\1+k=5\\K(6)=5-1\\K(6)=4\\K=2/3\\K=0.667/day$

That is the degradation constant of Z-contaminant

3 0

2 years ago

Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of

aleksandr82 [10.1K]

Answer:

Explanation:

Given

Temperature of solid $T=500\ K$

Einstein Temperature $T_E=300\ K$

Heat Capacity in the Einstein model is given by

$C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}$

$e^{\frac{3}{5}}=1.822$

Substitute the values

$C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})$

$C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}$

$C_v=0.97\times (3R)$

6 0

2 years ago

2. A F-22 Raptor has just climbed through an altitude of 9,874 m at 1,567 kph when a disk

BabaBlast [244]

The pressure difference across the sensor housing will be "95 kPa".

According to the question, the values are:

Altitude,

9874

Speed,

1567 kph

Pressure,

122 kPa

The temperature will be:

→ $T = 15.04-[0.00649(9874)]$

→ $= 15.04-64.082$

→ $= -49.042^{\circ} C$

now,

→ $P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}$

→ $= 27.074$

hence,

→ The pressure differential will be:

= $122-27$

= $95 \ kPa$

Thus the above solution is correct.

Learn more about pressure difference here:

brainly.com/question/15732832

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2 years ago

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alekssr [168]

Answer:

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Explanation:

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2 years ago