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svet-max [94.6K]
3 years ago
7

A strain gauge pressure sensor has the following specification. Will it be suitable for the measurement of pressure of the order

of 100 kPa to an accuracy of 5 kPa in an environment where the temperature is reasonably constant at about 20 degree C? Ranges: 2 to 70 MPa, 70 kPa to 1 MPa Excitation: 10 V d.c. or a.c. (r.m.s.) Full range output: 40 mV Non-linearity and hysteresis errors: 0.5% Temperature range: - 54 to +120 degree C
Engineering
1 answer:
brilliants [131]3 years ago
4 0

Answer: the measurement of pressure of the order of 100 KPa to an accuracy of +- 5Kpa is POSSIBLE with the sensor

Explanation:

Given that;

specifications include the range 70 kPa to 1 MPa which includes 100 kPa.

specification include non-linearity and hysteresis error of 0.5%

first we find the non-linear error

= 1 MPa * 0.5%

= 1 MPa * 0.005

= 0.005 MPa = 5 KPa

therefore, it includes the non-linearity error of +- 5 KPa.

the specification include the temperature range -54 to 120 degree Celsius which includes 20 degree Celsius

THEREFORE, the measurement of pressure of the order of 100 KPa to an accuracy of +- 5Kpa is POSSIBLE with the sensor

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How did the gene combination result in the different traits?
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5 0
3 years ago
A 40 mph wind is blowing past your house and speeds up as it flows up and over the roof. If the elevation effects are negligible
Finger [1]

The pressure at the point on the roof where the speed is 60mph is 14.66 psi

<u>Explanation:</u>

The question follows Bernoulli's equation

P1 + 1/2 ρ V1² = P2 + 1/2 ρ V2²

V1 = 40 X 5280 / 3600 = 58.7 ft/s

V2 = 60 X 5280 / 3600 = 88 ft/s

P2 = P1 + 1/2 ( 0.00238 ) [(58.7)² - (88)²]

P2 - P1 = -5.12lb/ft²

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We know,

P1 = 14.7 psi = 2116.8 lb/ft²

P2 = 2116.8 - 5.12 lb/ft²

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7 0
3 years ago
Consider a mild steel specimen with yield strength of 43.5 ksi and Young's modulus of 29,000 ksi. It is stretched up to a point
mezya [45]

Answer:

0.05%

Explanation:

From the question, we have;

The yield strength of the mild steel, \sigma _c = 43.5 ksi

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The inelatic strain \epsilon_c^{in} is given as follows;

\epsilon_c^{in} = \epsilon _c - \sigma _c/∈

Therefore, we have;

\epsilon_c^{in} = 0.002 - 43.5/(29,000) = 0.0005

Therefore, the inelastic strain, \epsilon_c^{in} = 0.0005 = 0.05%

Taking the inelastic strain as the residual strain, we have;

The residual strain = 0.05%

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