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3 years ago

11

Atmospheric pressure is measured to be 14.769 psia. a. What would be the equivalent reading of a water barometer (inches of H20)

? b. What would be the equivalent reading of a Mercury barometer (mm of Hg)?

1 answer:

Fofino [41]3 years ago

6 0

Answer:

(a) water height =408.66 in.

(b) mercury height=30.04 in.

Explanation:

Given: P=14.769 psi ( 1 psi= 6894.76 $\frac{N}{m^2}$ )

we know that $P=\rho\times g\timesh$

where $\rho =Density,g=9.81\frac{m}{s^2}$

h=height.

Given that P=14.769 psi ⇒P= 101828.6 7 $\dfrac{N}{m^2}$

(a) $P=\rho_{w}\times g\times h_{w}$

$\rho_{w}=1000\frac{Kg}{m^3}$

⇒101828.67= $1000\times 9.81\times h_{w}$

$h_{w}$ =10.38 m

So water barometer will read 408.66 in. (1 m=39.37 in)

(b) $P=\rho_{hg}\times g\times h_{hg}$

$\rho_{hg}$ =13600

So 101828.67= $13600\times 9.81\times h_{hg}$

$h_{hg}$ =0.763 m

So mercury barometer will read 30.04 in.

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Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

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True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .

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$Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n$

$\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }$

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$l_i$ is the instantaneous length

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$l_i = 470 * e^{0.0443}$

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