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bogdanovich [222]
3 years ago
13

Subcooled liquid water flows adiabatically in a constant diameter pipe past a throttling valve that is partially open. The liqui

d can be modeled as an incompressible flow (i.e., specific volume (v) is constant, u=CAT). The subscripts i and e refer to the conditions at the inlet and exit of the valve, respectively.
Indicate whether each difference below is positive, negative, or zero.


hi-he __________


pi-pe __________


ui-ue __________


ti-te ___________
Engineering
1 answer:
Llana [10]3 years ago
5 0

Answer:

hi-he = 0

pi-pe  = positive

ui-ue = negative

ti-te = negative

Explanation:

we know that fir the sub cool liquid water is

dQ = Tds = du +  pdv   ............1

and  Tds = dh - v dP         .............2

so now for process of throhling is irreversible when v is constant

then heat transfer is = 0 in irreversible process

so ds > 0

so here by equation 1 we can say

ds  > 0  

dv = 0 as v is constant

so that Tds = du    .................3

and du > 0

ue - ui > 0

and

now by the equation 2 throttling process  

here enthalpy is constant

so dh = 0

and Tds  = -vdP

so ds > 0  

so that -vdP > 0  

as here v is constant

so -dP =P1- P2

so P1-P2 > 0

so pressure is decrease here

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What is engineering?
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3 years ago
A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
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Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603                                         225

.203                                         368

.162                                          442

answer : 191 mPa

Explanation:

<u>Determine the stress present at the time of fracture for the original plate</u>

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603  bending stress = 225

at 0.203  bending stress = 368

at 0.162  bending stress = 442

<u>applying equation 1   determine the value of n for several combinations</u>

 ( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

 ( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) =  ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x =  0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

     = 225 * 0.84747

     <em>=  191 mPa </em>

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What standard feature is on all circular saws?

4. All of the above

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