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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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False

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An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

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The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

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Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origi
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Answer:

\vec F_{A} = -67500\,N\cdot (i + j)

Explanation:

The position of each point are the following:

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Answer:

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