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Aleonysh [2.5K]
4 years ago
12

3. Write down the total thermal resistance for a double-pipe heat exchanger. Show how to convert from total resistance to an ove

rall heat-transfer coefficient. Explain how flow regime may affect overall heat transfer rate.

Engineering
2 answers:
wlad13 [49]4 years ago
8 0

Answer: please check image for detailed calculation and explanation.

At the boundary between the pipes, there is discontinuity due to different regime; this gap can insulate heat flow causing a fluctuation in heat flow from the inner pipe to the outer pipe.

Nutka1998 [239]4 years ago
8 0

Answer:

\frac{1}{UA} =\frac{1}{A_{0}h_{0}  } +\frac{Rf_{0} }{A_{0} } +\frac{1}{2k\pi L} ln(\frac{di,0}{di,i} )+\frac{Rf_{i} }{A_{i} }

Explanation:

Global heat transfer determines the heat that is transferred from the inner tube to the outer tube. In the double tube heat exchanger, this global heat transfer coefficient is equal to:

\frac{1}{UA} =\frac{1}{A_{0}h_{0}  } +\frac{Rf_{0} }{A_{0} } +\frac{1}{2k\pi L} ln(\frac{di,0}{di,i} )+\frac{Rf_{i} }{A_{i} }

where subscript i refers to the inner surface and subscript 0 refers to the outer surface

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A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.
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Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

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Specific gravity G_{s} = \frac{P_s}{P_w} =  \frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)

% Saturation S = \frac{V_w}{Vv} × \frac{100}{1} =  \frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right} × \frac{100}{1}--------(3)

water content w =  \frac{M_w}{M_s} = \frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w =  \frac{M_w}{Ms}

Recall; mass = Density × volume

w = \frac{V_wP_w}{V_sP_s} ------(5)

From eqn. (2)  G_{s} = \frac{P_s}{P_w}

∴ \frac{1}{G_s} = \frac{P_w}{P_s} ------(6)

putting eqn. (6) into (5)

w = \frac{V_w}{V_sG_s} -----(7)

Again, from eqn (1)

V_s = \frac{V_v}{e}

substituting into eqn. (7)

w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v}  = S

∴ w = \frac{Se}{G_s} -----(8)

With eqn. (7), we can calculate

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when S = 100% = 1

Given, G_{s} = 2.7, e= 0.9

∴w= \frac{0.9*1}{2.7} = 0.333

∴ %w = 33.3%

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when S = 1% = 0.01

w= \frac{0.01*0.9 }{2.7} = 0.0033

∴ % w = 0.33%

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%S = \frac{V_w}{V_v} × \frac{100}{1} = \frac{V_w}{eV_s} × \frac{100}{1}

0.50 = \frac{V_w}{0.9* 100}  = 45cm^{3}

mass of water = P_wV_w= 1 * 45 = 45_{g}

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