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3 years ago

Which comment is someone who has a conventional personality type likely to make?

Engineering

2 answers:

Trava [24]3 years ago

4 0

Answer:

“How can I help?”

Though a second could be “Don’t tell me; show me.”, but that is very authoritative , while conventional personalities types don't typically like to be in an authoritative position.

Explanation:

svetlana [45]3 years ago

4 0

Answer:

just do it

Explanation:

I mean, its Nike's trademark.

You might be interested in

Decide whether the function is an exponential growth or exponential decay function, and find the constant percentage rate of gro

sasho [114]

Answer:

Just answered this to confirm my profile.

Explanation:

I dont have a clue, this is just to confirm my profile.

8 0

3 years ago

2. A counter flow tube-shell heat exchanger is used to heat a cold water stream from 18 to 78oC at a flow rate of 1 kg/s. Heatin

Anastaziya [24]

Answer:

a) $L = 220\,m$ , b) $U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

Explanation:

a) The counterflow heat exchanger is presented in the attachment. Given that cold water is an uncompressible fluid, specific heat does not vary significantly with changes on temperature. Let assume that cold water has the following specific heat:

$c_{p,c} = 4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C}$

The effectiveness of the counterflow heat exchanger as a function of the capacity ratio and NTU is:

$\epsilon = \frac{1-e^{-NTU\cdot(1-c)}}{1-c\cdot e^{-NTU\cdot (1-c)}}$

The capacity ratio is:

$c = \frac{C_{min}}{C_{max}}$

$c = \frac{(1\,\frac{kg}{s} )\cdot(4.186\,\frac{kW}{kg^{\textdegree}C} )}{(1.8\,\frac{kg}{s} )\cdot(4.30\,\frac{kW}{kg^{\textdegree}C} )}$

$c = 0.541$

Heat exchangers with NTU greater than 3 have enormous heat transfer surfaces and are not justified economically. Let consider that $NTU = 2.5$ . The efectiveness of the heat exchanger is:

$\epsilon = \frac{1-e^{-(2.5)\cdot(1-0.541)}}{1-(2.5)\cdot e^{-(2.5)\cdot (1-0.541)}}$

$\epsilon \approx 0.824$

The real heat transfer rate is:

$\dot Q = \epsilon \cdot \dot Q_{max}$

$\dot Q = \epsilon \cdot C_{min}\cdot (T_{h,in}-T_{c,in})$

$\dot Q = (0.824)\cdot (4.186\,\frac{kW}{^{\textdegree}C} )\cdot (160^{\textdegree}C-18^{\textdegree}C)$

$\dot Q = 489.795\,kW$

The exit temperature of the hot fluid is:

$\dot Q = \dot m_{h}\cdot c_{p,h}\cdot (T_{h,in}-T_{h,out})$

$T_{h,out} = T_{h,in} - \frac{\dot Q}{\dot m_{h}\cdot c_{p,h}}$

$T_{h,out} = 160^{\textdegree}C + \frac{489.795\,kW}{(7.74\,\frac{kW}{^{\textdegree}C} )}$

$T_{h,out} = 96.719^{\textdegree}C$

The log mean temperature difference is determined herein:

$\Delta T_{lm} = \frac{(T_{h,in}-T_{c, out})-(T_{h,out}-T_{c,in})}{\ln\frac{T_{h,in}-T_{c, out}}{T_{h,out}-T_{c,in}} }$

$\Delta T_{lm} = \frac{(160^{\textdegree}C-78^{\textdegree}C)-(96.719^{\textdegree}C-18^{\textdegree}C)}{\ln\frac{160^{\textdegree}C-78^{\textdegree}C}{96.719^{\textdegree}C-18^{\textdegree}C} }$

$\Delta T_{lm} \approx 80.348^{\textdegree}C$

The heat transfer surface area is:

$A_{i} = \frac{\dot Q}{U_{i}\cdot \Delta T_{lm}}$

$A_{i} = \frac{489.795\,kW}{(0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C} )\cdot(80.348^{\textdegree}C) }$

$A_{i} = 9.676\,m^{2}$

Length of a single pass counter flow heat exchanger is:

$L =\frac{A_{i}}{\pi\cdot D_{i}}$

$L = \frac{9.676\,m^{2}}{\pi\cdot (0.014\,m)}$

$L = 220\,m$

b) Given that tube wall is very thin, inner and outer heat transfer areas are similar and, consequently, the cold side heat transfer coefficient is approximately equal to the hot side heat transfer coefficient.

$U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

5 0

3 years ago

A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 5 ft from the right edge.

Anon25 [30]

That is too hard but u got that cuz i believe in you!!!

3 0

3 years ago

- WHEN YOU ARE TOWING A TRAILER:

zheka24 [161]

Answer:

And Im still going with B..

7 0

3 years ago

Marble A is placed in a hollow tube, and the tube is swung in a horizontal plane causing the marble to be thrown out. As viewed

Vilka [71]

Answer:

<em>The direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

Explanation:

The question is incomplete. So the picture, which is missing in question, is attached for your review.

As it can be seen in the picture, the ball coming out of the tube will have two components of velocity. One is along the length of tube (because ball is moving in that direction and is coming out from the hole), other is velocity component will be perpendicular to the tube (because the ball is made to move in that direction as the tube is rolling on the surface).

<em>So, taking the resultant of two vectors of velocity, the resultant direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

6 0

3 years ago