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TiliK225 [7]
3 years ago
12

The heat transfer coefficient decreases with increasing x for both the laminar and turbulent regions a. True b. False

Engineering
1 answer:
REY [17]3 years ago
3 0

Answer:

A) True  

Explanation:

Yes this is true when length is creases the heat transfer coefficient decease with length.

The heat transfer(h) coefficient is varying with x by given expression

For Laminar flow

h \alpha \dfrac{1}{x^{\frac{1}{2}}}

For turbulent flow

h \alpha \dfrac{1}{x^{\frac{1}{5}}}

But when flow is in transitional state the heat heat transfer(h) coefficient is increases with x.But for laminar as well as turbulent flow h is decrease when x increases.

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You have just finished your OST takeoffs for a single-story home and found 175 LF of interior walls with 2x6 studs 14" OC. The h
zimovet [89]

Answer:

Total BF for the interior wall is 7.50BD

Explanation:

Given Data:

· Size of stud = 2” x 6”

· Height of Wall = 8 ft

· Top plates = 2

· Bottom Plate = 1

BF stands for board feet in lumber/wood terminology. It is the unit of volume.

1 BF (Board feet) = 1 ft x 1 ft x 1 inch

Since there are total three plates at top and bottom, we have to deduct their thickness from wall height to calculate height of stud.

Height of stud = 8’ – 3 x 2” = 7’6” = 7.5 ft

Board feet of one stud = 7.50 6/12 x 2 = 7.50 BD

Total BF for the interior wall is 7.50BD

7 0
3 years ago
A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t
gayaneshka [121]

Answer:

Following are the proving to this question:

Explanation:

\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}

using the energy equation for entry and exit value :

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}

where

\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}

         = (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\  V^{2}_{1}\\\\

         = \frac{1}{(2f (\frac{l}{D})  )} \  V^{2}_{1}\\

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\

\to 0+0+Z_0 = 0  +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g}   \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} )  \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to  \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})

L.H.S = R.H.S

7 0
3 years ago
Define what a glass transition is and what happens to a polymer because of it
lesya692 [45]

Answer:

 The glass transition occur when the given temperature gets drop below the temperature of the glass transition for the polymers. During the creases of the long range of the given motion, the polymer start changing from the state of hard into the form of brittle.

The range of the glass transition temperature are between the 0 to -150°C. The basic use of the polymer is basically depend upon the main properties of the polymer.

7 0
3 years ago
An engineer lives in Hawaii at a location where the annual rain fall is 300 inches. She decides to use the rain to generate elec
Alex777 [14]

Answer:

80.7lbft/hr

Explanation:

Flow rate of water in the system = 3.6x10^-6

The height h = 100

1s = 1/3600h

This implies that

Q = 3.6x10^-6/[1/3600]

Q = 0.0000036/0.000278

Q = 0.01295

Then the power is given as

P = rQh

The specific weight of water = 62.3 lb/ft³

P = 62.3 x 0.01295 x 100

P = 80.675lbft/h

When approximated

P = 80.7 lbft/h

This is the average power that could be generated in a year.

This answers the question and also corresponds with the answer in the question.

4 0
3 years ago
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas
jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine (\eta), no unit, is defined by this formula:

\eta = \frac{\dot W}{\dot Q} (1)

Where:

\dot Q - Heat input, in watts.

\dot W - Power output, in watts.

If we know that \dot W = 600\,W and \eta = 0.3, then the heat input from the combustion phase is:

\eta = \frac{\dot W}{\dot Q}

\dot Q = \frac{\dot W}{\eta}

\dot Q = \frac{600\,W}{0.3}

\dot Q = 2000\,W

The heat input from the combustion phase is 2000 watts.

8 0
2 years ago
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