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3 years ago

5

Find the total amount of heat in Q lost through a wall 10' by 18' , with R value from q. 1. Inside temperature is 70 degrees F w

hile outside is 45 degrees F

2 answers:

marissa [1.9K]3 years ago

6 0

Answer:

Just think

Explanation:

gogolik [260]3 years ago

5 0

That is really easy just think about it

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The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

5 0

3 years ago

Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an

Svet_ta [14]

Answer:

2.46 * 10⁵ W/m³

Explanation:

See attached pictures for detailed explanation.

6 0

3 years ago

Read 2 more answers

A driver takes 3.5 s to react to a complex situation while traveling at a speed of 60 mi/h. How far does the vehicle travel befo

victus00 [196]

Answer:

x = 93.8 m.

Explanation:

During the entire the reaction time interval, the vehicle continues moving at the same speed that it was moving, i.e., 60 mi/hr.

In order to calculate the distance in meters, travelled at that speed, it is advisable first to convert the 60 mi/hr to m/seg, as follows:

$60 mi/hr = 60*\frac{1hr}{3,600s}*\frac{1,605m}{1mi} = 26.8 m/s$

Applying the definition of average velocity, we can solve for Δx, as follows:

Δx = 26.8 m/s* 3.5 s = 93.8 m

7 0

2 years ago

(c) As Engineering and Computing students, you must be familiar, with the respective professional bodies, as well as, Rules of P

guajiro [1.7K]

The professional ethics for computer engineers are:

They will Contribute to society and to human well-being.
They will Avoid harm.
Be honest and trustworthy.
They will be fair and take action that do to discriminate others.

<h3>What are the Characteristics of Code of Ethics?</h3>

The code of ethics are known to be a kind of a universal moral values, that is one that state that what a person expect of any given employee such as been trustworthy, respectful, responsible, and others.

Note that Rules of Practice, Professional Obligations and Codes of Ethics. are known to be put in place to avoid issues that may lead to conflict.

Therefore, i believe that As Engineering and Computing students, the respective professional bodies, Rules of Practice, Professional Obligations and Codes of Ethics are good and acts as a check and balance to us.

Therefore, The professional ethics for computer engineers are:

They will Contribute to society and to human well-being.
They will Avoid harm.
Be honest and trustworthy.
They will be fair and take action that do to discriminate others.

Learn more about Engineering rules from

brainly.com/question/17169621

#SPJ1

4 0

1 year ago

A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of

Burka [1]

Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end + area of the curve surface

$=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}$

= 2.136 *10^-4 $m^{2}$

Heat flux = $\frac{0.6}{2.136 * 10^{-4} }$ = 2808.99 $W/m^{2}$

fraction of heat dissipated from the top and bottom surface

$=\frac{\frac{2*\pi D^{2} }{4} }{\frac{2*\pi D^{2}}{4} + \pi DL } \\\\=\\\frac{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} }{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} +\pi *\frac{0.4}{100} *\frac{1.5}{100} } \\\\=\frac{2.51*10^{-5} }{2.136*10^{-4} } \\\\\= 0.1175$

=11.75%

8 0

3 years ago

Read 2 more answers