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3 years ago

Am I alive I really need to know?

Engineering

1 answer:

Nesterboy [21]3 years ago

5 0

Hell no,cause i’m not

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Thermal energy measured by?

Harlamova29_29 [7]

Answer:

Thermal energy is measured using a thermometer denominated in Fahrenheit, Celsius and Kelvin

3 0

2 years ago

g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th

xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

3 0

3 years ago

On a piece of paper, sketch the x-y stress state and the properly oriented principal stress state. Use the resulting sketch to a

stealth61 [152]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step and very detailed solution of the given problem

6 0

4 years ago

Plzzzz helppp design process in order

MA_775_DIABLO [31]

Answer:

generate

define

present

evaluate

develop

construct and test

7 0

3 years ago

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

4 years ago