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3 years ago

Why does the moon lack an atmosphere?

1 answer:

8 0

Because when it had any, the warmth of the sun heated the atmosphere to the point where the average speed of the gas molecules exceeded the escape velocity for a body with the moon's mass and size. So, little by little, they escaped.

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A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

3 years ago

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$

The time independent constant is 16.0787 s

$E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}$

At t = 6

$\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}$

From the first equation

$\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451$

$E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa$

$E_r(6)=4.35614\ MPa$

6 0

3 years ago

The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr

Mandarinka [93]

Answer:

The magnitude of minimum potential difference is 1800 V

Explanation:

Given:

Electric field $E = 3 \times 10^{6} \frac{V}{m}$

Gap between electrodes $d = 0.060 \times 10^{-2}$ m

For finding the minimum potential difference,

$\Delta V = E \times d$

$\Delta V = 3 \times 10^{6} \times 0.060 \times 10^{-2}$

$\Delta V = 1800$ V

Therefore, the magnitude of minimum potential difference is 1800 V

8 0

3 years ago

What are solar flares

Bad White [126]

It's a brief of intense high energy radiation from the sun's surface. When it happens it disrupts the earth's communication and power line transmissions

8 0

3 years ago

Read 2 more answers

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min

Yuki888 [10]

Answer:

A. To find the mass flow rate.

We use= 220 x 0.355/ 60

= 1.3kg/s

B. Volume flowrate is = mass flowrate / density

But density is 1000kg/m³

= 1.3kg/s/ 1000kg/m³

= 0.0013m³/s

C. Flow speead at 1

= 0.0013m³/s / (2 x 10-2m)²

= 6.5m/s

D.flow speed at 2

0.0013m³/s / (8x 10-2m)²

=1.63m/s

E. Gauge pressure at point 1

= 152+ 1/1000 ( 1.63)²- 6.5² + 1000( 9.8) ( 0-1.35)

= 119kpa

7 0

3 years ago