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svetlana [45]
3 years ago
12

If the radiant energy from the Sun comes in as a plane EM wave of intensity 1340 W/m2, calculate the peak values of E and B

Physics
1 answer:
Nastasia [14]3 years ago
5 0

Answer:

Given that

I = 1340 W/m²

µo = 4πx 10^-7 Tm/A

sc = 3 x 10^8 m/s

So to find the peak values of B we use

I = (c Bm²) / (2 µo)

1340 = (3 x 10^8 x Bm²) / (2 x 4πx 10^-7)

1340 = (3 x 10^8 x Bm²) / (25.12 x 10^-7)

(3 x 10^8 x Bm²) = 1340 x (25.12 x 10^-7)

(3 x 10^8 x Bm²) = 3.366 x 10^-3

Bm² = 3.366 x 10^-3 / 3 x 10^8

Bm² = 1.122 x 10^-11

Bm = 3.4 x 10^-6 T

Also to find the peak values of E we

use Em = c x Bm

Em = (3 x 10^8) (3.4 x 10^-6)

Em = 1.005 N/C

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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
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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

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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

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3 years ago
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A step index fiber has a numerical aperture of NA = 0.1. The refractive index of its cladding is 1.465. What is the largest core
Vladimir [108]

Answer:

diameter = 9.951 × 10^{-6} m

Explanation:

given data

NA = 0.1

refractive index = 1.465

wavelength = 1.3 μm

to find out

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solution

we know that for single mode v number is

V ≤ 2.405

and v = \frac{2*\pi *r}{ wavelength} NA

here r is radius    

so we can say

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so diameter is = 2  ×  4.975979 × 10^{-6} m

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