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Alex
3 years ago
13

A)

Physics
1 answer:
Savatey [412]3 years ago
7 0

Answer:

Distance covered is equal to all the distance traveled.

So for example, if you go from A to B, and then from B to C, the total distance covered is AB + BC.

Displacement is equal to the difference between the final position and the initial position.

So if we go from A to B, the displacement is simply the line AB.

While if we go from A to B, and then from B to C, the displacement will be a segment that directly connects A and C, such that:

displacement = √( (AB)^2 + (BC)^2)

Now, if we want to find the points such that the magnitude of the distance covered is equal to the magnitude of the displacement, we need to look at the pairs that are directly connected by a straight line.

Those are:

A to B  ( or B to A)

B to C  (or C to B)

C to D  (or D to 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

\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

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