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3 years ago

14

A plate and frame heat exchanger has 15 plates made of stainless steel that are 1 m tall. The plates are 1 mm thick and 0.6 m wi

de and are spaced 4.8 mm apart. The exchanger is used to cool water. The volume flow rate of this water is 0.011 m3/s, and its inlet temperature is 30 C. Raw water is the cooling medium. It enters the exchanger at 6 C with a volume flow rate of 0.02 m3/s. Calculate the outlet temperatures.

1 answer:

hodyreva [135]3 years ago

6 0

Answer:

14.506°C

Explanation:

Given data :

flow rate of water been cooled = 0.011 m^3/s

inlet temp = 30°C + 273 = 303 k

cooling medium temperature = 6°C + 273 = 279 k

flow rate of cooling medium = 0.02 m^3/s

Determine the outlet temperature

we can determine the outlet temperature by applying the relation below

Heat gained by cooling medium = Heat lost by water

= ( Mcp ( To - 6 ) = Mcp ( 30 - To )

since the properties of water and the cooling medium ( water ) is the same

= 0.02 ( To - 6 ) = 0.011 ( 30 - To )

= 1.82 ( To - 6 ) = 30 - To

hence To ( outlet temperature ) = 14.506°C

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b) Potential difference: -26,800 V

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Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

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r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

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$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

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The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

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$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

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So the potential difference is

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Answer:

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