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lapo4ka [179]
2 years ago
14

Two fluids, A and B exchange heat in a counter – current heat exchanger. Fluid A enters at 4200C and has a mass flow rate of 1 k

g/s. Fluid B enters at 200C and also has a mass flow rate of 1 kg/s, Effectiveness of heat exchanger is 75%. Determine the heat transfer rate and exit temperature of fluid B.
Engineering
1 answer:
Volgvan2 years ago
5 0

Answer:

Your question has some missing information below is the missing information

Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )

answer : 300 kW , 95°c

Explanation:

Given data:

Fluid A ;

Temperature of Fluid ( Th1 )  = 420° C

mass flow rate (mh)  = 1 kg/s

Fluid B :

Temperature ( Tc1) = 20° C

mass flow rate ( mc ) = 1 kg/s

effectiveness of heat exchanger = 75% = 0.75

<u>Determine the heat transfer rate and  exit temperature of fluid</u> <u>B</u>

Cph = 1000 J/kgk

Cpc = 4000 J/Kgk

Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B

exit temp of fluid  B = 95°C

heat transfer = 300 kW

attached below is a the detailed solution

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b) Potential difference: -26,800 V

c) Work: 4.3\cdot 10^{-15} J

Explanation:

a)

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

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

where

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q is the charge

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

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

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b)

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

(x = 0.40 m, y = 0.30 m)

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\Delta V is the potential difference

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q=-1.6\cdot 10^{-19}C is the charge of the electron

\Delta V=-26,800 V is the potential difference

Therefore, the work required on the electron is

W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J

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