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A heat exchanger is to heat water (Cp= 4.18 KJ/Kg.C ) from 25 to 60 C...

A heat exchanger is to heat water (Cp= 4.18 KJ/Kg.C ) from 25 to 60 C at rate of 0.2 kg/s. the heating is to be acomplished by geothermal water ( Cp= 4.31 KJ/Kg.c ) available at 140C at a mass flow rate of 0.3Kg/s. Determine the rate of heat transfer in the heat exchanger and the exit temprature of geothermal water.

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Concepts and reason

First Law of thermodynamics:

For a closed thermodynamic system, the variation in internal energy is equivalent to the difference in the heat energy supplied to the system and the work done by the system.

ΔΕ = Q-W

Internal energy:

The total amount of microscopic energies such as nuclear energy, bonding energy, kinetic energy of the molecules etc. possessed by a system itself without any external influences is termed as internal energy. It is denoted by U.

Heat exchanger:

It is a device which is used to transfer heat between two fluids operating at different temperature levels. In heat exchangers, work cannot be achieved (W =0)
.

Substituting the inlet and outlet temperatures of water in the expression for rate of heat transfer in constant pressure process, the rate of heat transferred to the water can be calculated. By thermal equilibrium, the rate of heat transferred to the water is said to be equivalent to the rate of heat supplied by the geothermal water. Then, exit temperature of the geothermal water is calculated by substituting the rate of heat transfer in heat exchanger in the expression to calculate the rate of heat transfer in constant pressure process.

Fundamentals

Write the expression to calculate the change in internal energy for a constant pressure process.

AE = mc, (Tz -)

Here, mass of the system is , specific heat constant for constant pressure process is , and the initial and final temperatures of the system are and respectively.

The relation for First law of thermodynamics is expressed as follows:

Since no work is said to be done in heat exchanger, substitute for .

ΔΕ = 0-0
0= ΔΕ

Substitute me,(T2 – T)
for ,

Q=mc, (T2 – T)

Rewrite the above expression to calculate the rate of heat transfer for fluid flowing through control volume.

Q=mc, (T2 – T)
…… (1)

Here, mass flow rate of the fluid is and temperatures of the fluid at inlet and outlet of control volume are and respectively.

The value of specific heat capacity of water at constant pressure is 4.18kJ/kg.°C
.

Calculate the rate of heat transferred to the water.

(am) – (22) *s*u=0

Here, mass flow rate of water is , specific heat capacity of water is , inlet and exit temperatures of water are and respectively.

Substitute 0.2 kg/s
for , 4.18 kJ/kg-°C
for , for and for .

Q = 0.2 kg/sx 4.18 kJ/kg. °C(60°C -25°C)
Q = 29.26 kJ/s

Write the expression to calculate the rate of heat transferred from geothermal water.

Q=mçc,(Tivo) -T20)

Here, mass flow rate of geothermal water is , inlet and exit temperatures of geothermal water are and respectively.

Substitute 0.3kg/s
for , 4.18 kJ/kg.C
for , 140°C
for and 29.26 kJ/s
for .

140°C - Tecido
29.26 kJ/s = 0.3kg/sx 4.18kJ/kg-°C(140°C-Teil)
29.26 kJ/s
0.3kg/sx 4.18 kJ/kg. °C
Teie = 140°C - 22.63°C
Tein

Ans:

The rate of heat transfer in the heat exchanger is 29.26 kJ/s
.

The exit temperature of geothermal water (720)
is 117.37°C
.

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