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PPT – Topic 2:Heat Exchanger Fundamentals, Recuperative Heat Exchangers PowerPoint presentation | free to download - id: 61dd52-YzAwZ

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Topic 2 Heat Exchanger Fundamentals,

Recuperative Heat Exchangers

- Heat Exchangers
- UA-LMTD Design Method
- e -NTU Design Method
- An Example

- Heat Exchangers

A heat exchanger is a device for transferring

heat from one fluid to another. There are three

main categories Recuperative, in which the two

fluids are at all times separated by a solid

wall Regenerative, in which each fluid transfers

heat to or from a matrix of material Evaporative

(direct contact), in which the enthalpy of

vaporization of one of the fluids is used to

provide a cooling effect.

Heat Exchanger (HX) Design Methods

HX designers usually use two well-known methods

for calculating the heat transfer rate between

fluid streamsthe UA-LMTD and the

effectiveness-NTU (number of heat transfer units)

methods. Both methods can be equally employed for

designing HXs. However, the ?-NTU method is

preferred for rating problems where at least one

exit temperature is unknown. If all inlet and

outlet temperatures are known, the UA-LMTD method

does not require an iterative procedure and is

the preferred method.

LMTD (Log Mean Temperature Difference)

The most commonly used type of heat exchanger is

the recuperative heat exchanger. In this type the

two fluids can flow in counter-flow, in

parallel-flow, or in a combination of these, and

cross-flow.

The true mean temperature difference is the

Logarithmic mean Temperature difference (LMTD),

is defined as

Heat Transfer Rate of a Heat Exchanger

The heat transferred for any recuperative heat

exchanger can be calculated as(refer to the

diagram shown on the previous slide)

Heat Exchanger UA-LMTD Design Method

Where U is the overall heat transfer coefficient

(and is assumed to be constant over the whole

surface area of the heat exchanger).

Heat Exchanger e-NTU Design Method

6-row, 6-pass plate fin-and-tube cross

counterflow HX

3-row, 3-pass plate fin-and-tube crossflow HX

Effectiveness of a 6-row, 6-pass plate

fin-and-tube cross counterflow HX

A HX Example

A schematic representation of a hybrid central

receiver is shown in the following slide (Slide

12). In this system, molten nitrate salt is

heated in a central receiver to temperature as

high as 1,050oF (565oC). The molten salt is then

passed through a heat exchanger, where it is used

to preheat combustion air for a combined-cycle

power plant. For more information about this

cycle, refer to Bharathan et. Al. (1995) and Bohn

et al. (1995). The heat exchanger used for this

purpose is shown in Slide 13. The plates of the

heat exchanger are made of steel and are 2 mm

thick. The overall flow is counter-flow

arrangement where the air and molten salt both

flow duct-shape passages (unmixed). The shell

side, where the air flow takes place, is baffled

to provide cross flow between the lateral

baffles. The baseline design conditions are

A hybrid central-receiver concept developed at

the NREL

Molten-salt-to-air HX used to preheat combustion

air

A HX Example (continued)

Air flow rate 0.503 kg/s per passage (250

lbm/s) Air inlet temperature 340oC

(650oF) Air outlet temperature 470oC

(880oF) Salt flow rate 0.483 kg/s per

passage (240 lbm/s) Salt inlet temperature

565oC (1050oF) Salt outlet temperature

475oC (890oF) Find the overall heat-transfer

coefficient for this heat exchanger. Ignore the

fouling resistances.

Solution

Solution (continued)

Solution (continued)

Solution (continued)

Solution (continued)

Solution (continued)

- Recuperative Heat Exchangers
- Definition of Recuperative HX
- Types of Recuperative HX
- Design Factors
- Examples

A Recuperative Heat Exchanger (HX) is one in

which the two fluids are separated at all times

by a solid barrier.

Waste-Heat water-Tube Boiler

Shell Boiler using Waste Gas

Furnace Gas Air Pre-Heater

Two-Pass Shell-and-Tube Heat Exchanger

Gas-to-Gas Heat Recovery with a Plate-Fin Heat

Exchanger

A

A

Liquid-to-Liquid Plate-Fin Heat Exchanger

Basic Equations

Heat Exchanger Configurations

Extended Surfaces Fins, fpi (fins per inch)

Example 5.4 (Eastop Croft) Fin Surface

Example 5.4

A flat surface as shown in the previous slide has

a base temperature of 90oC when the air mean bulk

temperature is 20oC. Air is blown across the

surface and the mean heat transfer coefficient is

30 W/m2-K. The fins are made of an aluminum

alloy the fin thickness is 1.6 mm, the fin

height is 19 mm, and the fin pitch is 13.5 mm.

Calculate the heat loss per m2 of primary surface

with and without the fins assuming that the same

mean heat transfer coefficient applies in each

case. Neglect the heat loss from the fin tips and

take a fin efficiency of 71.

Example 5.4 (continued)

?-NTU Method (Effectiveness Number of Thermal

Units Method)

?-NTU (Effectiveness against NTU) for

shell-and-tube heat exchangers

(with 2 shell passes and 4, 8, 12 tube passes)

Characteristics of ?-NTU Chart

- For given mass flow rates and specific heats of

two fluids the value of ? depends on the NTU and

hence on the product (UAo). Thus for a given

value of U the NTU is proportional to Ao. It can

then be seen from the ?-NTU chart that increasing

Ao increases ? and hence the saving in fuel. - The capital cost of the heat exchanger increases

as the area increases and ?-NTU chart shows that

at high values of ? large increase in area

produce only a small increase in ?. - The NTU and hence the effectiveness, ? can be

increased for a fixed value of area by increasing

the value of the overall heat transfer

coefficient, U.

Increasing HX ? with Fixed Ao (1)

The NTU, and hence ? can be increased for a fixed

value of the area by increasing the value of the

overall heat transfer coefficient, U, which can

be increased by increasing the heat transfer

coefficient for one or both of the individual

fluids.

The heat transfer coefficient can be increased by

reducing the tube diameter, and/or increasing the

mass flow rate per tube.

Increasing HX ? with Fixed Ao (2)

- Since , for a constant total mass

flow rate the number of tubes per pass must be

reduced correspondingly if the mass flow rate per

tube is increased. - Also, the heat transfer area is given by

, where n is the number of tubes per

pass, and p is the number of tube passes.

Therefore, to maintain the same total heat

transfer area for a reduced tube diameter in a

given type of heat exchanger, it is necessary to

increase the length of the tubes per pass, L,

and/or the number of tubes per pass (which will

reduce the heat transfer rate.) - The design process is therefore an iterative

process in order to arrive at the optimum

arrangement of tube diameter, tube length, and

number of tubes.

Overall HX Design Considerations

- Altering the inside diameter of a tube to

increase the heat transfer coefficient for flow

through the tube will alter the heat transfer on

the shell side. - A full economic analysis also requires

consideration of the pumping power for both

fluids. Pressure losses in fluid flow due to

friction, turbulence, and fittings such as

valves, bends etc. are proportional to the square

of the flow velocity. The higher the fluid

velocity and the more turbulent the flow the

higher is the heat transfer coefficient but the

greater the pumping power.

Example 5.5

(a) A shell-and-tube heat exchanger is used to

recover energy from engine oil and consists of

two shell passes for water and four tube passes

for the engine oil as shown diagrammatically in

the following figure. The effectiveness can be

calculated based on Eastop Equation (3.33). For a

flow of oil of 2.3 kg/s entering at a temperature

of 150oC, and a flow of water of 2.4 kg/s

entering at 40oC, use the data given to

calculate (i) the total number of tubes

required (ii) the length of the tubes (iii) the

exit temperatures of the water and oil (iv) the

fuel cost saving per year if water heating is

currently provided by a gas boiler of efficiency

0.8. (b) What would be the effectiveness and fuel

saving per year with eight tube passes?

Example 5.5 (continued)

Use EES for Eastop Example 5.5" hot

oil m_dot_H2.3 oil mass flow rate,

kg/s t_H1150 hot oil inlet temperature,

C cp_H2.19 mean spefici heat of oil,

kJ/kg-K rho_H840 mean oil density,

kg/m3 C_Hm_dot_Hcp_H hot fluid capacity,

kW/K cold water m_dot_C2.4 water mass flow

rate, kg/s t_C240 cold water inlet

temperature, C cp_C4.19 mean specific heat

of water, kJ/kg-K C_Cm_dot_Ccp_C cold fluid

capacity, kW/K data eta_boiler0.8 gas

boiler efficiency v_H0.8 oil velocity in the

tube, m/s eta_Hx0.7 require HX

effectiveness n_pass4 four pass heat

exchanger d_i0.005 tube inside diameter,

m d_o0.007 tube outside diameter,

m U0.400 overall heat transfer coefficient,

kW/m2-K t_hours4000 annual usage,

h cost1.2 cost of water heating, p/kWh

- a(i) calculate the total number of tubes

required - V_dot_Hm_dot_H/rho_H
- A_crossV_dot_h/v_H
- A_1PId_i2/4
- n_tubeA_cross/A_1n_pass
- n_tube697 tubes
- a(ii) calculate the length of the tubes
- Rmin(C_H,C_C)/max(C_H,C_C)
- eta_HX(1-exp(-NTU(1-R)))/(1-Rexp(-NTU(1-R)))
- NTUUA_o/min(C_H,C_C)
- A_oPId_on_tubeL_tube
- L_tube1.27 m
- a(iii) calculate the exit temperature of oil

and water - Eta_HXC_H(t_H1-t_H2)/(min(C_h,C_C)(t_H1-t_C2))
- C_H(t_H1-t_H2)C_C(t_C1-t_C2)
- t_C178.6 C t_H273.0 C

- a(iv) calculate the total heat transfer and

fuel cost saving per year - Q_dotC_C(t_C1-t_C2)
- Fuel_savingQ_dott_hourscost/(eta_boiler100)
- Fuel_saving23271 British Pounds
- b(v) calculate eta2_hx if double Ao
- NTU22NTU
- eta2_HX(1-exp(-NTU2(1-R)))/(1-Rexp(-NTU2(1-R))

) - eta2_HX0.881
- b(vi) calulate t_H2, t_C2 and fuel_saving
- eta2_HXC_H(t_H1-t2_H2)/(min(C_h,C_C)(t_H1-t_C2)

) - C_H(t_H1-t2_H2)C_C(t2_C1-t_C2)
- Q2_dotC_C(t2_C1-t_C2)
- Fuel2_savingQ2_dott_hourscost/(eta_boiler100)
- Fuel2_saving29279 British Pounds