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SUSTAINABLE ENERGY UTILISATION – SEU-LAB4

DETERMINING HEAT TRANSFER COEFFICIENTS FOR A FINNED-TUBE HEAT EXCHANGER WITH FORCED

CONVECTION

Before the lab lesson, you should be able to answer the following questions:

1. What is meant by U and αf in eq.3?

2. In eqn.4, which is the main contribution to the heat transfer coefficient? Is it from convection, diffusion or radiation? On what do you base your answer?

3. Is the length used to calculate the average and the inner tube areas, Am and A1,

in eqn. 3, the same as the one used to calculate the boiling number in eq. 10? Explain.

4. Is the refrigerant flow in the condenser the same as in the finned coil

evaporator? Explain.

5. How is the fin efficiency ξ determined? It is strongly recommended that you make necessary calculations and draw the line to determine fin efficiency before you come to the lab lesson.

6. How is the heat transfer coefficient αf calculated according to 8.55 in [1]1?

1 Granryd et al., 2008, Refrigerating Engineering, Dept of Energy Technology, KTH, Stockholm, Sweden

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SUSTAINABLE ENERGY UTILISATION – LAB LESSON NO. 4

DETERMINING HEAT TRANSFER COEFFICIENTS FOR A

FINNED-TUBE HEAT EXCHANGER WITH FORCED CONVECTION

OBJECTIVE: The objective of this lab lesson is to determine the overall heat transfer coefficient and the heat transfer coefficient related to the fin surface of a heat exchanger placed in an air duct with forced convection and cooled on the inside by boiling refrigerant. The theory below is mainly based on Chapter 8 in the textbook Refrigerating Engineering by Granryd et al., 2002 [1]. Bring the book to the lab lesson.

THEORY: The following equations describe the heat transfer for a finned-tube heat exchanger (compare with figure 2):

Q.

= U. A. ϑ ϑ = Q.

/(U. A) (1a)

Q.

= α1. A1.ϑ 1 ϑ 1= Q.

/(α1. A1) (1b)

Q.

= (λt/ δt). Am.ϑ t ϑ t= Q.

.δt /(λt. Am) (1c)

Q.

= (α2. A2 + αf. ξ. Af).ϑ 2 ϑ 2 = Q.

/(α2. A2 +αf . ξ. Af) (1d)

The overall heat transfer is given by eq.1a, while eqns. 1b-1d give the heat transfer on the inside tube, through the tube wall and on the finned outside. The temperature difference over the heat exchanger can be expressed as:

ϑ = ϑ 1+ϑ t+ϑ 2 (2)

The heat transfer for a finned-tube heat exchanger can then, if we assume that α2=αf, be expressed as:

AU .1

= )..(

1

2 AA ffξα +

+ Amt

t

.λδ +

A11.1

α (3)

The surfaces are as follows: Af is the finned surface. A2 is the free outer tube area between the fins. Am and A1 are the average and inner tube areas. Finally, A = (A2+ Af) is the overall surface area (for values, see page 4). The overall heat transfer coefficient

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U is referred to A. The other heat transfer coefficients α1, α2 and αf are referred to A1, A2 and Af. Further, is the thermal conductivity of the tube; δt = (d2-d1)/2 is the tube wall thickness and ξ is the fin efficiency (compare with figure 2).

The heat transfer coefficient between the air and the outer surface of the finned-tube heat exchanger, Af, refer to the total heat transfer coefficient, αf, which is the sum of heat transfer coefficients from convection (αc), diffusion (αd), and radiation (αr) where,

αf = αc +αd + αr (4)

The contribution from radiation in this forced convection finned-tube heat exchanger is very small. It can be neglected since the active area for radiation is only the circumferential area of the heat exchanger.

A contribution from diffusion is obtained when dew or frost appears on the tube or finned-tube surface. This occurs when the surface temperature is lower than the dew point of the surrounding air (see also 8.50 in [1]). As we have a closed air duct and add no water, we do not expect much contribution from diffusion in steady state condition.

What temperature difference should be used when calculating the overall heat transfer coefficient (U)? The logarithmic mean temperature difference (ϑ m) is commonly used in parallel and counter-flow heat exchangers and is defined as:

ϑ m = (ϑ 1 -ϑ 2)/ ln(ϑ 1/ϑ 2) (5)

Where ϑ 1 and ϑ 2 are temperature differences between the fluids at he two ends of the heat exchanger.

The overall heat transfer coefficient of finned-tube heat exchanger with forced convection is also usually based on the logarithmic mean temperature difference (ϑ m). The values for U and αf, calculated from test results using the logarithmic mean temperature difference as a base, represent surface mean values of the local values along the heat exchanger.

Some equations and data for heat transfer coefficients of finned-tube heat exchanger with forced convection can be found in 8.23, 8.54 –55 in [1].

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TEST EQUIPMENT:

The test equipment consists of a test rig with a finned-coil evaporator. On the airside, the circulating air is heated and cooled in a duct. The airflow and thereby the air velocity is controlled by a speed regulator connected to the fan (see figure 1).

The cooling is provided with a flooded finned-coil evaporator with five parallel tubes and a low-pressure receiver. This free-convection flooder evaporator is connected to a compressor refrigeration unit with a rotary compressor and a brazed plate heat exchanger as a condenser (see figure 4 and compare with 8.04-05 in [1]).

TEST PROCEDURE:

The lab assistant has usually started the lab lesson machinery and adjusted the equipment for the first point of measuring. This includes following steps:

Open the water faucet to the condenser and adjusting some valves in the refrigeration circuit. Start the fan and the compressor. The speed regulator connected to the fan and the heater is adjusted for the first measuring point. The hand-regulated valve between the receiver and the finned-coil evaporator should be adjusted to give a good circulation of refrigerant in the evaporator tubes. This can be done by first adjusting the valve to give a small superheat after the evaporator (t16) and then open it more to get he desired circulation and a mean vapour quality x≈0.4-0.6 at the exit of the evaporator tubes (consult the Lab Assistant). When temperatures are stable for each measuring point, the measured values should be noted in Table 1.

Adjust the speed regulator connected to the fan for the next measuring point. Adjust also the power to the heater so that the temperature of the circulated air at the inlet of the finned coil is maintained constant. Adjust the hand-regulated valve between the receiver and the evaporator as mentioned above.

When all measurements have been taken, the Assistant will show you how to close the machinery.

CALCULATIONS AND REPRESENTATION:

To calculate the heat transfer coefficient on the airside (αf) from eq.3, we first need to know the overall heat transfer coefficient (U), the heat transfer coefficient on the inside (α1), and the fin efficiency (ξ). For the representation, we also need to know the front velocity of the air into the fin battery (Wfr).

Calculation of the overall heat transfer coefficient (U):

The overall heat transfer coefficient (U) is determined in the following way: Through

an energy balance, over the condenser we determine the condenser capacity (Q.

1), the

circulated refrigerant flow ( I,Rm& ) in the condenser and then finally the refrigerating

capacity of the evaporator (Q.

2). To get an even temperature in the evaporator, we use

the system with flooded finned-coil evaporator with five parallel tubes and a low-

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pressure receiver, which gives incomplete evaporation. We get he following equations (see also figure 3):

Q.

1= ( m& ·cp)w·∆tW (6)

I,Rm& = Q.

1/∆h1 = Q.

1/(h1k-hs) = Q.

1/(h14-h11) (7)

Q.

2 = ERm ,& . (h16-h15)

Q.

2= I,Rm& . ∆h2 = I,Rm& . (h2k-hs) = I,Rm& . (h13-h11) (8)

U= 2Q& /(A.ϑ m) (9)

The logarithmic mean temperature difference (ϑ m) is determined from eq.5. Assuming incomplete evaporation (x<1), we can choose ϑ 1 = ta,in - t15 and ϑ 2 = ta,out - t16, where, ta,in = (t1+t2+t3+t4)/4 and ta,out = (t5+t6+t7+t8)/4. Symbols are defined in Table 1.

The following values for the finned-coil surfaces, distances and thermal conductivity are valid with symbols according to eqn.3.

A1 = 0.198 m2 A = A2 + Af =2.65 m2

Am = 0.221 m2 d1 = 0.0078 m

A2 = 0.232 m2 d2 = 0.0095 m

Af = 2.42 m2 λt = 380 W/(m.°C)

Calculation of the heat transfer coefficient on the inside (α1):

The heat transfer coefficient on the inside α1 (which also could be defined as αboil), can be determined from a relation given by Pierre for boiling refrigerants in tubes, which gives the average coefficient along the tube.

For incomplete evaporation (vapor quality at exit of tube <1), Pierre equation gives;

Nu = C1. Re. Kf ½ (10)

Where, Nu = α d/λl’ is the Nusselt number; Re = 4 IIRm ,& / (πdµl

’) is the Reynold’s number; Kf = ∆h2,II / (g.L) is Pierre’s “boiling number”;

The refrigerant flow in each of the parallel evaporator tube is II,Rm& = Q.

2 / (5. ∆h2,II)

and ∆h2,II =r.∆ x where r = latent heat of vaporization and ∆ x = change in vapour quality between outlet and inlet of the evaporator (a circulation number n can be defined; n=1/∆ x ).

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By adjusting the evaporator in the way discussed before, we aim to achieve x≈0.4-0.6 (the circulation number can be assumed in between, n=1/∆ x ; 1,7-2,5) (see t16 in Figure 3). To calculateα1 from eq.10 (incomplete evaporation), the heat of vaporization as well as thermal conductivity (λl’) and the dynamic viscosity (µl

’) of the liquid at the evaporating temperature can be obtained from Tables A: 10-12 for R134a in [1]. The length (L) and the inner diameter (d) of each of the parallel tubes are 6·0.270 and 0.0078 m respectively. Acceleration due to gravity (g) is assumed to be 9.81 m/s2.

The constant (C1) Pierre used in eq.10 was 1.1·10-3. Note that the thermal properties have been revised since Pierre established his equation and it is now suggested that the constant C1 should be reduced by 15% (see 8.32 in [1]).

Calculations of the heat transfer coefficient related to the finned surface:

To determine the heat transfer coefficient (αf) we calculate the product (αf ξ) with the aid of eq.3, first with the simplification that we put A2 = 0, so that we get (αf ξ Af) = αf ξ Af. Then αf ξ is easily calculated from:

)..1

Affξα

= AU .

1 -

Amt

t

.λδ -

A11.1

α

Only the air velocity and the heat transfer coefficient varies for the finned coil during the lab lesson and as a result, we get the fin efficiency ξ = f(αf). Hence, a given heat transfer coefficient αf, determines the fin efficiency.

The following procedure to determine the fin efficiency (preferably done as a before coming to the lab lesson) results in an almost straight line in Figure 5.

Start by using Figure 6 “Fin efficiency for different types of fins”. The fin efficiency is a function of αf. The following dimensions of the fins that can be viewed as hexagonal: 2.r = 9.6 mm, 2.H = 24.8 mm, 2.R = 21.6 mm, δf = 0.2 mm, λf = 210 W/(m.K). Calculate ρ for hexagonal fins and draw ρ = const in Figure 6. Calculate L (m) and then (α/λ.δ)L2 for example for αf = 50, 100, 150 and 200 W/(m.K) and determine ξ and then αf ξ for these values.

These ξ and αf ξ values can now be plotted in Figure 5. A comparison with (αf) from the simplified eq.3 (mentioned above) gives the fin efficiencyξ. This ξ value is now put into the complete eq.3 to give a more accurate value forαf.

The front velocity of the air into the finned-coil Wfr = V.

/Afr is obtained as follows: The front area of the air duct can be measured with a ruler as Afr = 0.270*0.275 m2.

The relation between the volume flow (V.

) and the pressure drop (∆P) for the

calibrated orifice (see Figure 2) is V.

= 0.0305 (∆P)0.5.

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The heat transfer coefficient αf is finally also determined with the methods described under 8.54 and 8.55 in [1]. The finned-coil in the laboratory rig has approximately the following dimensions (with symbols as appearing in Table 8.55a): s = 3.8 mm, δf = 0.2 mm, L = 65 mm, z = 1, d = 9.6 mm, ST = 25.5 mm, SL = 21.6 mm, zr = 3, staggered tubes. You can for air, use the thermal conductivity λ=0.024 W/m.K and kinematic viscosity ν = 1.47*10-5 m2/s.

PRESENTATION IN DIAGRAM:

- Plot in Diagram 1, the calculated values for the overall heat transfer coefficient U and the heat transfer coefficient related to the fin surface αf (calculated from the complete eq.3) as a function of the air velocity Wfr. Draw straight lines through these sets of values.

- Compare the U-values with the corresponding values in Table 8.23a

- Draw as a line in Diagram 1, the appropriate equation from 8.54 in [1]. Determine for αf (calculated from eq.3), constants and exponents according to αf = C.(ρ)n.(Wfr)n. Compare.

- Plot αf also from appropriate equations in 8.55 in [1]. Symbols are given above.

Discuss the results.

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TABLE 1: MEASURED VALUES TEST NO:

1 2 3 4

Cooling water mass flow rate (kg/s) &mw

Compressor inlet pressure (bar) p2k

Compressor outlet pressure (bar) p1k

Air temp before finned battery (°C) t1

- " - t2

- " - t3

- " - t4

Air temp after finned battery t5

- " - t6

- " - t7

- " - t8

Cooling water inlet temp t9

Cooling water outlet temp t10

Condenser outlet temp t11 (ts)

Receiver inlet temp t12

Compressor inlet temp t13 (t2k)

Compressor outlet temp t14 (t1k)

Evaporator inlet temp t15

Evaporator outlet temp t16

Ambient air temp t17

Pressure drop (Pa) ∆p

Estimated change in vapor quality ∆x

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TABLE 2: CALCULATIONS

TEST NO:

1 2 3 4

Transferred condenser power &Q1

Refri. mass flow rate in condenser & ,mR I

Cooling capacity (eqn 8) &Q2

Temp difference between media ϑ1

- " - ϑ2

Mean temp difference ϑm

Overall heat transfer coefficient U

Dynamic viscosity, R134a (liquid) µ’l

Enthalpy change in evaporator ∆h2,II

Refrigerant flow in one tube & ,mR II

Reynold's number (one tube) ReII

"Boiling number" (one tube) Kf,II

Nusselt number NuII

Thermal conductivity, R134a λ’l

Inside heat transfer coefficient α1

Value in simplified equation 1/(αf·ξ·Af)

Value of simplified eqn product αf·ξ

Fin efficiency ξ

Heat transfer coefficient (Eqn 3) αf

Pressure drop (orifice) ∆p

Volumetric Air flow rate &V

Front velocity of air wfr

Heat transfer coefficient (8.54) αf

Factor kRe

Factor Ch

Heat transfer coefficient (8.55) αf