Mass Flow Rate Calculation
For the present example, care must be taken with the calculation of the Reynolds number for the coolant. While the Reynolds number of the coolant is varied, the Reynolds number is specific to flow through a pinfin array. As shown in Equation 5.9, the pin diameter is used as the characteristic length and the coolant velocity is taken as the maximum velocity of the coolant occurring as the coolant travels through the minimum crosssectional area (between two pins).
At a specific Reynolds number, the maximum velocity can be calculated by rearranging Equation (5.9). However, an average channel velocity is needed to determine the required mass flow rate through the channel. Based on the pinfin spacing and mass conservation, the average velocity through the channel can be determined.
Rearranging the Equation 5.10 provides the average coolant velocity through the channel.
With the given pinfin arrangement, the crosssectional area can be calculated as A_{c} = h(2.5D), and the minimum area is A,_{mi}„ = h(l.5D).
Combining Equations 5.9 through 5.11 the mass flow rate desired for a specific Reynolds number can be calculated, and this is shown in Equation (5.12).
TABLE 5.1
Calculated Mass Flow Rates over Range of Reynolds Numbers
Nominal Re 
Actual Re 

6000 
5779 
0.00592 
10.000 
10.088 
0.0103 
15.000 
14.127 
0.0145 
20.000 
20.160 
0.0206 
25,000 
25.558 
0.0262 
Table 5.1 summarizes the coolant mass flow rates over the range of Reynolds numbers considered in this example.
Heat Loss Calibration (Raw Data)
The copper test section was insulated to maximize the resistance to heat transfer through the support material. Using a low conductivity, insulating material will deter the transfer of heat through the support material by conduction. However, the test section cannot be perfectly insulated, and therefore, the amount of heat transferred by modes other than convection must be quantified. For the present experiment, supplementary steadystate tests were performed to establish a matrix of heat loss data. Two heat loss tests were completed in which the wall temperature within the duct was heated to 11 °C and 50°C above the ambient room temperature. It is necessary to record the ambient room temperature during both the actual tests and the heat loss tests because the temperature difference between the test section and the room is the driving temperature difference for heat transfer away from the test section. In addition to the wall and ambient temperatures, the power supplied to the heater during these tests must be recorded. The raw data (including the average wall temperature taken from the array of thermocouples) acquired during the heat loss tests is shown in Table 5.2.
Heat Transfer Enhancement (Raw Data)
During each heat transfer experiment, it is necessary to measure the wall temperature, inlet air temperature, ambient room temperature, and the power supplied to the
TABLE 5.2
Raw Data from Supplementary Heat Loss Tests
G о й 
L_{mb} (°C) 
Heater Voltage (V) 
Heater Current (Amp) 

Low data 
32.1 
21.0 
4.5 
0.256 
High data 
71.1 
21.0 
15.2 
0.865 
TABLE 5.3
Raw Data Acquired over Range of Reynolds Numbers for Overall Heat Transfer Experiment
Re 

5778 
10,088 
14,127 
20,160 
25,558 

W°C) 
21.1 
21.1 
21.1 
21.1 
21.1 
G о 5 K^{S} 
21.1 
21.2 
21.0 
20.7 
21.0 
G о K‘ 
41.2 
40.4 
40.6 
38.4 
40.3 
T_{Kl}(°C) 
41.0 
40.1 
40.3 
38.1 
39.8 
T_{KJ} (°C) 
41.3 
40.6 
40.8 
38.7 
40.6 
G о КГ 
40.7 
39.7 
39.8 
37.5 
39.2 
т_{к} 5 со 
41.6 
40.8 
41.2 
39.1 
41.3 
G о КГ 
40.8 
40.0 
40.6 
37.9 
39.7 
G о КГ 
41.1 
40.2 
40.4 
38.2 
40.1 
G о КГ 
41.1 
40.2 
40.4 
38.2 
40.1 
G о КГ 
41.2 
40.4 
40.7 
38.6 
40.4 
Heater voltage (V) 
31.9 
38.3 
43.3 
46.6 
52.9 
Heater current (Amp) 
1.81 
2.17 
2.46 
2.65 
3.00 
heater. For this overall heat transfer experiment, multiple thermocouples are embedded into the copper surface. However, as one can see in Table 5.3, the variation in these thermocouples is minimal.
Comparing Tables 5.2 and 5.3 it is obvious the wall temperatures maintained during the actual test fall between those of the “high” and “low” heat loss tests. It is a good practice to have the heat loss data bracket the actual test data, so linear interpolation can be used to estimate the actual heat losses during the steadystate experiment.
Heat Transfer Enhancement Data Reduction
The authors of the current experimental study sought to compare overall Nusselt numbers obtained in rectangular cooling channels under variety of flow conditions and channel configurations. Before acquiring the nondimensional Nusselt number, the convective heat transfer coefficient must first be calculated. As the wall temperature of the channel is constant, it is necessary to modify the traditional form of the convective heat transfer Equation to incorporate the constant wall temperature boundary condition. Equation (5.13) shows the heat transfer coefficient definition with the driving temperature difference taken as the log mean temperature difference (which is defined in Equation (5.14)).
In order to calculate the log mean temperature difference, and thus the convective heat transfer coefficient, it is necessary to know the outlet bulk temperature. While the outlet temperature could be measured with thermocouples, the outlet temperature can also be calculated using the first law, energy balance. The calculation of the outlet temperature serves as a check for the estimation of the heat losses; if the heat losses have been properly taken into the account, the calculated outlet temperature should be close to the measured outlet temperature. The outlet temperature can be calculated using Equation (5.15).
As shown above in Equation (5.15), the rate of heat loss, fi_{oss}, must be known prior to calculating the outlet temperature. Linear interpolation can be applied to determine the actual rate of heat loss during the test using the supplementary heat loss calibration data. Equation (5.16) shows the basic linear interpolation, and this can be rearranged to explicitly solve for the rate of heat loss, as shown in Equation (5.17).
As shown in Equation (5.17), the ambient, room temperature is needed to determine the rate of heat loss, as the heat is being lost to the room. Also, as this is an overall heat transfer experiment, the wall temperature always represents the average of the nine thermocouples embedded in the channel wall. The rates of heat transfer can be calculated using the measured voltages and currents supplied to the heaters (fi = VI). Using this equation, and Equation (5.17), the power input and heat loss for each Reynolds number is summarized in Table 5.4.
With the calculated heat loss, and the known power input, the outlet temperature of the air can be calculated using Equation (5.15). The mass flow rate was previously determined, and is shown in Table 5.1 for each Reynolds number. Taking the specific heat of air to be c_{p} = 1005 J/kgK, the outlet air temperature was calculated and shown in Table 5.5.
TABLE 5.4
Power Input and Lost over Reynolds Number Range
Actual Re 
Q,np_{U}t (W) 
Qbis (W) 
Qnet — Qinput Qlms (A0 
5779 
57.79 
3.93 
53.87 
10.088 
83.18 
3.66 
79.52 
14.127 
106.3 
3.74 
102.6 
20.160 
123.3 
3.05 
120.2 
25,558 
158.6 
3.62 
155.0 
TABLE 5.5
Calculated Outlet Temperatures
Re 
^outlet ( О 
5779 
30.2 
10.088 
28.8 
14.127 
28.1 
20.160 
26.5 
25.558 
26.9 
TABLE 5.6
Calculated Log Mean Temperature Differences
Re 
A7im (°C) 
5779 
15.0 
10.088 
14.9 
14.127 
15.8 
20.160 
14.5 
25.558 
16.0 
With knowledge of the outlet bulk temperatures, the log mean temperature difference for each Reynolds number can be calculated using Equation (5.14), and these values are shown for the current case in Table 5.6.
Finally, the overall Nusselt number for the channel can be determined. For the calculation of the Nusselt number, the heat transfer coefficient shown in Equation (5.13), should be nondimensional. For the present experiment, the characteristic length is taken as the pin diameter. With this knowledge, Equation (5.18) can be used to determine the channel Nusselt numbers with the thermal conductivity of air, к = 0.0254 W/mK.
The surface area shown in Equation (5.18) is the total area of copper exposed to the cooling air. For the current case where air only travels streamwise through the channel (no lateral ejection), the total surface area includes the endwalls on the top and bottom walls, the two sidewalls, and the area of the pins. Given the channel and pinfin dimensions, the total area of exposed copper is A_{s} = 0.0215 m^{2}. With this information, the channel Nusselt numbers can be calculated, and for the given range of Reynolds numbers, they are shown in Table 5.7.
Plotting the data in Table 5.7 on a loglog plot replicates a single set of data shown in Figure 5.3.
TABLE 5.7
Channel Averaged (Overall) Nusselt Numbers
Re 
Nu 
5779 
42.0 
10.088 
62.3 
14.127 
76.3 
20.160 
96.9 
25.558 
113.3 
FIGURE 5.3 Overall Nusselt number as a function of Re.