# 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 pin-fin 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 cross-sectional 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 pin-fin 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 pin-fin arrangement, the cross-sectional area can be calculated as Ac = 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 steady-state 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 Lmb (°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 KS 21.1 21.2 21.0 20.7 21.0 G о K‘ 41.2 40.4 40.6 38.4 40.3 TKl(°C) 41.0 40.1 40.3 38.1 39.8 TKJ (°C) 41.3 40.6 40.8 38.7 40.6 G о КГ 40.7 39.7 39.8 37.5 39.2 тк 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 steady-state 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 non-dimensional 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, fioss, 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 cp = 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,npUt (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 non-dimensional. 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 pin-fin dimensions, the total area of exposed copper is As = 0.0215 m2. 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 log-log 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.