Measurement of Conductivity of Liquids
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1 Name: Lab Section: Date: ME4751, Energy Systems Laboratory Measurement of Conductivity of Liquids Objective: The objective of this experiment is to measure the conductivity of fluid (liquid or gas) and compare the results with available data if possible. Theory: Conduction heat flux (energy transfer rate per unit area) through any material is expressed by the Fourier s law, (1) where κ is the conductivity and dt/dx is the temperature gradient at the location. Minus sign indicates that the energy flows from the high temperature to the low temperature. Consider a material with a thickness of L with the surface temperatures at T 1 and T 2 (Fig. 1). If there is no energy generation in the medium temperature profile is linear (constant slope) and Eqn. (1) can be approximated by (2) where the thermal conductivity is evaluated at the mean temperature of the fluid, (T 1 +T 2 )/2. T 1 q T 2 L Figure 1. Linear temperature distribution in a slab x If the material is fluid, in particular a gas, the effects of convection heat transfer can be significant unless the thickness is very small. 1
2 Experimental Set-Up: In the present experiment, P. A. Hilton unit for the measurement of fluid conductivity is used. Fig. 2 shows the conductivity measurement unit (cylinder) connected to the heat transfer control module. Power (0-240 V, 0-2 Amp) outlet (1) supplies energy to the heater and raises the temperature of the cylindrical plug. Heat transfer occurs from the surface of the plug through the thin layer of testing fluid to the cooling water channel. Testing fluid is charged by a syringe at the charging port (8) until fluid is visible at the vent port (9). Cooling water enters (6) and leaves (7) at a fixed flow rate. Conduction heat transfer rate through the thin layer of testing fluid is determined by measuring the power input, and the temperature difference across the fluid layer (2 & 3, or 4 & 5). At the steady state, energy balance requires that the electrical input power (Q e ) should be equal to the energy transfer through the thin layer of fluid by conduction (Q c ) plus incidental energy loss (Q i ). Input power and the total conduction heat transfer rate are V (3) Q κa T/ r (4) where A is the surface area of the cylinder ( m 2 ), r is the thickness of the fluid layer (0.3 mm) and T is the temperature difference across the fluid layer (T 1 -T 2 ). Eqn. (4) is justified since the clearance between the plug and the cooling water jacket is very small compared to plug diameter (39 mm). (Proof is in Appendix) The incident energy loss accounts for energy loss from the sides of the unit and radiation heat transfer across the fluid in the clearance. By using a fluid with well known conductivity (air), the conduction heat transfer rate can be easily calculated and the incident energy loss is then estimated by Q % Q & Q (5) The approximate magnitude of the loss is correlated to the temperature difference and the results are shown in Fig. 3. Experimental Procedures: Prepare the unit as follows (Figure 4); (1) Ensure that the main switch is off. (2) Remove the hexagonal bolt from the end of the plug and the fluid injection connection. (3) Push plug out of the jacket. Be careful not to damage the o-rings. (4) Clean the plug and the jacket with paper towel. (5) Re-assemble plug, jacket and end cap, with o-rings seal fitted at each end. 2
3 (6) Pass water through the jacket at about 3 liters per minute (actual quantity is not critical but a copious supply is necessary so that the jacket will operate at a sensibly constant temperature). (7) Connect the small flexible tube to the charging and vent unions at either end of the plug and introduce the liquid or gas to the radial clearance between the plug and jacket. Note (a) When using a liquid it is important that sufficient is passed through the clearance space to ensure no air pockets in the clearance. When the fluid has high viscosity it is difficult to push the fluid in. In that case turn on the power to increase the temperature of the fluid to decrease the viscosity. When the fluid has low viscosity (water) it is essential the liquid is injected very slowly with the vent plug held higher than the charging port. (b) When using a gas, the space must be thoroughly purged with the gas, and the flexible tube at both ends must be closed off to prevent leakage. Data collection & Analysis: (i) Switch on the electrical supply and turn the voltage control clockwise to give approximately 170 volts for liquids and 110 volts for gases. Actual power supply can be varied. The results are more accurate with larger temperature differences. (ii) At intervals check the temperature of the plug and the jacket surface until they reach a steady state. Fill in table 1. Table 1. Collected Data Time (min) T 1 ( 0 C) T 2 ( 0 C) Current (amp) Voltage (Volt) 0 (iii) Plot T1 & T 2 vs. time (min). (Attach the plot) (iv) At the steady state, calculate Q e using Eqn. (3) and find Q i from Fig. 3. 3
4 (v) Calculate the conductivity by using Eqn. (4) and compare with available data. Figure 2. Heat transfer control module and fluid conductivity unit assembly (top) and Plug /jacket assembly details (bottom) 4
5 Figure 3. Incident energy loss as a function of temperature difference 5
6 Figure 4. plug/ jacket details (top) and charging process (bottom) 6
7 Appendix: Consider conduction heat transfer from the inner surface to the outer surface as shown in the figure. r 2 r 1 L The inner surface and outer surface temperatures are T s1 and T s2, respectively and the medium in the concentric volume has a conductivity κ. The length of the cylinder is L. Without energy generation in the medium, it can be shown that the total conduction heat transfer rate from the inner surface to the outer surface is Q '()*+, -, -. / a. The thickness of the medium is r= r 2 -r 1. By using binomial expansion ln1 8 9:8 8 3 ln+1<ε. ' Substituting +b. into +a., we get <A? +b. Q '(8 )*+, -, -. 8 κa T/ r (c) 7
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