--- title: "Power measurement with circulating calorimeter" subject: "RF and Microwave Engineering" chapter: 8 key: "semester_7/rf_and_microwave_engineering:circulating-calorimeter-power" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/circulating-calorimeter-power site: BCT Engineering --- # Power measurement with circulating calorimeter RF and Microwave Engineering · Chapter 8: RF and Microwave Measurements - Asked in: 2082 Baisakh, 4 marks; 2075 Bhadra, 5 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/circulating-calorimeter-power - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-8 ## Question Discuss in detail the power measurement using circulating calorimeter. ## Short answer A circulating calorimeter measures high continuous microwave power by flowing water through a matched load and measuring temperature rise. Power is P = rho v c DeltaT, with DC substitution to remove errors. Example: 0.5 L/min flow and 2.87 C rise gives 100.1 W. ## Answer A **circulating (flow) calorimeter** measures high continuous microwave power by passing a fluid (usually water) through the load at a steady rate and measuring the temperature rise of the fluid. All the absorbed power is carried away by the flowing fluid. #### Principle ```text P = ρ·v·c·ΔT (W) ρ = density of fluid (kg/m³) v = volume flow rate (m³/s) c = specific heat (J/kg·°C) ΔT = T_out − T_in (°C) (ρ·v = mass flow rate ṁ, so P = ṁ·c·ΔT) ``` #### Construction ```text water in (T_in) | RF in ====[ water load: glass/ ] [ dielectric tube in ] [ tapered waveguide ] | water out (T_out) | [flow meter] [ΔT sensor: thermocouples or thermopile] ``` 1. The **load** is a tube of water (or lossy liquid) placed across or along a tapered waveguide so that it is well matched and absorbs all incident power. 2. **Flow system:** a pump circulates water at a constant, measured rate. 3. **Temperature sensors** (thermocouples or thermopile) measure inlet and outlet temperatures. 4. **Heat exchanger** (closed loop) cools the water before it returns. #### Measurement procedure 1. Start the flow and wait for steady inlet temperature. 2. Apply microwave power and wait until the outlet temperature becomes steady. 3. Read the flow rate and ΔT and compute P. 4. **Substitution check:** switch off RF and pass a known DC/50 Hz power through a heater in the same load to get the same ΔT; the RF power equals this known power. This removes errors from flow measurement and heat leakage. #### Example Water flow 0.5 L/min (ρv = 0.5/60 = 8.33×10⁻³ kg/s), c = 4186 J/kg·°C, ΔT = 2.87 °C: ```text P = 8.333×10⁻³ × 4186 × 2.87 = 100.1 W ``` #### Features | Advantage | Limitation | |---|---| | Measures high power (100 W to MW) | Slow response | | Absolute, accurate (±1–2 %) | Needs pump, flow meter | | Continuous measurement | Heat loss and flow errors | | Load handles high power safely | Bulky set-up | Compared with a static calorimeter, it can run continuously because heat is removed all the time, so it suits transmitters, radars and magnetrons. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In a circulating calorimeter, how is the microwave power found? A. From the voltage and current across the water load B. From the mass flow rate, specific heat and the temperature rise of the flowing fluid C. From the time the water takes to boil D. From the change in resistance of a platinum wire Answer: B. All absorbed power is carried away by the fluid, so P = rho*v*c*dT = m_dot*c*dT. (Key point: Principle of flow calorimeter) 2. Water flows at 8.33x10^-3 kg/s (c = 4186 J/kg°C) and warms by 2.87 °C. What is the power? A. About 10 W B. About 50 W C. About 100 W D. About 1000 W Answer: C. P = 8.333e-3 x 4186 x 2.87 = about 100 W. (Key point: Power formula) 3. Why is a substitution check with a known heater power done? A. To increase the water flow rate B. To match the load to the waveguide C. To remove errors due to flow measurement and heat leakage D. To make the water boil faster Answer: C. Producing the same dT with known DC/50 Hz power makes RF power equal to that known power, cancelling flow and leakage errors. (Key point: Substitution check) 4. Why can a circulating calorimeter suit transmitters and radars better than a static one? A. Heat is removed continuously, so it can run continuously at high power B. It needs no pump or flow meter C. It responds instantly D. It measures only very low power Answer: A. The flowing fluid removes heat all the time, allowing continuous high-power measurement, though response is still slow and needs a pump. 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