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Power measurement with 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.

Discuss in detail the power measurement using circulating calorimeter.

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

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

           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:

P = 8.333×10⁻³ × 4186 × 2.87 = 100.1 W

Features

AdvantageLimitation
Measures high power (100 W to MW)Slow response
Absolute, accurate (±1–2 %)Needs pump, flow meter
Continuous measurementHeat loss and flow errors
Load handles high power safelyBulky 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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