Fluid coupling and torque converter

Why it matters

A conventional automatic transmission has no clutch pedal: the engine is joined to the gearbox through oil. A fluid coupling lets the engine idle with the vehicle stationary and take up drive smoothly; a torque converter does the same and also multiplies torque when the vehicle is pulling away. Understanding their torque, speed and efficiency relations explains automatic-transmission behaviour (creep, stall speed, lock-up, heat) and is standard exam material.

Key ideas

• Fluid coupling. Two facing bladed half-tori sit in an oil-filled casing: the impeller (pump), driven by the engine, and the turbine (runner), connected to the gearbox input shaft. The impeller flings oil outward; the oil crosses to the turbine at the outer radius, gives up its angular momentum, flows inward and returns to the impeller at the inner radius. This vortex flow exists only if the impeller turns faster than the turbine, so there is always some slip.
  • With only two elements and no reaction member, the torque on the turbine equals the torque on the impeller (neglecting small bearing and windage losses). A coupling can never multiply torque.
  • Efficiency therefore equals the speed ratio: about 0.95–0.98 at cruising (2–5% slip), falling to zero at stall (turbine stationary), when all input power becomes heat in the oil.
  • Torque capacity rises roughly with the square of speed and the fifth power of diameter, so at idle very little torque is transmitted (slight creep) and the drive takes up smoothly as the engine speeds up. Couplings also damp torsional vibration and protect against shock loads.
• Torque converter. A third bladed element, the stator (reactor), sits between turbine outlet and impeller inlet, mounted on a one-way (sprag or roller) clutch on a fixed sleeve.
  • At low turbine speed, oil leaving the turbine would strike the backs of the impeller blades and oppose it. The curved stator blades turn this flow so it enters the impeller in the helping direction. The reaction torque taken by the stator adds to the turbine torque: T_t = T_p + T_s (magnitudes).
  • Stall: turbine stationary, maximum torque ratio, typically about 1.8–2.5 for car converters (take the actual value from the maker's characteristic). Stall speed is the engine speed at which the converter absorbs full engine torque with the turbine held.
  • As turbine speed rises, the oil leaving the turbine strikes the stator blades from the back; at the coupling point (speed ratio typically about 0.8–0.9) the stator torque falls to zero, the one-way clutch overruns and the stator freewheels. Above that point the converter behaves as a fluid coupling, torque ratio about 1, efficiency close to the speed ratio.
  • Efficiency η = TR × SR is zero at stall, peaks below the coupling point (around 0.85–0.9), dips near it and then rises along the coupling line. A converter whose stator is locked would see efficiency fall steeply above the peak; the freewheel avoids that.
• Lock-up clutch. A friction clutch (often with a torsional damper) inside the converter locks the turbine to the casing at cruising speeds, eliminating slip losses and heat; it is controlled (sometimes with deliberate small slip) by the transmission controller.
• Multi-stage and variable-pitch converters with two stators or several turbines give higher stall ratios for buses and heavy machinery.
• Couplings and converters lose efficiency in slip, so the oil needs a cooler; a stuck stator clutch or slipping lock-up shows up as overheating and poor fuel economy.

Formulas

SR = N_t / N_p, s = (N_p − N_t) / N_p = 1 − SR

• SR: speed ratio (–); N_t, N_p: turbine and impeller speeds (rev/min); s: slip (–, multiply by 100 for %).

TR = T_t / T_p

• TR: torque ratio (–); T_t, T_p: turbine (output) and impeller (input) torques (N·m). Fluid coupling: TR = 1.

T_t = T_p + T_s

• T_s: stator reaction torque (N·m), equal and opposite to the torque the stator exerts on the casing; zero above the coupling point.

η = P_t / P_p = TR × SR

• η: efficiency (–); fluid coupling η = SR = 1 − s.

P = 2π · N · T / 60, heat generated = P_p − P_t

• P: power (W); N: speed (rev/min); T: torque (N·m).

K = N_p / √T_p

• K: capacity factor (rev/min per √(N·m)), constant for a given converter at a given speed ratio. It follows from T_p ∝ N_p² (with T ∝ ρ · N² · D⁵ for geometrically similar units), so at the same speed ratio T_p2 = T_p1 · (N_p2 / N_p1)².

Worked examples

Example 1 (standard). A fluid coupling transmits 150 N·m from an engine running at 2400 rev/min with 4% slip. Find turbine speed, output power, efficiency and heat generated.

1. N_t = N_p · (1 − s) = 2400 × 0.96 = 2304 rev/min.
2. T_t = T_p = 150 N·m (fluid coupling).
3. P_p = 2π × 2400 × 150 / 60 = 37 699 W; P_t = 2π × 2304 × 150 / 60 = 36 191 W.
4. η = P_t / P_p = 0.96 (equal to the speed ratio).
5. Heat = 37 699 − 36 191 = 1508 W.

Answer: N_t = 2304 rev/min, P_t ≈ 36.2 kW, η = 96%, about 1.51 kW of heat into the oil.

Example 2 (GATE level). A torque converter is driven at 2000 rev/min with an impeller torque of 200 N·m. At this instant the turbine turns at 800 rev/min and the torque ratio is 1.9. Find (a) turbine torque and stator torque, (b) efficiency and power lost, (c) the capacity factor, and (d) the impeller torque if the engine speed rises to 2500 rev/min at the same speed ratio.

1. T_t = TR · T_p = 1.9 × 200 = 380 N·m; T_s = T_t − T_p = 380 − 200 = 180 N·m.
2. SR = 800 / 2000 = 0.40; η = TR × SR = 1.9 × 0.40 = 0.76.
3. P_p = 2π × 2000 × 200 / 60 = 41 888 W; P_t = 2π × 800 × 380 / 60 = 31 835 W; loss = 10 053 W.
4. K = N_p / √T_p = 2000 / √200 = 141.4 rev/min per √(N·m).
5. Same SR, so same K: T_p = (2500 / 141.4)² = 200 × (2500/2000)² = 312.5 N·m.

Answer: T_t = 380 N·m, T_s = 180 N·m; η = 76%, about 10.1 kW lost as heat; K ≈ 141; T_p ≈ 313 N·m at 2500 rev/min.

Common mistakes

• Claiming a fluid coupling multiplies torque. With no reaction member, output torque equals input torque.
• Writing converter efficiency as the torque ratio, or forgetting to multiply by the speed ratio.
• Thinking the stator rotates with the impeller; it is held by a one-way clutch at low speed ratio and only freewheels above the coupling point.
• Taking slip as (N_p − N_t)/N_t; it is referred to the impeller (input) speed.
• Forgetting that all lost power appears as heat in the oil – stall tests must be short.
• Assuming the lock-up clutch multiplies torque; it removes slip and gives 1:1 drive only.

For GATE ME

Expect numericals on slip, efficiency and heat generated in a fluid coupling; torque ratio, stator torque and efficiency of a torque converter from given speeds and torques; and scaling of transmitted torque with speed (and occasionally diameter). Conceptual questions ask why a coupling cannot multiply torque, what the stator and its one-way clutch do, what the coupling point is, and why lock-up clutches are fitted. Practise reading torque-ratio and efficiency curves against speed ratio.

Quick check

1. A coupling runs with impeller at 3000 rev/min and turbine at 2880 rev/min. What are the slip and efficiency?
2. A converter has T_p = 150 N·m and T_t = 330 N·m. What is the stator torque?
3. What is the efficiency of any converter at stall?
4. What happens to the stator above the coupling point?
5. If a coupling transmits 100 N·m at 2000 rev/min, about what does it transmit at 3000 rev/min at the same slip?

Answers: 1. 4% slip, 96% efficiency. 2. 180 N·m. 3. Zero (turbine speed is zero). 4. Its one-way clutch overruns and it freewheels, so the converter acts as a coupling. 5. About 100 × (3000/2000)² = 225 N·m.