Stream Duty from Mass Flow (Q = ṁcΔT)

Every exchanger calculation has two halves that must agree. The transfer side says Q̇ = UAF·$\Delta T_{\mathrm{lm}}$; the process side says Q̇ = ṁcₚΔT for each stream. Write both, set them equal, and the whole problem closes. Because the heat leaving the hot stream must arrive in the cold one, ṁcₚΔT for the hot side equals ṁcₚΔT for the cold side, so the stream with the smaller ṁcₚ, the smaller heat capacity rate, always shows the larger temperature swing. That single observation lets you sanity-check a datasheet from across the room: if both streams change by the same amount, their capacity rates are equal.

Worked example: 2.5 kg/s of water (cₚ = 4186 J/(kg·K)) heated 12 K takes 2.5 × 4186 × 12 = 125.6 kW. In North American units the same physics reads 500,000 BTU/h into 20,000 lb/h of oil at cₚ = 0.5 across 50 °F. Traps: cₚ is not constant. Water is flat enough to ignore, but oils and glycols vary 10–20% over a working range, so use the value at the mean temperature. And this equation is sensible heat only; the moment anything boils or condenses, the temperature stops moving and you need ṁ times the latent heat instead.