Library / Chemical / CHE-03
Chemical
Sensible heating of a stream
Heat needed to warm a flowing fluid by a given temperature rise.
- Ref
- CHE-03
- Rev
- v1
- Field
- Chemical
- Result
- Q̇ [kW]
Q̇ heat rate in kW, ṁ mass flow in kg/s, cp specific heat in kJ/(kg·K), Tout and Tin outlet and inlet temperatures. Steady, single-phase, constant pressure, no shaft work.
The heat needed to warm a flowing fluid equals the mass flowing each second, times how much heat each kilogram needs per degree, times the temperature rise.
Calculator
- Temperature rise40 K
- Energy per hour1,204 MJ
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Water’s high specific heat makes it an efficient coolant: it carries four times the heat of air per kilogram per degree.
- The formula covers temperature change only. A change of phase (boiling, condensing) needs latent heat as well.
- The same equation gives the cooling duty with the sign reversed.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kW) | Local step | Change in result |
|---|---|---|---|---|---|---|
| ṁ Mass flow | 0.05 to 50 kg/s | 8.36 | 8,360 | 8,352 | +10% | +10 % |
| Tout Outlet temperature | 50 to 100 °C | 250.8 | 668.8 | 418 | +10% | +15 % |
| cp Specific heat | 0.5 to 5 kJ/(kg·K) | 40 | 400 | 360 | +10% | +10 % |
| Tin Inlet temperature | 0 to 40 °C | 167.2 | 501.6 | 334.4 | +10% | −5 % |
Sweep of Outlet temperature
| Outlet temperature (°C) | Heat duty (kW) |
|---|---|
| 50 | 250.8 |
| 56.25 | 303 |
| 62.5 | 355.3 |
| 68.75 | 407.5 |
| 75 | 459.8 |
| 81.25 | 512 |
| 87.5 | 564.3 |
| 93.75 | 616.5 |
| 100 | 668.8 |
Sweep one variable
Variables and how each one contributes
Heat duty
What the equation gives you. Everything else feeds it.
Mass flow
Mass of fluid per second. Directly proportional.
Specific heat
Water 4.18, air 1.0, oil about 2. Directly proportional.
Inlet temperature
Temperature entering. Lowers the duty as it rises (smaller rise).
Outlet temperature
Temperature leaving. Raises the duty as it rises.
Worked example with the values above
- Temperature rise: 60 − 20 = 40 K.
- Q = ṁcpΔT = 2 × 4.18 × 40.
- Q = 334.4 kW.
Assumptions and limits
- Steady flow, single phase, constant cp.
- Kinetic and potential energy changes neglected.
- No shaft work.
Common mistakes
- Forgetting latent heat if a phase change occurs.
- Mixing kJ with J.
- Using volume flow without converting to mass flow.
Statistics and uncertainty
Temperature rise is the weakest term when the difference is small: ±0.5 K on both thermometers gives ±0.7 K on a 40 K rise (1.8%). The specific heat varies by a few per cent with temperature and composition.
Where it is used
- Heater and cooler sizing.
- Engine and electronics coolant loops.
- Heat exchanger duty checks.
Maths you need
- Specific heat: energy per kg per degree
- Mass flow: kg per second