Heat a steady liquid stream of 10,000 lb/h from 70°F to 150°F. Use a constant specific heat of 4.18 kJ/(kg·K) and no reaction, phase change, or heat exchange with the surroundings.
1. Calculator Inputs
Select the units below before entering the numbers. These are fixed benchmark inputs for checking the calculator.
| Calculator field | Value | Unit / setting |
|---|---|---|
| Calculation basis | Continuous process: heat rate | |
| Mass flow | 10,000 | lb/h |
| Specific heat | 4.18 | kJ/(kg·K) |
| Temperature unit | °F | |
| Inlet temperature | 70 | °F |
| Outlet temperature | 150 | °F |
| Reaction, phase change, surroundings | All disabled |
2. Step-by-step Math
Convert mass flow
ṁ = 10,000 × 0.45359237 / 3,600 ṁ = 1.259978806 kg/s
Choose the steady-flow basis first, then select lb/h and enter 10,000. Changing the basis resets the mass-unit choices.
Convert the temperature difference
ΔT = 150 − 70 = 80°F ΔT = 80 × 5 / 9 ≈ 44.4 K
For a temperature difference, multiply by 5/9; do not subtract 32. The calculator accepts the two absolute temperatures in °F. Its specific-heat input remains in kJ/(kg·K).
Calculate sensible heat duty
Q̇ = ṁCpΔT Q̇ = 1.259978805556 × 4.18 × 44.444444444444 Q̇ = 234.076062543 kJ/s ≈ 234 kW
A positive result means external heating is required. All other terms are zero because their checkboxes are off.
Convert the duty to Btu/h
1 Btu/h ≈ 2.93e-04 kW Q̇ = 234.076062543 / 0.0002930710701722222 Q̇ ≈ 798,700.678 Btu/h
This uses the same Btu conversion as the calculator. Rounding specific heat to 1 Btu/(lb·°F) would give 800,000 Btu/h, a slightly different result because it changes the property input.
3. Compare Your Answers
Keep full precision during the calculation. The calculator may display fewer digits; compare answers rounded to three significant figures; retain unrounded intermediate values.
| Result | Expected answer | Your answer |
|---|---|---|
| Signed process duty | 234 kW | __________ |
| Sensible heat contribution | 234 kW | __________ |
| Reaction, phase change, surroundings | 0 kW each | __________ |
| Service | Heating required | __________ |
Second check
For a cooling check, swap the inlet and outlet temperatures while keeping all other inputs unchanged. The signed duty becomes −234 kW and the service becomes Cooling required.
4. Assumptions and Method
Specific heat is held constant and the material stays in one phase. This is process duty, not heater electrical demand or fuel input. Equipment efficiency, heat-transfer area, startup, and utility approach temperatures are outside this example.
Equation definitions and implementation: Heat Duty Calculator. See also calculation methodology and other worked examples.