HEAT EXCHANGER FORMULAS & EXAMPLE Heat Exchanger Example: Heating Water with Steam using a Modulating Control Valve
PUMP-TRAP SIZING K) If the condensate return line has a total back pressure of 10 PSIG, can a steam trap be used to drain the heat exchanger? At full-load conditions, the steam pressure is 50 PSIG and the condensate load is ~5,000 lbs/hr. Since the total back pressure of the return line is 10 PSIG, the differential pressure across the steam trap is 40 PSI. An appropriately sized steam trap can handle this situation. However, when the steam pressure is reduced to 10 PSIG or lower, due to lower heat demand, the differential pressure across the steam trap will be 0 PSIG or less. Without positive differential pressure across the trap, the condensate cannot drain from the HX. During this situation, the condensate will back up into the HX shell. Therefore, a steam trap will not be effective in discharging condensate from the HX under all conditions. L) If the water flow rate is reduced, less heat energy per unit of time is needed to heat the water and therefore the heat load will also reduce. This will cause a reduction in steam flow and pressure in the heat exchanger. If the steam pressure falls to or below the system back pressure, the condensate will begin to back up into the heat exchanger, causing the system to stall. Why is it important to prevent stall from occurring? Condensate flooding the heat exchanger space will cause poor temperature control, accelerated corrosion and potentially damaging waterhammer. These factors can cause rapid or premature failure of the unit, leading to costly repairs and downtime. M) For the heat exchanger size (surface area) calculated in Part D, what is the flow rate of water at which stall will occur? We will use two methods to calculate the water flow rate at stall and then compare the two methods. Method 1: Based on Heat Exchanger Size Stall occurs at the point where the steam pressure equals the back pressure. The steam pressure at stall is therefore 10 PSIG. From the saturated steam table, this is equivalent to a steam temperature ( Ts ) of 239°F. Formula 3 can now be used to calculate the mean temperature difference between the steam and water:
(T S - T o ) + (T S - T i ) = (239 - 140) + (239 - 50) = 99 + 189 = 288 = 144°F 2 2 2 2
ΔT M =
The heat load at stall is then calculated from Formula 1 : E stall = U A Δ T M = 120 Btu/(hr-ft 2 -°F) x 185 ft 2 x 144°F = 3,196,800 Btu/hr Finally, the volumetric flow rate of water at stall is calculated from Formula 8: Q w-stall = E stall / [500 x (T o - T i )] = 3,196,800 = 3,196,800 = 3,196,800 [500 x (140 – 50)] (500 x 90) 45,000 = 71 GPM Method 2: Based on % Stall Load Formula T S = Steam temperature at full-load = 298°F (50 PSIG steam) T B = Back pressure equivalent saturated steam temperature = 239°F (10 PSIG steam) T WM = Mean water temperature = T o + T i = 140 + 50 = 95°F 2 2 Using Formula 9 : % Stall Load = T B – T WM x 100 = 239 – 95 x 100 = .71 x 100 = 71% T S – T WM 298 – 95 The water flow rate at stall is then calculated using Formula 10 : Q stall = Q w-full load x (% Stall Load)/100 = 100 GPM x 71/100 = 71 GPM
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