Page 221 - Handbook of Energy Engineering Calculations

0, 0.2, 0.4, 0.6, and 0.8 load on the engine. Plot the result as curve B, Fig. 9.
Perform a similar calculation for condition c—full flow through the bypass
circuit. Plot the results as curve C, Fig. 9.

4. Compute the actual cooling-water flow rate
Find the points of intersection of the pump total-head curve and the three

system head-loss curves A, B, and C, Fig. 9. These intersections occur at 314,
325, and 330 gal/min (19.8, 20.5, and 20.8 L/s), respectively.
The initial design assumed a 10°F (5.6°C) temperature rise through the
engine with a water flow rate of 281 gal/min (17.7 L/s). Rearranging the

equation in step 1 gives Δ t = H/(400G). Substituting the flow rate for
condition a gives an actual temperature rise of Δt = (402)(3500)/[(500)(314)]
= 8.97°F (4.98°C). If a 180°F (82.2°C) rated thermostatic element is used in
the three-way valve, holding the outlet temperature t to 180°F (82.2°C), the
o
inlet temperature t will be Δt = t − t = 8.97; 180 − t = 8.97; t = 171.03°F
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(77.2°C).
5. Determine the required surge-tank capacity
The surge tank in a cooling system provides storage space for the increase in
volume of the coolant caused by thermal expansion. Compute this expansion
from E = 62.4 gΔ V, where E = expansion, gal (L); g = number of gallons

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3
required to fill the cooling system: Δ V = specific volume, ft /lb (m /kg) of
the coolant at the operating temperature − specific volume of the coolant,
3
3
ft /lb (m /kg) at the filling temperature.
The cooling system for this engine must have a total capacity of 281 gal
(1064 L), step 1. Round this to 300 gal (1136 L) for design purposes. The
system operating temperature is 180°F (82.2°C), and the filling temperature

is usually 60°F (15.6°C). Using the steam tables to find the specific volume
of the water at these temperatures, we get E = 62.4(300)(0.01651 − 0.01604)
= 8.8 gal (33.3 L).
Usual design practice is to provide two to three times the required

expansion volume. Thus, a 25-gal (94.6-L) tank (nearly three times the
required capacity) would be chosen. The extra volume provides for excess
cooling water that might be needed to make up water lost through minor
leaks in the system.