I will assume your heat exchanger uses the common counter flow principle - that is, the direction of flow of the "cold" water opposes that of the "hot" water, so the hottest water in the heating loop (entering the heat exchanger) is in touch with the hottest water of the pool loop (just before exiting the heat exchanger).
The heat flow across the exchanger is proportional to the temperature difference. Since the "input temperature" is fixed at 180 F, the only variable is the temperature of the "sink" - the pool water. The colder the pool water, the greater the heat flow.
At the input of the heat exchanger, the temperature is the temperature of the pool; at the exit, it will be somewhat warmer. The slower the water flows, the more heat it will pick up, and the hotter the water that re-enters the pool. However, the hotter the pool water in the exchanger, the smaller the thermal gradient, and therefore the smaller the heat flux into the pool water.
The water will heat most rapidly if the pool water runs quickly - this keeps the temperature difference greatest.
There is just one caveat: the power of the pump moving the water. If the pump is working harder to move water through a constricted valve, it would generate a little bit more power; if the water flow is set up so heat from the pump is dumped to the water, you will get a small amount of additional heating; but I don't believe that would ever offset the benefit of the faster gradient.
One other consideration: what happens to the surface of your pool. This relates to the way the output of your heat exchanges returns to the pool. If you have a jet that dumps deep inside the pool, there will be little disturbance at the surface; if it's aimed at the surface, you will cause some "stirring". As you may know, the greatest heat loss from a pool happens through evaporation - so if there is anything in your setup that increases evaporation as a function of flow rate through the heat exchanger, that will affect the total heating time.
If it were my pool, I would probably rig up a thermocouple and a data logger, and look at the evolution of temperature. Turn the flow rate up and down every two hours or so, and see if you can observe a change in heating rate on the temperature trace.
I am sorry - according to the laws of physics, your pool guy is wrong. Open that valve, and let the water flow!
Best Answer
Changing the Product Temperature - Heating up with Steam The amount of heat required to raise the temperature of a substance can be expressed as:
$Q = m C_p dT$ (1)
where
$Q$ = quantity of energy or heat (kJ)
$m$ = mass of the substance (kg)
$C_p$ = specific heat capacity of the substance ($kJ/kg C$ ) - Material Properties and Heat Capacities for several materials
$dT$ = temperature rise of the substance ( C)
Preferring Imperial Units - Use the Units Converter!
This equation can be used to determine a total amount of heat energy for the whole process, but it does not take into account the rate of heat transfer which is:
amount of heat energy per unit time In non-flow type applications a fixed mass or a single batch of product is heated. In flow type applications the product or fluid is heated when it constantly flows over a heat transfer surface.
Non-flow or Batch Heating In non-flow type applications the process fluid is kept as a single batch within a tank or vessel. A steam coil or a steam jacket heats the fluid from a low to a high temperature.
The mean rate of heat transfer for such applications can be expressed as:
$q = m C_p dT / t$ (2)
where
$q$ = mean heat transfer rate (kW (kJ/s))
$m$ = mass of the product (kg)
$C_p$ = specific heat capacity of the product (kJ/kg C) - Material Properties and Heat Capacities for several materials
$dT$ = Change in temperature of the fluid ( C)
$t$ = total time over which the heating process occurs (seconds)
Example - Time required to Heat up Water with direct Injection of Steam
The time required to heat 75 kg of water ($C_p = 4.2 kJ/kg C$) from temperature 20 C to 75 C with steam produced from a boiler with capacity $200 kW (kJ/s)$ can be calculated by transforming eq. 2 to
$t = m C_p dT / q$
$ = (75 kg) (4.2 kJ/kg C) ((75 C) - (20 C)) / (200 kJ/s) $
$ = 86 s$
Note! - when steam is injected directly to the water all the energy from the steam is transferred instantly.
When heating through a heat exchanger - the heat transfer coefficient and temperature difference between the steam and the heated fluid matters. Increasing steam pressure increases temperature - and increases heat transfer. Heat up time is decreased.
Overall steam consumption may increase - due to higher heat loss, or decrease - due to to shorter heat up time, depending on the configuration of the actual system.
Flow or Continuous Heating Processes In heat exchangers the product or fluid flow is continuously heated.
The mean heat transfer can be expressed as
$q = C_p dT m / t$ (3)
where
$q$ = mean heat transfer rate (kW (kJ/s))
$m / t$ = mass flow rate of the product (kg/s)
$C_p$ = specific heat capacity of the product (kJ/kg C) - Material Properties and Heat Capacities for several materials
$dT$ = change in temperature of the fluid (C)
Calculating the Amount of Steam If we know the heat transfer rate - the amount of steam can be calculated:
$ms = q / he$ (4)
where
$ms$ = mass of steam (kg/s)
$q$ = calculated heat transfer (kW)
$he$ = evaporation energy of the steam (kJ/kg)
The evaporation energy at different steam pressures can be found in the SteamTable with SI Units or in the Steam Table with Imperial Units.
Example - Batch Heating by Steam A quantity of water is heated with steam of 5 bar (6 bar abs) from a temperature of 35 C to 100 C over a period of 20 minutes (1200 seconds). The mass of water is 50 kg and the specific heat capacity of water is 4.19 kJ/kg C.
Heat transfer rate:
$$q = (50 kg) (4.19 kJ/kg C) ((100 C) - (35 C)) / (1200 s)$$
$$ = 11.35 kW$$
Amount of steam:
$$ms = (11.35 kW) / (2085 kJ/kg)$$
$$ = 0.0055 kg/s$$
$$ = 19.6 kg/h$$
Example - Continuously Heating by Steam Water flowing at a constant rate of 3 l/s is heated from 10 C to 60 C with steam at 8 bar (9 bar abs).
The heat flow rate can be expressed as:
$$q = (4.19 kJ/kg C) ((60 C) - (10 C)) (3 l/s) (1 kg/l)$$
$$ = 628.5 kW$$
The steam flow rate can be expressed as:
$$ms = (628.5 kW) / (2030 kJ/kg)$$
$$ = 0.31 kg/s$$
$$ = 1115 kg/h$$