Temperature Rise for Electrically Heated Power Washers

The equations used by #Riveer

Temperature rise formula for non-direct immersion style electrically heated units.

limitations

On our non-direct immersion style heaters, expect -5°F for temperature loss across the coil.
Because our heaters are non-direct immersion style heat exchangers, the maximum temperature output you can get is 180°F-185°F.
When are heaters are in the automatic position and have reached system temperature setting, the initial start-up temperature will be 180°F-185°F.
- Once the trigger gun is activated, the temperature will begin to decrease to your continuous temperature output.
- Continuous temperature output is usually reached after approximately 1 minute of constant running.
- Any time the trigger is released the temperature starts to recover (back to 185°F set point).

equations for continuous temperature output

Start from the kW rating of the heaters and the mass flow rate of the pumps in gal per min:

the easy way

\begin{aligned} \dot{Q} & = m C Δ T \\ Δ T & = \frac{\dot{Q}}{m C} \end{aligned}

where $\dot{Q} =$ the rate of heat flow in $k W \cdot s$
$m =$ the mass of water
$C = 4.184 \frac{J}{g \cdot ° C}$ , the specific enthalpy of water.

Given the conversion factors for water at $20 ° C$ :

\frac{8.435 l b_{m}}{1 gal H_{2} O} \cdot \frac{1 k g}{2.2 l b_{m}} = 3.793 \frac{k g}{gal}

Then the heat equation yields:

\begin{aligned} Δ T & = \frac{\dot{Q} kW}{m \frac{gal}{min} (\frac{1 min}{60 \sec}) (3.793 \frac{k g}{gal}) 4.184 \frac{k J}{k g \cdot ° C}} \\ = \frac{\dot{Q} kW}{m \frac{gal}{m i n}} \cdot \frac{60 (\frac{\sec}{m i n})}{3.793 (\frac{k g}{gal}) \cdot 4.184 (\frac{k W \cdot \sec}{k g \cdot ° C})} \\ Δ T = \frac{\dot{Q} k W}{m gpm} (3.78 \frac{° C \cdot gal}{k W \cdot min}) \end{aligned}

$∴$ the minimum temperature that the system will output during continuous operation is $Δ T$ plus the temperature of the incoming water.
For ground water, the temperature will be $\approx$ the air temperature, averaged over the entire year, for that clime and place.

numerical example (easy)

Example: Calculate temperature output of a 24 kw heater with an output of 4 gpm.

\begin{aligned} Δ T & = \frac{24 k W}{4 gpm} (3.78 \frac{° C \cdot gal}{k W \cdot min}) = 22.68 ° C \\ in freedom units: \\ Δ T & = 22.68 ° C (\frac{9 ° F}{5 ° C}) = 40.83 ° F \end{aligned}

rough estimate

for a rough estimate:

$1 gal \approx 3.8 L$ and $1 L \approx 1 k g$

kW of heat divided by gallons per min of flow rate
multiplied by 60 $(\frac{\sec}{min})$
divided by 3.8 $(\frac{k g}{g a l})$
divided by 4.184 $(\frac{k W \cdot s}{k g \cdot ° C})$

This gives the result in science units $(° C)$ . To find $Δ T$ in ~~fake units~~ freedom units $(° F)$ multiply the result by 9/5 (or 1.8).

the hard way

1 BTU = heat required to raise 1 lb of water by 1°F
$1 B T U \approx 1055 J$

kW to BTU/hr conversion

Total kW \cdot \frac{3412 \frac{B T U}{h r}}{1 kW} = Total \frac{B T U}{h r}

BTU/hr to temperature rise

Total BTUH divided by GPM
Divided by 60 (minutes in 1 hour)
Divided by 8.34 (lbs per gallon)
= Temperature rise in Fahrenheit

x \frac{B T U}{h r} \cdot \frac{1 min}{y gal} \cdot \frac{1 hr}{60 min} \cdot \frac{1 gal H_{2} O}{8.34 l b_{m}} = Δ T ° F

numerical example (hard)

Example: Calculate temperature output of a 24 kw heater with an output of 4 gpm.

\begin{aligned} 24 kW \cdot \frac{3412 \frac{B T U}{h r}}{1 kW} & = 81, 888 \frac{B T U}{h r} \\ 81, 888 \frac{° F \cdot l b_{m} H_{2 O}}{hr} \cdot \frac{1 min}{4 gal} \cdot \frac{1 hr}{60 min} \cdot \frac{1 gal H_{2} O}{8.34 l b_{m}} & = Δ T ° F \\ \frac{24 kw}{4 \frac{gal}{min}} \cdot \frac{3412}{60 \cdot 8.34} & = Δ T = 40.91 ° F \end{aligned}

Add the temperature rise $(Δ T)$ to the incoming water temperature to get your continuous temperature output.