How to calculate fluid's temperature change in a pump? To calculate the pressure at the outlet of a pump we use pump performance characteristics i.e. charts giving pump head as a function of volumetric flow. When the fluid flows through a pump, it's temperature slightly rises.

Is there a formula or other method to calculate the temperature rise in a pump?

 A: Starting with the definition of specific enthalpy $$h=u+pv,$$ we apply the differential operator to give $$dh=du+pdv+vdp.$$ Now, we know that the $du$ term can be expanded using fundamental thermodynamics. Namely, the change in the internal energy of the system $(du)$ is equal to the heat transfer to the system minus the work done by the system. In equation form, this means that $$du=Tds-pdv.$$ Plugging this result into the above expression for enthalpy gives $$dh=(Tds-pdv)+pdv+vdp=Tds+vdp.$$ Ignoring irreversibilities for the time being, the $Tds$ term will be equal to zero. Thus, for an ideal incompressible pump the differential change in stagnation enthalpy is simply $$dh=vdp=\frac{dp}\rho$$ Expanding the differential to a finite difference (which is licit because the flow is incompressible) gives $$\Delta h=v\Delta p=\frac{\Delta p}\rho=\frac{\dot{W}_{in}}{\dot{m}}=w_{in}$$ As was pointed out in the comments, we cannot explicitly solve for the temperature increase using algebra because we are not working with a calorically perfect fluid. However, the new temperature can be solved for implicitly because the outlet specific enthalpy can be written as a state function of temperature and pressure: $$h(p_{out},T_{out})=h(p_{in},T_{in})+\frac{(p_{out}-p_{in})}{\rho},$$ and a simple interpolation of table values for enthalpy will yield the corresponding temperature. Adding an isentropic pump efficiency $\eta_p$ to account for irreversibilities changes matters, but not in any revolutionary way. We simply use its definition to calculate the actual enthalpy rise in relation to the ideal enthalpy rise. Defining the isentropic pump efficiency as $$\eta_p=\frac{\Delta h_i}{\Delta h}\leq 1,$$ our equation for the exit enthalpy becomes $$\boxed{h(p_{out},T_{out})=h(p_{in},T_{in})+\frac{(p_{out}-p_{in})}{\eta_p \rho}},$$ which predictably reduces to the ideal equation in the limit as $\eta_p\rightarrow 1$. Finally, notice that because $\eta_p\leq 1$ the actual enthalpy rise is always greater than the ideal enthalpy rise for a given $\Delta p$ across the pump. Thus the actual pump requires greater power input than an ideal pump to achieve the same increase in pressure.
A: While perhaps a bit pedantic depending on the original poster's background, I thought I'd outline the considerations in some detail to answer this question generally.
In general, you need to model the pump using a steady-state energy balance across the pump control volume (flange to flange):
$$\Delta H+\Delta KE+\Delta PE=Q-W_s$$
The only work assumed is shaft work.  $PV$ work isn't allowed by the definition of the fixed control volume, and other exotic types of work (such as electromagnetic, surface tension, etc.) aren't typically seen in most situations in a pump. A typical simplifying assumption for most pumps is that the center line of the suction and the  discharge are level ($\Delta PE=0$) and the velocities of the suction and discharge are the same ($\Delta KE=0$).  However, if you want to include these small effects, you certainly can w/o much added difficulty.
Less intuitively, the heat transfer across the pump control volume is usually assumed to be zero.  For fluids near ambient temperatures, this is certainly negligible, and even for fluids at temperature, the heat transfer is usually low relative to the enthalpy/work changes.  However, you should keep it in the back of your mind.  Again, assuming you can quantify it (which would take some intricate heat transfer modeling work), you can include it fairly easily.
So, we have the pretty simple appearing equation:
$$\Delta H=-W_s$$
I'm sure you're wondering where the "mechanical energy balance equation" is that explicitly contains a pressure rise term in it.  The short answer is that real flowsheeting tool pump rating calculations don't use it.  They use the above.  Because the "mechanical energy balance" makes the enthalpy term disappear (via simplifying assumptions), it fundamentally can't conceive of a temperature rise in the fluid.  Real flowsheeting pump algorithms have an exact (if somewhat complex) function of enthalpy vs temperature and pressure.  So, if I specify how much shaft work I put into the pump, I automatically know (w/ assumptions above) what my enthalpy rise is.
A quick aside.  It should be obvious that temperature across a pump has to go up (w/ above assumptions) since the effect on entropy of the fluid will be decreased via compression (more ordered system).  For entropy of the fluid to remain the same (again, no external heat transfer), which is the best case possible (i.e reversible), the temperature of the fluid must go up to offset the lower entropy due to pressure rise.  So, there is a minimum temperature rise required for pumping.  Inefficiency in the pumping only adds more fluid entropy which must mean higher outlet temperatures.
Now, I also know what my outlet pressure is (say by the pump head curve for a given volumetric rate).  So, I'm searching for the temperature that corresponds to the outlet enthalpy $H_2=H_1+W_s$ at the given outlet temperature.  $W_s$ is read off the pump curve as "bhp" (brake horsepower).  Note this is different than the hydraulic horsepower you would calculate from pressure rise term in the "mechanical energy balance" since it incorporates the reversibility that manifests itself in enthalpy.  So, in the end, its just finding out what outlet temperature gives enthalpy $H_2$.
In general, there is little/no data on enthalpy of liquids as a function of temperature and pressure (water being a notable exception).  Almost always, the enthalpy is calculated via an ideal gas enthalpy with departure functions derived from an equation of state (Soave-Redlich-Kwong, Peng-Robinson, etc.)
This almost always requires a flowsheeting tool as the calculations are just too tedious to do by hand.
A: First Law of Thermodynamics :
Q - W = Δ E system
Assuming the Q = 0 and W of the pump negative and Cp of water = 1 Btu/lb-F:
W = 500 x GPM  x v x (P2 – P1)  = 500 x GPM x (T2 – T1)
W = 500 x GPM x v x (P2 – P1)  = 500 x GPM x (T2 – T1)
W = v x (P2 – P1)  = (T2 – T1)
T2 = v x (P2 – P1) + T1
(Where v is the average specific volume of the water)
