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Pumps Temperature Increase

Reference data and engineering information about pumps temperature increase for fluid mechanics applications.

pumpstemperatureincrease

Overview

Engineering reference data for Pumps Temperature Increase in fluid mechanics.

Key Formulas

Reynolds Number

Re=ρvDμRe = \frac{\rho v D}{\mu}

Ratio of inertial to viscous forces — determines flow regime.

Bernoulli's Equation

P+12ρv2+ρgh=constP + \frac{1}{2}\rho v^2 + \rho g h = \text{const}

Conservation of energy for steady, inviscid, incompressible flow.

Continuity Equation

A1v1=A2v2A_1 v_1 = A_2 v_2

Conservation of mass for incompressible flow.

Darcy-Weisbach

ΔP=fLDρv22\Delta P = f \frac{L}{D} \frac{\rho v^2}{2}

Pressure drop due to friction in a pipe.

Variables

SymbolDescriptionUnit
ReReReynolds number
ρ\rhoFluid densitykg/m³
vvFlow velocitym/s
DDCharacteristic dimensionm
μ\muDynamic viscosityPa·s
PPPressurePa
ffDarcy friction factor

Calculation Examples

The temperature rise (ΔT\Delta T) in a pump can be calculated using the following formula:

ΔT=Ps(1μ)cpqρ\Delta T = \frac{P_s (1 - \mu)}{c_p \cdot q \cdot \rho}

Where ΔT\Delta T is the temperature rise (C^\circ\text{C}), PsP_s is the brake power (kW), μ\mu is the pump efficiency (dimensionless), cpc_p is the specific heat capacity of the fluid (kJ/kgC\text{kJ/kg}^\circ\text{C}), qq is the volume flow rate (m3/s\text{m}^3/\text{s}), and ρ\rho is the fluid density (kg/m3\text{kg/m}^3).

Example 1: Normal Operating Conditions

For a water pump with a flow rate of 6 m³/h (0.0017 m³/s), brake power of 0.11 kW, and an efficiency of 28% (0.28), the temperature rise is:

ΔT=(0.11 kW)(10.28)(4.2 kJ/kgC)(0.0017 m3/s)(1000 kg/m3)=0.011C\Delta T = \frac{(0.11\ \text{kW})(1 - 0.28)}{(4.2\ \text{kJ/kg}^\circ\text{C})(0.0017\ \text{m}^3/\text{s})(1000\ \text{kg/m}^3)} = 0.011^\circ\text{C}

The specific heat of water (cpc_p) is taken as 4.2 kJ/kg°C.

Example 2: Throttled Flow Condition

If the flow is reduced to 2 m³/h (0.00056 m³/s) by throttling the discharge valve, with a corresponding brake power of 0.095 kW and a reduced efficiency of 15% (0.15), the temperature rise increases significantly:

ΔT=(0.095 kW)(10.15)(4.2 kJ/kgC)(0.00056 m3/s)(1000 kg/m3)=0.035C\Delta T = \frac{(0.095\ \text{kW})(1 - 0.15)}{(4.2\ \text{kJ/kg}^\circ\text{C})(0.00056\ \text{m}^3/\text{s})(1000\ \text{kg/m}^3)} = 0.035^\circ\text{C}

Key Insight: Reducing pump flow via throttling decreases pump efficiency and dramatically increases the fluid's temperature rise. This relationship, linking temperature rise to flow rate, is typically characterized by specific pump curves provided in manufacturer documentation.

References