Flow Rates in Heating Systems

Calculate flow rates in heating systems

       The volumetric flow rate in a heating system can be expressed as

                                                q = h / (c_p ρ dt)         (1)

where

q = volumetric flow rate, h = heat flow rate , c_p = specific heat capacity , ρ = density ,

dt = temperature difference 

This generic equation can be modified for the actual units - SI or imperial - and the liquids in use.

                       Volumetric Water Flow Rate in Imperial Units

For water with temperature 60 ^oF flow rate can be expressed as

q = h (7.48 gal/ft³) / ((1 Btu/lbm.^oF) (62.34 lb/ft³) (60 min/h) dt)        

    = h / (500 dt)         (2)

where

q = water flow rate (gal/min), h = heat flow rate (Btu/h), ρ = density (lb/ft³),

dt = temperature difference (^oF)

For more exact volumetric flow rates the properties of hot water should be used.

                           Water Mass Flow Rate in Imperial Units

Water mass flow can be expressed as:

                                m = h / ((1.2 Btu/lbm.^oF) dt)        

                                    =  h / (1.2 dt)     (3)

where

m = mass flow (lb_m/h)

                        Volumetric Water Flow Rate in SI-Units

Volumetric water flow in a heating system can be expressed with SI-units as

                      q = h / ((4.2 kJ/kg^oC) (1000 kg/m³) dt)        

                         = h / (4200 dt)        (4)

where

q = water flow rate (m³/s), h = heat flow rate (kW or kJ/s), dt = temperature difference (^oC)

For more exact volumetric flow rates the properties of hot water should be used.

 

                             Water Mass Flow Rate in SI-units

Mass flow of water can be expressed as:

                                          m = h / ((4.2 kJ/kg.^oC) dt)

                                              = h / (4.2 dt)        (5)

where

m = mass flow rate (kg/s)

                          Example - Flow Rate in a Heating System

A water circulating heating systems delivers 230 kW with a temperature difference of 20^oC.

The volumetric flow can be calculated as:

q = (230 kW) / ((4.2 kJ/kg ^oC) (1000 kg/m³) (20 ^oC))

    = 2.7 10^-3 m³/s

The mass flow can be expressed as:

m = (230 kW) / ((4.2 kJ/kg ^oC) (20^oC))

    = 2.7 kg/s

                            Example - Heating Water with Electricity

10 liters of water is heated from 10^oC to 100^oC in 30 minutes. The heat flow rate can be calculated as 

h = (4.2 kJ/kg ^oC) (1000 kg/m³) (10 liter) (1/1000 m³/liter) ((100^oC) - (10^oC)) / ((30 min) (60 s/min))

  =  2.1 kJ/s (kW)

The 24V DC electric current required for the heating can be calculated as

I = (2.1 kW) (1000 W/kW)/ (24 V)

   = 87.5 Amps

Hot Water Boiler - Water Circulation Rates

           

   Boiler power and water flow through boilers - Imperial and SI-units

    The relation between

•  Boiler capacity (or power)
•  Temperature drop of the water flow in a system serviced by boiler
•  Circulated water flow

               are indicated in the diagrams below:

Water Circulation - Boiler Capacity in BHP - Temperature drop in degrees Fahrenheit

• 1 gal (US)/min = 6.30888x10^-5 m³/s = 0.0227 m³/h = 0.06309 dm³(liter)/s = 2.228x10^-3 ft³/s = 0.1337 ft³/min = 0.8327 Imperial gal (UK)/min
• 1 hp (English horse power) = 745.7 W = 0.746 kW = 550 ft lb/s = 2,545 Btu/h

      Water Circulation - Boiler Capacity in MBtu/h - Temperature drop in degrees Fahrenheit

       

• 1 Btu/h = 0.293 W

                 Water Circulation - Boiler Capacity in kW - Temperature drop in degrees Celsius

• 1 kW = 3,412 Btu/h = 1.341 British hp

Example - Water flow through Boiler

A boiler with capacity 50 kW increases the temperature of the circulated water with 20^oC - the water flow through the boiler according the diagram above is 0.6 kg/s

Heat Transfer Formula ,some problems and solution

  Heat Transfer Formula

                                Heat transfer is all about the transfer of heat from one point to another. If we consider any system which will be at higher temperature compared to surroundings, there will be transfer of heat from system to the surroundings.

                            Heat transfer is given by

Where,   m is the mass,
             C is the specific heat and
             ΔT is the temperature difference in K.


There are three types of Heat transfer

1. Conduction
2. Convection
3. Radiation.

                         Heat transfer (Q) by Conduction formula is given by

Where, k is the thermal conductivity of the material,
           A is the cross sectional area,
           T_Hot is the higher temperature,
           T_Cold is the cooler temperature,
           t is the time taken,
           d is the thickness of the material.

                             Heat transfer by Convection is given by
                 Where H_c is the heat transfer coefficient,

                        The Heat transfer by radiation is given by
              Here σ is the Stefan Boltzmann Constant.

                 Heat transfer is usually expressed in Joules. Heat transfer formula is used in problems to find the heat transfer taking place for any given material.

                  

                               Heat Transfer Problems

Below are some problems on Heat transfer which may be helpful for you.

Solved Examples

Question 1: A room is maintained at 20^oC where as temperature outside the room is 10^oC. There is one of the window in the corner having area 1m by 2m. Calculate the heat transfer?
Thermal Conductivity of glass is 1.4 W/mK.
Solution:

Thermal Conductivity of glass is 1.4 W/mK.
Initial temperature Tcold = 10^oC,
Final temperature THot = 20^oC,
Area A = 1m ×$\times$ 2m = 2 m²,

The Heat transfer Q = kA(THot−TCold)d$\frac{kA(T_{Hot}-T_{Cold})}{d}$
                             = 1.4W/mK×2m2×10oC0.003m$\frac{1.4W/mK\times 2m^2\times {10}^{o}C}{0.003m}$
                   

                           = 9333.33 W.

Question 2: Calculate the Heat lost by the block when iron block decreases its temperature from 60^oC to 40^oC if the mass of the body is 2 Kg.
Specific heat of iron C = 0.45 kJ/kg K.
Solution:

Given: Initial temperature T_i = 60^oC,
          Final temperature T_f = 40^oC,
          Mass of the body m = 2 kg,
The Heat lost is given by Q = m c Δ$\mathrm{\Delta }$ T
                                        = 2 Kg ×$\times$ 0.45 kJ/kg K ×$\times$ 293 K
                                        = 263.7 J.