HEAT EXCHANGER / CROSS FLOW RADIATOR (1KW) PROTOTYPE DESIGN

Introduction

A radiator is a heat exchanger designed to transfer heat from a hot fluid to a colder fluid without direct contact. In drying, we envisage applications where hot water flows through tubes while ambient air passes over finned surfaces. Heat is transferred from the water to the tube walls, then through the fins, and finally to the air.

The use of fins significantly increases the available heat transfer area and arrangement of tubes breaks laminar resulting in higher thermal efficiency. The design of a radiator involves selecting appropriate tube dimensions, fin geometry, tube arrangement, and flow rates to achieve the desired outlet air temperature

week 1 

Study the woeking principle of the cross flow radiator (Heat Exchanger)

• Hot water from the solar water-heating system flows through 11 staggered copper tubes.
• The copper tubes have an outer diameter of 10 mm and an inner diameter of 8 mm.
• A fan/blower forces air across the tubes in a cross-flow direction.
• Heat is transferred from the hot water to the flowing air through the copper tube walls.
• The tubes are arranged in a staggered pattern with 30 mm horizontal spacing and 25 mm vertical spacing.
• The staggered arrangement increases air–tube contact and improves heat transfer.
• The heated air is then supplied to the drying chamber.
• Tube diameter, wall thickness, and tube length affect water flow and heat transfer.
• Number of tubes, tube pitch, and tube rows determine the total heat-transfer area and water distribution.
• Fin thickness, fin spacing, and fin height increase the heat-transfer surface area and allow effective airflow.
• Proper selection of all design parameters improves heat-transfer efficiency and produces a higher air outlet temperature.

Geometrical parameters

• Tube diameter
• Tube wall thickness
• Tube length
• Number of tubes
• Tube pitch
• Number of tube rows
• Fin thickness
• Fin spacing
• Fin height

Design parameters fianilised, material selection and fabrication

week 2

Assembly preparation stage

• Design parameters finalized for a 1 kW cross-flow radiator.
• Copper tubes cut to the required 200 mm length.
• Aluminium fins prepared with 0.16 mm thickness.
• Holes made according to the required tube pitch.
• Tubes arranged in a staggered pattern.
• Tube–fin assembly completed with 4 mm fin spacing.
• Radiator core fabrication is currently in progress.
• Next step: Header connection.

header connection

Preparation of fins and assembly of the copper tubes with fins and headers.

• Header Preparation: The reshaped copper header pipe was prepared for connecting the copper tubes.
• Hole Making: Holes were made in the header according to the required tube arrangement.
• Copper Tube Insertion: The copper tubes were inserted into the prepared holes of the header.
• Fin Assembly: Aluminium fins were placed over the copper tubes to increase the heat-transfer surface area.
• Core Assembly: The tubes, fins, and headers were assembled to form the radiator core.
• Brazing Process: The tube-to-header joints are being brazed to make them strong and leak-proof.
• Present Status: The radiator core fabrication is in progress, and the main components have been successfully assembled.
• Next Step: Complete brazing, check for water leakage, align the fins, and perform thermal performance testing.

week 3

 Header & Fin Preparation

• Copper header plates were cut.
• Header plates were welded.
• A plywood template was prepared.
• 4 mm fin spacing was maintained

 Assembly & Sealing

• The radiator was assembled.
• M-Seal was applied to all joints.
• Joints were checked for proper sealing.

Leakage Testing

• The first leakage test was conducted.
• Leakage points were identified.
• Araldite was applied to repair the leakage.
• The radiator was left for 24 hours to cure.
• A second leakage test showed improved sealing.

Third leakage test

• The third leakage test was conducted.
• The leakage points was corrected.
• After cure for 24 hours.
• leakages are now negligible.

Final Status

• The radiator assembly was completed.
• All leakage tests are conducted.
• The leakages are negligible.
• The radiator will be ready for performance evaluation.

Week 4

Heat Exchanger Assembly completed 

• Completed the fabrication of the cross-flow heat exchanger. 
• Connected the water inlet and outlet pipes.
• Connection of the fan(90 cfm) is done.
• Checked the overall assembly and structural alignment.

WORKING ON PROJECT (SOLAR DRYER)

WEEK 1 (from 02/09/26 to 06/09/26)

1. Project Discussion and Aim:
During the initial discussion with Dixit Sir, it was decided to develop a system to improve the efficiency of the solar drying process. The system will use solar energy during sunny conditions and provide an alternative heating method when sunlight is not available.

2. Alternative Heating System:
For cloudy weather or low-sunlight conditions, a commercial radiator with hot-water circulation and a fan will be used to produce hot air inside the dryer. This will help maintain the drying process even without direct sunlight.

CYCLIC TIMER IS ADDED  

• A cyclic timer is added to the system to automatically switch the water heating rod ON and OFF.

week 2  (from 07/09/26 to 13/09/26)

calculations (for system efficiency)

Formula             

$$Q=\dot m C_p\Delta T$$

Where:

• Flow rate = 13 L/min
• Water density approximately 1 kg/L
• 
  $C_{\mathrm{p <span> \; \; of\; water\; \approx \; 4.186\; kJ/kg·°C </span>}}$

   

   

   

   

• $\dot{m}=13\mathrm{/}60=\mathrm{0.2167<span>kg</span>}$

   

   

   

   

   

1. Fan OFF

Water:

• Inlet = 60°C
• Outlet = 48°C
• Temperature drop = 12°C

Therefore,

$$Q=0.2167\times4.186\times12$$

$$Q=\mathbf{10.88\ kW}$$

So, based on 13 L/min, the calculated heat transfer from the water is  approx 10.88 kW.

Your air temperature increased from 24°C to 27°C.

Temperature effectiveness:

$$\frac{27-24}{60-24}\times100 =\mathbf{8.33\%}$$

2. Fan ON

Water:

• Inlet = 60°C
• Outlet = 45°C
• Temperature drop = 15°C

$$Q=0.2167\times4.186\times15$$

$$Q=\mathbf{13.61\ kW}$$

So, based on 13 L/min, the calculated heat transfer from the water is ≈13.61 kW.

Air temperature:

• Initial = 24°C
• Final = 38°C
• Air temperature rise = 14°C

Temperature effectiveness:

60−2438−24​×100=38.89%

These all calculations are on the basis of rated flow rate of water of pump which is 13 lit/min.

After the actual clculation of the flow rate of water we determine the acutal heat transfer rate of the system.

week 03  (14/09/26 to 20/09/26)

testing of radiator’s efficiency in close loop oven cabinate.

When fan is OFF :-
Water temperature
Initial – 60•c
Final – 47°c
Air temperature
Ambient – 29°c
Final – 44.5°c
After 44.5°c the temperature is constant.

When Fan is ON :-
Water temperature
Initial – 60°c
Final – 45°c
Air temperature
Ambient – 29°c
Final – 58.5°c
Fan air flow rate – 3.3m/s
After 1hrs the temperature is constant at 58.5°c

calculation of system efficiency.

Water flow rate: 13 L/min
Water density = 1 kg/L; specific heat, Cp = 4.186 kJ/kg·°C.

1. Water Mass Flow Rate
m■ = 13 ÷ 60 = 0.2167 kg/s

2. Fan OFF
Water inlet = 60°C
Water outlet = 47°C
Air inlet = 29°C
Air outlet = 44.5°C
Water temperature drop: ∆T = 60 − 47 = 13°C

Heat transfer:
Q = m■ × Cp × ∆T
Q = 0.2167 × 4.186 × 13 = 11.8 kW

Temperature effectiveness:
ε = [(44.5 − 29) / (60 − 29)] × 100 = 50.0%

3. Fan ON
Water inlet = 60°C
Water outlet = 45°C
Air inlet = 29°C
Air outlet = 58.5°C
Air velocity = 3.3 m/s
Air temperature became constant at 58.5°C after about 1 hour.
Water temperature drop: ∆T = 60 − 45 = 15°C

Heat transfer:
Q = 0.2167 × 4.186 × 15 = 13.6 kW

Temperature effectiveness:
ε = [(58.5 − 29) / (60 − 29)] × 100 = 95.2%

Study of Heat Pump Dryer System                   (ANAND MOHITE, from DHAMARI) 

• During this week, the heat pump dryer system was studied with the help of the given system diagram.
• The major components such as the fan, condenser, evaporator, heaters, exhaust fan and drying chambers were identified.     
• The airflow arrangement and working principle of the dryer were studied.
• It was understood that the fan circulates air through the heating and heat-pump sections before supplying hot air to the drying chamber.
• The hot air removes moisture from the material placed inside the chamber. The moist air is then passed through the evaporator for dehumidification and is reheated for further circulation.               
• The role of the condenser, evaporator and heaters in maintaining the required drying conditions was also studied.                     
• The study helped in understanding the air circulation, heating, dehumidification and drying process of the heat pump dryer.
• Outcome:
  The basic components, airflow path and working principle of the heat pump dryer were studied and understood.

week 4  (from 21/09/26 to 26/09/26

Temperature vs Time chart (by readings of datalogger)

cacluation of the losses from the pump inlet to radiator inlet.

• After discussion with Dixit Sir, we discussed that we need to determine the actual water flow rate through the system. 
• It was decided to experimentally measure the water collected over a known time period and calculate the actual flow rate.
• Measurement of actual water flow rate
  During the experiment, 1 litre of water was collected in 14 seconds.                               
• The calculated actual flow rate was:
  4.29 L/min
  0.0715 kg/s mass flow rate.
• Measurement of temperature drop before radiator inlet

1.    The water temperature was observed to decrease from 60°C to 58°C before reaching the radiator.                         
2. Therefore, the temperature drop in the section between the pump/system and radiator inlet was 2°C.

• Calculation of heat loss
  Using the measured mass flow rate and specific heat of water, the heat loss was calculated.
  Calculated heat-loss rate: ≈ 598.5 W ≈ 599 W.

• This indicates that approximately 599 J of thermal energy is lost per second in the section corresponding to the measured 60°C → 58°C temperature drop.

• Energy-loss calculation
• Heat loss per second :599 J
• Heat loss per minute: 35,940 J
• Heat loss per hour: 2.1564 MJ
• Equivalent energy loss: 0.599 kWh/hour ≈ 0.599 unit/hour.

calculation of actual heat transfer rate of the system with by using actual flow rate of water in radiator outlet.

Water-Side Heat Transfer Rate
Actual water flow rate = 4.29 L/min
Mass flow rate: mnw = 4.29/60 = 0.0715 kg/s
Specific heat of water: Cp,w = 4.186 kJ/kg·K

Equation: Q = mn × Cp × ∆T

Fan OFF
∆T = 60 − 47 = 13°C
Q = 0.0715 × 4.186 × 13 = 3.89 kW

Fan ON
∆T = 60 − 45 = 15°C
Q = 0.0715 × 4.186 × 15 = 4.49 kW

Increase in water-side heat transfer = 4.49 − 3.89 = 0.60 kW (600 W).
Percentage increase = (0.60/3.89) × 100 ≈ 15.4%.

 Air-Side Heat Transfer Rate
For the fan-ON condition, the calculation uses the blower’s rated air flow of 90 CFM. 

90 CFM = 0.04248 m³/s
At about 29°C, air density ≈ 1.165 kg/m³
Air mass flow rate: mna = 0.04248 × 1.165 ≈ 0.0495 kg/s
Specific heat of air: Cp,a ≈ 1.005 kJ/kg·

Fan ON: ∆Ta = 58.5 − 29 = 29.5°C
Qair = 0.0495 × 1.005 × 29.5 ≈ 1.47 kW

week 5  (from 27/09/26 to 04/10/26)

Tray Design and Fabrication

• The existing metal drying tray was further developed for use in the dryer.
• Grinding and finishing work was carried out on the tray/frame to improve its shape and fitting.
• The tray was checked for proper placement and handling inside the drying chamber.
• The existing solid tray surface was studied for modification to improve hot-air circulation.
• Tray dimensions.

Saw dust drying calculation

Weight Loss During Drying
Weight loss = Initial weight − Final weight
= 350 − 315
= 35 g

Percentage Weight Loss
Percentage weight loss = (Weight loss / Initial weight) × 100
= (35 / 350) × 100
= 10%
4. Drying Rate
Drying rate = Weight loss / Drying time
= 35 / 1
= 35 g/h
In minutes:
Drying rate = 35 / 60 = 0.583 g/min