Car radiator cools the engine by cooling the anti-freeze coolant using air. The radiator is essentially a crossflow heat exchanger with single-pass and only coolant is unmixed. The coolant enters and exits the radiator at 210F and 150F and at a mass flow rate of 0.35kg/s. Its heat capacity is Cp = 3.5 J/g.K. Air, at 25C flows through the radiator at a rate of 600g/s. Size the radiator knowing that the overall 2 heat-transfer coefficient of the radiator is about 8000 W/m K 

 

?Size = 0.285 m²

Step-by-step explanation

First, we need to find for the value of Q of the antifreeze using the formula:

Q= mCp(T1-T2)

Where:

Q = heat released by the antifreeze

m = mass flow rate of antifreeze

T1= entering temperature of antifreeze

T2 = leaving temperature of the antifreeze

Next we need to analyze the given and convert them if needed:

• m = 0.35 kg/s = 350 g/s
• Cp = 3.5 J/g K
• T1 = 210F = 98.89°C
• T2 = 150 F = 65.56.°C

Therefore, substituting the given into the equation:

Q = 350 g/s × 3.5 J/ g C × (98.89-65.56)°C

Q = 40 829.25 J

Note that based on the conservation of heat, the heat released by the the antifreeze is equal to the heat absorded by the air. This is shown using the equation :

Q antifreeze = Q air

Q air = mCp(T2 - T1)

Next we need to analyze the given and convert them if needed:

• m air = 600 g/s
• Cp = 1 J/g K
• T1 = 25°C

Substituting the given into the equation:

40 829.25 J = 600 g/s × 1 J/g K × (T2 - 25)

T2 = 93.05 °C

To solve for the size of the radiator, we need to use the formula:

Q = UAdT

Where:

Q = heat released or absorbed

U = heat transfer coefficient

A = area of radiator

dt =Log mean temperature

dT is solved as:

dt = [(65.56-25)-(98.89-93.05)] ÷ ln[(65.56-25)÷(98.89-93.05)]

dt = 17.91°C

Substituting the given into the equation:

Q = UAdT

40829.25 J = 8000 W/m K × A × 17.9 °C

A = 0.285 m²

Therefore the size of the radiator is 0.285 m^2.