Calculating Heating Time of Lube Oil SAE 40 - 19000 Liters at 20° C to 40° C

[amit4644]

Homework Statement

I have problem in calculating heating time of Lube Oil ( SAE 40 ). There is a tank of Lube Oil ( SAE 40) of 19000 liters at 20° C. this oil has to be heated up from 20° C to 40° C with help of a heater of 85 kW through which oil flows at 4800 liter/hour and back to tank.

How much time it will take to heat up complete tank from 20° C to 40° C. it is a closed loop circuit and small distance. just oil from tank flows at 4800 liter/hour through 85kW heater and then goes back to tank.

any suggestions for finding time needed ??

Thanks

Homework Equations

The Attempt at a Solution

 

[Valkarie]

To calculate the time needed to heat up the lube oil, you need to use the equation: Time = (Volume of Oil x Specific Heat Capacity x Delta T) / (Power of Heater x Efficiency) where: Volume of Oil = 19000 liters Specific Heat Capacity = 1.84 kJ/kgK Delta T = 20° C Power of Heater = 85 kW Efficiency = 0.8 Therefore, the time needed to heat up the lube oil is: Time = (19000 x 1.84 x 20) / (85 x 0.8) = 456.2 hours

 

[Mene]

To calculate the heating time of Lube Oil SAE 40, we need to use the formula Q = mcΔT, where Q is the heat energy required, m is the mass of the oil, c is the specific heat capacity of the oil, and ΔT is the change in temperature. In this case, we can assume that the specific heat capacity of Lube Oil SAE 40 is approximately 2.1 J/g°C.

First, we need to calculate the mass of the oil in the tank. Given that the tank has a volume of 19000 liters, we can convert this to mass using the density of Lube Oil SAE 40, which is approximately 0.88 g/mL. This gives us a mass of 19000 liters x 0.88 g/mL = 16720 kg.

Next, we need to calculate the heat energy required to raise the temperature of the oil from 20°C to 40°C. This can be done by substituting the values into the formula: Q = (16720 kg) x (2.1 J/g°C) x (40°C - 20°C) = 7027200 J.

Since we know that the heater has a power of 85 kW and the oil flows at a rate of 4800 liters/hour, we can calculate the heating time using the formula t = Q/P, where t is the time, Q is the heat energy, and P is the power. Substituting the values, we get t = (7027200 J) / (85,000 W) = 82.6 seconds.

Therefore, it will take approximately 82.6 seconds to heat up the complete tank of Lube Oil SAE 40 from 20°C to 40°C. However, this calculation assumes ideal conditions and does not take into account any heat loss or other factors that may affect the heating time. It is always important to consider these factors and make adjustments accordingly in real-world situations.