Question

Engine oil \(\left(c_{p}=2100 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be heated from \(20^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) in a \(2-\mathrm{cm}\)-diameter thinwalled copper tube by condensing steam outside at a temperature of \(130^{\circ} \mathrm{C}\left(h_{f g}=2174 \mathrm{~kJ} / \mathrm{kg}\right)\). For an overall heat transfer coefficient of \(650 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the length of the tube required to achieve it.

Short answer

Answer: The required length of the copper tube is approximately 7.045 meters.

Step-by-step solution

Calculate the mass flow rate of engine oil.

Given, the flow rate of engine oil is \(0.3\ kg/s\).

Determine the heat transfer required by the engine oil to raise its temperature.

To find the heat transfer required to raise the temperature of engine oil from \(20^\circ C\) to \(60^\circ C\), we can use the formula: \(Q=mc_p\Delta T\) Where \(Q\) is the heat transfer, \(m\) is the mass flow rate of engine oil, \(c_p\) is the specific heat capacity and \(\Delta T\) is the change in temperature. Given \(m=0.3\ kg/s\), \(c_p=2100\ J/kg\cdot K\), and \(\Delta T = 60-20 = 40^\circ C\). Therefore, \(Q = (0.3)(2100)(40)= 25200\ W\).

Calculate the rate of heat transfer between the steam and engine oil.

We are given the overall heat transfer coefficient U as \(650\ W/m^2\cdot K\). Also, we know \(Q=U\cdot A\cdot\Delta T_{lm}\) where the area A is given by \(A = \pi d L\) and \(\Delta T_{lm}\) is the log mean temperature difference. First, we need to calculate the log mean temperature difference. The outlet temperature of the engine oil is given as \(60^\circ C\). Since the steam is at a constant temperature, the inlet temperature difference can be calculated as \(\Delta T_1 = 130 - 60 = 70^\circ C\). Similarly, the outlet temperature difference is \(\Delta T_2 = 130 - 20 = 110^\circ C\). Now, we can find the log mean temperature difference: \(\Delta T_{lm} = \frac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2/\Delta T_1)}=\frac{110-70}{\ln(110/70)} = 87.25^\circ C\)

Find the length of the copper tube required for the desired heat transfer.

We have the rate of heat transfer \(Q=25200\ W\) and \(\Delta T_{lm}=87.25^\circ C\). We can now solve for the area, \(A\) \(A = \frac{Q}{U\cdot \Delta T_{lm}}=\frac{25200}{650\cdot 87.25}=0.443\ m^2\) Now, we can find the length of the copper tube. We know the diameter of the tube is \(2\ \mathrm{cm} = 0.02\ \mathrm{m}\) and the area \(A=0.443\ \mathrm{m}^2\). \(L = \frac{A}{\pi d}=\frac{0.443}{\pi\cdot 0.02}=7.045\ m\) So, the length of the copper tube required to achieve the desired rate of heat transfer is approximately \(7.045\ \mathrm{m}\).

Key concepts

Engine Oil Heating

Heating engine oil involves raising its temperature to meet operational or process needs. In this exercise, the engine oil is heated from 20°C to 60°C, and understanding this process requires examining the energy required to achieve this temperature increase. This process occurs within a thin-walled copper tube where the oil is heated using condensing steam.

To heat the oil effectively, steam at a temperature of 130°C is utilized, condensing on the exterior of the tube. The heat from the steam transfers through the copper tube wall into the flowing engine oil, thus increasing its temperature. The specific parameters such as flow rate and diameter of the tube influence the heating process and determine the efficiency of heat transfer.

Specific Heat Capacity

Specific heat capacity, denoted as \( c_p \), is an important property that defines how much energy is required to change the temperature of a substance. For engine oil in this scenario, \( c_p = 2100 \ \mathrm{J/kg\cdot K} \). This value indicates how much heat energy is needed to raise the temperature of 1 kilogram of oil by 1 Kelvin.

To calculate the total amount of heat transfer \( Q \) needed to elevate the oil temperature from 20°C to 60°C, the formula used is:

• \( Q = mc_p\Delta T \)

where \( m \) is the mass flow rate, \( \Delta T \) is the temperature change (which in this case, is 40°C), and \( c_p \) is the specific heat capacity. Hence, understanding the specific heat capacity allows us to quantify the amount of heat energy required for the oil's temperature increase.

Log Mean Temperature Difference

The log mean temperature difference (LMTD) is a crucial concept in determining the efficiency of heat exchangers, like the one used to heat engine oil. LMTD provides a weighted average temperature difference between the hot and cold fluids across the heat exchanger. In this setup, it considers the steam at 130°C and the engine oil, varying from 20°C to 60°C.

LMTD is calculated as:

• \( \Delta T_{lm} = \frac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2/\Delta T_1)} \)

where \( \Delta T_1 \) is the temperature difference at the inlet, and \( \Delta T_2 \) at the outlet. For this example, \( \Delta T_1 = 70 \ \mathrm{°C} \) (130-60) and \( \Delta T_2 = 110 \ \mathrm{°C} \) (130-20), resulting in an LMTD of approximately 87.25°C. A higher LMTD indicates a greater potential for heat transfer, thus more efficient heating.

Heat Transfer Coefficient

The heat transfer coefficient \( U \) is a vital factor in determining how effectively heat is transferred through the heat exchanger. In this example, the heat transfer coefficient is given as \( 650 \ \mathrm{W/m^2\cdot K} \). This coefficient quantifies the heat transfer per unit area per degree of temperature difference between the fluids.

The formula for heat transfer \( Q \) including \( U \) and the LMTD is:

• \( Q = U\cdot A\cdot\Delta T_{lm} \)

where \( A \) is the surface area of the tube. The heat transfer coefficient helps us calculate the tube's required length to achieve the desired heating, ensuring the engine oil reaches the correct temperature with optimum efficiency.