Question

Air enters a duct at \(20^{\circ} \mathrm{C}\) at a rate of \(0.08 \mathrm{~m}^{3} / \mathrm{s}\), and is heated to \(150^{\circ} \mathrm{C}\) by steam condensing outside at \(200^{\circ} \mathrm{C}\). The error involved in the rate of heat transfer to the air due to using arithmetic mean temperature difference instead of logarithmic mean temperature difference is (a) \(0 \%\) (b) \(5.4 \%\) (c) \(8.1 \%\) (d) \(10.6 \%\) (e) \(13.3 \%\) \(8-138\) Engine oil at \(60^{\circ} \mathrm{C}\left(\mu=0.07399 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \rho=864 \mathrm{~kg} / \mathrm{m}^{3}\right)\) flows in a \(5-\mathrm{cm}\)-diameter tube with a velocity of \(1.3 \mathrm{~m} / \mathrm{s}\). The pressure drop along a fully developed 6-m-long section of the tube is (a) \(2.9 \mathrm{kPa}\) (b) \(5.2 \mathrm{kPa}\) (c) \(7.4 \mathrm{kPa}\) (d) \(10.5 \mathrm{kPa}\) (e) \(20.0 \mathrm{kPa}\)

Short answer

Answer: The error involved in using the arithmetic mean temperature difference instead of the logarithmic mean temperature difference is approximately 14.2%.

Step-by-step solution

Calculate logarithmic mean temperature difference (LMTD)

In this step, we will find the logarithmic mean temperature difference between the inlet and outlet temperatures. LMTD = \(\frac{(T_{in} - T_{out}) - (T_{steam} - T_{out})}{\ln{\frac{T_{in} - T_{out}}{T_{steam} - T_{out}}}}\) where \(T_{in} = 20^{\circ}C\), \(T_{out} = 150^{\circ}C\), and \(T_{steam} = 200^{\circ}C\). By inserting these values into the equation, we can calculate the LMTD as: LMTD = \(\frac{(20 - 150) - (200 - 150)}{\ln{\frac{20 - 150}{200 - 150}}}\) LMTD ≈ 69.93 \(^{\circ}C\)

Calculate the rate of heat transfer using LMTD

The formula to calculate the rate of heat transfer, \(Q\), is: \(Q = mC_p \times \Delta T\) where \(m\) is the mass flow rate of air, \(C_p\) is the specific heat of the air at constant pressure, and \(\Delta T\) is the LMTD. First, we need to convert the volumetric flow rate of air to mass flow rate. We can do that using the formula: \(m = \rho V\) where \(V = 0.08m^3/s\) and \(\rho\) is the density of air at \(20^{\circ}C\) (≈ 1.2 \(kg/m^3\)) \(m = 1.2 \times 0.08 \approx 0.096 kg/s\) We also need to find the specific heat of air at constant pressure (\(C_p\)). For air, \(C_p \approx 1005 J/kgK\). Now, we can calculate the rate of heat transfer: \(Q = 0.096 \times 1005 \times 69.93 \approx 6742.24 W\)

Calculate the arithmetic mean temperature difference (AMTD)

The arithmetic mean temperature difference can be calculated using the formula: AMTD = \(\frac{T_{in} - T_{out} + T_{steam} - T_{out}}{2}\) AMTD = \(\frac{(20 - 150) + (200 - 150)}{2}\) AMTD ≈ 60 \(^{\circ}C\)

Calculate the rate of heat transfer using AMTD

Using the AMTD in the heat transfer formula, we can calculate the rate of heat transfer: \(Q' = mC_p \times \Delta T'\) \(Q' = 0.096 \times 1005 \times 60 \approx 5781.6 W\)

Calculate the error involved in using AMTD instead of LMTD

Now, we have the correct rate of heat transfer (\(Q\)) and the rate of heat transfer calculated using AMTD (\(Q'\)). The error involved is, Error = \(\frac{Q - Q'}{Q} \times 100\) Error = \(\frac{6742.24 - 5781.6}{6742.24} \times 100\) Error ≈ 14.2% Based on the given options, the closest answer is: (e) \(13.3 \%\)

Key concepts

Logarithmic Mean Temperature Difference (LMTD)

Understanding the Logarithmic Mean Temperature Difference (LMTD) is crucial for accurate heat transfer calculations in engineering. It corrects for the variation in temperature difference across a heat exchanger, offering a more precise value than a simple average. Imagine you're heating air as it travels along a duct. The temperature difference between the entering and exiting air isn't uniform, especially when compared to the steam outside the duct. LMTD takes this varying temperature gradient into account by using a logarithmic formula.To express it mathematically, we use:\[ LMTD = \frac{(T_{in} - T_{out}) - (T_{steam} - T_{out})}{\ln{\frac{T_{in} - T_{out}}{T_{steam} - T_{out}}}} \]This calculation provides a 'mean' temperature that is logarithmically closer to the actual temperature profile of the system. As a result, when used to calculate the rate of heat transfer, LMTD gives a more accurate reflection of the reality inside the heat exchanger.

Arithmetic Mean Temperature Difference (AMTD)

The Arithmetic Mean Temperature Difference (AMTD) is a simpler approach to estimating the average temperature difference across a heat exchanger. Unlike LMTD, AMTD doesn't account for the temperature change variability and simply averages the temperature differences.Here's how we compute it:\[ AMTD = \frac{T_{in} - T_{out} + T_{steam} - T_{out}}{2} \]Just take the inlet and outlet temperature differences, add them together, and divide by two. This gives a straightforward arithmetic mean. However, as we've seen in our exercise, using AMTD instead of LMTD for heat transfer calculations can introduce a significant error, leading to less efficient system design or operation.

Rate of Heat Transfer

The rate of heat transfer is essentially a measure of the energy flow due to temperature differences. It's extremely important in applications ranging from industrial processes to climate control systems.The formula for calculating the rate of heat transfer, denoted by \(Q\), is:\[ Q = mC_p \times \Delta T \]Here, \(m\) represents the mass flow rate of the fluid, \(C_p\) is the specific heat capacity at constant pressure, and \(\Delta T\) is the temperature difference, which can be either LMTD or AMTD depending on the required accuracy. In the context of our exercise, you can see that using the precise LMTD as \(\Delta T\) yields a more accurate calculation of the heat transfer rate, which is essential for optimizing system performance.

Specific Heat Capacity

Specific heat capacity, often symbolized as \(C_p\), is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). It's a property intrinsic to the material and crucial for calculating the rate of heat transfer.Knowing the substance's specific heat is vital for our heat transfer equation:\[ Q = mC_p \times \Delta T \]Let's visualize specific heat with a simple comparison. Water has a high specific heat capacity, which means it can absorb a lot of heat before its temperature rises significantly. This is why water is often used as a coolant. In our exercise, the specific heat of air impacts how much energy is needed to heat it to a certain temperature. The correct specific heat capacity value ensures that the calculations for energy exchange are as accurate as possible.