All Exams Test series for 1 year @ ₹349 only
Question

A balanced counter flow heat exchanger has a surface area of 20 m2 and overall heat transfer coefficient of 20 W/m2–K. Air (CP = 1000 J/kg - K) entering at 0.4 kg/s and 280 K is to be preheated by the air leaving the system at 0.4 kg/s and 300 K. The outlet temperature (in K) of the heated air is ___

The correct answer is

300

To determine the outlet temperature of the heated air in a balanced counter flow heat exchanger, we will use the principles of heat transfer, specifically focusing on the properties of balanced counter flow heat exchangers and the overall heat transfer rate. We will calculate the heat capacity rates, identify the constant temperature difference, compute the total heat transferred, and finally find the desired outlet temperature.

Heat Capacity Rates Calculation

First, let's calculate the heat capacity rate for both the cold air (to be heated) and the hot air (leaving the system). The heat capacity rate (\(C\)) is given by the product of the mass flow rate (\(\dot{m}\)) and the specific heat (\(C_p\)).

  • Mass flow rate of cold air, \(\dot{m}_c = 0.4 \text{ kg/s}\)
  • Mass flow rate of hot air, \(\dot{m}_h = 0.4 \text{ kg/s}\)
  • Specific heat of air, \(C_p = 1000 \text{ J/kg – K}\)

For the cold air:

\({C_c = \dot{m}_c \times C_p = 0.4 \text{ kg/s} \times 1000 \text{ J/kg – K} = 400 \text{ W/K}}\)

For the hot air:

\({C_h = \dot{m}_h \times C_p = 0.4 \text{ kg/s} \times 1000 \text{ J/kg – K} = 400 \text{ W/K}}\)

Since \(C_c = C_h = 400 \text{ W/K}\), this confirms that it is a balanced counter flow heat exchanger.

Counter Flow Heat Exchanger Property

A key property of a balanced counter flow heat exchanger is that the temperature difference between the hot and cold fluids remains constant throughout the heat exchanger. This means the temperature difference at one end is equal to the temperature difference at the other end. Let this constant temperature difference be \(\Delta T\).

\(\Delta T = (T_{h,in} - T_{c,out}) = (T_{h,out} - T_{c,in})\)

We are given the following temperatures:

  • Inlet temperature of cold air, \(T_{c,in} = 280 \text{ K}\)
  • Outlet temperature of hot air, \(T_{h,out} = 300 \text{ K}\)

Temperature Difference Calculation

We can calculate the constant temperature difference using the known inlet and outlet temperatures:

\(\Delta T = T_{h,out} - T_{c,in} = 300 \text{ K} - 280 \text{ K} = 20 \text{ K}\)

This means the temperature difference between the hot and cold fluid is \(20 \text{ K}\) at all points along the heat exchanger.

Heat Transfer Rate Determination

Now, we can calculate the total heat transfer rate (\(Q\)) using the overall heat transfer coefficient (\(U\)), the surface area (\(A\)), and the constant temperature difference (\(\Delta T\)).

  • Overall heat transfer coefficient, \(U = 20 \text{ W/m}^2\text{–K}\)
  • Surface area, \(A = 20 \text{ m}^2\)

\({Q = U \times A \times \Delta T}\)

\({Q = 20 \text{ W/m}^2\text{–K} \times 20 \text{ m}^2 \times 20 \text{ K} = 8000 \text{ W}}\)

Outlet Temperature Calculation

Finally, we can use the calculated heat transfer rate and the heat capacity rate of the cold air to find its outlet temperature (\(T_{c,out}\)). The heat transferred to the cold fluid is given by:

\({Q = C_c (T_{c,out} - T_{c,in})}\)

Rearranging the formula to solve for \(T_{c,out}\):

\({T_{c,out} - T_{c,in} = \frac{Q}{C_c}}\)

\({T_{c,out} = T_{c,in} + \frac{Q}{C_c}}\)

Substitute the known values:

\({T_{c,out} = 280 \text{ K} + \frac{8000 \text{ W}}{400 \text{ W/K}}}\)

\({T_{c,out} = 280 \text{ K} + 20 \text{ K}}\)

\({T_{c,out} = 300 \text{ K}}\)

Therefore, the outlet temperature of the heated air is \(300 \text{ K}\).

To summarize the parameters and results:

Parameter Symbol Value Unit
Surface Area \(A\) 20 \(\text{m}^2\)
Overall Heat Transfer Coefficient \(U\) 20 \(\text{W/m}^2\text{–K}\)
Specific Heat of Air \(C_p\) 1000 \(\text{J/kg – K}\)
Mass Flow Rate (Cold/Hot) \(\dot{m}\) 0.4 \(\text{kg/s}\)
Inlet Temperature (Cold) \(T_{c,in}\) 280 \(\text{K}\)
Outlet Temperature (Hot) \(T_{h,out}\) 300 \(\text{K}\)
Heat Capacity Rate (Cold/Hot) \(C_c, C_h\) 400 \(\text{W/K}\)
Constant Temperature Difference \(\Delta T\) 20 \(\text{K}\)
Total Heat Transfer Rate \(Q\) 8000 \(\text{W}\)
Outlet Temperature (Cold) \(T_{c,out}\) 300 \(\text{K}\)

Was this answer helpful?

Important Questions from Heat Exchanger Analysis

  1. The fin effectiveness can be enhanced by selecting _____ value of heat transfer co-efficient.

  2. NTU, which is a measure of effectiveness of heat exchanger, stands for _________.

  3. LMTD stands for _______.

  4. Water (Cp = 4.18 kJ/kg.K) at 80°C enters a counter flow heat exchanger with a mass flow rate of 0.5 kg/s. Air (Cp = 1 kJ/kg.K) enters at 30°C with a mass flow rate of 2.09 kg/s. If the effectiveness of the heat exchanger is 0.8, the LMTD (in °C) is

  5. For a heat exchanger, ΔTmax is the maximum temperature difference and ΔTmin is the minimum temperature difference between the two fluids. LMTD is the log mean temperature difference. Cmin and Cmax are the minimum and the maximum heat capacity rates. The maximum possible heat transfer (Qmax) between the two fluids is

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App