Parallel Flow Heat Exchanger Exit Temperature Calculation

This problem involves determining the exit temperature of the cold fluid in a parallel flow heat exchanger. We will utilize key concepts such as heat capacity rates, Number of Transfer Units (NTU), and heat exchanger effectiveness to solve this problem. The specific relation for effectiveness provided in the question will be crucial for our calculations.

Heat Capacity Rate Determination

The heat capacity rate (C) is defined as the product of the specific heat at constant pressure (\(C_p\)) and the mass flow rate (\(\dot{m}\)). We are given that the heat capacity rates of the hot and cold fluids are equal.

• Mass flow rate of hot fluid, \(\dot{m}_h\) = 1 kg/s
• Specific heat of hot fluid, \(C_{p,h}\) = 4 kJ/kgK

The heat capacity rate for the hot fluid (\(C_h\)) is calculated as:

\(\text{C}_h = \dot{m}_h \times C_{p,h}\)

\(\text{C}_h = 1 \text{ kg/s} \times 4 \text{ kJ/kgK} = 4 \text{ kJ/sK} = 4 \text{ kW/K}\)

Since the heat capacity rates of the hot and cold fluid are equal:

\(\text{C}_h = \text{C}_c = 4 \text{ kW/K}\)

In heat exchanger analysis, the minimum heat capacity rate (\(\text{C}_{\text{min}}\)) is used for calculating NTU and effectiveness. In this case, since \(\text{C}_h = \text{C}_c\), we have:

\(\text{C}_{\text{min}} = \text{C}_{\text{max}} = 4 \text{ kW/K}\)

Number of Transfer Units (NTU) Calculation

The Number of Transfer Units (NTU) is a dimensionless parameter that indicates the size of the heat exchanger. It is calculated using the overall heat transfer coefficient (U), the heat transfer surface area (A), and the minimum heat capacity rate (\(\text{C}_{\text{min}}\)).

• Overall heat transfer coefficient, U = 1 kW/m²K
• Heat transfer surface area, A = 5 m²
• Minimum heat capacity rate, \(\text{C}_{\text{min}}\) = 4 kW/K

The formula for NTU is:

\(\text{NTU} = \frac{\text{U} \times \text{A}}{\text{C}_{\text{min}}}\)

\(\text{NTU} = \frac{1 \text{ kW/m}^2\text{K} \times 5 \text{ m}^2}{4 \text{ kW/K}}\)

\(\text{NTU} = \frac{5}{4} = 1.25\)

Heat Exchanger Effectiveness Calculation

The effectiveness (\(\varepsilon\)) of a heat exchanger is the ratio of the actual heat transfer rate to the maximum possible heat transfer rate. The problem provides a specific relation for the effectiveness:

\(2\varepsilon = 1 - \exp (-2\text{NTU})\)

Substitute the calculated NTU value into this relation:

\(2\varepsilon = 1 - \exp (-2 \times 1.25)\)

\(2\varepsilon = 1 - \exp (-2.5)\)

Now, calculate the value of \(\exp (-2.5)\):

\(\exp (-2.5) \approx 0.082085\)

Substitute this value back:

\(2\varepsilon = 1 - 0.082085\)

\(2\varepsilon = 0.917915\)

\(\varepsilon = \frac{0.917915}{2}\)

\(\varepsilon \approx 0.4589575\)

Cold Fluid Exit Temperature Determination

The effectiveness (\(\varepsilon\)) can also be expressed in terms of fluid temperatures. For a heat exchanger where \(\text{C}_{\text{min}} = \text{C}_{\text{max}}\) (which is the case here as \(\text{C}_h = \text{C}_c\)), the effectiveness is given by:

\(\varepsilon = \frac{\text{Actual Heat Transfer Rate}}{\text{Maximum Possible Heat Transfer Rate}} = \frac{\text{C}_c (T_{c,\text{out}} - T_{c,\text{in}})}{\text{C}_{\text{min}} (T_{h,\text{in}} - T_{c,\text{in}})}\)

Since \(\text{C}_c = \text{C}_{\text{min}}\), the expression simplifies to:

\(\varepsilon = \frac{T_{c,\text{out}} - T_{c,\text{in}}}{T_{h,\text{in}} - T_{c,\text{in}}}\)

We are given the inlet temperatures:

• Hot fluid inlet temperature, \(T_{h,\text{in}}\) = 102°C
• Cold fluid inlet temperature, \(T_{c,\text{in}}\) = 15°C

Now, substitute the known values into the effectiveness equation:

\(0.4589575 = \frac{T_{c,\text{out}} - 15}{102 - 15}\)

\(0.4589575 = \frac{T_{c,\text{out}} - 15}{87}\)

Multiply both sides by 87:

\(T_{c,\text{out}} - 15 = 0.4589575 \times 87\)

\(T_{c,\text{out}} - 15 \approx 39.99\)

Finally, solve for \(T_{c,\text{out}}\):

\(T_{c,\text{out}} = 15 + 39.99\)

\(T_{c,\text{out}} \approx 54.99\)°C

Rounding to the nearest whole number, the exit temperature for the cold fluid is approximately 55°C.