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Question

In a heat exchanger, the hot liquid enters with a temperature of 180°C and leaves at 160°C. The cooling fluid enters at 30°C and leaves at 110°C. The capacity ratio of the heat exchanger is

The correct answer is

0.25

Understanding Heat Exchanger Capacity Ratio

This problem asks us to determine the capacity ratio ($C_r$) of a heat exchanger given the inlet and outlet temperatures of both the hot liquid and the cooling fluid. The capacity ratio is a crucial parameter in heat exchanger analysis, defined as the ratio of the smaller heat capacity rate to the larger heat capacity rate.

Key Definitions

  • Heat Capacity Rate: It is the product of the mass flow rate ($m$) and the specific heat capacity ($c_p$) of a fluid. It's often denoted by $C$. $$C = m \cdot c_p$$ The units are typically W/°C or kW/K.
  • Capacity Ratio ($C_r$): It is the ratio of the minimum heat capacity rate ($C_{min}$) to the maximum heat capacity rate ($C_{max}$) among the two fluids. $$C_r = \frac{C_{min}}{C_{max}}$$ The capacity ratio is a dimensionless quantity, and its value is always between 0 and 1 ($0 \le C_r \le 1$).

Given Data

Let's list the provided temperatures:

Hot Liquid Inlet Temperature $T_{h,in}$ 180°C
Hot Liquid Outlet Temperature $T_{h,out}$ 160°C
Cooling Fluid Inlet Temperature $T_{c,in}$ 30°C
Cooling Fluid Outlet Temperature $T_{c,out}$ 110°C

Calculating Heat Transfer Rate ($Q$)

We can calculate the heat transfer rate ($Q$) based on the temperature change of either the hot liquid or the cooling fluid. Assuming no heat loss to the surroundings, the heat lost by the hot liquid equals the heat gained by the cold fluid.

  • Heat lost by hot liquid: $$Q = m_h \cdot c_{ph} \cdot (T_{h,in} - T_{h,out})$$ Where $m_h$ is the mass flow rate of the hot liquid and $c_{ph}$ is its specific heat capacity. We can represent the heat capacity rate of the hot fluid as $C_h = m_h \cdot c_{ph}$. $$Q = C_h \cdot (180^\circ C - 160^\circ C) = C_h \cdot 20^\circ C$$
  • Heat gained by cooling fluid: $$Q = m_c \cdot c_{pc} \cdot (T_{c,out} - T_{c,in})$$ Where $m_c$ is the mass flow rate of the cooling fluid and $c_{pc}$ is its specific heat capacity. We can represent the heat capacity rate of the cold fluid as $C_c = m_c \cdot c_{pc}$. $$Q = C_c \cdot (110^\circ C - 30^\circ C) = C_c \cdot 80^\circ C$$

Determining Heat Capacity Rates Relationship

Since the heat transfer rate ($Q$) must be the same for both fluids:

$$C_h \cdot 20^\circ C = C_c \cdot 80^\circ C$$

Now, we can find the ratio of the heat capacity rates:

$$\frac{C_h}{C_c} = \frac{80^\circ C}{20^\circ C} = 4$$

This tells us that the heat capacity rate of the hot fluid ($C_h$) is 4 times the heat capacity rate of the cold fluid ($C_c$).

Calculating the Capacity Ratio ($C_r$)

To find the capacity ratio $C_r$, we need to identify the minimum ($C_{min}$) and maximum ($C_{max}$) heat capacity rates.

  • From the ratio $\frac{C_h}{C_c} = 4$, we know $C_h > C_c$.
  • Therefore, $C_{min} = C_c$ and $C_{max} = C_h$.
  • The capacity ratio is calculated as: $$C_r = \frac{C_{min}}{C_{max}} = \frac{C_c}{C_h}$$
  • Using the inverse of the ratio we found earlier ($\frac{C_h}{C_c} = 4$): $$C_r = \frac{1}{4} = 0.25$$

The capacity ratio of the heat exchanger is 0.25.

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Important Questions from Heat Exchanger Analysis

  1. The logarithmic mean temperature difference for parallel flow heat exchangers is

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

  3. A cold liquid enters a counter flow heat exchanger at 15 deg at a rate of 8 kg/s. A hot stream of the same liquid enters the heat exchanger at 75 deg at 2 kg/s. Assuming the specific heat of the fluid as 4 kJ/kg.°C, determine the maximum heat transfer rate.

  4. For the same inlet and exit temperatures of the hot and cold fluids, the Log mean temperature difference (LMTD) is

  5. NTU effectiveness method for the analysis of heat exchanger is used when:

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