A small shell and tube exchanger with one tube pass [A = 4.64 m2 and U = 280 W/m2. OC] is to be used to heat high pressure water initially at 20OC with hot air initially at 260OC. If the exit water temperature is not to exceed 93OC and the air flow rate to 0.45 kg/s, calculate the water flow rate.
The Correct Answer and Explanation is :
To calculate the water flow rate in this heat exchanger, we need to use the heat exchanger equation that relates heat transfer, temperature change, heat capacity, and mass flow rates:
[
Q = U \cdot A \cdot \Delta T_m
]
Where:
- ( Q ) is the heat transfer rate (W)
- ( U ) is the overall heat transfer coefficient (W/m²·°C)
- ( A ) is the heat transfer area (m²)
- ( \Delta T_m ) is the mean temperature difference (°C)
The heat transfer rate, ( Q ), is the same for both the hot and cold fluids, so it can also be expressed in terms of the mass flow rate of water and its specific heat capacity, ( C_p ), as:
[
Q = m_{\text{water}} \cdot C_p \cdot \Delta T_{\text{water}}
]
Where:
- ( m_{\text{water}} ) is the mass flow rate of the water (kg/s)
- ( C_p ) is the specific heat capacity of water (approximately 4.18 kJ/kg·°C)
- ( \Delta T_{\text{water}} ) is the temperature change of the water (°C)
Step 1: Calculate the heat transfer rate, ( Q )
We start by calculating the temperature difference between the hot air and water. The mean temperature difference, ( \Delta T_m ), is estimated using the following approximation:
[
\Delta T_m = \frac{(T_{\text{hot, in}} – T_{\text{cold, out}}) + (T_{\text{hot, out}} – T_{\text{cold, in}})}{2}
]
Where:
- ( T_{\text{hot, in}} ) is the inlet temperature of the hot air (260°C)
- ( T_{\text{cold, out}} ) is the outlet temperature of the cold water (93°C)
- ( T_{\text{hot, out}} ) is the outlet temperature of the hot air (this needs to be calculated)
- ( T_{\text{cold, in}} ) is the inlet temperature of the cold water (20°C)
First, let’s assume the outlet temperature of the hot air is about 100°C (a rough estimate).
Then:
[
\Delta T_m = \frac{(260 – 93) + (100 – 20)}{2} = \frac{167 + 80}{2} = 123.5°C
]
Now, the heat transfer rate, ( Q ):
[
Q = U \cdot A \cdot \Delta T_m = 280 \cdot 4.64 \cdot 123.5 = 160,000 \, \text{W} = 160 \, \text{kW}
]
Step 2: Calculate the water mass flow rate
Now, we can use the equation for the water flow rate:
[
Q = m_{\text{water}} \cdot C_p \cdot \Delta T_{\text{water}}
]
[
160,000 = m_{\text{water}} \cdot 4180 \cdot (93 – 20)
]
[
160,000 = m_{\text{water}} \cdot 4180 \cdot 73
]
[
m_{\text{water}} = \frac{160,000}{4180 \cdot 73} = \frac{160,000}{305,140} \approx 0.524 \, \text{kg/s}
]
Conclusion:
The water flow rate required to achieve the desired temperature rise is approximately 0.524 kg/s.
Explanation:
In this calculation, we used heat exchanger theory to determine the heat transfer rate, which must be the same on both the air and water sides of the exchanger. By calculating the mean temperature difference and applying the heat transfer coefficient and area, we can find the heat transfer rate. Then, using the specific heat capacity of water and the temperature change in the water, we calculated the required flow rate. This solution assumes steady state conditions and that the outlet temperature of the air is roughly 100°C.