Saturated steam at 100°C is condensing on shell side of a shell and tube heat exchanger. The cooling water enters the tube at 30°C and leaves at 70°C. Calculate arithmetic mean temperature difference in counter flow arrangement.
50°C
The problem asks us to calculate the Arithmetic Mean Temperature Difference (AMTD) for a shell and tube heat exchanger operating in a counter-flow arrangement. We are given the conditions for the condensing steam (hot fluid) and the cooling water (cold fluid).
Understanding Heat Exchanger Temperature Differences
In heat exchangers, the temperature difference between the hot and cold fluids is crucial for determining the heat transfer rate. Since this temperature difference varies along the length of the exchanger, we often use a mean temperature difference. The Arithmetic Mean Temperature Difference (AMTD) is a simpler way to approximate this, calculated as the average of the temperature differences at the two ends of the heat exchanger.
Identifying Fluid Temperatures
We need to identify the inlet and outlet temperatures for both the hot and cold fluids.
- Hot Fluid (Steam): It is condensing at a constant temperature, $T_h = 100^\circ\text{C}$.
- Cold Fluid (Cooling Water):
- Inlet Temperature, $T_{c,in} = 30^\circ\text{C}$.
- Outlet Temperature, $T_{c,out} = 70^\circ\text{C}$.
Calculating Temperature Differences at Ends (Counter Flow)
For a counter-flow arrangement, the temperature difference at one end ($\Delta T_1$) is calculated using the hot fluid's outlet temperature and the cold fluid's inlet temperature. The temperature difference at the other end ($\Delta T_2$) is calculated using the hot fluid's inlet temperature and the cold fluid's outlet temperature.
Since the steam is condensing, its inlet and outlet temperatures are the same ($100^\circ\text{C}$).
- End 1: Difference between hot fluid outlet and cold fluid inlet. $ \Delta T_1 = T_h - T_{c,out} $ $ \Delta T_1 = 100^\circ\text{C} - 70^\circ\text{C} = 30^\circ\text{C} $
- End 2: Difference between hot fluid inlet and cold fluid outlet. $ \Delta T_2 = T_h - T_{c,in} $ $ \Delta T_2 = 100^\circ\text{C} - 30^\circ\text{C} = 70^\circ\text{C} $
Calculating Arithmetic Mean Temperature Difference (AMTD)
The formula for the Arithmetic Mean Temperature Difference is:
$$ AMTD = \frac{\Delta T_1 + \Delta T_2}{2} $$Substituting the calculated values:
$$ AMTD = \frac{30^\circ\text{C} + 70^\circ\text{C}}{2} $$ $$ AMTD = \frac{100^\circ\text{C}}{2} $$ $$ AMTD = 50^\circ\text{C} $$Conclusion
The Arithmetic Mean Temperature Difference (AMTD) for this counter-flow heat exchanger is $50^\circ\text{C}$.
Important Questions from Heat Exchanger Analysis
The fin effectiveness can be enhanced by selecting _____ value of heat transfer co-efficient.
NTU, which is a measure of effectiveness of heat exchanger, stands for _________.
LMTD stands for _______.
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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
