Question 17.4: Performance of a Wet Cooling Coil A testing and balancing co......
Performance of a Wet Cooling Coil
A testing and balancing contractor tests a coil to determine if its performance is as specified. The coil specified by the designer had the following characteristics (refer to Figure 17.21 for nomenclature) at the design point:
1. Entering air: T_{db,i} = 95°F (35°C), T_{wb,i} = 75°F (23.9°C)
2. Leaving air: T_{db,o} = 66°F (18.9°C), T_{wb,o} = 61°F (16.1°C)
3. Entering coil water temperature: T_{w,i} = 55°F (12.7°C)
4. Airflow rate: 20,000 ft³/min (9,440 L/s)
During the coil test, the contractor measures the following data. Since test conditions rarely are the same as design conditions, the operating condition is different. Use these measured data to determine whether the coil has the specified cooling capacity.
• Entering air: T_{db,i} = 90°F (32.2°C), T_{wb,i} = 73°F (22.8°C)
• Leaving air: T_{db,o} = 63°F (17.2°C), T_{wb,o} = 59°F (15.0°C)
• Entering coil water temperature: T_{w,i} = 52°F (11.1°C)
• Airflow rate: 20,000 ft³/min (9,440 L/s)
The air side is the minimum-capacitance side of the coil, and the airflow during the test is the same as that at design conditions.
Given: Design and measured data as given earlier
Figure: See Figure 17.21.
Assumptions: The coil is tested at sea level. Its surface is wet—conrmed by plotting the process on a psychrometric chart.
Find: Cooling capacity of coil
Lookup values: Enthalpies are read from the psychrometric chart as follows:
h_{a,i,des} =38.7 Btu/lb_{m}
h_{a,o,des} = 27.2 Btu/lb_{m}
h_{a,i,test} = 36.8 Btu/lb_{m}
h_{a,o,test} = 25.8 Btu/lb_{m}
h_{w,i,des} = 23.2 Btu/lb_{m} (saturated at 55°F)
h_{w,i,test} = 21.5 Btu/lb_{m} (saturated at 52°F)
The way to compare the performance of a coil to the design specifications is to calculate the effectiveness for both cases. For the design conditions, the enthalpy form of Equation 12.34 is
Effectiveness \varepsilon = \frac{Actual heat transfer rate}{Maximum possible heat transfer rate}
= \frac{\dot{Q}}{\dot{Q}_{max}} = \frac{(T_{h,i} – T_{h,o})}{(T_{h,i} – T_{c,i})} \times \frac{\dot{C}_{h}}{\dot{C}_{min}} (12.34)
\varepsilon_{des} = \frac{\dot{Q}_{des}}{\dot{m}_{a}(h_{a,i} – h_{w,i})} (17.11)
\dot{m}_{a} = 20,000 ft^{3}/min \times 0.075 lb_{m}/ft^{3} \times 60 min/h= 90,000 lb_{m}/h
\dot{Q}_{des} = \dot{m}_{a} \Delta h_{a} = 90,000 lb_{m}/h
\times (38.7 Btu/lb_{m} – 27.2 Btu/lb_{m})
= 1,035 kBtu/h
The design effectiveness is
\varepsilon = \frac{1,035,000 Btu/h}{90,000 lb_{m}/h \times (38.7 – 23.2) Btu/lb_{m}} = 0.742Under test conditions, the same calculations give these results:
\dot{Q}_{des} = \dot{m}_{a} \Delta h_{a} = 90,000 lb_{m}/h \times (36.8 – 25.8) Btu/lb_{m}= 990 kBtu/h
The calculated effectiveness is
\varepsilon_{test} = \frac{990,000 Btu/h}{90,000 lb_{m}/h \times (36.8 – 21.5) Btu/lb_{m}} =0.719The measured effectiveness is 97% of that specified by the designer, a small difference within the tolerance of the measurement errors.
Comments
This type of calculation is necessary whenever results from a test and balance contractor are being evaluated. Testing will almost never be done at design conditions for either heating or cooling coils. The concept of heat exchanger effectiveness provides a convenient method for using the results at test conditions to determine coil performance at the design condition.