Question 14.4: Analysis of Chiller Systems Consider a large chiller using R......
Analysis of Chiller Systems
Consider a large chiller using R-134a with watercooled condenser and evaporator. The effectiveness values of the condenser and evaporator heat exchangers are 0.90 and 0.75, respectively. The water inlet temperature to the condenser is 25°C while the leaving water temperature in the evaporator is controlled at 2°C. The cooling capacity is 500 kW. The compressor isentropic efficiency is given as 0.9, with no vapor superheating. The water flow rate through the evaporator is 8.0 kg/s while that through the condenser is 12.0 kg/s. Find the power input and COP of this system.
Given: T_{cd,in} = 25°C, T_{ch,out} = 2°C, \dot{Q}_{cool} = 500 kW, \dot{m}_{cd} = 12.0 kg/s, \dot{m}_{ch} = 8.0 kg/s, \eta_{isen} = 0.90, \varepsilon_{evap} = 0.75, \varepsilon_{cond} = 0.9
Figure: See Figure 14.10.
Assumptions: All processes are steady state. The cycle follows the standard VC cycle while considering the fact that the compression is not isentropic. Specific heat of water c_{p} = 4.186 kJ/kg .
Find: \dot{W}_{comp}, COP
This problem cannot be solved sequentially like in the two previous examples. One could adopt an iterative approach or better still use an equation solver software program. There are different ways by which an iterative procedure can be followed. One approach is to assume an initial guess value for the refrigerant condensing temperature T_{3} =T_{cond} =40°C.
(i) Rewrite Equation 14.25 in terms of the evaporator temperature:
= \frac{(\dot{m}_{ch} \cdot c_{p}) \varepsilon_{evap}}{(1 – \varepsilon_{evap})} (T_{ch,out} – T_{evap}) (14.25)
T_{evap} \equiv T_{1} = T_{ch,out} – \dot{Q}_{evap} \frac{(1 – \varepsilon_{evap})}{(\dot{m}_{ch} \cdot {c_{p}) \varepsilon_{evap}}}= 2 – 500 \times \frac{(1 – 0.75)}{(8 \times 4.186 \times 0.75)} = -3.0°C
(ii) The entropy at state 1 is found to be s_{1} = 1.7255 kJ/kg, which is also equal to s_{2}.
This allows the enthalpies at states 1 and 2 to be found as h_{1} = 396.84 kJ/kg and h_{2} = 425.04 kJ/kg.
(iii) At state 3, the refrigerant is saturated vapor with h_{3} = 256.41 kJ/kg = h_{4}.
(iv) The refrigerant mass flow rate:
= 3.56 kg/s
(v) As in Example 14.2, we use Equation 14.14 to get the enthalpy of the vapor leaving the compressor:
\eta_{isen} = \frac{h´_{2} – h´_{1}}{h"_{2} – h´_{1}} (14.14)
h´_{2} = h_{1} – \frac{(h_{2} – h_{1})}{\eta_{isen}} = 396.84 + \frac{(425.04 – 396.84)}{0.9}= 428.17 kJ/kg
(vi) The compressor power input (from Equation 14.7)
\dot{W}_{comp} = \dot{m}_{r} (h_{2} – h_{1}) (14.7)
\dot{W}_{comp} = 3.56 kg/s \times (428.17 – 396.84) kJ/kg= 111.55 kW
(vii) The heat rejection rate at the condenser can be found very simply from an energy balance (this will give identical results as using Equation 14.8)
\dot{Q}_{cond} = \dot{m}_{r} (h_{2} – h_{3}) (14.8)
\dot{Q}_{cond} = \dot{Q}_{cool} + \dot{W}_{comp} = 500 + 111.55 = 611.55 kW(viii) We can now use Equation 14.24 to verify our initial guess of T_{3} = 40°C
\dot{Q}_{cond} = (\dot{m}_{cd} \cdot c_{p}) \varepsilon_{cond}(T_{cond} – T_{cd,in})
= \frac{(\dot{m}_{cd} \cdot c_{p}) \varepsilon_{cond}}{(1 – \varepsilon_{cond})} (T_{cond} – T_{cd,out}) (14.24)
T_{cond} = T_{cd,in} + \frac{\dot{Q}_{cond}}{(\dot{m}_{cd} \cdot c_{p}) \varepsilon_{cond}}= 25 + \frac{611.55}{(12 \times 4.186 \times 0.9)} = 38.5°C
(ix) The difference between the aforementioned value and the initial guess value is small; a second iteration would likely provide the necessary convergence. If the present difference is acceptable, the chiller actual COP is easily determined:
COP = \frac{500 kW}{111.55 kW} = 4.48