Answer:
a) W_cycle = 200 KW , n_th = 33.33 % , Irreversible
b) W_cycle = 600 KW , n_th = 100 % , Impossible
c) W_cycle = 400 KW , n_th = 66.67 % , Reversible
Explanation:
Given:
- The temperatures for hot and cold reservoirs are as follows:
TL = 400 K
TH = 1200 K
Find:
For each case W_cycle , n_th ( Thermal Efficiency ) :
(a) QH = 600 kW, QC = 400 kW
(b) QH = 600 kW, QC = 0 kW
(c) QH = 600 kW, QC = 200kW
- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
Solution:
- The work done by the cycle is given by first law of thermodynamics:
W_cycle = QH - QC
- For categorization of cycle is given by second law of thermodynamics which states that:
n_th < n_max ...... irreversible
n_th = n_max ...... reversible
n_th > n_max ...... impossible
- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:
n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %
And, n_th = W_cycle / QH
a) QH = 600 kW, QC = 400 kW
- The work done by cycle according to First Law is:
W_cycle = 600 - 400 = 200 KW
- The thermal efficiency of the cycle is given by n_th:
n_th = W_cycle / QH
n_th = 200 / 600 = 33.33 %
- The type of process according to second Law of thermodynamics:
n_th = 33.333 % n_max = 66.67 %
n_th < n_max
Hence, Irreversible Process
b) QH = 600 kW, QC = 0 kW
- The work done by cycle according to First Law is:
W_cycle = 600 - 0 = 600 KW
- The thermal efficiency of the cycle is given by n_th:
n_th = W_cycle / QH
n_th = 600 / 600 = 100 %
- The type of process according to second Law of thermodynamics:
n_th = 100 % n_max = 66.67 %
n_th > n_max
Hence, Impossible Process
c) QH = 600 kW, QC = 200 kW
- The work done by cycle according to First Law is:
W_cycle = 600 - 200 = 400 KW
- The thermal efficiency of the cycle is given by n_th:
n_th = W_cycle / QH
n_th = 400 / 600 = 66.67 %
- The type of process according to second Law of thermodynamics:
n_th = 66.67 % n_max = 66.67 %
n_th = n_max
Hence, Reversible Process
