Answer:fan blades, rotating , abrasive wheel machinery
Explanation:
Answer: P = 0.416 kW
Explanation:
taken a step by step process to solving this problem.
we have that from the question;
the amount of heat rejected Qn = 4800 kJ/h
the cooling effect is Ql = 3300 kJ/h
Applying the first law of thermodynamics for this system gives us
Шnet = Qn -Ql
Шnet = 4800 - 3300 = 1500 kJ/h
Next we would calculate the coefficient of performance of the refrigerator;
COPr = Desired Effect / work output = Ql / Шnet = 3300/1500 = 2.2
COPr = 2.2
The Power as required gives;
P = Qn - Ql = 4800 - 3300 = 1500 kJ/h = 0.416
P = 0.416 kW
cheers i hope this helps!!!!1
Answer: 67.392s
Explanation: detailed calculation is shown below
Answer:
second-law efficiency = 62.42 %
Explanation:
given data
temperature T1 = 1200°C = 1473 K
temperature T2 = 20°C = 293 K
thermal efficiency η = 50 percent
solution
as we know that thermal efficiency of reversible heat engine between same temp reservoir
so here
efficiency ( reversible ) η1 = 1 -
............1
efficiency ( reversible ) η1 = 1 -
so efficiency ( reversible ) η1 = 0.801
so here second-law efficiency of this power plant is
second-law efficiency =
second-law efficiency =
second-law efficiency = 62.42 %