Answer:
110.87 dB
Explanation:
(I got it right on Acellus)
I= P/4(pi)r^2 = 60/4(pi)6.25^2
60/4(pi)6.25^2=0.12223
B=10log(I/Io)
B=10log(0.12223/1*10^-12) = 110.87 dB
111 in sigfigs
Answer: a) 0.04kW = 40W
b) 0.05
Explanation:
A)
Thermal efficiency of the power cycle = Input / output
Input = 10 kW + 14,400 kJ/min = 10 kW + 14,400 kJ/(60s) = 10 kW + 14,400/60 kW.
Output = 10 kW
Thermal Efficiency = Output / Input = 10kW / 250kW = 0.04KW = 40W
B)
Maximum Thermal Efficiency of the power cycle = 1 - T1/T2
Where T1 = 285kelvin
And T2 = 300kelvin
Maximum Thermal Efficiency = 1 - T1/T2 = 1 - 285/300 = 0.05
