Please show step by step Question 1 Determine the COP of a refrigerator that rem
ID: 2995566 • Letter: P
Question
Please show step by step
Question 1 Determine the COP of a refrigerator that removes heat from teh food compartment at a rate ot 4,691 k h tor each kW of power it consumes. (Hint: watch your units) 10 points Save Answer Question 2 Steam enters an adiabatic turbine with a pressure of 2 MPa, temperature of 400 C, and velocity ot 50 m/s. The steam leaves the turbine at a pressure of 20 kPa, specific volume of 6.885 m3/kg, and velocity of 180 m/s. The mass flow rate of the steam is 6 kg/s. Determine the work output of the turbine in kw.Explanation / Answer
1.
COP = Heat removed / Power input
Heat removed = 4691 kJ/hr = 4691 / (60*60) kW = 1.303 kW
COP = 1.303 / 1 = 1.303
2.
From steam properties at P1 = 2 MPa and T1 = 400 deg C we get h1 = 3250 kJ/kg
From steam properties at P2 = 20 kPa and v2 = 6.885 m^3/kg we get h2 = 2370 kJ/kg
W = m*(h1 - h2 + V1^2 / 2 - V2 ^2 /2)
W = 6*(3250*10^3 - 2370*10^3 + 50^2 /2 - 180^2 / 2)
W = 5190300 Watts
W = 5190.3 kW
3.
Electric power = VI
= 10*120
= 1200 Watts
In 30 minutes, energy drawn = 1200*(30*60) = 2160000 J = 2160 kJ
Specific energy drawn = 2160 / 2.5 = 864 kJ/kg
Initial specific volume v1 = V1 / m
= 1.5 / 2.5
= 0.6 m^3/kg
From water properties at P1 = 1.5 bar and v1 = 0.6 m^3/kg we get h1 = 1620 kJ/kg
Q - W = h2 - h1
0 - (-864) = h2 - 1620
h2 = 2484 kJ/kg
From water properties at P2 = P1 = 1.5 bar and h2 = 2484 kJ/kg we get v2 = 1.05 m^3/kg
4.
From R134a properties at P1 = 1 MPa and T1 = 12 C we get h1 = 68.3 kJ/kg
From R134a properties at P2 = 1 MPa and T2 = 60 c we get h2 = 293 kJ/kg
(2*m)*68.3 + m*(293) = (2m + m)*h3
h3 = 143.2 kJ/kg
From R134a properties at P3 = P2 = P1 = 1 MPa and h3 = 143.2kJ/kg, we get v3 = 0.00513 m^3/kg
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.