Answer:
Actual COP = 5.368
Maximum theoretical COP = 6.368
Explanation:
Given - An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to the compressor is -5oC.
To find - If this device operates using R134a as the working fluid. Calculate the actual COP of this device as well as the maximum theoretical COP.
Proof -
Given that,
An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa.
From Refrigerant 134-a Table
At T1 = -5°C
h1 = 247.505 KJ/kg
S1 = 0.93434 KJ/kg
At P2 = 900 KPa
S1 = S2
h2 = 274.679 Kj/Kg
h3 = h4 = 101.61 KJ/g
So,
Compressor work (Wc) = h2 - h1
= 274.679 - 247.505
= 27.174
⇒Compressor work (Wc) = 27.174 KJ/kg
Now,
Heat out (Qout) = h2 - h3
= 274.679 - 101.61
= 173.069
⇒Heat out (Qout) = 173.069 KJ/kg
Now,
Heat input (Qin) = h1 - h4
= 274.505 - 101.61
= 145.895
⇒Heat input (Qin) = 145.895 KJ/kg
So,
Actual COP at the refrigerator is -
(COP)R = (Qin)/(Wc)
= (145.895)/ (27.174)
= 5.368
⇒Actual COP = 5.368
Now,
Maximum theoretical COP is -
(COP) = (Qout)/(Wc)
= (173.069)/ (27.174)
= 6.368
⇒Maximum theoretical COP = 6.368
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Answer:
The answer is
C. Split phase motor
Explanation:
Clamp meters rely on the principle of magnetic induction to make non contact AC current measurements. Electric current flowing through a wire produces a magnetic field.
Which is similar to basic mode of operation of electric motor and split phase motor is a type of electric motor.
What is a a clamp on meter?
Clamp meters are electrical testers which have wide jaws that are able to clamp around an electrical conductor. Originally designed as a single purpose tool for measuring AC current, clamp meters now include inputs for accepting test leads and other probes that support a wide range of electrical measurements, the jaws of a clamp meter permit work in tight spaces and permits current measurements on live conductors without circuit interruption.
Answer:
Heat required =7126.58 Btu.
Explanation:
Given that
Mass m=20 lb
We know that
1 lb =0.45 kg
So 20 lb=9 kg
m=9 kg
Ice at -15° F and we have to covert it at 200° F.
First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.
We know that
Specific heat for ice 
Latent heat for ice H=336 KJ/kg
Specific heat for ice 
We know that sensible heat given as
Heat for -15F to 32 F:
Q=858.69 KJ
Heat for 32 Fto 200 F:
Q=6330.74 KJ
Total heat=858.69 + 336 +6330.74 KJ
Total heat=7525.43 KJ
We know that 1 KJ=0.947 Btu
So 7525.43 KJ=7126.58 Btu
So heat required to covert ice into water is 7126.58 Btu.
