Tan Wee Ser:

It depends on what kind of system you are looking at. For engines, water-steam cycle, there can be isothermal process. Try looking at Rankine, Brayton and Otto Cycles.

Tan Wee Ser:

It depends on what kind of system you are looking at. For engines, water-steam cycle, there can be isothermal process. Try looking at Rankine, Brayton and Otto Cycles.

Lynton Mogridge:

It sounds like you are running a sub-critical two phase refrigeration cycle. This is similar to your domestic refrigerator. If you look up refrigeration cycle on Wikepedia it will give you the details and the normal performance measure COP (coefficient of performance), more detail in Cengel and Boles Chapter 10.

Lynton.

Lynton Mogridge:

It sounds like you are running a sub-critical two phase refrigeration cycle. This is similar to your domestic refrigerator. If you look up refrigeration cycle on Wikepedia it will give you the details and the normal performance measure COP (coefficient of performance), more detail in Cengel and Boles Chapter 10.

Lynton.

Lynton Mogridge:

It sounds like you are running a sub-critical two phase refrigeration cycle. This is similar to your domestic refrigerator. If you look up refrigeration cycle on Wikepedia it will give you the details and the normal performance measure COP (coefficient of performance), more detail in Cengel and Boles Chapter 10.

Lynton.

Weizhi Yao:

Lynton Mogridge:

It sounds like you are running a sub-critical two phase refrigeration cycle. This is similar to your domestic refrigerator. If you look up refrigeration cycle on Wikepedia it will give you the details and the normal performance measure COP (coefficient of performance), more detail in Cengel and Boles Chapter 10.

Lynton.

Thanks, I found that the COP in air conditioner is about 3.x,


If we use 1km copper line, radius is 900 mm2,

At 20 deg c

R = 0.0172*10^-6 * 1000 / (900*10^-6) = 0.0191(ohm)

At 90 deg c, temperature coefficient of resistivity is 0.00393 at 20 deg c

R = 0.0191*1.00393^70 = 0.02513(ohm)


The current is 1000A

P=1000*1000*0.0191=19100(W)

P90 = 1000*1000*0.02513 = 25130(W)

Psav = 25130-19100 = 6030(W)


Per year, the energy saving is

W = Psav* 24*365 = 52822.8(kw*h)   -- per kilometer


To keep the line 20 deg c, the input power of compressor should be

19100/3.5 = 5457.14(W)


It seems it is worth implementing the solution.

Am I right? Could anybody working in electricity domain point out the rough difference between the real value and the ideal value?

 

Weizhi Yao:

Lynton Mogridge:

It sounds like you are running a sub-critical two phase refrigeration cycle. This is similar to your domestic refrigerator. If you look up refrigeration cycle on Wikepedia it will give you the details and the normal performance measure COP (coefficient of performance), more detail in Cengel and Boles Chapter 10.

Lynton.

Thanks, I found that the COP in air conditioner is about 3.x,


If we use 1km copper line, radius is 900 mm2,

At 20 deg c

R = 0.0172*10^-6 * 1000 / (900*10^-6) = 0.0191(ohm)

At 90 deg c, temperature coefficient of resistivity is 0.00393 at 20 deg c

R = 0.0191*1.00393^70 = 0.02513(ohm)


The current is 1000A

P=1000*1000*0.0191=19100(W)

P90 = 1000*1000*0.02513 = 25130(W)

Psav = 25130-19100 = 6030(W)


Per year, the energy saving is

W = Psav* 24*365 = 52822.8(kw*h)   -- per kilometer


To keep the line 20 deg c, the input power of compressor should be

19100/3.5 = 5457.14(W)


It seems it is worth implementing the solution.

Am I right? Could anybody working in electricity domain point out the rough difference between the real value and the ideal value?