1259 AM 0 94 KT OO Done thermo prac chapter5pdf any such cyc

12:59 AM 0 94% KT OO Done thermo prac chapter5.pdf any such cycle. s59 As shown in an airconditioner operating ical power in kw kW required when dwelling a Car a day when the and T found in part the dwelling through the wal and roof klh might anet power input to the air conditioner ampresor of 2kW yes determine the of perform no determine the minimum theoretical power inpuL in kW. see A heat pump cycle is used to maintais the interior of building at At steady state, the heat pump enero by heat traneder from well waser at l and di- at a rate of charges hy heat tramlar to the building 120000 klan over a period of days an electric meter record that h of clectricity provided hea pump Determine (a) the amount of energs that the bear pump over the I4dy Period fromithe well water by heat traerler. (b) the heat pump\'s coefficient of performanae revenible heat pump the coefficient cycle operating betweenhet and cold reservoin refrigeration cycle has a of Fig. P5.63 equal to Ts of the value for a revenile Problems: Developing Engineering Skills 251 the energy rejected in kw. ebh the lowest theoretical temperature the the maimum theoretical power in kw.that would be by a power operatingbetweenthe cols the same aaellaient of and the surroundings would you renommend making use of this opportunity for developing power? arranged in s64 At steady state, a heat pump provides NIA50 kuh to dwelling at 3rCon a day when the outside Te and energy by heat TC The power input to the heat pump kw. electricity cousseents per kW-h compare the actual trander to a energy by heat operating with the minimum theoretical operating cont T and rejects transfer from the for each day of operation.

Solution

Solution 5.63

We know that COP is equal to the ratio of desired effect to the work input in a refrigeration cycle.

COP = 4.5 given

Input given, Q= 0.8KW

COP= Desired effect/ input given

= Q1/Q

4.5 = Q1/Q

Q1= 0.8 x 4.5

=3.6KW

= Q1/(Q2-Q1)              as Q1+Q = Q2

4.5 = 3.6/(Q2-3.6)

Q2= 4.4 KW ans.

T1, ambient temp= 20 C = 273+20=293 K

T2, condenser temperature = 28 C= 273+28 = 301 K

T3 = evaporator temperature = ?

COP in terms of temperature can be given as

COP = T3/(T2-T3)

i.e, 4.5 = T3 / (301-T3)

So,

4.5x(301-T3) = T3

T3 = 246.27 K ans.

This is the lowest possible temp. in the given refrigerator which will be at the evaporator.

I would surely suggest to utilize every bit of power available to the maximum extent we can.

Solution 5.63

We know that COP is equal to the ratio of desired effect to the work input in a refrigeration cycle.

COP = 4.5 given

Input given, Q= 0.8KW

COP= Desired effect/ input given

= Q1/Q

4.5 = Q1/Q

Q1= 0.8 x 4.5

=3.6KW

= Q1/(Q2-Q1)              as Q1+Q = Q2

4.5 = 3.6/(Q2-3.6)

Q2= 4.4 KW ans.

T1, ambient temp= 20 C = 273+20=293 K

T2, condenser temperature = 28 C= 273+28 = 301 K

T3 = evaporator temperature = ?

COP in terms of temperature can be given as

COP = T3/(T2-T3)

i.e, 4.5 = T3 / (301-T3)

So,

4.5x(301-T3) = T3

T3 = 246.27 K ans.

This is the lowest possible temp. in the given refrigerator which will be at the evaporator.

I would surely suggest to utilize every bit of power available to the maximum extent we can.

Solution 5.63

We know that COP is equal to the ratio of desired effect to the work input in a refrigeration cycle.

COP = 4.5 given

Input given, Q= 0.8KW

COP= Desired effect/ input given

= Q1/Q

4.5 = Q1/Q

Q1= 0.8 x 4.5

=3.6KW

= Q1/(Q2-Q1)              as Q1+Q = Q2

4.5 = 3.6/(Q2-3.6)

Q2= 4.4 KW ans.

T1, ambient temp= 20 C = 273+20=293 K

T2, condenser temperature = 28 C= 273+28 = 301 K

T3 = evaporator temperature = ?

COP in terms of temperature can be given as

COP = T3/(T2-T3)

i.e, 4.5 = T3 / (301-T3)

So,

4.5x(301-T3) = T3

T3 = 246.27 K ans.

This is the lowest possible temp. in the given refrigerator which will be at the evaporator.

I would surely suggest to utilize every bit of power available to the maximum extent we can.

Solution 5.63

We know that COP is equal to the ratio of desired effect to the work input in a refrigeration cycle.

COP = 4.5 given

Input given, Q= 0.8KW

COP= Desired effect/ input given

= Q1/Q

4.5 = Q1/Q

Q1= 0.8 x 4.5

=3.6KW

= Q1/(Q2-Q1)              as Q1+Q = Q2

4.5 = 3.6/(Q2-3.6)

Q2= 4.4 KW ans.

T1, ambient temp= 20 C = 273+20=293 K

T2, condenser temperature = 28 C= 273+28 = 301 K

T3 = evaporator temperature = ?

COP in terms of temperature can be given as

COP = T3/(T2-T3)

i.e, 4.5 = T3 / (301-T3)

So,

4.5x(301-T3) = T3

T3 = 246.27 K ans.

This is the lowest possible temp. in the given refrigerator which will be at the evaporator.

I would surely suggest to utilize every bit of power available to the maximum extent we can.