# Absolute Zero on Thermodynamic Temperature Scale

If more and more engines are added in series to Fig. 7.7, the total work output (W + W  + W + … ) cannot exceed Q1 (according to 1st law).

In the limiting case when total output is equal to Q1, then _the heat rejection to sink by the last engine will be zero. This will violate the Kelvin-Plank’s 2nd law (some heat is to be rejected to sink). Therefore as the number of engines increases the heat rejection of the last engine decreases but it can never be zero. We can imagine that in the limiting case (no. of engines tends to infinity) the heat rejection is zero and the sink temperature in this case is also zero. Thus absolute zero is a conceptual lower limit on thermodynamic scale of temperature. ·

SOLVED EXAMPLES

Example 7.1. Fig. 7.8 shows a heat pump (HP), a heat engine (HE). The heat pump is driven by heat engine. The work out from HE is also driving one machine giving 30KW work input. Find.

(a) the rate of heat supply from the 840°C source i.e., Q1E = ?

(b) the rate of total heat rejection to 60°C heat reservoir (sink).

i.e.,                                          QlP + Q2E = ?

Solution.

ɳ HP = Q1P – Q2P/ Q1P = Q1P – 17 / Q1P

or                                 1 – 273 + 5 / 273 + 60 = Q1P – 17 / Q1P

Q1P = 20.36 KW

[ W = Q1P x ɳ HP = 20.36 x 0.165 = 3.36 kw]

W net for HE = 30 + W = 30 + 3.36

= 33.36 KW

For                   HE,1 –Q2E /Q1E = 1 – 60 + 273 / 840 + 273

or                                 Q2E / Q1E = 0.3

again                                        Wnet = Q1E – Q2E

33.36 = Q1E – Q2E

Solving equations {1) and (2) we get Q1E = 47.66KW. Ans. (a)

Q2E = 0.3 x 47.66 = 14.3 KW

Thus total heat rejection to sink at 60°C

= QIP + Q2E = (20.36 + 14.3) KW

= 34.66 KW. Ans. (b)

Example 7.2. Fig. 7.9 shows a reversed Carnot heat engine (RCHE) (a refrigeration plant).

Calculate the power required to drive the refrigeration plant.

Solution.

We have Q1 /Q2 = T1 /T2

Q1 / 5 = 273 + 25 / 273 -5

Q1 = 5.56 KW

W = Q1 – Q2 = ( 5.56 – 5 ) = 0.56 KW

= Power required.

Example 7.3. A heat engine is used to drive a heat pump. The heat transfers from the heat engine and from the heat pump are used to heat the water circulating through the radiators of a building. The thermal efficiency of the heat engine is 30% and the COP of the heat pump is 5. Evaluate the ratio of the total heat transfers to the circulating water to the heat input to heat engine.

Solution.                          ɳ HE = 0.3

1 – Q2E / Q1E = 0.3

Q2E / Q1E = 0.7

COP HP = Q1P / Q1P – Q2P =5

Q2P / Q1P = 1/5

1- Q2P / Q1P = 1/5

1 – 1/5 = Q2P / Q1P , Q2P / Q1P = 4 /5

We also have                           W = Q1E  – Q2E = Q1P – Q2P

Q1E – Q1E x 0.7 = Q1P – 4 /5 Q1P

0.3 Q1E = Q1P /5

Q1P / Q1E = 1.5

To find               Q2E + Q1P / Q1E = Q2E / Q1E + Q1P / Q1E = 0.7 + 1.5

Example 7.4. If 30 KJ are added to Car not cycle at a temperature of 100°C and 20 KJ are rejected at 00c, determine the location of absolute zero on the Celsius scale.

Solution.              Q2 /Q1 = T2 / T1 = x + 0 / x + 100

20 /30 = x / x + 100

3x = 2x + 200

x  = 200

The location of absolute zero on the Celsius scale will be – 200°C.