Question

Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that the maximum absolute temperature in the cycle is twice the minimum absolute temperature, and the net work output of the cycle is 60 kJ. If the steam changes from saturated vapor to saturated liquid during heat rejection, determine the temperature of the steam during the heat rejection process. The temperature of the steam during the heat rejection process is

Answer 1

The question requests the temperature of steam during the heat rejection process of a Carnot cycle. The efficiency of the Carnot engine is 50%, but without data like the latent heat of vaporization, we cannot find the exact temperature. Additional information such as values from steam tables would be needed to calculate the temperature.

Step-by-step explanation:

The question asks to determine the temperature of the steam during the heat rejection process in a Carnot cycle when the steam changes from saturated vapor to saturated liquid. It is given that the maximum temperature in the cycle is twice the minimum temperature, and the net work output of the cycle is 60 kJ.

According to the Carnot cycle principles, the efficiency (e) of a heat engine is given by:

e = 1 - (Tc / Th)

However, in this case, we know that Th = 2 * Tc, thus:

e = 1 - (Tc / 2 * Tc) = 1 - 1/2 = 1/2

This means the efficiency of the cycle is 0.5 (or 50%). We also know the net work output which is the product of efficiency and the heat input Qh:

W = e * Qh = 0.5 * Qh

The net work is given as 60 kJ, which implies:

60 kJ = 0.5 * Qh
Qh = 120 kJ

During the heat rejection process at the colder temperature (Tc), the heat rejected (Qc) is the difference between the heat input and the net work output:

Qc = Qh - W
Qc = 120 kJ - 60 kJ
Qc = 60 kJ

Since the efficiency is 50%, the heat rejected and heat added are equal in magnitude:

Qc = Qh / 2

Therefore, using the relation Qc = m * hfg, where m is the mass of the steam (0.025 kg) and hfg is the latent heat of vaporization (specific to the temperature Tc), we can find Tc. This value for hfg would need to be looked up in a steam table corresponding to the temperature Tc where the steam changes from saturated vapor to saturated liquid.

However, due to insufficient data provided in the question (such as latent heat values or specific temperatures), we cannot determine the exact temperature Tc using the provided information alone.

Answer 2

The temperature of the steam during the heat rejection process is 42.5°C

Step-by-step explanation:

Given the data in the question;

the maximum temperature T
`_H` in the cycle is twice the minimum absolute temperature T
`_L` in the cycle

T
`_H` = 0.5T
`_L`

now, we find the efficiency of the Carnot cycle engine

η
`_{th` = 1 - T
`_L`/T
`_H`

η
`_{th` = 1 - T
`_L`/0.5T
`_L`

η
`_{th` = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η
`_{th` = 1 - W
`_{net`/Q
`_H`

where W
`_{net` is net work done, Q
`_H` is is the heat supplied

we substitute

0.5 = 60 / Q
`_H`

Q
`_H` = 60 / 0.5

Q
`_H` = 120 kJ

Now, we apply the first law of thermodynamics to the system

W
`_{net` = Q
`_H` - Q
`_L`

60 = 120 - Q
`_L`

Q
`_L` = 60 kJ

now, the amount of heat rejection per kg of steam is;

q
`_L` = Q
`_L`/m

we substitute

q
`_L` = 60/0.025

q
`_L` = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q
`_L` = h
`_{fg` = 2400 kJ/kg

now, at h
`_{fg` = 2400 kJ/kg from saturated water tables;

T
`_L` = 40 + ( 45 - 40 ) (
`(2400-2406.0)/(2394.0-2406.0)\\}` )

T
`_L` = 40 + (5) × (0.5)

T
`_L` = 40 + 2.5

T
`_L` = 42.5°C

Therefore, The temperature of the steam during the heat rejection process is 42.5°C