## What Is Qh In Thermodynamics

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This portion assumes that the temperature of cooling water for the condenser is 25oC and the high-temperature supply from the boiler/steam generator is 320oC

Substituting the component entropy gains into Eq (4.14) gives the entropy production per kg of water circulated:

The largest contribution is due to heat transfer over non-zero temperature differences between the cycle components and their associated thermal reservoirs.

### The Brayton cycle - a gas turbine

The Brayton cycle utilizes a gas (e.g., air, He) as the working fluid which, unlike the water Rankine cycle, is directly heated by the primary energy source. The latter can be either a combustor wherein hydrocarbon fuel burns in air, or a nuclear reactor, in which fission energy is transformed into heat that is removed by the flowing coolant gas. These two variations are shown in Fig. 4.13 along with the T-S process diagram for each. Fig. 4.13 Two types of Brayton cycles. Left: combustion heat source; Right; nuclear heat source (HTGR means High Temperature Gas-cooled Reactor)

The combustion version is not a true cycle in that the coolant air is not recycled to the inlet of the compressor, which draws in fresh air. With the nuclear reactor heat source, the coolant is high-purity helium, which is recycled to the compressor after passing through a cooler. In the combustion version, the hot exhaust simply dissipates heat to the atmosphere. In both cases, states 1 and 2 are at a high pressure pH and states 3 and 4 are at a low pressure pL. The output of the turbine (WT) is divided between the net work produced by the cycle (W) and the work required to operate the compressor (WC), or W = WT - WC. The 1st Law for the cycle is W = QH - QL, where QH is the energy delivered to the gas by the nuclear reactor and QL is the heat rejected to the low-temperature reservoir.

Focusing on the right hand diagram of Fig. 4.13, the cycle efficiency is:

Qh Qh Qh

Flow through the nuclear reactor and the cooler is (approximately) isobaric, so for an ideal gas,

Substituting these into Eq (4.15) yields:

Assuming that flow through the compressor and the turbine is isentropic, Eq (3.14) gives:

Zl = I± = | PL T2 T1 I Ph which leads to: T3/T4 = T2/T1 and to the final result:

Contrary to the Carnot cycle, the Brayton cycle efficiency is determined by the high and low pressures but is independent of the temperature of the low-temperature thermal reservoir (there is no high temperature thermal reservoir)

4.5.2 Refrigeration cycle (heat pump)

This cycle utilizes a condensable fluid such as ammonia (NH3) or freon and resembles a reverse Rankine cycle. A typical refrigeration cycle is shown in Fig. 4.14.

The heat extracted from a low-temperature reservoir converts a two-phase mixture to saturated vapor, which is then compressed adiabatically and (nearly) reversibly. On the T-s diagram, the line from state 1 to state 2 is shown as isentropic. The condenser rejects heat to a high-temperature reservoir and in the process converts the superheated vapor to saturated liquid. The final unit is an expansion valve which, by reducing the pressure of the flow, evaporates some of the liquid to produce the two-phase mixture for the cycle to repeat. This component is highly irreversible, which accounts for the tilt of the 3-4 arrow in the T-s diagam.

A numerical example of the conditions at various states is shown in Table 4.3. The underlined numbers are unit specifications and correspond to the values on the process/EOS diagram in Fig. 4.14. The horizontal lines point to values obtained from the freon equivalent of the steam tables - for example, at - 20oC, the saturation pressure is 1.5 atm and the enthalpy and entropy of the saturated vapor are 179 kJ/kg and 0.71 kJ/kg-K, respectively. The vertical arrows are the result of process restraints (constant pressure, enthalpy, or entropy). Thus the arrow between Fig. 4.14 A refrigeration cycle

states 2 and 3 approximate an isobar. Similarly, the enthalpies at states 3 and 4 are equal because of the 1s Law for a flow device (the expansion valve) with no heat or work exchanged with the surroundings. Finally, the compressor is assumed to be both adiabatic and reversible, so the entropy at states 1 and 2 are the same. T2 is fixed by intersection of the isobar originating at state 3 and the isentrope starting from state 1.

Additional applications of the First Law yield the following information: Condenser: QH = h2 - h3 = 136 kJ/kg

Evaporator: QL = hi - h4 = 104 "compressor (shaft )work = Qh - Ql = h2 - hi = 32 kJ/kg

Instead of an efficiency, the quality of the refrigeration operation is described by the coefficient of performance

The power requirements for a typical household refrigerator with this COP, a cooling capacity (QL) of 3.1 kW and a freon flow rate of 0.03 kg /s is 3.1/3.2 ~ 1 kW.

designed to speed up the flow of a gas. These devices are represented in Fig. 4.15 Fig. 4.15 Flow devices that are both adiabatic and do not perform or accept work

The two classes of flow devices differ in two principal ways: fluid velocity changes and reversibility. In both cases, the open system consists of the section of pipe containing the constriction and imaginary surfaces perpendicular to the flow direction that are sufficiently far upstream and downstream to escape perturbations of the flow by the device. The inlet fluid condition is completely fixed (thermodynamic state and velocity) and the outlet condition is partially specified. The First law, and if applicable, the form of the Second law for reversible processes, are used to calculate the remaining downstream conditions. The mass flow rate is constant in time, so the analysis is on the basis of a unit of flowing mass.

4.6.1 Orifice or valve - water

For an orifice or a valve, the First law reduces to:

Example: What is the outlet steam quality in Fig. 4.15(a)? The assumption that condensation occurs in the outlet stream needs to be verified. Using the upstream pressure and temperature in the superheated steam table A.3, the entrance and exit enthalpies are equal: hi = he = 2785 kJ/kg. This enthalpy is slightly less than the enthalpy of the saturated vapor at the downstream pressure of 1.6 MPa, which is obtained from Table A.2. Because the outlet water is in the two-phase region, the temperature is Te = 201oC, corresponding to saturation at 1.6 MPa. The quality of the downstream steam is:

Suppose, however, that the downstream pressure is specified as 1.0 MPa, In this case, he is greater than the enthalpy of the saturated vapor, and the downstream steam remains superheated. To determine the final state of the gas, steam table A.3 is entered with the combination pe =1.0 MPa, he =2785 kJ/kg, which yields a downstream temperature of 183oC.

That flow through an orifice is irreversible seems intuitive because of fluid turbulence (or at least laminar friction) created by the abrupt reduction in flow cross section. This qualitative assessment can be verified by comparing the entropies of the gas before and after passage through the orifice. At 8 MPa and 300oC, steam table A.3 gives si = 5.79 kJ/kg-K. For the downstream condition (pe = 1.0 MPa, he = 2785 kJ/kg), se = 6.60 kJ/kg-K. Thus, se > s;, proving that gas flow through an orifice is indeed an irreversible process.

4.6.2 Nozzle - ideal gas

The First law applied to a nozzle is:

Because of the smooth shape of the walls, the flow is approximately reversible. Since the system is also adiabatic, the Second law provides the additional relation:

In order to relate the constant-entropy condition of this particular of open system to the changes in pressure and temperature of the ideal gas, Eq (3.14) is employed. Even though this equation was derived for changes in a closed system, it is applicable to open systems as well. The reason is that this equation involves only thermodynamic properties, and so depends only on the initial and final states of the fluid, not on the process that caused the change. Another way of viewing isentropic flow is to imagine that the flow consists of small packets of fluid acting as closed systems undergoing reversible adiabatic expansion or compression as they move from the inlet to the outlet.

Once Te is determined from Eq (3.14), the First law can be applied to complete the solution of the problem.

Example: What are the exit temperature and velocity of steam flowing through the nozzle of Fig. 4.15(b)? The specific heat of steam (Cp) is 33 J/mole-K (Fig. 2.4) or 1833 J/kg-K. The heat capacity ratio CP/CV = 1.337.

Applying Eq (3.14) yields:

+1 -1 ## Getting Started With Solar

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• veera
How to determine qh and ql in diesel cycle?
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