# THERMAL PART

Q1: (60 marks)

#### A piston-cylinder device initially contains air at 200 kPa and 127 0C. At this stage, the piston is resting on a pair of stops, as shown in the figure below and the enclosed volume is 200 litres. The mass of the cylinder is such that a 300 kPa inside pressure is required to move it. The air is now heated until its volume has doubled. Determine (a) the final temperature, (b) the work done by the air, and (c) the total heat transferred to the atmosphere.

**Solutions **

In such problem, volume, temperature, mass, pressure, and friction are dependable to access the solution. In the assumption that, there is no friction amid the hedges of cylinder and piston, and by assuming the volume employed by the stops. Then, the solution of final temperature, work was done by air and amount of heat transferred to the air should are calculated in the following ways.

a) Temperature = initial temperature multiply by final pressure divided by the initial pressure inside the container. That is: T2 = T1(P2/P1)

Implying that, T2 = 400.5*(300/200) 0C = 537.25 0K

T3 = 537.25 0K * 2 = 947.5 0C

Therefore, the final temperature produced by heating the air in the container is twice the amount of initial temperature. The result is because of direct proportionality of temperature to pressure and volume. Therefore, when the air in the container heated, the volume doubles with relation to pressure hence the temperature could also multiply twice.

#### b) Work done by the air equals the integration of the change of pressure concerning the change of volume in the heated air. The change of pressure in the heated air could affect the energy to be produced by the piston while pressing the volume in the air downwards. Therefore, the work was done by pressing the piston solved by applying the following method. Work is done given by,

Then, letting P2 = final pressure, V3 = final temperature after heating the air in the container and V2 = the initial temperature before heating the air in the container.

Therefore, work done (W) = P2(V3-V2).

Again, letting V1 = mass multiplied by the rate of temperature and by initial temperature, then altogether divided by the initial pressure of air experienced from a piston.

That is V1 = mRT/P = [1*0.56702*(273.15+127)] / 200 = 1.13446m3

Therefore, the work done (W) = 300kPa (2*1.13446m3 – 1.341448m3) =278.2416Kj

The amount of work done on pressing the piston in the container depends on the pressure, and the volume produced when the air was heated. Thereby, the production of energy depending on the work manufactured during pressing of the piston in the container could also affect the amount of pressure resulted.

The total amount of heated transferred from the piston to the container could result from the change of energy plus the work done in the container. The change of energy produced is thereby, the mass of air resulted multiplied by a change of temperature and finally multiplied by the volume of capacity. Therefore, ∆E = M * CV (T3-T1)

Again, CV @750 0K = 0.8

The total of heat transferred (Q) = 1*0.8*(947.5 – 400.5) + 278.2416 kJ = 715.8416Kj

The sum of heat produced when the air in the container heated depends on the mass of air resulted and since the volume at the final stops ignored, then the amount of mass obtained could be less compared to when the volume at last press could consider.

#### Experiment analysis

The change of temperature in the air makes an enormous difference of work done during the pressing on of the piston to the container. The result is due to the more significant gap created by the change of volume when the air was heated. Regarding the assumptions made during the experiment, the friction amid to cylinder hedge and the piston could not account for the calculation which in real life situation the conduction of the two parts will produce frictional force. This resultant to the minimum sum of the work done and the amount of energy produced when the air is heated to double the volume in the container.

Again, the cross-sectional area of the springe could also affect the amount of energy and work done from the pressed and heated air. The occupation of air in the container could determine the total volume of the air to be heated. Therefore, the larger the cross-sectional area, the more significant the amount of air to be burned and vice-versa. This resultant is due to the dependent of volume to the field across the piston area covering the air. Similarly, the cross-sectional area affects the pressure exerted by the piston in the container. The initial volume should equal the capacity of the system since there was no force used at the equilibrium position of a cylinder and the spring is just touching the walls of the container.

Thereby, the dislocation occurred in the container is directly proportional to force of piston, then, the pressure in spring is given by relating piston displacement to change of volume in the bottle.

The amount of work done depends on the sum of pressure produced when the piston pressed assuming no frictional forces obtained from the container and the springe. Again, the more substantial amount of volume attained affects the total work done by pressing of the piston into the bowl.

The sum of pressures in integration depended on the components of work produced. The initial load is constant since there was no change of volume.

Therefore, calculating total work done in spring applies the following formula.

Thus, the amount of heat transferred from the heated container to the springe depends on the amount of energy resulted from pressing the piston plus total amount of obtained work produced.

According to the assumptions made initially, the final volume in the spring could account the amount work done to the springe. Again, the frictional forces deduct the amount of energy released piston in conduct to the container. These results affect volume and pressure exerted by the piston. Finally, the assumption of an ideal gas to be pressed, there was the considerable change of temperatures and pressure obtained, (Kurozawa, Oliveira & Hubinger, 2015).

Q2: (40 marks)

#### A mixture of water and steam has the pressure of 0.8 bar and dryness of 0.9. Calculate its specific volume and specific enthalpy of the steam. (Use the Thermodynamic Properties of Water to do manual calculations.)

The boiling point of water been up to 100%, it produces drenched steams which are due to the small foams contravening the water shells and pull the small water droplets in accompanied by steam. Therefore, the lack of superheater in use, the steam source central to partly wet from the summed of liquid.

**The portion of steam dehydration **

Steam dehydration fraction is used to enumerate the total amount of water within the vapour. That is if the gas contains 19% water by mass, then, its dehydration is 81% dry, naturally, have a dryness fraction of 0.81. The primary importance of steam dehydration is that it has an unswerving outcome on the total quantity of transportable energy confined within the vapour. The result affects the heating quality and competent especially in water.

Calculation of total latent heat

The specific volume of latent heat of water is obtained using the following formula

V = x * v_{g}+ (1- x) * v_{f}

Whereby, x is dryness in percentage or fraction, v_{f} is the specific volume of full water, and vg is the specific volume of saturated steam. Thus, the value of the specific volume of saturated water is negligible because it is likely similar to the specific volume of saturated vapour.

Therefore, specific volume of wet steam (V) = 0.9*0.1 + (1- 0.9) = 0.19cm3/g

Calculation of specific enthalpy of wet vapour

Specific enthalpy of wet steam is outputted: h = h_{f} + x* h_{fg}, whereby

X = dryness of steam

hf = specific Enthalpy of saturated water

hfg = specific enthalpy of saturated steam minus specific enthalpy of saturated water.

According to the steam table, 0.8 bar ranged the hf and hfg at 781kJ/kg and 2000kJ/kg respectively. Then, h = 781 + (0.9 * 2000) =2581kJ

#### Analysis of changes in steam

Evaporation process takes place when the liquid is changing into steam. Similarly, the process of reversing the steam state to liquid is called liquefaction or known as condensing process. If the fluid is heated, the temperature increases in proportionality to heat shifted. Therefore, the heat transferred is outputted by, Q = mc∆T. The great change of temperature and pressure affect the change of specific heat of the liquid.

The transfer of energy and heat in the liquid is more affected by the internal energy and enthalpy. Enthalpy and the specific internal energy are well known, hence recorded in a particular table. When the fluid heated to boiling point, it gains sensible power represented as uf or hf. When the liquid becomes saturated with heat and no more absorption of heat without changing state (from liquid to vapour and gaseous form), the process is called evaporation. The boiling point of liquid shows the saturation temperature of the fluid hence such liquid is called saturated liquid.

The super heat supplied to the liquid makes the liquid to evaporate which in turn drives the vapour. The resultant gas at saturation temperature could term as dry saturated vapour. Therefore, by definition, the steam is gaseous state adjoining to the temperature at which it determinates condensation of such fluid. H_{fg} and u_{fg} symbolise the quantity of enthalpy and inner energy needed to evaporate 1 kg. This h_{fg} and u_{fg} of liquid are called latent enthalpy and latent internal energy fluid correspondingly.

The evaporation process depends on temperature and atmospheric pressure. Boling point of the liquid is directly proportional to the force exerted on it. Therefore, the higher the pressure, the higher the boiling point. Again, the lower the pressure, the less the boiling point of the fluid. Keeping the temperature of a molten constant, the change of pressure could also make boil. The end at which the water boils due to change of force when temperatures are kept constant is called saturation pressure, (Mehta, More, & Malek, 2015).

In increasing the heat on vapour, the latent water becomes hotter compared to boiling point, and if the temperature incremented, then the steam turns to gas, hence is known as superheated vapour.

#### The specific volume of latent water

The specific capacity of full water(v_{f}) and the particular quantity of a dehydrated saturated vapour (v_{g}) determine the amount of specific volume of latent water. Again, v_{f} is negligible because v_{f} is too small and also slightly similar to vg. All changes of state are measured under mass of water. The saturation curve explains the rate of temperature against the change of pressure when the pool is heated. Amount of potential internal energy depends on superheat provided to the liquid. The curve for the civil power shows increment as the pressure increases. Such reason is due to direct proportionality of the pressure to the latent internal heat of the liquid. All the changes in fluids depend on a shift in temperature and change of force exerted to the liquid. Superheated vapours which result in gas could also tabulate under heat and pressure graph. In such case, pressure and temperature graph accommodate all the changes in the gaseous and in liquid states, (Pourshaghaghy, 2015).

The amount of energy produced during the evaporation and condensation process depends on the pressure and the amount of temperature applied to the liquid. Again, the evaporation method requires higher temperature and less weight compared to condensation process which acts vice-versa.

**References**

Kurozawa, L.E., Oliveira, R.A., Hubinger, M.D. and Park, K.J., 2015. Thermodynamic Properties of Water Desorption of Papaya. *Journal of food processing and preservation*, *39*(6), pp.2412-2420.

Mehta, R.N., More, U., Malek, N., Chakraborty, M. and Parikh, P.A., 2015. Study of stability and thermodynamic properties of water-in-diesel nanoemulsion fuels with nano-Al additive. *Applied Nanoscience*, *5*(8), pp.891-900.

Pourshaghaghy, A., 2015. CALCULATION OF SOME THERMODYNAMIC PROPERTIES OF WATER FOR THE BOUNDARY POINTS BETWEEN REGIONS 3 AND 4 OF INDUSTRIAL FORMULATION IAPWS-IF97. *Journal of Engineering Studies and Research*, *21*(1), p.66.