Here questions and answers both are available I need your handwriting and also please don't copy and paste chegg answers Thermodynamics of Fluid Flow 59EXAMPLE 2.5Combustion gases flow into a gas turbine stage at a rate of 1,4 kg/s. At the inlet, the static pressure andtemperature are 550 kPa and 950 K, respectively, and the velocity is 180 m/s. At the outlet of the stage,the static pressure is 250 kla and the velocity is 75 m/s. If the power output of the stage is 250 kW, findthe efficiency of the stage when(a) the stage is the first one.(b) the stage is one of the middle stages.(c) the stage is the last one.Take ,- 1.004 kJ/kg K, y= 1.41.Solution: The gas constant is given byR-7-1 = 291.9J/kg KThe density is550 x1000RT= 1.9834 kg/m291.9 x950The total temperature at inlet is given by(180)Toi = 950 += 966.14 K2x1004The total pressure at inlet is given by(180)Po =550 +19834 x= 582.13 kPa2x1000Now by isentropic relation between p, and p, we haveア-INYT,250 41.4= 0.7983550Therefore, the static temperature at the end of isentropic process is given byT= 950 x 0.7983 = 758.385 KAnd the corresponding total temperature is given by(75)T= 758.385 += 761.1862x1004Actual power developed is(7- T) it = 1.004 (966.14 - 761.186)1.4 = 288.1 kW60 Turbomachines(a) and (b)If the stage is the first one or any one of the middle stages, the efficiency may be taken as total-to-total.Therefore,250Efficiency== 86.78%288.1(c)If the stage is the last stage, the efficiency to be considered is total-to-static. Therefore,250Eficiency =2501.004(966.14 - 758.385)250292= 85.62%Comment: In an expansion process or a turbine process, a "static" state at the inlet to the first stagecannot be generally considered because the turbine cannot have a steady work output with the inletfrom a stagnant source (atmosphere or a storage tank). Hence, for the expansion process, "static-to-static" or "static-to-total" efficiency need not be considered, unless it is some sort of "instantaneous"efficiency.2.7 Overall Isentropic Efficiency versus Stage EfficiencyThe overall isentropic efficiency in the present context refers to the thermodynamic efficiency of multi-stageturbomachines. (In Chapter 1, the overall efficiency was identified as the net effect of mechanical, hydraulic,and volumetric efficiencies. The contexts were different there.) A multi-stage machine comprises severalstages. A typical multi-stage air compressor may have about eight or ten stages. However, a multi-stage steamturbine may have even 18 or more stages. In a given machine, generally, each stage is designed to have thesame pressure ratio and the same eficienty. The method of manufacturing the components is not different

Question

Here questions and answers both are available I need your handwriting and also please don't copy and paste chegg answers Thermodynamics of Fluid Flow 59EXAMPLE 2.5Combustion gases flow into a gas turbine stage at a rate of 1,4 kg/s. At the inlet, the static pressure andtemperature are 550 kPa and 950 K, respectively, and the velocity is 180 m/s. At the outlet of the stage,the static pressure is 250 kla and the velocity is 75 m/s. If the power output of the stage is 250 kW, findthe efficiency of the stage when(a) the stage is the first one.(b) the stage is one of the middle stages.(c) the stage is the last one.Take ,- 1.004 kJ/kg K, y= 1.41.Solution: The gas constant is given byR-7-1 = 291.9J/kg KThe density is550 x1000RT= 1.9834 kg/m291.9 x950The total temperature at inlet is given by(180)Toi = 950 += 966.14 K2x1004The total pressure at inlet is given by(180)Po =550 +19834 x= 582.13 kPa2x1000Now by isentropic relation between p, and p, we haveア-INYT,250 41.4= 0.7983550Therefore, the static temperature at the end of isentropic process is given byT= 950 x 0.7983 = 758.385 KAnd the corresponding total temperature is given by(75)T= 758.385 += 761.1862x1004Actual power developed is(7- T) it = 1.004 (966.14 - 761.186)1.4 = 288.1 kW60 Turbomachines(a) and (b)If the stage is the first one or any one of the middle stages, the efficiency may be taken as total-to-total.Therefore,250Efficiency== 86.78%288.1(c)If the stage is the last stage, the efficiency to be considered is total-to-static. Therefore,250Eficiency =2501.004(966.14 - 758.385)250292= 85.62%Comment: In an expansion process or a turbine process, a &#34;static&#34; state at the inlet to the first stagecannot be generally considered because the turbine cannot have a steady work output with the inletfrom a stagnant source (atmosphere or a storage tank). Hence, for the expansion process, &#34;static-to-static&#34; or &#34;static-to-total&#34; efficiency need not be considered, unless it is some sort of &#34;instantaneous&#34;efficiency.2.7 Overall Isentropic Efficiency versus Stage EfficiencyThe overall isentropic efficiency in the present context refers to the thermodynamic efficiency of multi-stageturbomachines. (In Chapter 1, the overall efficiency was identified as the net effect of mechanical, hydraulic,and volumetric efficiencies. The contexts were different there.) A multi-stage machine comprises severalstages. A typical multi-stage air compressor may have about eight or ten stages. However, a multi-stage steamturbine may have even 18 or more stages. In a given machine, generally, each stage is designed to have thesame pressure ratio and the same eficienty. The method of manufacturing the components is not different

Accepted Answer

Given:-m=1.4kg/sAt inlet:T=950 KP=550 kPaV=180m/sAt outlet:P=250 kPaV=75m/sP=250 kWTo…