for P= 36.29 bar,. h₁ = hg =2801.7 h4 = hf = 1057.5 k31kg hfg 1744.2 kJ/kg 43149 T = 244.2 C = 517.2K. .5,' = sg = 6.111 kJ/kg K '= 2.740 kJ/kg. 54= •We have s₁ =52. • 7.080 = S,' + Cvapor in (I+). 7.080 = 6.111+ 2.1 In (STA-2) T= 820.45 K. hh,+Cvapor CT, T₁) hi = 2801.7 + 2.1 (820.45-517.2) h₁ =3438.53 kJ/kg CS Scanned with CamScanner Cp water = 2.1 kJ/kg k 171 for hy Wp = √3 (P4-P3) 0.001065 (36.29-2.26) 30% To produce more power the upper management decided to modify the power plant by placing the stage 2 as a saturated liquid stage. The condenser site was planned to leave as it is with the same parameters. The flow rates stays the same. See the figure below for the data. Again, this an ideal case: S2-S1; S3=54. Boiler P1=36.29bar lin Draw: The T-S diagram of the cycle Turbine Condenser Pump P2=2.26 bar -2 Saturated vapor T6=? 6 Lour T5=15C 5 Cooling water 3 saturated liquid

Solve the following completely. 1. In a boiler observation made were as follows: Rate of feed water per hour = 1,000 kg Temperature of feed water = 30°C Steam pressure = 10 bar Quality of steam = 0.9 Coal consumption rate per hour = 100 kg Calorific value of coal = 33,000 kJ/kg Mass of ash and unburnt coal= 10 kg/h Calorific value of ash and unburnt coal = 3,000 kJ/kg Quantity of flue gas/kg of coal = 20 kg %3D Discharged gas temperature = 300°C Atmospheric temperature = 20°C Specific heat of flue gas = 1.03 kJ/kg K Calculate: (i) Boiler efficiency. (ii) Percentage of heat loss in flue gas. (iii) Percentage of heat loss to ash and unburnt coal. (iv) Percentage of heat loss unaccounted for.

Important to plot the data in the given diagram

Derive the seven general property equation of the following thermodynamics processes: a. ISOMETRIC PROCESS - Any Process relation - Work Non-flow - Internal Energy - Heat Transferred - Enthalphy - Entropy - Work Steady flow refer with the picture. Show the detailed explanation please

THERMODYNAMICS

1.

a) To bring your methane (CH4) to market you have to compress it to a pressure of 1 MPa. You are producing methane from a reservoir at a rate of 2 kg/sec and a pressure and temperature of 0.4MPA and 280K. This methane is run through an adiabatic compressor to bring it up to a pressure of 1 MPa. Using the appropriate balance equations and the chart given, find the minimum amount of power (work per time) required to compress the methane. b) To put methane on a ship for export, the methane needs to be liquefied. This is done by compressing and cooling a methane stream to P = 10MPa and T = 200K. It is then run through a choke valve operating adiabatically with an outlet pressure of P = 0.1 MPa. If the upstream methane stream consists of a mass flow rate of 10 kg/sec, using the appropriate balance equations and the attached chart, find rate at which liquid methane is produced downstream of the valve.

Please answer these 2 questions

A:YV 36 ll 46.ll Test 3.pdf F1 F2 a F3 A d. C a F4 E Consider the following values: - F5 a = 12 m; b = 10 m; c = 10 m; d = 8 m; F1 = 6 kN; F2 = 7 kN; F3 = 8 kN; F4 = 9 kN; F5 = 10 kN; a = 30° 0 = 45° B = 60° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 111.6 kN.m b) 112.6 kN.m c) - 184.6 kN.m d) - 109.6 kN.m e) - 104.1 kN.m f) None of them 2] What is the resultant moment of the five forces acting on the rod about point B? a) -95.8 kN.m b) 42.7 kN.m c) 80.9 kN.m d) 10.9 kN.m e) 51.8 kN.m f) None of them 3] What is the resultant moment of the five forces acting on the rod about point C? a) 32.6 kN.m b) - 21.1 kN.m c) 17.4 kN.m d) - 16.5 kN.m e) 15.6 kN.m f) None of them 4] What is the resultant moment of the five forces acting on the rod about point D? a) 37.6 kN.m b) 73.8 kN.m c) 71.6 kN.m d) 52.1 kN.m e) 21.1 kN.m f) None of them 5] What is the moment of the force F2 about point E? a) 90.7 kN.m b) - 88.1 kN.m c) 54.6 kN.m d) 103.1 kN.m e)…

Drag and Drop the correct answers in the blanks. A mass mis hanging from the end of a vertical spring that has a spring constant k. The equilibrium position of the mass, yo, is The total potential energy of the mass when it is displaced by s from the equilibrium position is given by Y. is zero + U(s) = ks + positive * negative 1 1 ks+ 21 + U(s) ks² + mgyo %3D U(s) mgyo