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.

3G 9:.. %79 Final 1st attem... F2 У - ахis Fs F4 F1 х - ахis X- axis y - axis F3 F2 F1 Figure (1) Figure (2) 250 N 36.87 D F2 60° 0.5 m 1 m o! 0.2 m 1 m 300 NE F1 1 m Figure (3) Figure (4) Q1] Answer the following questions: For figure (1): The object is subjected to two forces F1 = 25 N, and F2 = 50 N. Set x = 30°, 0 = 45° , and ß = 30°. 1- What is the magnitude of the resultant of these two forces? a) 39.4 N b) 80.1N c) 41.6 N d) 69.8 N e) 71.4 N f) 46.1 N g) None of them 2 - What is the direction of the resultant force measured counterclockwise from the positive x-axis? а) 14.30 b) 392.6° c) 9.2° d) 322.9° e) 6.8° f) 344.2° g) None of them For figure (2): The arm in is subjected to five forces as follows: F1 = 135 N, F2 = 98 N, F3 = 210 N, F4 = 56 N. If the y – direction resultant force is 144 N, and x = 40°, 0 = 50°, B = 70°: 3 – What is the magnitude of the force F5? a) 318.4 N b) 99.1N c) 151.6 N d) 36.5 N e) 111.2 N f) 18.8 N g) None of them 4 - What is the magnitude of the…

1-find the internal energy in KJ of 2 kg water contained in 1120 liter at P-1 bar & v= 0.56 m^3/kg.

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

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)…

An open system is often referred to as control volume, which is a properly selected region in space in which mass and energy can flow across the boundaries as figure 1.2. The boundary of a open thermodynamic system is called the control surface Across the Boundaries E = Yes F 0 = Yes w =Yes Control surface ass YES W CONTROL VOLUME energy YES Figure 1.2. A cooling/heating radiator is an example of such a system – give two more examples of such a system.

From 20 °C to 90 °C, 0.5 kg of an ideal gas is heated at a constant pressure of 1 bar. What is the value of the work done by the gas? Assume: C= 1.005 KJ/Kg K and C= 0.718 KJ/Kg K. Select one: O a. 100.45 K] O b. 1004.5J Oc 10.045 KJ O d. 800 J

7 Table 1. Saturation: Temperatures Sperie Velume Internal Energy Esthalyy Eatregy 65.37 100as 1192 20 I 101342 190.5 24042 64.20 idie s4 I6 2604I 10I15 1858 24 1021 0 ITRIO 28041 61 102386 1502 2604I 1927 2 I776.2 28040 1028 54 13755 2040 10257 171T1.4 18009 3000 12187 2455E 3.523 4179 240 35071 6170 237 3.171 63.0 1019 2ó išs49 1224 122개 232 3A146 6.159 24 3484 6108 239 386 19. 10332I IST0 26040 10132 I10.5 2803. 58.7 103T90 IM60 260 1042.09 1 7616 280 T 37M 104259 i5612 2603A 1046 756.7 205 S67 1047 2 iss4 2603T 10S 64 1751.7 28014 272)3936 61225 55. 1051.99 1555 260 S 1056.43 1346 2832 3344 240 241 242 243 2.1015 34422 6.14 27101 34260 6.1366 2.7I 340 61296 12291 3463 3.534 3.56 12334 12155 12177 244 27380 3374 61154 54.11 105.7I 15467 2603.4 106.23 L741.7 2803.0 53.75 64 154 2603.2 52.2 6 16 1S3 26030 100 RS 1731 225 S1.0 T08 1339 202 101 6 i1765 28022 S1 0 1075 64 15210 26026 1080 51 7214 28019 245 24 247 24 3.648 3.11 3.776 340 1239 12422 13444 1246 i 249 2.1412 33612 6108)…

%.00 @ 4G JO 100 I. Hw 1 Name: How many minutes you took ID. to solve this problem? Section: HW-1 Determine the magnitude and coordinate direction angles of the resultant force acting on the bracket. F = 450 N 45 60 F, = 600 N