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- Two gaseous streams containing the same fluid enter a mixing chamber and leave as single stream. For the first gas the entrance condition are A1 = 500 cm, vj = 130 m/s, pi = 1.60 kg/m3. For the second gas the entrance condition are A2 = 400 cm², v2 = 0.503 m³/kg. The exit stream condition is A3 = 650 cm' v3 = 130 m/s and v3 = 0.437 m/kg. Determine (a) the mass flow rate and (b) the velocity of gas 2.(a) A substance flows through a turbine at the rate of 100 lb./min with ΔKE = 0 and Q = 0. At entry, its pressure is 175 psia, its volume is 3.16 ft.3/lb., and its internal energy is 1166.7 BTU/lb. At exit, its pressure is 0.813 psia, its volume is 328 ft.3/lb., and its internal energy is 854.6 BTU/lb. What horsepower is developed? (b) The same as (a) except that the heat loss from the turbine is 10 BTU/lb. of steam.A thermodynamic steady flow system receives 4.56kg per min of fluid where P1=137.90 KPa,V1=0.0388M^3kg, V2=122M/s, and U1=17.16KJ/Kg. The fluid leaves the system at a boundary where P2=551.6KPa, V2=0.193 M^3/kg, V2=183M/s and U2=52.80 KJ/kg. During passage through the system the fluid receives 3,000J/s of heat, determine the work.
- A thermodynamic steady flow system receives a 4.56 kgm per min of fluid where p1=137.90 kPa, v1=0.0388 m^3/kgm, ϑ1=122 m/s, and u1= 17.16 KJ/kgm. The fluid leaves the system at a boundary where p2= 551.6 kPa, v2= 0.193 m^3/kgm, ϑ2= 183 m/s, and u2= 52.80 KH/kgm. During pasage through the system the fluid receives 3000 J/s of heat. Determine the work.Two gaseous streams containing the same fluid enter a mixing chamber and leave as a single stream. For the first gas, the entrance conditions are Ai 500 cm2, vi 130 m/s, pi 1.60 kg/m3. For the second gas, the entrance conditions are A2 400 cm2, 7h2 8.84 kg/s, u2 z0.502 m3/kg. The exit stream condition is v3 130 m/s and v3 0.437 m3/kg. Determine (a)the total mass leavingThe volume of a compressible fluid system changes from 1 ft^3 to 5 ft^3 during an internally reversible process in which the pressure varies as P=[(100/V) + 50] psia when V is in ft^3. A. For the process, find the area behind the P-V curve. B. If the process is steady flow with AKE=5 BTU, APE=-2 BTU and AH=120 BTU, Find the work. C. If the process is steady flow with AKE=5 BTU, APE=-2 BTU and AH=120 BTU, Find the heat. D. If the process is nonflow, find Work. E. If the process is nonflow, find Heat. F. If the process is nonflow, find AU.
- Consider water (density = 1000 kg/m^3) flow in a piping network. The pressure, velocity, and elevation at a specified point (point 1) of the flow rate are 155 kPa, 1.8 m/s, and 14 m. The pressure and velocity at point 2 are 170 kPa and 3.5 m/s. Neglecting frictional effects, the elevation at point 2 is3. In a refinery, certain oil with sp.gr 0.9 flows with a velocity of 2 m/s in a pipe of diameter 30cm. Along the flow, the pipe diameter gets reduced to 25 cm. Determine the velocity and mass flow rate of oil at this section. 4. A circular pipe of uniform diameter 500mm carries water under pressure 30 N/cm2 . The mean velocity of water at the inlet (at the datum) is 2.0 m/s. Find the total head or total energy per unit weight of the water at a cross-section, which is 5 m above the datum line. 5. A tapered pipe, through which water is flowing, is having diameter, 30cm and 20 cm at the cross-sections 1 and 2 respectively. The velocity of water at section 1 is given as 3.5m/s. Find the velocity head at section 1 and 2 and also rate of discharge. 6. Water is flowing through a pipe having diameter 300mm and 200 mm at the bottom and upper end respectively. The pressure at the lower(bigger) end is 25 N/cm2 and the pressure at the upper end is 10 N/cm2. Determine the difference in datum…1.) While the pressure remains constant at 100 psia the volume of a system of air changes from20 ft3to 10 ft3. What are (a) ∆U in Btu, (b) Q in Btu, (c) Wn, Btu2.) Four pounds of air gain 0.491 Btu/R of entropy during a nonflow isothermal process. If p1= 120 psia and V2 = 42.5 ft3find (a) V1 and T1 (b) Q, and (c) Ws with υ1 = 100 ft/sec and υ2 =500 ft/sec.3.) A 1 hp stirring motor is supplied to a tank containing 22.7 kg of air. The stirring action isapplied for 2 hours and the tank losses 850 kJ/hr of heat. Calculate (a) the paddle work inkJ, (b) change of internal in kJ, and (c) the change in temperature of air after 2 hours,assuming that the process occurs at constant volume.4.) A gas with cp = 5.2 kJ/kg.K and R = 2.056 kJ/kg.K initially at 2000 kPa, 3m3and 115oCundergoes an isentropic process to 105 kPa. Find (a) V2 and T2. What is the work (b) if theprocess is nonflow, (c) if the process is steady flow with ΔKE = 10 kJ?
- 6. A system receives 10,320 Ib/hr of fluid at the following initial properties: u, = 2,731 kJ/kg; P1 = 16 kPa; v1 = 0.081 m/kg; and u1 = 15 m/s. Exit properties are as follows: uz=2,795 kJ/kg, P2= 37 kPa, v2=0.176 m³/kg; and v2 = 17.1 m/s. The fluid rejects 0.76 kJ/s of heat as it goes through the system. Find the following: a. Work in kJ/kg. b. Work produced in 3 minutes. c. Power in MW.Water flows at steady state. It is siphoned from an open container as shown below. The height of the water from the ground is 18.6 feet while the height of point 3 from the ground is 4.2 feet. The atmospheric pressure is 14.7 psia. Determine the velocity at point 2. Assume that the velocity in the tank is 0 since it is very large and the pressure at point 3 is 0 since it is a free jet. Round your answers to 1 decimal place only. No need to write the units. 1 1 1 Hint: Use the formula P, + -=-= PV ₁² ₁ + 8²₁ = P₂+ = Pv₂ + 8 ² ₂ = P3+ -Pv 3+8²3₁ '1 2 WaterWater flows at steady state. It is siphoned from an open container as shown below. The height of the water from the ground is 18.6 feet while the height of point 3 from the ground is 4.2 feet. The atmospheric pressure is 14.7 psia. Determine the velocity at point 2. Assume that the velocity in the tank is 0 since it is very large and the pressure at point 3 is 0 since it is a free jet. Round your answers to 1 decimal place only. No need to write the units. 1 1 1 Hint: Use the formula P₁ + pv²₁ + gz₁ = = P₂ + ½-½pv²₂ + gz₂ = P3 + ² pv²3 + gz3 2 Water 2