thermorevsteamQ_A

thermorevsteamQ_A - 98 Brett Road Piscataway NJ 08854-8058...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
98 Brett Road Piscataway NJ 08854-8058 Phone: (732) 445-3655 FAX: (732) 445-0730 email: muller@caes.rutgers.edu Center for Advanced Energy Systems Michael R. Muller, Director Problem Set: Thermodynamics Review - Steam 1. An adiabatic piston-cylinder system contains a 1000-W immersion heater and 4 kg of H 2 0 initially at 1 atm and 96 percent quality. The heater is operated for 7 min., during which the pressure is held constant. Find the final volume of the system. Solution: Process is constant pressure. It is a control mass problem. Recall with control mass you can determine changes in internal energy and work separately. Normally this is the simplest way if you have the final conditions. In this problem you are only told that it is a constant pressure process. For this case, the heat input is equal to the change in enthalpy which fixes the final state. M Q h h Q H H V V P Q U U PdV Q dW dQ dE initial final initial final initial final initial final / ) ( = = = = + = δ Find the energy added by heat transfer into the system by the immersion heater (a Watt is a joule/sec). The energy added is 420 kJ or 105 kJ/kg The original state is fixed by the pressure and quality of the steam: The original enthalpy - 2585.71 kJ/kg The original specific volume is – 1.61 m 3 /kg The original volume is - 6.44 m 3 Get the second enthalpy by using the first law: New enthalpy – 2690.71 kJ/kg (just add the heat in) Final specific volume – 1.72 m 3 /kg Final volume = 4 kg * 1.72 m 3 /kg = 6.88 m 3 Note, even though you equated enthalpy change with heat transfer to solve this problem, there is work done. Determine the amount of work out of the system:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Office of Industrial Productivity and Energy Assessment 2 Michael R. Muller, Director W = P(V final – V initial ) = 100000 N/m 2 x (0.44) or 44 kJ And the internal energy change can then be determined from the first law 2. Steam flows through a small steam turbine at the rate of 10,000 kg/h, entering at 600 0 C and 2.0 MPa and emerging at 0.01 MPa with 4 percent moisture. The flow enters at 50 m/s at a point 2 m above the discharge and leaves at 80 m/s. Compute the shaft-power output, assuming that the device is adiabatic but considering kinetic and potential energies. How much error would be made were these secondary terms neglected? What are the areas of the inlet and discharge pipes? Solution: Note – 4% moisture is the same as a quality of 96% Find inlet states: h = 3690 kJ/kg v = 0.1996 m 3 /kg mass flow rate = 10000 kg/h yields volume flow rate = 1996 m 3 /hr. With velocity of 50 m/s obtain inlet area = 0.011 m 2 Find outlet states: h = 2489 kJ/kg v = 14.09 m 3 /kg mass flow rate = 10000 kg/h yields volume flow rate = 140,900 m 3 /hr. With velocity of 80 m/s obtain
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

thermorevsteamQ_A - 98 Brett Road Piscataway NJ 08854-8058...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online