Assignment__4 - Assignment#4 PROBLEM 4.1 PROBLEM STATEMENT...

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Assignment #4 9/20/04 PROBLEM 4.1 PROBLEM STATEMENT: Water (2 kg) is contained in the vertical frictionless piston/ cylinder device shown below. The mass of the piston is such that the H 2 O exists as a saturated liquid at 10.0 MPa. There is a heat transfer to the cylinder until the piston reaches some stops, at which point the total volume is 0.02 m 3 . There is an additional heat is transfer to the water until the H 2 O exists as a saturated vapor. Determine the total work and the total heat transfer, and show the process on a T-v diagram. DIAGRAM DEFINING SYSTEM AND PROCESS: GIVEN: Water, m=2 kg, frictionless piston State 1: Saturated liquid (x 1 = 0), P 1 =10.0 MPa State 2: V 2 =0.02 m 3 , P 2 =P 1 =10.0 MPa State 3: Saturated vapor (x 3 = 1), V 3 =V 2 =0.02 m 3 FIND: Total work and total heat transfer (kJ) ASSUMPTIONS: 1. The water in the container is a closed system. 3. No net change in kinetic or potential energy. GOVERNING RELATIONS: First Law, IN,a-b OUT,a-b b a Q - W = U - U QUANTITATIVE SOLUTION: Once the piston reaches the stops, there is no change in volume, so OUT,1-3 OUT,1-2 OUT,2-3 Total work : W = W + W 1 2 1 = mP (v - v ) From Table 11s, v 1 = v L (10 MPa) = 0.001452 m 3 /kg and u 1 =1392.8 kJ/kg. 3 3 2 0.02(m ) v = 0.01m /kg 2(kg) = Thus, 3 OUT,1-2 1000kPa W = 2(kg) 10.0(MPa) (0.01 0.001452)(m /kg) 1MPa =171.0 kJ × × × - - IN,1-3 IN,1-2 IN,2-3 IN,1-2 2 1 OUT,1-2 IN,2-3 OUT,2-3 Q = Q + Q Q = U - U + W Q W 3 2 IN,1-3 3 1 OUT,1-2 = U - U Q = U - U + W Since state 3 is a saturated vapor, u 3 =u V , which from Table 11s, corresponds to v 3 =v V =0.01 m 3 /kg. This value falls between 15 MPa and 16 MPa on Table 11s. Q IN,1-3
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Assignment #4 9/20/04 By interpolation we find the value of u 3 where v 3 =v V =0.01 m 3 /kg. 3 3 3 3 (0.01034-0.01)(m /kg) u = 2455.0(kJ/kg) (2431.3-2455)(kJ/kg) (0.01034-0.00931)(m /kg) u =2447.2 kJ/kg + × Hence, the total heat transfer is IN,1-3 3 1 OUT,1-2 IN,1-3 Q = m(u -u )+ W = 2(kg) (2447.2-1392.8)(kJ/kg)+171.0(kJ) Q =2280 kJ × P-v Diagram DISCUSSION OF RESULTS: In applying the energy balance principle (First Law), it is good practice to show Q and W terms and thin eliminate terms for a given process as appropriate. In this case, for example, - = OUT,2 3 W 0 because the voume stays constant. By explicitly showing this in the equation, it is clear that this has been accounted for. 0.01 m 3 /kg v T 1 3 2 P 3 P 1 = P 2 =10.0 MPa
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Assignment #4 9/20/04 PROBLEM 4.4 PROBLEM STATEMENT: Water (3 kg) in the frictionless piston-cylinder device shown is initially at a pressure of 1 MPa and a quality of 75%. There is a heat transfer until the temperature is 500°C. It takes a pressure of 2 MPa to lift the piston. What is the total heat transferred in the process? DIAGRAM DEFINING SYSTEM AND PROCESS:
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This note was uploaded on 04/20/2010 for the course M E 320 taught by Professor Deinert during the Spring '08 term at University of Texas.

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Assignment__4 - Assignment#4 PROBLEM 4.1 PROBLEM STATEMENT...

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