HW 2-10 Soln by langkunxg

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									Homework #2 Solutions
Fall 2010 - Solution

1. Warmer. Because energy is added to the room air in the form of electrical work.



2. A water pump increases water pressure. The power input is to be determined.
Analysis The power input is determined from
W  V ( P  P )
 
          2    1                                      50 psia
                                      1 Btu               1 hp     
    (1.2 ft 3 /s)(50  10 )psia                     
                                                       0.7068 Btu/s 
                                                                                                   Water
                                 5.404 psia  ft 3                   
                                                                                                 10 psia
    12.6 hp
The water temperature at the inlet does not have any significant effect on the required power.



3. The available head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to
be determined.
Assumptions 1 The flow is steady. 2 Water levels at the reservoir and the discharge site remain constant. 3 Frictional
losses in piping are negligible.
Properties We take the density of water to be           = 1000
kg/m3 = 1 kg/L.                                                    1
Analysis The total mechanical energy the water in a dam
possesses is equivalent to the potential energy of water
at the free surface of the dam (relative to free surface of
discharge water), and it can be converted to work
entirely. Therefore, the power potential of water is its        120 m             overall = 80%
                                                     
potential energy, which is gz per unit mass, and mgz for
a given mass flow rate.
                                             1 kJ/kg                       Turbin         Generator
     emech  pe  gz  (9.81 m/s 2 )(120 m)                 1.177 kJ/kg   e
                                             1000 m 2 /s2                           2
The mass flow rate is
        m  V  (1000 kg/m3 )(100 m3/s)  100,000 kg/s
         
Then the maximum and actual electric power generation become

                                                               1 MW 
          Wmax  Emech  memech  (100 ,000 kg/s)(1.17 7 kJ/kg) 
                                                                             117 .7 MW
                                                                 1000 kJ/s 
                               
          Welectric   overallWmax  0.80(117.7 MW)  94.2 MW
Discussion Note that the power generation would increase by more than 1 MW for each percentage point
improvement in the efficiency of the turbine–generator unit.



4. Superheated water vapor cools at constant volume until the temperature drops to 250°F. At the final state, the
pressure, the quality, and the enthalpy are to be determined.
Analysis This is a constant volume process (v = V/m = constant), and the initial specific volume is determined to be
         P1  180 psia 
                        v  3.0433 ft /lbm
                                      3
                                                                  (Table A-6E)
         T1  500  F  1
At 250°F, vf = 0.01700 ft3/lbm and vg = 13.816 ft3/lbm. Thus                                     H2O
at the final state, the tank will contain saturated liquid-vapor                               180 psia
mixture since vf < v < vg , and the final pressure must be the                                  500F
saturation pressure at the final temperature,
          P  Psat@250 F  29.84 psia

(b) The quality at the final state is determined from                                     T               1
                 v 2 v f             3.0433  0.01700
          x2                                           0.219
                      v fg            13 .816  0.01700
(c) The enthalpy at the final state is determined from
                                                                                                          2
         h  h f  xh fg  218.63  0.219 945.41  426.0 Btu/lbm
                                                                                                                  v

5. One side of a two-sided tank contains an ideal gas while the other side is evacuated. The partition is removed and
the gas fills the entire tank. The gas is also heated to a final pressure. The final temperature is to be determined.
Assumptions The gas is specified as an ideal gas so that ideal gas relation can be used.
Analysis According to the ideal gas equation of state,
         P2  P1
         V 2  V1  2V1  3V 1                                                    Ideal
                                                                                              Evacuated
                                                                                   gas
Applying these,                                                                                  2V1
                                                                                 927°C                        Q
            m1  m1                                                                V1
          P1V 1 P2V 2
               
           T1    T2
           V1         V2
                  
            T1        T2
                           V2     3V
            T2  T1            T1 1  3T1  3927  273 ) K   3600 K  3327 C
                           V1     V1
6. Two rigid tanks that contain hydrogen at two different states are connected to each other. Now a valve is opened,
and the two gases are allowed to mix while achieving thermal equilibrium with the surroundings. The final pressure
in the tanks is to be determined.
Properties The gas constant for hydrogen is 4.124 kPa·m3/kg·K (Table A-1).
Analysis Let's call the first and the second tanks A and B. Treating H 2 as an ideal gas, the total volume and the total
mass of H2 are
         V  V A  V B  0.5  0.5  1.0 m 3                                                        A                  B
               PV         (400 kPa)(0.5 m )                3
         mA   1  
               RT                                   0.1655 kg                                    H2                     H2
               1  A (4.124 kPa  m /kg  K)(293 K)
                                    3
                                                                                                V = 0.5 m3           V = 0.5 m3
               PV             (150 kPa)(0.5 m 3 )                                              T=20C                T=50C
         mB   1  
               RT                                   0.0563 kg                                P=400 kPa             P=150 kPa
               1  B (4.124 kPa  m /kg  K)(323 K)
                                        3

         m  m A  m B  0.1655  0.0563  0.2218 kg
Then the final pressure can be determined from
                 mRT 2           (0.2218 kg)(4.124 kPa  m 3 /kg  K)(288 K)
          P                                                                   264 kPa
                  V                                1.0 m 3
7. A rigid tank contains 6 kg of an ideal gas at 3 atm and 40°C. Now a valve is opened, and half of
mass of the gas is allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the
tank is
(a) 186°C (b) 59°C (c) -43°C (d) 20°C (e) 230°C

Answer (a) 186°C

For R = constant and V = constant

P1 m1 T1
  
P2 m2 T2

       m1 P2       6  2.2 
T2          T1          40  273  459 K (186C )
       m2 P1       3  3 




8. A 1-m3 rigid tank contains 10 kg of water (in any phase or phases) at 160°C. The pressure in the
tank is
(a) 738 kPa (b) 618 kPa (c) 370 kPa (d) 2000 kPa (e) 1618 kPa

Answer (b) 618 kPa



Specific volume ν = 1 (m3) /10 (kg)/ = 0.01 m3/kg.

At this specific volume, νf < ν < νfg

so the water is saturated and, from Table A-4, P = Psat = 618 kPa.

								
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