Assignment6_solutions
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ENSC 388 Assignment #6 Assignment date: Wednesday Oct. 21, 2009 Due date: Wednesday Oct. 28, 2009 Problem 1 A turbine operating at steady state receives air at a pressure of π1 = 3.0 πππ and a temperature of π1 = 390πΎ. Air exits the turbine at a pressure of π2 = 1.0 πππ. The work developed is measured as 74 ππ½ per ππ of air flowing through the turbine. The turbine operates adiabatically, and changes in kinetic and potential energy between inlet and exit can be neglected. Using the ideal gas model for air, determine the turbine efficiency. πΜπΆπ = 1.0 πππ πΜ 2 π2 = 1.0 πππ π1 = 390πΎ 1 π2 = 1.0 πππ TURBINE Problem 2 Components of a heat pump for supplying heated air to a dwelling are shown in the schematic below. At steady state, Refrigerant 134a enters the compressor at 6 β, 3.2 πππ and is compressed adiabatically to 75 β, 14 πππ. From the compressor, the refrigerant passes through the condenser, where it condenses to liquid at 28 β, 14 πππ. The refrigerant then expands through a throttling valve to 3.2 πππ. The states of the refrigerant are shown on the accompanying T-s diagram. Return air M. Bahrami ENSC 388 Assignment # 6 1 from the dwelling enters the condenser at 20 β, 1 πππ with a volumetric flow rate of 0.42 π3 /π and exits at 50 β with a negligible change in pressure. Using the ideal gas model for the air and neglecting kinetic and potential energy effects, (a) determine the rates of entropy production, in ππ/πΎ, for control volumes enclosing the condenser, compressor, and expansion valve, respectively. (b) Discuss the sources of irreversibility in the components considered in part (a). Indoor return air π5 = 20β π5 = 1 πππ (π΄π)5 = 0.42 π3 /π Condenser π3 = 28β 3 π3 = 14 πππ Expansion Valve π4 = 3.2 πππ 4 Supply air π6 = 50β π6 = 1 πππ π2 = 75β π2 = 14 πππ 2 3 14 πππ 28β Turbine 3.2 πππ 1 Evaporator M. Bahrami 75β 2 T π1 = 6 β π1 = 3.2 πππ 1 6β 4 s Outdoor air ENSC 388 Assignment # 6 2 Problem 1: Known: ππ‘ππ‘ππ ππππππππ‘πππ πππ π‘βπ ππππππ π ππ . Find: - Turbine efficiency. π 3 πππ 1 π1 = 390πΎ Actual expansion Isentropic expansion 2s 2 1 πππ π Assumptions: - Problem is steady state. - The expansion is adiabatic and the changes in kinetic and potential energy between inlet and exit can be neglected. - The air is modeled as an ideal gas. Analysis: Isentropic turbine efficiency can be obtained from πΜπΆπ πΜ ππ‘ = πΜ ( ππΆπ Μ )π The numerator of the efficiency is known. The denominator is evaluated as follows. M. Bahrami ENSC 388 Assignment # 6 3 The work developed in an isentropic expansion from the given inlet state to the specified exit pressure is πΜπΆπ ( ) = β1 − β2π πΜ π From Table A-17, at 390πΎ, β1 = 390.88 ππ½/ππ. To determine β2π , applying an exact solution leads to π2 ππ (π2π ) = ( ) ππ (π1 ) π1 With π1 = 3 πππ, π2 = 1 πππ and ππ (π1 ) = 3.481 form Table A-17 at 390πΎ 1 ππ (π2π ) = ( ) × 3.481 = 1.1603 3 Interpolating in Table A-17 gives β2π = 285.27 ππ½/ππ. Thus πΜπΆπ ππ½ ππ½ ππ½ ( ) = 390.88 [ ] − 285.27 [ ] = 105.6 [ ] πΜ π ππ ππ ππ Hence isentropic turbine efficiency will be ππ½ 74 [ ] ππ ππ‘ = = 0.7 ππ 70% ππ½ 105.6 [ ] ππ M. Bahrami ENSC 388 Assignment # 6 4 Problem 2: Known: ππ‘ππ‘ππ ππππππππ‘πππ πππ π‘βπ ππππππ π ππ . Find: - The entropy production rates for control volumes enclosing the condenser, compressor, and expansion valve, respectively and discuss the sources of irreversibility in these components. Assumptions: - Each component is analyzed as a control volume at steady state. - The compressor operates adiabatically, and the expansion across the valve is a throttling process. - For the control volume enclosing the condenser, πΜπΆπ = 0 and πΜπΆπ = 0. - Kinetic and potential energy effects can be neglected. - The air is modeled as an ideal gas with constant ππ = 1.005 ππ½/ππ. πΎ. Analysis: (a) Lets us begin by obtaining property data at each of the principal refrigerant states located on the T-s diagram. At the inlet to the compressor, the refrigerant is a superheated vapour at 6 β, 3.5 πππ, so from Table A-13, π 1 = 0.9415 ππ½/ππ. πΎ. Similarly, at state 2, the refrigerant is superheated vapour at 75 β, 14 πππ, so interpolating in Table A-13 gives π 2 = 0.9561 ππ½/ππ. πΎ and β2 = 291.3 ππ½/ππ. State 3 is compressed liquid at 28 β, 14 πππ. From Table A-11, π 3 ≈ π π@28 β = 0.33846 ππ½/ππ. πΎ and β3 ≈ βπ@28 β = 90.69 ππ½/ππ. The expansion through the valve is a throttling process, thus β4 = β3 . Using data from Table A-11, the quality at state 4 is π₯4 = M. Bahrami β4 − βπ@3.2πππ βππ@3.2πππ ππ½ ππ½ 90.69 [ ] − 55.16 [ ] ππ ππ = = 0.18 ππ½ 196.71 [ ] ππ ENSC 388 Assignment # 6 5 And the specific entropy is π 4 = π π@3.2πππ + π₯4 π ππ@3.2πππ = 0.21637 [ ππ½ ππ½ ] + 0.18 × 0.71369 [ ] ππ. πΎ ππ. πΎ ππ½ → π 4 = 0.3448 [ ] ππ. πΎ Condenser: Consider the control volume enclosing the condenser. With the first and third assumptions, the entropy rate balance reduces to 0 = πΜπππ (π 2 − π 3 ) + πΜπππ (π 5 − π 6 ) + π Μπππ,ππππ To evaluate π Μπππ requires the two mass flow rates, πΜπππ and πΜπππ , and the change in specific entropy for the air. These are obtained next. Evaluating the mass flow rate of air using the ideal gas model πΜπππ π5 π3 100[πππ] ππ = (π΄π)5 = 0.42 [ ] × = 0.5 [ ] ππ½ π π π π β5 0.286 [ ] × 293 [πΎ] ππ. πΎ π The refrigerant mass flow rate is determined using an energy balance for the control volume enclosing the condenser together with the 1 st, 3rd and 4th assumptions to obtain πΜπππ = πΜπππ (β6 − β5 ) β2 − β3 With the 5th assumption, β6 − β5 = ππ (π6 − π5 ). Inserting values πΜπππ ππ ππ½ 0.5 [ ] × 1.005 [ ] × (323[πΎ] − 293[πΎ]) ππ π ππ. πΎ = = 0.075 [ ] ππ½ ππ½ π 291.3 [ ] − 90.69 [ ] ππ ππ The change in specific entropy of the air is M. Bahrami ENSC 388 Assignment # 6 6 π 6 − π 5 = ππ ππ π6 π6 ππ½ 323[πΎ] 1[πππ] − π ππ = 1.005 [ − π ππ ] × ππ π5 π5 ππ. πΎ 293[πΎ] β 1[πππ] 0 → π 6 − π 5 = 0.098 [ ππ½ ] ππ. πΎ Finally, solving the entropy balance for π Μπππ and inserting values π Μπππ = πΜπππ (π 3 − π 2 ) + πΜπππ (π 6 − π 5 ) ππ ππ½ ππ½ ππ ] × (0.33846 [ ] − 0.9561 [ ]) + 0.5 [ ] π ππ. πΎ ππ. πΎ π ππ½ ππ × 0.098 [ ] = 2.677 × 10−3 [ ] ππ. πΎ πΎ π Μπππ,ππππ = 0.075 [ Compressor: For the control volume enclosing the compressor, the entropy rate balance reduces with the first and third assumptions to 0 = πΜπππ (π 1 − π 2 ) + π Μπππ,ππππ or ππ ππ½ ππ½ π Μπππ,ππππ = πΜπππ (π 2 − π 1 ) = 0.075 [ ] × (0.9561 [ ] − 0.9415 [ ]) π ππ. πΎ ππ. πΎ ππ = 1.095 × 10−3 [ ] πΎ Valve: Finally, for the control volume enclosing the throttling valve, the entropy rate balance reduces to 0 = πΜπππ (π 3 − π 4 ) + π Μπππ,π£ππ Solving for π Μπππ,π£ππ and inserting values M. Bahrami ENSC 388 Assignment # 6 7 π Μπππ,π£ππ = πΜπππ (π 4 − π 3 ) = 0.075 [ ππ ππ½ ππ½ ] × (0.3448 [ ] − 0.33846 [ ]) π ππ. πΎ ππ. πΎ ππ = 4.755 × 10−4 [ ] πΎ (b) The following table summarize, in rank order, the calculated entropy production rates ππΎ ] π² compressor 1.095 × 10−3 valve 4.755 × 10−4 condenser 2.677 × 10−3 Component πΜ πππ [ Entropy production in the compressor is due to fluid friction, mechanical friction of the moving parts, and internal heat transfer. For the valve, the irreversibility is primarily due to fluid friction accompanying the expansion across the valve. The principal source of irreversibility in the condenser is the temperature difference between the air and refrigerant streams. In this problem, there are no pressure drops for either stream passing through the condenser, but slight pressure drops due to fluid friction would normally contribute to the irreversibility of condensers. The evaporator lightly has not analyzed in this solution. Note (1): Since the temperature difference is large in problem one, applying approximate solution may leads to considerable error. Try to apply the approximate solution and see how different the result is. Note (2): Due to relatively small temperature change of the air, the specific heat ππ can be taken constant at the average of the inlet and exit air temperature. Note (3): Temperature in πΎ are used to evaluate πΜπππ , but since a temperature difference is involved the same result would be obtained if temperature in β were used. Temperatures in πΎ must be used when a temperature ratio is involved. M. Bahrami ENSC 388 Assignment # 6 8
