ME 354 Tutorial, Week#5 Refrigeration Cycle A large refrigeration plant is to be maintained at -15 o C, and it requires refrigeration at a rate of

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1 ME 54 Tutorial, Week#5 Refrigeration Cyle A large eration plant is to be maintained at -5 o C, and it requires eration at a rate of 00 kw. The ondenser of the plant is to be ooled by liquid water, whih experienes a temperature rise of 8 o C as it flows over the oils of the ondenser. The plant uses erant-4a between the pressure limits of 0 kpa and 700 kpa. Assuming the ompressor has an isentropi effiieny of 75%, determine: a) The mass flow rate of the erant b) The power input to the ompressor ) The mass flow rate of the ooling water d) The rate of exergy destrution assoiated with the ompression proess (assume T 0 5 o C)

2 Step : Write out what is required to solve for a) The mass flow rate of the erant b) The power input to the ompressor ) The mass flow rate of the ooling water d) The rate of exergy destrution assoiated with the ompression proess (assume T 0 5 o C) Step : Property table Step 4: Assumptions Assumptions: ) Δke, Δpe 0 ) SSSF ) Throttling proess is adiabati 4) State is a saturated P0 kpa 5) State is a saturated P700 kpa Step 5: Solve Part a)

3 State Table MPa h h g@p0 kpa kj/ s s g@p0 kpa.7487 kj/k Performing an energy balane on the throttling valve using the assumptions Δke, Δpe 0 and the proess is adiabati gives h 4 h. So h 4 will be known if h is known. State is saturated liquid at a pressure of 700 kpa, therefore h an be determined using Table A-. State Table A-@P0.7 MPa h h 4 kpa 6.90 kj/ Substituting these values into Eq the value of m& 00 Q& L s (orreted) ( h h4) ( ) 0.68 /s Answer a) Part b) The power input to the ompressor, on the ompressor as shown in Eq. an be determined. W, an be found by performing an energy balane W h h (Eq) was alulated in part a) but (h -h ) must still be determined. The problem statement provides an isentropi effiieny for the ompressor. Rearranging the definition of the isentropi effiieny, as shown in Eq, an expression for (h -h ) in terms of known quantities, h, and h s, whih an be determined, is found. ( h s h ) ( h h ) s η ( hs h ) η (Eq) For state s, s s kj/*k. Using this information with Table A-, it is found that state s is outside of the vapor dome and into the superheated region i.e. s >s g@p700 kpa. Looking in Table P0.7 MPa, it is found that s lies between the entropies with orresponding temperatures of 0 o C and 40 o C. Using s kj/*k to interpolate between the entropies at 0 o C and 40 o C, we an find the enthalpy at the state s.

4 ( h s hs 0.9) s s h s h s s s 0.9 h s (0.507)( ) (Not Corret) h s 4.97 Note: h s is what the enthalpy at state would be if the ompression proess from was isentropi i.e. s s. Substituting Eq into Eq, the value of W an be determined. ( ) ( hs h) & & (orreted) W m 0.68 η s kj/s Answer b) Part ) Part ) asks for the mass flow rate of the ooling water, w, used in the ondenser. The rate of heat transfer out of the ondenser must be equal to the rate of heat transfer into the ooling water. The rate of heat transfer out of the ondenser, Q, an be determined by H performing an energy balane on the ondenser as shown in Eq4. Q H h h (Eq4) The rate of heat transfer into the ooling water an be expressed as a funtion of the temperature rise the ooling water experienes using Eq5. Q ΔT (Eq5) H w Combining Eq4 and Eq5, and rearranging to isolate for the mass flow rate of the ooling water, Eq6 is obtained. Q H w ΔT w ΔT (Eq6) 4

5 h and have been previously determined, so only h,, and Δ T need to be found. Rearranging Eq, the enthalpy at state, h, an be found. h s η + h h From Table A-, for water is given as 4.8 kj/* o C. The problem statement gives the temperature rise in the ooling water as 8 o C i.e. Δ T 8 o C. Substituting the values of h,, Δ T, h and into Eq6, we an determine the mass flow rate of the ooling water. w ΔT 0.68 s kj 4.8 o C ( ) o ( 8)[ C] 4.00 /s Answer ) Part d) The rate of exergy destrution an be expressed in terms of the surroundings temperature, T 0, and the rate of entropy generation, S GEN, as shown in Eq7. T (Eq7) 0S GEN We an determine the entropy generation term, S the ompressor as shown in Eq.8 S in out GEN system GEN, by applying an entropy balane on S + S ΔS (Eq8) Sine there is no heat transfer in or out of the ompressor, the only mehanism for entropy transfer into and out of the ompressor is by the mass flow. Therefore S and S are equal to sand s respetively. Using the steady state assumption, out Δ S is zero. Applying the above information to Eq8, Eq9 is obtained. system Substituting Eq9 into Eq7, Eq0 is obtained. S GEN s s (Eq9) ( s ) X T (Eq0) 0m s in 5

6 s an be determined from Table P0.7 MPa using h 85.7 kj/ to interpolate between the entropies orresponding to temperatures of 40 o C and 50 o C. ( s sh ) ( s s ) h 88.5 h s s h ( s s ) h 88.5 h s (0.79)( ) Substituting the known variables, assuming T 0 5 o C, into Eq0, the rate of exergy destrution during the ompression proess an be determined. kj K kj ( )[ K]( 0.68) ( ) s K 7.87 /s Answer d) Alternative: The exergy an also be found by performing an exergy balane on the ompressor. in out system Δ (Eq) Sine there is no heat transfer in or out of the ompressor the only mehanism for exergy transfer into and out of the ompressor is by the mass flow and the work oming in the ompressor. Using the steady state assumption, X is zero. Applying the above information to Eq, Eq is obtained. Replaing Ψ and Ψ by their value, ( Ψ Ψ ) + W (Eq) ( h h T ( s s )) + W 0 (Eq) From Eq, W h h. Substituting this information in Eq, Eq4 is obtained. ( h h T ( s s )) + 0 (Eq4) 6

7 By simplifying the previous equation, an equation idential to Eq0 for obtained. X is ( s ) X T (Eq4) 0m s Step 5: Conluding Remarks & Disussion The mass flow rate of the erant was found to be 0.68 /s. The power input to the ompressor was found to be.6 kj/s. The mass flow rate of the ooling water was found to be 4.0 /s. The rate of exergy destrution assoiated with the ompression proess was found to be 7.87 kj/s assuming a dead state temperature, T 0 5 o C. 7