Exp 8 practice solution

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School

University of Wisconsin, Madison *

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Course

370

Subject

Mechanical Engineering

Date

Dec 6, 2023

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pdf

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Uploaded by MinisterHippopotamusPerson1488

1) The values contained in the table below are gathered from operation of the ME 370 refrigerator experiment with the compressor running at 1450 rpm and with ambient conditions of T amb = 22 ˚C, P amb = 745 mmHg. The temperature and pressure values are the raw data provided directly from Labview. State s: 2. Comp out, 3. Cond out, 4. Throttle valve in, 5. Evap in, 6. Evap out, 8. Comp in Time T 2 T 3 T 4 T 5 T 6 T 8 T evap,ave T cond,ave (s) (C) (C) (C) (C) (C) (C) (C) (C) 130 71.9 29.9 24.6 -4.7 22.4 27.7 22.3 23.5 860 98.2 43.5 33.9 -4.4 12.4 32.4 10.2 38.1 P cond,in P cond,out P evap,in P evap,out ! Q cool , r ! W s COP cool ! m r ! V (kPa) (kPa) (kPa) (kPa) (kW) (kW) - (kg/s) (L/min) 721 707 144 75.5 4.09 3.37 0.022 1084 1051 146 82.6 1.24 X 0.022 X a) Find the shaft power at time = 130 s ! W s (kW) = ____1.21______ COP = ! Q cool ! W s ⇒ ! W s = ! Q cool COP = 4.09 kW 3.37 = 1.21 kW b) Determine the cooling power of the refrigerant in the evaporator at time = 860 s. Properties may be obtained from the attached chart. Show all your work. Identify the locations on the chart of each state for which you use the properties in your calculation. Label these with the corresponding state number and the subscript c (for example 2 c for state 2). ! Q cool , r (kW) = ___3.56_______ ! Q cool = ! m r h 6 − h 5 ( ) . h 5 = h 4 The enthalpy at state 6 c and state 4 c are given from the attached chart: h 6 = 262 kJ/kg and h 5 = h 4 = 100 kJ/kg. Then ! Q cool = 0.022 kg s 162 kJ kg = 3.56 kW c) Find the rate at which heat is exchanged between the evaporator tank and the surroundings at time = 130 s. Assume that the sum of m*c P (the mass times specific heat) for the evaporator tank and its contents is 175 (kJ/K). ! Q surr , evap (kW) = ___1.19____

from equation (11) in the lab manual, the heat transferred to the water from the surroundings is given by: ! Q surr , evap = ! Q cool , r − ! Q cool , w where, from equation (10) we have that ! Q cool , w = − mc P ∑ ( ) dT dt # $ % & ' ( evap . From the data provided above, dT dt = 10.2 − 22.3 860 − 130 = − 0.01658 ° C s , so that ! Q cool , w = − 175 kJ K " # $ % & ' − 0.01658 K s " # $ % & ' = 2.9 kW Then ! Q surr , evap = 4.09 kW − 2.90 kW = 1.19 kW . Note that we have assumed that the slope of dT/dt was constant through the run, which was true from all the data sets gathered in experiment 8. d) Describe and explain what will happen to the time required to cool the water in the evaporator tank from 22 ˚C to 10 ˚C, and the energy consumed in the process, if the compressor speed is reduced to 500 rpm. As observed in the data from experiment 8, when the compressor speed is reduced from 1450 rpm to 500 rpm, the cooling rate is reduced as well and it will therefore take a longer time to cool from 22 ˚C to 10 ˚C. However, because the system is operating closer to ideal situations when it is running slower, less entropy will be generated in the process, and the COP will be higher. That means that although the same energy is removed from the water, it will take less energy to accomplish that task, a fact that was also observed in the data gathered in experiment 8.

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