# Calculo de calor

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ME 40 Thermodynamics Spring 2009

Quiz #3

March 16, 2009

Name: SID:

Instructions: Read each question carefully. Take into consideration the point values for each question. Write your nameand SID on each page. You have roughly 45 minutes. One double-sided reference sheet is permitted. Good luck!

Name: SID:

Question 1: [10 points]
Consider a device consisting of two cycles assketched below. The heat engine cycle (the power cycle) operates between two thermal reservoirs at 500 K and 300 K respectively. The refrigeration cycle interacts with reservoirs at 300 K and 275 Krespectively. The power from the heat engine cycle is used entirely to run the refrigeration cycle. Both cycles are assumed to be ideal Carnot cycles. Determine the ratio of heat inputs into these twocycles,
Q in , refrigeration/Q in , power

Q in,power

Q in,refrigeration

solution: The work from Carnot cycle Wengine=(1-300/500) Qin,power=0.4 Qin,power Work required from the Carnotrefrigeration cycle= Qin,refrigeration/COP Carnot where COP Carnot= 1/(300/275-1)= 11 Work required from the Carnot refrigeration cycle= Qin,refrigeration /11 Setting Work required from the Carnot refrigerationcycle= Wengine Qin,refrigeration/11= 0.4 Qin,power Qin,refrigeration/ Qin,power=4.4

Name: SID:

Question 2: [5 points]
An ideal gas initially at 300 K and 500 kPa (State 1) passes through anadiabtic throttling device. The pressure of this ideal gas drops to 100 kPa (State 2). Determine 1) the change of entropy per unit mass of this ideal gas (2 points), and 2) the minimum work required toreverse the throttling process from State 2 to State 1 (3 points). The gas constant of this ideal gas is R=0.3 kJ/kg-K. 1st law gives T2=T1 as h2=h1. Second law: the entropy change for an ideal gasis ∆s= Cp ln(T2/T1) –R ln (P2/P1)= -R ln (P2/P1)= -0.3 kJ/kg-K ln (0.2)=0.48 kJ/kg-K The minimum work required is equal to the loss of potential work= T*∆s= 300 K * 0.48 kJ/kg-K = 144 kJ/kg....