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Michael D.

Biomedical Engineer with Graduate Level Education

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Physics (Thermodynamics)

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Question:

An ideal vapor-compression refrigeration cycle that uses R-134a as its working fluid maintains a condenser at 1,000 kPA and the evaporator at $$4^{\circ}C$$. Determine this system's Coefficient of Performance (COP) and the amount of power required to service a 400 kW cooling load. $$4^{\circ}$$

Michael D.

Answer:

I used http://www.pfri.uniri.hr/~pkralj/R134a_thermo_prop_si.pdf for my R-134a saturation tables. State 1, Saturated Vapor $$T_{1}=4^{\circ}C,\, P_{1}=337.85 kPa,\,h_{1}=401.1\,kJ/kg,\,s_{1}=1.7257\,kJ/(kg\cdot K)$$ Temperature known, saturation properties found in saturation table. State 2, Superheated Vapor $$P_{2}=1,000 kPa,\,s_{2}=1.725\,kJ/(kg\cdot K),\,h_{2}=423.7\,kJ/kg$$ Pressure given. Movement from state one to state 2 is isentropic, so entropy (s) remains the same. Used linear interpolation of the two closest values of the "superheated vapor at constant pressure, 1,000 kPa" chart to calculate $$h_{2}$$ State 3, Saturated Liquid $$P_{3}=1,000 kPa,\,h_{3}=255.6\,kJ/kg$$ Used linear interpolation of the two closest values of the "superheated vapor at constant pressure, 1,000 kPa" chart to calculate $$h_{2}$$ State 4, Liquid Vapor Mixture $$P_{3}=1,000 kPa,\,s_{2}=1.725\,kJ/(kg\cdot K),\,h_{2}=255.6\,kJ/kg$$ Movement from State 3 to State 4 is isenthalpic, meaning there is no change in enthalpy $$COP=\frac{Q_{L}}{\left | W_{c} \right |}=\frac{m\cdot\left ( h_{1}-h_{4} \right )}{m\cdot\left ( h_{1}-h_{2} \right )}=\frac{401.1-255.6}{\left | 401.1-423.7 \right |}=6.44$$ $$Q_{L}=400\,kW=m\cdot(h_{1}-h_{4})$$ $$m=\frac{400\,kW}{(h_{1}-h_{4})}=\frac{400\,kW}{(401.1-255.6)kJ/kg}=2.75\,kg/s$$ $$W_{c}=m\cdot(h_{1}-h_{2})=2.75\cdot(401.1-423.7)=-62.15 kW$$

Biochemistry

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Question:

Calculate the free energy change for glucose entry into cells when the extracellular concentration is 5 mM and the intracellular concentration is 3 mM.

Michael D.

Answer:

$$\Delta G = RT\ln\frac{\left [ glucose \right ]_{in}}{\left [ glucose \right ]_{out}}$$ $$= \left ( 8.3145 J\cdot K^{-1}\cdot mol^{-1} \right )\left ( 298 K \right )\ln\frac{0.003}{0.005}$$ $$= -1270J\cdot mol^{-1} =-1.27kJ\cdot mol^{-1}$$

Statistics

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Question:

An experiment tosses a coin 4 times. Assume the coin is fair. Given: A = At least three consecutive heads or tails B = At least two heads from the 4 tosses Calculate $$P\left ( A \cup B \right )$$

Michael D.

Answer:

A sample space with 2 outcomes for each event is $$2^{n}=2^{4}=16$$ $$P(A) =\frac{(HHHT, THHH, HTTT, TTTH)}{16}=\frac{4}{16}=0.25=25\%$$ P(B) can be calculated by finding the probabilities that it does not happen (the probabilities that only 1 head or 0 heads are tossed) $$P(B) =1-P(1 \,head)-P(2\,heads)=1-0.5^{4}-4\cdot0.5^{4}=0.6875=68.75\%$$ The probability of the union of A and B can be calculated by summing the probability of A happening and the probability of B happening and subtracting from the probability that both events occur at the same time. The probability that both A and B occur at the same time is calculated by multiplying P(A) by the probability that B will occur, given A occurring, which in this case is 50% or 0.5. $$P\left ( A \cup B \right )=P(A)+P(B)-P\left ( A \cap B \right )=P(A)+P(B)-P(A)\cdot P\left ( A \mid B \right )$$ $$P\left ( A \cup B \right )=0.25+0.6875-0.25 \cdot 0.5=0.8125=81.25\%$$

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