Bioengineering Application The heart pumps blood at an average rate of 5 L/min.

Question

Bioengineering Application

The heart pumps blood at an average rate of 5 L/min. The gauge pressure on the venous (intake) side is 0 mm Hg and that on the arterial (discharge) side is 100 mm Hg. Energy is supplied to the heart as heat released by the absorption of oxygen in the cardiac muscles: 5 mL (STP)O2/min is absorbed, and 20.2 J is released per mL of O2 absorbed. Part of this absorbed energy is converted to flow work (the work done to pump blood through the circulatory system), and the balance is lost as heat transferred to the tissues surrounding the heart.

(a)  

Simplify Equation 7.4-12 for this system, assuming (among other things) that there is no change in internal energy from inlet to outlet.

(b)  

What percentage of the heat input to the heart (Qin) is converted to flow work? (The answer may be thought of as the efficiency of the heart as a pump.)

Explanation / Answer

(a) Equation 7.4-12 is not given in the question

(b) 5 mL (STP)O2/min is absorbed, and 20.2 J is released per mL of O2 absorbed

   So, energy input = 5 mL/min * 20.2 J/mL = 101 J/min

To calculate flow work, the pressure used can be obtained from the pressure difference for venous and arterial side i.e .(100-0) mm Hg= 100 mm Hg and volume change is the blood pumped i.e. 5 L/min

So, heat input to the heart = PV = 100 mm Hg * 5L/min = 100 * 0.00132 * 5 L atm / min [1 mm Hg = 0.00132 atm]

                                                                                        = 100 * 0.00132 * 5 * 101.3 J/min [1 L-atm = 101.3 J]

                                                                                         = 66.858 J/min

Hence, percentage of the heat input to the heart (Qin) is converted to flow work = (66.858/101) * 100% = 66.2%

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