The end diastolic volume of a heart is 140 mL Assume that it is a sphere. At end diastole, the intraventricular pressure is 7mmIIg. The wall thickness at this time is 1.1 cm. At the end of isovolumetric contraction, the intraventricular pressure is 80 mml Ig.

A. What is the wall tension at end diastole?

B. What is the wall tension at the end of the isovolumetric contraction?

C. At the end of systole, the intraventricular volume is 65 mL, the pressure is 100 mml Ig, and its wall thickness is 1.65 cm. What is the wall tension at this time?

D. The wall stress is related to tension by sigma = T/w, where a is the wall stress, T is the tension, and w is the wall thickness. Calculate the wall stress from A, B, and C.

We know that, V = 140 mL = 0.00014 m3

Assume that it is a sphere. so, we have

V = (4/3) / pir3

r3 = (0.00014 m3) (3) / (4) (3.14)

r = / sqrt[3]{}/sqrt[3]{}3/sqrt{}3.34 x 10-5 m3

r = 1.93 x 10-7 m

(a) The wall tension at end diastole will be given as:

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end diastole = 7 mmHg = 933.2 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (933.2 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 8.18 x 10-3 N

(b) The wall tension at the end of isovolumetric contraction will be given as:

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end of isovolumetric contraction = 80 mmHg = 10665.7 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (10665.7 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 9.35 x 10-2 N

(d) The wall stress from A and B which will be given as:

we know that, / sigma = T / w

For part A, we have

/sigmaA = (8.18 x 10-3 N) / (0.011 m)

/sigmaA = 0.743 N/m

For part B, we have

/sigmaB = (9.35 x 10-2 N) / (0.011 m)

/sigmaB = 8.5 N/m