Question

The blood speed in a normal segment of a horizontal artery is 0.16 m/s. An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to one-eighth the normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery?

Answer 1

The difference in blood pressures is
`P_1 -P_2 = 854.8 \  Pa`

Step-by-step explanation:

From the question we are told that

The blood speed in a normal segment is
`v_1 = 0.16 \ m/s`

The area of the normal segment is
`A_1`

The area of the abnormal segment is
`A_2 = (A_1)/(8)`

Generally from continuity equation we have that

`v_1 A_1 = v_2 A_2`

=>
`v_1 A_1 = v_2 * (A_1)/(8)`

=>
`0.16 = v_2 * (1)/(8)`

=>
`v_2 = 1.28 \ m/s`

Generally from Bernoulli's equation the difference in pressure is mathematically represented as

`P_1 -P_2 = (1)/(2) \rho (v_2^2 - v_1^2 )`

Here
`\rho` is the density of blood with value
`\rho = 1060 \ kg /m^3`

`P_1 -P_2 = (1)/(2) (1060) ( 1.28 ^2 - 0.16 ^2 )`

=>
`P_1 -P_2 = 854.8 \  Pa`