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InterviewSolution
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Many types of race cars depend on negative lift (or down-force) to push them down against the track surface so they can take turns quickly without sliding out into the track wall . Part of the negative lift is the ground force , which is a force due to the airflow beneath the car . As the race car in Fig moves forward at 27.25 m/s , air is forced to flow over and under the car . The air forced to flow under the car . The air forced to flow under the car enters through a vertical cross-sectional area is A_(1) = 0.0310 m^(2) . Treat this flow as steady flow through a stationary horizontal pipe that decreases in cross-sectional area from A_(0) to A_(1) (Fig) (a) At the moment it passes through A_(0) , the air is at atmospheric pressure p_(0) . At what pressure p_(0) . At what pressure p_(1) is the air as it moves through A_(1) ? |
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Answer» <P> Solution :From Eq. , we can WRITE`A_(0) v_(0) = A_(1) v_(1)` , or `v_(1) = v_(0) - (A_(0))/(A_(1)) "" (14-52)` Substituting Eq.14-52 into Eq.14-51 and rearranginggive us `p_(1) = p_(0) - (1)/(2) RHO v_(0)^(2) ((A_(0)^(2))/(A_(1)^(2)) - 1) "" (14-53)` The speed of the air as it enters `A_(0)` at the front of the CAR is equal to 27.25 m/s , the speed of the car as it moves forward through the air . Substituting this speed , the air density `rho = 1.21 KG//m^(3)` , and the values for `A_(0)` and `A_(1)` into Eq. , we find `p_(1) = p_(0)- (1)/(2) (1.21 kg //m^(3)) (27.25 m//s)^(2) [((0.0330 m^(2))^(2))/((0.0310 m^(2))^(2)) - 1]` `p_(1) = p_(0) - 59.838 Pa = p_(0) - 59.8 Pa` |
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