Consider two-dimensional flow over the diamond airfoil shown...

Consider two-dimensional flow over the diamond airfoil shown in Problem 2 except with only the bottom half of the symmetric airfoil. So, the upper surface is the original chord line (dashed line). a. Do you expect the lift to be positive/negative/zero? Explain briefly. b. Do you expect the drag to be larger/smaller/the same as the complete airoil evaluated in Problem 1? Briefly explain. c. Do you expect the airfoil with pitch nose up/nose down/not at all? Briefly explain.

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00:01 Hi, here in this question we are given that the value of p is equal to 30 feet which can be also written as 9 .144 meter whereas the value of velocity v equals to 300 feet per second which is equal to 91 .44 meter per second whereas the value of p is equal to 114 .7 pisa which can be also written as 101 .352 pascal.

00:41 So here further alpha is equal to 2 degrees and v equals to 60 degree fahrenheit which is equal to 15 degree celsius.

00:51 So here first of all we need to calculate the cross section of the air which is affected.

00:57 So here the cross section of the air can be calculated.

00:59 A equals to b square pi divided by substituting the above value.

01:05 So we have area a equals to 91 .44 square multiplied with pi divided by 4.

01:12 So here on simplifying this we have area covered equals to 65 .7 meter square.

01:19 Now here we need to calculate the mass.

01:21 So m equals to rho kv.

01:23 Substituting the values we have mass m equals to 1 .22 multiplied with 65 .7 multiplied with 91 .42.

01:35 So here on calculating this mass affected would be equal to 7329 kg per second.

01:45 So here this is mass value, this is rate of change of mass with respect to time.

01:50 Now here we need to calculate the value of force.

01:56 So here summation of fx is equal to m dot multiplied with vx outside minus vx inside...

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