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A thin airfoil is mounted in a low-speed, subsonic wind tunnel, in which the Mach number is 0.1. At a point on the airfoil, the pressure coefficient is measured to be -1.2. If the flow velocity is increased such that the free-stream Mach number is 0.6, the pressure coefficient at the same point on the airfoil will approximately be:
- a)-3.5
- b)-2.9
- c)-1.5
- d)-0.75
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A thin airfoil is mounted in a low-speed, subsonic wind tunnel, in wh...
Explanation:
The pressure coefficient is defined as:
Cp = (P - P_inf) / (1/2 * rho * V_inf^2)
where P is the local pressure, P_inf is the free-stream pressure, rho is the air density, and V_inf is the free-stream velocity.
Given:
Mach number in the low-speed wind tunnel, M_1 = 0.1
Pressure coefficient at a point on the airfoil, Cp_1 = -1.2
At Mach number 0.1:
From the isentropic relations, we know that:
P_1 / P_inf = (1 + (gamma-1)/2 * M_1^2)^(gamma/(gamma-1))
where gamma is the ratio of specific heats.
For air, gamma = 1.4
Therefore, P_1 / P_inf = 0.990
Cp_1 = -1.2
At Mach number 0.6:
Using the same relations, we can find the pressure coefficient at Mach number 0.6:
P_2 / P_inf = (1 + (gamma-1)/2 * M_2^2)^(gamma/(gamma-1))
where M_2 = 0.6 is the new Mach number.
Therefore, P_2 / P_inf = 1.244
Using the definition of Cp:
Cp_2 = (P_2 - P_inf) / (1/2 * rho * V_inf^2)
Cp_2 = (1.244 * P_inf - P_inf) / (1/2 * rho * V_inf^2)
Cp_2 = 0.488 * Cp_1
Cp_2 = 0.488 * (-1.2)
Cp_2 = -0.586
Approximately, Cp_2 = -1.5 (option C is correct)
Note:
The change in Mach number causes changes in the flow properties, including pressure, density, and temperature. The pressure coefficient is a dimensionless quantity that relates the local pressure to the free-stream pressure and velocity. A negative value of Cp indicates that the local pressure is lower than the free-stream pressure.
The pressure coefficient is defined as:
Cp = (P - P_inf) / (1/2 * rho * V_inf^2)
where P is the local pressure, P_inf is the free-stream pressure, rho is the air density, and V_inf is the free-stream velocity.
Given:
Mach number in the low-speed wind tunnel, M_1 = 0.1
Pressure coefficient at a point on the airfoil, Cp_1 = -1.2
At Mach number 0.1:
From the isentropic relations, we know that:
P_1 / P_inf = (1 + (gamma-1)/2 * M_1^2)^(gamma/(gamma-1))
where gamma is the ratio of specific heats.
For air, gamma = 1.4
Therefore, P_1 / P_inf = 0.990
Cp_1 = -1.2
At Mach number 0.6:
Using the same relations, we can find the pressure coefficient at Mach number 0.6:
P_2 / P_inf = (1 + (gamma-1)/2 * M_2^2)^(gamma/(gamma-1))
where M_2 = 0.6 is the new Mach number.
Therefore, P_2 / P_inf = 1.244
Using the definition of Cp:
Cp_2 = (P_2 - P_inf) / (1/2 * rho * V_inf^2)
Cp_2 = (1.244 * P_inf - P_inf) / (1/2 * rho * V_inf^2)
Cp_2 = 0.488 * Cp_1
Cp_2 = 0.488 * (-1.2)
Cp_2 = -0.586
Approximately, Cp_2 = -1.5 (option C is correct)
Note:
The change in Mach number causes changes in the flow properties, including pressure, density, and temperature. The pressure coefficient is a dimensionless quantity that relates the local pressure to the free-stream pressure and velocity. A negative value of Cp indicates that the local pressure is lower than the free-stream pressure.
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A thin airfoil is mounted in a low-speed, subsonic wind tunnel, in which the Mach number is 0.1. At a point on the airfoil, the pressure coefficient is measured to be -1.2. If the flow velocity is increased such that the free-stream Mach number is 0.6, the pressure coefficient at the same point on the airfoil will approximately be:a)-3.5b)-2.9c)-1.5d)-0.75Correct answer is option 'C'. Can you explain this answer?
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