10 points 1. Answer the following questions If you extract heat from a rigid gas

Question

10 points 1. Answer the following questions If you extract heat from a rigid gas container, the gas change from one thermodynamic state to other one. What thermodynamic process is performed? will the pressure and temperature increase? Why? Define what is the stagnation property If you have an isentropic flow, what happen with the stagnation properties at several nozzle sections? In a supersonic flow, if the area decrease, the velocity increase or decrease? At what Mach number values will you have a Mach cone? a. b. c. d. e.

Explanation / Answer

1. (a)

Accordingto first law of thermodynamics-

dQ = dU + dW ....... Eq. (1)

(where, dQ = heat transfer, dW= work transfer and dU= change in internal energy)

Now, for an ideal gas - dU= m*(Cv)*dT

dW = P*dV (for closed system)

Here, Cv= calorific value at constant volume,

dT = change in temperature,

P = pressure,

dV = change in volume

Now, Eq. (1) becomes-

dQ = m*(CV)*dT + P.dV

Since it is a rigid container so no change in volume of gas. So dV = 0 (This process is constant volume or isochoric process)

So,

dQ= m*CV*dT

Since heat is extracted so L.H.S. of above equation is negative. So R.H.S. must be negative so dT must be negative. So change in temperature is negative. So temperature will decrease so pressure will also decrease (since for constant volume process pressure is proportional to temperature).

(b)

Stagnation property at a point is obtained if the local flow velocity at that point is imagined as zero isentropically).

(d)

dA/A = (M2 -1) (dV/V)

Where, A = area, M= Mach number, V= velocity

Since flow is supersonic so M>1 and A is decreasing i.e. dA = negative . So dV= negative.

So velocity decreases in this area.

(e)

For Mach cone, Mach number should be greater than 1.

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