Problems of Chapter 5

Problem 5.1

The ratio of specific heats reads

Подпись:Подпись: f/ + 2

/

From this we obtain 2

Y – 1 and

/ I -/ —r 1 . I -/ r 1 t QO*

Y-1

Problem 5.2

a) From Section 6.2 we obtain the relation for the maximum speed

tm = V – І’I-

which yields Vm = 6324.5 m/s. Because the static temperature T is zero, the speed of sound is zero and the Mach number is infinitely large.

b) The specific heat at constant pressure of the Lighthill gas is cPL =4 R = 1148.24 m2/s2 K.

The relation for the exit speed is

vexit 2(ht CPL Texit)?

which yields vexit = 6140.3 m/s.

For the Lighthill gas the ratio of the specific heats is yl = 1.333. The speed of sound aexu = a/yl RTexu = 608.9 m/s and the Mach number Мєхц = 10.08.

c) The relation for the exit speed with 20 per cent of the reservoir enthalpy frozen is

Vexit — J 2(0.8 ht c. pL Texit),

which yields vexit = 5656.8 m/s. With the same speed of sound assumed as for c), the exit Mach number is Mexit = 9.29.

d) The general result is that both the nozzle exit speed and Mach number are affected by the different thermodynamic conditions. Read again Sub-Section 5.5.2.

Problem 5.3

Подпись: cp Problems of Chapter 5

The static temperature at H = 30 km altitude is = 226.509 K. The specific heat at constant pressure is

In terms of у this reads

cP = -^-R.

Y – 1

a) The ratio of specific heats at Tx = 226.509 K is y = 1.4. The specific heat at constant pressure then is cp = 1004.71 m2/s2K.

With that we obtain

ht = CpT + [177]— = 2, 227, 575.86 nr/s2.

From this the total temperature is found to be Tt = ht/cp = 2,217.1 K.

b) With y =1.3 the specific heat at constant pressure is cp = 1243.92 m2/s2K. The total temperature then is Tt = ht/cp = 1,790.76 K.

c) With y = 1.1 the specific heat at constant pressure is cp = 3157.66 m2/s2K. The total temperature then is Tt = ht/cp = 705.45 K.

d) The value y =1.3 appears to lie in a realistic range, see Fig. 5.2.

Problem 5.4

We measure the heat flux at the two locations and find q = 0.0617 MW/m2 at x = 1 m and q = 0.0247 MW/m2 at x = 6 m.

Eq. (3.27) reduces in our case to

At x = 6 m we then obtain the scaled value q = 0.0617/605 = 0.0252 MW/m2.

The difference to the measured value is Aq = + 0.0005 MW/m2, which is about 2 per cent difference. With regard to the distance x, eq. (3.27) scales the computed heat flux quite well.

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