Allen/Eggers Hypersonic Drag, 1957

In METRIC!


See: http://hdl.handle.net/2060/19930091020

Symbols:

T_r

Kelvins

Recovery temperature

T_w

Kelvins

Wall temperature (relatively small, will be ignored)

T

Kelvins

Temperature at altitude

M

unitless

Mach number at altitude

H

J / m2

Heat transferred per unit area

h

J / m2 - K

Heat transfer coefficient

C_v

J / kg K

Specific heat capacity at constant volume

C_p

J / kg K

Specific heat capacity at constant pressure

C_f

?

Skin effect coefficient

\gamma

C_p / C_v

Specific heat capacity ratio, typically 1.4

\sigma

meters

nose radius

k_r

?

Thermal conductivity at the recovery temperature

Nu_r

unitless

Nusselt number

Re_{\sigma}

unitless

Reynolds number for nose cone radius \sigma

Pr

unitless

Prandtl number assumed 1

\mu_r

?

coefficient of viscosity at the recovery temperature

Assuming that the Prandtl number is unity.

Allen/Eggers Eq 23: T_r = T \left( 1 + { \Large { { \gamma-1 } \over 2 } } M^2 \right) \approx { \Large { { \gamma-1 } \over 2 } } M^2 T ~ ~ for high mach numbers

Allen/Eggers Eq 25: ( T_r - T_w ) = V^2 / 2 C_p ... since V_{sound} = \sqrt{ ( \gamma - 1 ) C_p T } ~ ~ at altitude.

Allen/Eggers Eq 26: h = { 1 \over 2 } ~ C_f ~ C_p ~ \rho ~ V Heat transfer coefficient (all subcripted ._l in the original) .

Page 17: (modified to ditch minus sign)

Allen/Eggers Eq 42a(?): \Large { { d H_s } \over { d t } } = { { Nu_r k_r ( T_r - T_w ) } \over \sigma } ~ ~ Heat transfer rate per unit area at the stagnation point

Page 18:

Allen/Eggers Eq 42b(?): Nu_r = 0.934 ~ Re_{\sigma}^{0.5} ~ Pr^{ 0.4 } ~ ~ Nusselt number at recovery temperature (unitless)

"note that???" Re_{\sigma} ~ = ~ \rho ~ V \sigma / \mu_r

The Prandtl number is assumed to be unity, so

Eq KHL1: ~ Nu_r = 0.934 ~ Re_{\sigma}^{0.5} = 0.934 ~ \Large \sqrt{ { \rho ~ V ~ \sigma } \over \mu_r }

Eq KHL2: ~ { \Large { { d H_s } \over { d t } } } = 0.934 ~ { \Large \sqrt{ { \rho ~ V ~ \sigma } \over \mu_r } ~ { { k_r V^2 } \over { 2 C_p \sigma } } } ~ = ~ 0.467 ~ { \Large \sqrt{ { \rho ~ V } \over { \mu_r ~ \sigma } } ~ { { k_r V^2 } \over { C_p } } }

From the definition of the Prandtl number in Hypersonic and High Temperature Gas Dynamics by Anderson (1989)

Eq 16.42: ~ Pr ~=~ \mu C_p / k_T ~ ( assumed 1 ), we can infer

Eq KHL3: ~ k_r = k_T ~=~ \mu C_p ~=~ \mu_r C_p

Eq KHL4: ~ { \Large { { d H_s } \over { d t } } } = 0.467 ~ { \Large \sqrt{ { \rho ~ V } \over { \mu_r ~ \sigma } } ~ { {\mu_r C_p V^2 } \over { C_p } } } ~ = ~ 0.467 ~ { \Large \sqrt{ { \rho ~ V \mu_r } \over { \sigma } } } ~ V^2 ~ ~ This is identical to the Allen/Eggers equation 43. OK so far!

The Allen/Eggers paper gives a fps expression for the coefficient of thermal velocity \mu_r without citing a source. The book Hypersonic and High Temperature Gas Dynamics by Anderson (1989) has provides this equation on page 605 in section 16.6 Transport Properties for High Temperature Air:

Eq And1: ~ \mu_0 = \left( 1.462e-5 { gm \over { cm ~ s ~ K^{1/2} } } \right) ~ { \Large { T^{1/2} \over { 1 + 112/T } } }

Let's assume \mu_0 = \mu_r (hm...) and simplify the demominator (making the viscosity a wee bit larger) and convert it to mks:

Eq KHL5: ~ \mu_r = \left( 1.462e-6 { kg \over { m ~ s ~ K^{1/2} } } \right) ~ T^{1/2}

Now let's assume T \approx ~ T_r - T_w and use Eq 25:

Eq KHL6: ~ \mu_r = \left( 1.462e-6 { kg \over { m ~ s ~ K^{1/2} } } \right) ~ V / \sqrt{ 2 C_p } ~= \left( 1.03e-6 { kg \over { m ~ s ~ K^{1/2} } } \right) ~ V / \sqrt{ C_p }

Combining KHL4 and KHL6:

Eq KHL7: ~ { \Large { { d H_s } \over { d t } } } = 0.467 { \Large \sqrt{ { \rho ~ \left( 1.03e-6 { kg \over { m ~ s ~ K^{1/2} } } \right)~ { V^2 } } \over { \sqrt{ C_p } \sigma } } } ~ V^2 ~ = \left( 4.74e-4 { kg^{1/2} \over { m^{1/2} s^{1/2} K^{1/4} } } \right) ~ C_p^{-1/4}~ { \Large \sqrt{ \rho \over \sigma } } ~ V^3

Except for the factor of C_p^{-1/4} , this resembles equation 44 in the Allen/Eggers paper:

Allen/Eggers Eq 44: ~ { \Large { { d H_s } \over { d t } } } ~=~ 6.8e-6 ~ { \Large \sqrt{ \rho \over \sigma } } ~ V^3 in foot-slug-second-BTU units.

So, where is the discrepancy? The units are correct for equation KHL7, both sides work out to kg \over { s^3 } . I presume the authors assumed that air has a fairly constant specific heat capacity, until it disassociates at high temperature, which was not well characterised in 1957. Here is a table up to 1500 Kelvin - heat capacity increases from 1005 J / kg K at 300K up to 1200 J / kg K. Our gas is much thinner and hotter (8000 Kelvin?) than that table is intended for. As a wild guess, assume 1300 J / kg K - if it is larger, the heating is lower.

That results in C_p^{1/4} = 6 m^{1/2} s^{-1/2} K^{-1/4} so our equation simplifies to:

Eq KHL8: ~~~~~~~~~~~ { \Large { { d H_s } \over { d t } } } ~=~ 8e-5 \left( kg^{1/2} \over m \right)~ { \Large \sqrt{ \rho \over \sigma } } ~ V^3 ~~~~~ W/m2

This is the heat flux at the stagnation point in the center of the round nose of a vehicle; the heating is less over the rest of the nose.

If \sigma = 1 m, \rho = 2.22e-8 kg/m3, and V = 11 km/s, the heat flux at the nose is 160 kW/m2.

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DragAllenEggers1957 (last edited 2016-12-13 04:48:32 by KeithLofstrom)