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← Revision 16 as of 2016-12-13 04:48:32 ⇥
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Deletions are marked like this. | Additions are marked like this. |
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See: http://hdl.handle.net/2060/19930091020 | |
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|| $ T_r || Kelvins || Recovery temperature || | || $ T_r $ || Kelvins || Recovery temperature || |
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|| $ Pr $ || unitless || Prandtl number = 1 || | || $ Pr $ || unitless || Prandtl number assumed 1 || |
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Eq 23: $ T_r = T \left( 1 + { { \gamma-1 } \over 2 } M^2 \right) \approx { { \gamma-1 } \over 2 } M^2 T $ | 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 |
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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 25: $ ( T_r - T_w ) = V^2 / 2 C_p $ ... since $ V_{sound} = \sqrt{ ( \gamma - 1 ) C_p T } ~ ~ $ at altitude. |
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Eq 26: $ h = { 1 \over 2 } ~ C_f ~ C_p ~ \rho ~ V $ Heat transfer coefficient (all subcripted $._l$ in the original) . | Allen/Eggers Eq 26: $ h = { 1 \over 2 } ~ C_f ~ C_p ~ \rho ~ V $ Heat transfer coefficient (all subcripted $._l$ in the original) . |
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Eq 42a(?): $ \Large { { d H_s } \over { d t } } = { { N_{ur} k_r ( T_r - T_w ) } \over \sigma } \approx { { N_{ur} k_r T_r } \over \sigma } $ Heat transfer rate per unit area at the stagnation point | 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 |
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Eq 42b(?): $ Nu_r = 0.934 ~ Re_{\sigma}^{0.5} ~ Pr^{ 0.4 } $ Nusselt number at recovery temperature (unitless) | Allen/Eggers Eq 42b(?): $ Nu_r = 0.934 ~ Re_{\sigma}^{0.5} ~ Pr^{ 0.4 } ~ ~ $ Nusselt number at recovery temperature (unitless) |
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"note that???" $ Re_{\sigma} ~ = ~ \rho ~ V ~ \sigma / \mu_r $ | "note that???" $ Re_{\sigma} ~ = ~ \rho ~ V \sigma / \mu_r $ |
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The Prandtl number is assumed to be unity, so | |
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Eq KHL1: $ ~ Nu_r = 0.934 ~ Re_{\sigma}^{0.5} $ = 0.934 $ ~ \Large \sqrt{ { \rho ~ V ~ \sigma } \over \mu_r } $ | |
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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 } } } $ | |
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From the definition of the Prandtl number in ''Hypersonic and High Temperature Gas Dynamics'' by Anderson (1989) | |
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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. [[ https://www.ohio.edu/mechanical/thermo/property_tables/air/air_Cp_Cv.html | 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/m^2^ || 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/m^3^, and V = 11 km/s, the heat flux at the nose is 160 kW/m^2^. |
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 |
note 1: in the original document, many variables have subscript ._l indicating "local" or at altitude, a complication not needed here
note 2: \gamma can be higher for diatomic or ionized gasses.
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) .
- .. much omitted ...
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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
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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.