航空发动机原理复习之计算题总结(四)

文章目录

类型六：调节规律
- 06 知识回顾
- - 稳态下压气机和涡轮共同工作条件
  - 涡轮落压比
  - 两种调节规律

类型六：调节规律

06 知识回顾

稳态下压气机和涡轮共同工作条件

1.转速一致

对于单转子涡喷发动机，压气机转子转速等于涡轮转子的转速，即
n c = n T n_c=n_T nc=nT
2.流量连续

流过涡轮的燃气流量 q m , g q_{m,g} qm,g等于流入压气机的空气流量 q m , a q_{m,a} qm,a加上进入燃烧室的燃油流量 q m , f q_{m,f} qm,f，再减去引气系统引走的空气流量 q m , c o l q_{m,col} qm,col,即
q m , g = q m , a + q m , f − q m , c o l q_{m,g}=q_{m,a}+q_{m,f}-q_{m,col} qm,g=qm,a+qm,f−qm,col
此式可以写成：
q m , g = β q m , a q_{m,g}=\beta q_{m,a} qm,g=βqm,a
其中：
β = q m , a + q m , g − q m , c o l q m , a = 1 + f − v c o l \beta =\frac{q_{m,a}+q_{m,g}-q_{m,col}}{q_{m,a}}=1+f-v_{col} β=qm,aqm,a+qm,g−qm,col=1+f−vcol
3.压力平衡

涡轮进口处燃气总压等于压气机进口处空气总压乘以燃烧室总压恢复系数，即
p 3 ∗ = σ b p 2 ∗ p_{3}^{*}=\sigma _bp_{2}^{*} p3∗=σbp2∗
4.功率平衡

涡轮提供给压气机的功率与压气机压缩空气所消耗的功率平衡，即
N c = N T η m N_c=N_T\eta _m Nc=NTηm

考虑到：
N c = q m , a w c , N T = q m , g w T = β q m , a w T N_c=q_{m,a}w_c,N_T=q_{m,g}w_T=\beta q_{m,a}w_T Nc=qm,awc,NT=qm,gwT=βqm,awT
所以：
w c = w T β η m w_c=w_T\beta \eta _m wc=wTβηm

涡轮落压比

π T ∗ = [ σ e A 5 q ( λ 5 ) σ t A t q ( λ t ) ] 2 n ′ n ′ + 1 \pi _{T}^{*}=\left[ \frac{\sigma _eA_5q\left( \lambda _5 \right)}{\sigma _tA_tq\left( \lambda _t \right)} \right] ^{\frac{2n^{\prime}}{n^{\prime}+1}} πT∗=[σtAtq(λt)σeA5q(λ5)]n′+12n′

几何不可调时，当涡轮导向器最小截面和喷管都处于临界或超临界状态,落压比为常数。
π T ∗ n ′ + 1 2 n ′ = σ e A 5 σ t A t = c o n s t {\pi _{T}^{*}}^{\frac{n^{\prime}+1}{2n^{\prime}}}=\frac{\sigma _eA_5}{\sigma _tA_t}=const πT∗2n′n′+1=σtAtσeA5=const

两种调节规律

1. n = n m a x = c o n s t n=n_{max}=const n=nmax=const和 A 5 = c o n s t A_5=const A5=const的最大工作状态调节规律

2. n = n m a x = c o n s t n=n_{max}=const n=nmax=const和 T 3 ∗ = c o n s t T_3^*=const T3∗=const的最大工作状态条件概率

1.某单轴涡喷发动机，在飞行高度 H = 0 H=0 H=0,马赫数 M a = 0 Ma=0 Ma=0以最大转速工作时，压气机的 π c ∗ = 12.0 \pi_c^*=12.0 πc∗=12.0，效率 η c ∗ = 0.85 \eta_c^*=0.85 ηc∗=0.85，涡轮后燃气总温 T 4 ∗ = 1000 K T_4^*=1000\ \mathrm{K} T4∗=1000 K，若该发动机采用转速 n = n= n=常数，喷管喉部面积 A 5 = A_5= A5=常数的调节规律，当 H = 11 k m H=11\ \mathrm{km} H=11 km（ T 0 ′ = 216.5 K T_{0}^{\prime}=216.5 \mathrm{K} T0′=216.5K）, M a = 2.0 Ma=2.0 Ma=2.0时， π c ∗ ′ = 8.0 {\pi _{c}^{*}}^{\prime}=8.0 πc∗′=8.0， η c ∗ ′ = 0.82 {\eta _{c}^{*}}^{\prime}=0.82 ηc∗′=0.82，试计算:当飞行条件从 H = 0 H=0 H=0， M a = 0 Ma=0 Ma=0变到 H = 11 k m H=11\ \mathrm{km} H=11 km, M a = 2.0 Ma=2.0 Ma=2.0时，涡轮前、后燃气总温 T 3 ∗ ′ {T_{3}^{*}}^{\prime} T3∗′、 T 4 ∗ ′ {T_{4}^{*}}^{\prime} T4∗′各为多少?若发动机采用转速 n = n= n=常数，涡轮前燃气总温 T 3 ∗ = T_3^*= T3∗=常数的调节规律，试计算: H = 11000 m H=11000\ \mathrm{m} H=11000 m， M a = 2.0 Ma=2.0 Ma=2.0时，涡轮后燃气总温 T 4 ∗ ′ ′ {T_{4}^{*}}^{^{''}} T4∗′′为多少?当飞行条件从 H = 0 H=0 H=0, M a = 0 Ma=0 Ma=0变到 H = 11000 m H=11000\ \mathrm{m} H=11000 m， M a = 2.0 Ma=2.0 Ma=2.0时，喷管喉部面积 A 5 A_5 A5变化了多少?（假设涡轮导向器，喷管均处于超临界状态，空气和燃气的流量相同,定压比热容分别为 1.005 k J / ( k g ⋅ K ) 1.005\ \mathrm{kJ/(kg\cdot K)} 1.005 kJ/(kg⋅K)、 1.158 k J / ( k g ⋅ K ) 1.158\ \mathrm{kJ/(kg\cdot K)} 1.158 kJ/(kg⋅K);绝热指数分别为 1.40 1.40 1.40、 1.33 1.33 1.33，其他各种损失系数和部件效率等于 1.0 1.0 1.0）

解:空气和燃气的流量相同，即 β = 1 \beta=1 β=1

H = 0 H=0 H=0、 M a = 0 Ma=0 Ma=0，进气道内为绝能流动，可以得到： T 1 ∗ = T 0 ∗ = T 0 = 288.15 K T_{1}^{*}=T_{0}^{*}=T_0=288.15 \mathrm{K} T1∗=T0∗=T0=288.15K

功率平衡： w c = w T β η m w_c=w_T\beta \eta _m wc=wTβηm，由于 β = 1 \beta=1 β=1、 η m = 1 \eta_m=1 ηm=1,可以得到 w c = w T w_c=w_T wc=wT,即：
c p T 1 ∗ η c ∗ ( π c ∗ γ − 1 γ − 1 ) = c p ′ ( T 3 ∗ − T 4 ∗ ) \frac{c_pT_{1}^{*}}{\eta _{c}^{*}}\left( {\pi _{c}^{*}}^{\frac{\gamma -1}{\gamma}}-1 \right) =c_{p}^{\prime}\left( T_{3}^{*}-T_{4}^{*} \right) ηc∗cpT1∗(πc∗γγ−1−1)=cp′(T3∗−T4∗)
可以算出涡轮前总温：
T 3 ∗ = c p T 1 ∗ c p ′ η c ∗ ( π c ∗ γ − 1 γ − 1 ) + T 4 ∗ = 1005 × 288.15 1158 × 0.85 ( 1 2 1.4 − 1 1.4 − 1 ) + 1000 = 1304.19 K T_{3}^{*}=\frac{c_pT_{1}^{*}}{c_{p}^{\prime}\eta _{c}^{*}}\left( {\pi _{c}^{*}}^{\frac{\gamma -1}{\gamma}}-1 \right) +T_{4}^{*}=\frac{1005\times 288.15}{1158\times 0.85}\left( 12^{\frac{1.4-1}{1.4}}-1 \right) +1000=1304.19\,\,\mathrm{K} T3∗=cp′ηc∗cpT1∗(πc∗γγ−1−1)+T4∗=1158×0.851005×288.15(121.41.4−1−1)+1000=1304.19K
由于涡轮效率为 η T ∗ = 1 \eta_T^*=1 ηT∗=1，所以有：
π T ∗ = ( T 3 ∗ T 4 ∗ ) γ ′ γ ′ − 1 = ( 1304.19 1000 ) 1.33 1.33 − 1 = 2.916 \pi _{T}^{*}=\left( \frac{T_{3}^{*}}{T_{4}^{*}} \right) ^{\frac{\gamma ^{\prime}}{\gamma ^{\prime}-1}}=\left( \frac{1304.19}{1000} \right) ^{\frac{1.33}{1.33-1}}=2.916 πT∗=(T4∗T3∗)γ′−1γ′=(10001304.19)1.33−11.33=2.916
H = 11 k m H=11\ \mathrm{km} H=11 km、 M a = 2 Ma=2 Ma=2，进气道内为绝能流动，可以得到：
T 1 ∗ ′ = T 0 ∗ ′ = T 0 ′ ( 1 + γ − 1 2 M a 2 ) = 216.5 × ( 1 + 1.4 − 1 2 × 2 2 ) = 389.7 K {T_{1}^{*}}^{\prime}={T_{0}^{*}}^{\prime}={T_0}^{\prime}\left( 1+\frac{\gamma -1}{2}Ma^2 \right) =216.5\times \left( 1+\frac{1.4-1}{2}\times 2^2 \right) =389.7\,\,\mathrm{K} T1∗′=T0∗′=T0′(1+2γ−1Ma2)=216.5×(1+21.4−1×22)=389.7K
若该发动机采用转速 n = n= n=常数，喷管喉部面积 A 5 = A_5= A5=常数的调节规律,涡轮导向器，喷管均处于超临界状态,可知涡轮落压比 π T ∗ = 2.916 \pi_T^*=2.916 πT∗=2.916不变。

根据功率平衡：
c p T 1 ∗ ′ η c ∗ ′ ( π c ∗ ′ γ − 1 γ − 1 ) = c p ′ T 3 ∗ ′ ( 1 − 1 π T ∗ γ ′ − 1 γ ′ ) η T ∗ ′ \frac{c_p{T_{1}^{*}}^{\prime}}{{\eta _{c}^{*}}^{\prime}}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right) =c_{p}^{\prime}{T_{3}^{*}}^{\prime}\left( 1-\frac{1}{{\pi _{T}^{*}}^{\frac{\gamma ^{\prime}-1}{\gamma ^{\prime}}}} \right) {\eta _{T}^{*}}^{\prime} ηc∗′cpT1∗′(πc∗′γγ−1−1)=cp′T3∗′ 1−πT∗γ′γ′−11 ηT∗′
可以计算出 H = 11 k m H=11\ \mathrm{km} H=11 km, M a = 2.0 Ma=2.0 Ma=2.0时，涡轮前燃气总温为：
T 3 ∗ ′ = c p T 1 ∗ ′ ( π c ∗ ′ γ − 1 γ − 1 ) c p ′ η c ∗ ′ η T ∗ ′ ( 1 − 1 π T ∗ γ ′ − 1 γ ′ ) = 1005 × 389.7 × ( 8 1.4 − 1 1.4 − 1 ) 1158 × 0.82 × 1 × ( 1 − 1 2.91 6 1.33 − 1 1.33 ) = 1435.12 K {T_{3}^{*}}^{\prime}=\frac{c_p{T_{1}^{*}}^{\prime}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right)}{c_{p}^{\prime}{\eta _{c}^{*}}^{\prime}{\eta _{T}^{*}}^{\prime}\left( 1-\frac{1}{{\pi _{T}^{*}}^{\frac{\gamma ^{\prime}-1}{\gamma ^{\prime}}}} \right)}=\frac{1005\times 389.7\times \left( 8^{\frac{1.4-1}{1.4}}-1 \right)}{1158\times 0.82\times 1\times \left( 1-\frac{1}{2.916^{\frac{1.33-1}{1.33}}} \right)}=1435.12\,\,\mathrm{K} T3∗′=cp′ηc∗′ηT∗′(1−πT∗γ′γ′−11)cpT1∗′(πc∗′γγ−1−1)=1158×0.82×1×(1−2.9161.331.33−11)1005×389.7×(81.41.4−1−1)=1435.12K
由于涡轮效率为 η T ∗ ′ = 1 {\eta _{T}^{*}}^{\prime}=1 ηT∗′=1，所以有：
T 4 ∗ ′ = T 3 ∗ ′ π T ∗ γ ′ − 1 γ ′ = 1435.12 2.91 6 1.33 − 1 1.33 = 1100.44 K {T_{4}^{*}}^{\prime}=\frac{{T_{3}^{*}}^{\prime}}{{\pi _{T}^{*}}^{\frac{\gamma ^{\prime}-1}{\gamma ^{\prime}}}}=\frac{1435.12}{2.916^{\frac{1.33-1}{1.33}}}=1100.44\,\,\mathrm{K} T4∗′=πT∗γ′γ′−1T3∗′=2.9161.331.33−11435.12=1100.44K
若发动机采用转速 n = n= n=常数，涡轮前燃气总温 T 3 ∗ = T_3^*= T3∗=常数的调节规律

根据功率平衡：
c p T 1 ∗ ′ η c ∗ ′ ( π c ∗ ′ γ − 1 γ − 1 ) = c p ′ ( T 3 ∗ − T 4 ∗ ′ ′ ) \frac{c_p{T_{1}^{*}}^{\prime}}{{\eta _{c}^{*}}^{\prime}}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right) =c_{p}^{\prime}\left( T_{3}^{*}-{T_{4}^{*}}^{^{''}} \right) ηc∗′cpT1∗′(πc∗′γγ−1−1)=cp′(T3∗−T4∗′′)
可以计算出 H = 11 k m H=11\ \mathrm{km} H=11 km, M a = 2.0 Ma=2.0 Ma=2.0时，涡轮后燃气总温为：
T 4 ∗ ′ ′ = T 3 ∗ − c p T 1 ∗ ′ c p ′ η c ∗ ′ ( π c ∗ ′ γ − 1 γ − 1 ) = 1304.19 − 1005 × 389.7 1158 × 0.82 ( 8 1.4 − 1 1.4 − 1 ) = 969.51 K {T_{4}^{*}}^{^{''}}=T_{3}^{*}-\frac{c_p{T_{1}^{*}}^{\prime}}{c_{p}^{\prime}{\eta _{c}^{*}}^{\prime}}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right) =1304.19-\frac{1005\times 389.7}{1158\times 0.82}\left( 8^{\frac{1.4-1}{1.4}}-1 \right) =969.51\,\,\mathrm{K} T4∗′′=T3∗−cp′ηc∗′cpT1∗′(πc∗′γγ−1−1)=1304.19−1158×0.821005×389.7(81.41.4−1−1)=969.51K
此时涡轮落压比：
π T ∗ ′ = ( T 3 ∗ T 4 ∗ ′ ′ ) γ ′ γ ′ − 1 = ( 1304.19 969.51 ) 1.33 1.33 − 1 = 3.304 {\pi _{T}^{*}}^{\prime}=\left( \frac{T_{3}^{*}}{{T_{4}^{*}}^{^{''}}} \right) ^{\frac{\gamma ^{\prime}}{\gamma ^{\prime}-1}}=\left( \frac{1304.19}{969.51} \right) ^{\frac{1.33}{1.33-1}}=3.304 πT∗′=(T4∗′′T3∗)γ′−1γ′=(969.511304.19)1.33−11.33=3.304
由于：
π T ∗ = [ σ e A 5 q ( λ 5 ) σ t A t q ( λ t ) ] 2 γ ′ γ ′ + 1 \pi _{T}^{*}=\left[ \frac{\sigma _eA_5q\left( \lambda _5 \right)}{\sigma _tA_tq\left( \lambda _t \right)} \right] ^{\frac{2\gamma ^{\prime}}{\gamma ^{\prime}+1}} πT∗=[σtAtq(λt)σeA5q(λ5)]γ′+12γ′
涡轮导向器，喷管均处于超临界状态,其他各种损失系数和部件效率等于 1.0 1.0 1.0，上式可改写为：
π T ∗ = ( A 5 A t ) 2 γ ′ γ ′ + 1 \pi _{T}^{*}=\left( \frac{A_5}{A_t} \right) ^{\frac{2\gamma ^{\prime}}{\gamma ^{\prime}+1}} πT∗=(AtA5)γ′+12γ′
当飞行条件从 H = 0 H=0 H=0, M a = 0 Ma=0 Ma=0变到 H = 11000 m H=11000\ \mathrm{m} H=11000 m， M a = 2.0 Ma=2.0 Ma=2.0时，喷管喉部面积 A 5 A_5 A5变化为：
A 5 ′ A 5 = ( π T ∗ ′ π T ∗ ) γ ′ + 1 2 γ ′ = ( 3.304 2.916 ) 1.33 + 1 2 × 1.33 = 1.116 \frac{A_{5}^{\prime}}{A_5}=\left( \frac{{\pi _{T}^{*}}^{\prime}}{\pi _{T}^{*}} \right) ^{\frac{\gamma ^{\prime}+1}{2\gamma ^{\prime}}}=\left( \frac{3.304}{2.916} \right) ^{\frac{1.33+1}{2\times 1.33}}=1.116 A5A5′=(πT∗πT∗′)2γ′γ′+1=(2.9163.304)2×1.331.33+1=1.116

2.已知某一单转子涡喷发动机在 H = 0 H=0 H=0， M a = 0 Ma=0 Ma=0（标准大气压条件）工作时，压气机出口总温为 T 2 ∗ = 642 K T_2^*=642\ \mathrm{K} T2∗=642 K，涡轮后燃气总温 T 4 ∗ = 1100 K T_4^*=1100\ \mathrm{K} T4∗=1100 K，涡轮效率 η T ∗ = 0.98 \eta_T^*=0.98 ηT∗=0.98，若发动机采用转速 n = n m a x = n=n_{max}= n=nmax=常数，涡轮前燃气总温 T 3 ∗ = T 3 , m a x ∗ = T_3^*=T_{3,max}^*= T3∗=T3,max∗=常数的调节规律，当 H = 11 k m H=11\ \mathrm{km} H=11 km（ T 0 ′ = 216.5 K T_{0}^{\prime}=216.5 \mathrm{K} T0′=216.5K）, M a = 2.0 Ma=2.0 Ma=2.0时， π c ∗ ′ = 8.0 {\pi _{c}^{*}}^{\prime}=8.0 πc∗′=8.0， η c ∗ ′ = 0.81 {\eta _{c}^{*}}^{\prime}=0.81 ηc∗′=0.81，涡轮效率 η T ∗ ′ = 0.95 {\eta _{T}^{*}}^{\prime}=0.95 ηT∗′=0.95，当飞行条件从 H = 0 H=0 H=0, M a = 0 Ma=0 Ma=0变到 H = 11000 m H=11000\ \mathrm{m} H=11000 m， M a = 2.0 Ma=2.0 Ma=2.0时，涡轮前、后燃气总温 T 3 ∗ T_3^* T3∗、 T 4 ∗ T_4^* T4∗各为多少?落压比为多少？（假设涡轮导向器，喷管均处于超临界状态，各种机械损失效率为 0.98 0.98 0.98，两种情况下油气比 f = 0.02 f=0.02 f=0.02，引气冷却系数 v c o l = 0.01 v_{col}=0.01 vcol=0.01，定压比热容分别为 1.005 k J / ( k g ⋅ K ) 1.005\ \mathrm{kJ/(kg\cdot K)} 1.005 kJ/(kg⋅K)、 1.158 k J / ( k g ⋅ K ) 1.158\ \mathrm{kJ/(kg\cdot K)} 1.158 kJ/(kg⋅K);绝热指数分别为 1.40 1.40 1.40、 1.33 1.33 1.33）

解： β = 1 + f − v c o l = 1 + 0.02 − 0.01 = 1.01 \beta =1+f-v_{col}=1+0.02-0.01=1.01 β=1+f−vcol=1+0.02−0.01=1.01

H = 0 H=0 H=0、 M a = 0 Ma=0 Ma=0，进气道内为绝能流动，可以得到： T 1 ∗ = T 0 ∗ = T 0 = 288.15 K T_{1}^{*}=T_{0}^{*}=T_0=288.15 \mathrm{K} T1∗=T0∗=T0=288.15K

采用转速 n = n m a x = n=n_{max}= n=nmax=常数，涡轮前燃气总温 T 3 ∗ = T 3 , m a x ∗ = T_3^*=T_{3,max}^*= T3∗=T3,max∗=常数的调节规律

功率平衡： w c = w T β η m w_c=w_T\beta \eta _m wc=wTβηm，即：
c p ( T 2 ∗ − T 1 ∗ ) = c p ′ ( T 3 ∗ − T 4 ∗ ) β η m c_p\left( T_{2}^{*}-T_{1}^{*} \right) =c_{p}^{\prime}\left( T_{3}^{*}-T_{4}^{*} \right) \beta \eta _m cp(T2∗−T1∗)=cp′(T3∗−T4∗)βηm
可以计算出涡轮前总温：
T 3 ∗ = c p c p ′ β η m ( T 2 ∗ − T 1 ∗ ) + T 4 ∗ = 1005 1158 × 1.01 × 0.98 × ( 642 − 288.15 ) + 1100 = 1410.26 K T_{3}^{*}=\frac{c_p}{c_{p}^{\prime}\beta \eta _m}\left( T_{2}^{*}-T_{1}^{*} \right) +T_{4}^{*}=\frac{1005}{1158\times 1.01\times 0.98}\times \left( 642-288.15 \right) +1100=1410.26\,\,\mathrm{K} T3∗=cp′βηmcp(T2∗−T1∗)+T4∗=1158×1.01×0.981005×(642−288.15)+1100=1410.26K
H = 11 k m H=11\ \mathrm{km} H=11 km、 M a = 2 Ma=2 Ma=2，进气道内为绝能流动，可以得到：
T 1 ∗ ′ = T 0 ∗ ′ = T 0 ′ ( 1 + γ − 1 2 M a 2 ) = 216.5 × ( 1 + 1.4 − 1 2 × 2 2 ) = 389.7 K {T_{1}^{*}}^{\prime}={T_{0}^{*}}^{\prime}={T_0}^{\prime}\left( 1+\frac{\gamma -1}{2}Ma^2 \right) =216.5\times \left( 1+\frac{1.4-1}{2}\times 2^2 \right) =389.7\,\,\mathrm{K} T1∗′=T0∗′=T0′(1+2γ−1Ma2)=216.5×(1+21.4−1×22)=389.7K
由功率平衡：
c p T 1 ∗ ′ η c ∗ ′ ( π c ∗ ′ γ − 1 γ − 1 ) = c p ′ ( T 3 ∗ ′ − T 4 ∗ ′ ) β η m \frac{c_p{T_{1}^{*}}^{\prime}}{{\eta _{c}^{*}}^{\prime}}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right) =c_{p}^{\prime}\left( {T_{3}^{*}}^{\prime}-{T_{4}^{*}}^{\prime} \right) \beta \eta _m ηc∗′cpT1∗′(πc∗′γγ−1−1)=cp′(T3∗′−T4∗′)βηm
可以计算出:
T 4 ∗ ′ = T 3 ∗ ′ − c p T 1 ∗ ′ c p ′ η c ∗ ′ β η m ( π c ∗ ′ γ − 1 γ − 1 ) = 1410.26 − 1005 × 389.7 1158 × 0.81 × 1.01 × 0.98 ( 8 1.4 − 1 1.4 − 1 ) = 1067.95 K {T_{4}^{*}}^{\prime}={T_{3}^{*}}^{\prime}-\frac{c_p{T_{1}^{*}}^{\prime}}{c_{p}^{\prime}{\eta _{c}^{*}}^{\prime}\beta \eta _m}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right) =1410.26-\frac{1005\times 389.7}{1158\times 0.81\times 1.01\times 0.98}\left( 8^{\frac{1.4-1}{1.4}}-1 \right) =1067.95\,\,\mathrm{K} T4∗′=T3∗′−cp′ηc∗′βηmcpT1∗′(πc∗′γγ−1−1)=1410.26−1158×0.81×1.01×0.981005×389.7(81.41.4−1−1)=1067.95K
其中： T 3 ∗ ′ = T 3 ∗ = 1410.26 K {T_{3}^{*}}^{\prime}=T_{3}^{*}=1410.26 \mathrm{K} T3∗′=T3∗=1410.26K

功率平衡还可以写为：
c p T 1 ∗ ′ η c ∗ ′ ( π c ∗ ′ γ − 1 γ − 1 ) = c p ′ T 3 ∗ ′ ( 1 − 1 π T ∗ γ ′ − 1 γ ′ ) η T ∗ ′ β η m \frac{c_p{T_{1}^{*}}^{\prime}}{{\eta _{c}^{*}}^{\prime}}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right) =c_{p}^{\prime}{T_{3}^{*}}^{\prime}\left( 1-\frac{1}{{\pi _{T}^{*}}^{\frac{\gamma ^{\prime}-1}{\gamma ^{\prime}}}} \right) {\eta _{T}^{*}}^{\prime}\beta \eta _m ηc∗′cpT1∗′(πc∗′γγ−1−1)=cp′T3∗′ 1−πT∗γ′γ′−11 ηT∗′βηm
解出涡轮落压比：
π T ∗ = ( 1 − c p T 1 ∗ ′ ( π c ∗ ′ γ − 1 γ − 1 ) c p ′ T 3 ∗ ′ η c ∗ ′ η T ∗ ′ β η m ) γ ′ 1 − γ ′ = ( 1 − 1005 × 389.7 ( 8 1.4 − 1 1.4 − 1 ) 1158 × 1410.26 × 0.81 × 0.95 × 1.01 × 0.98 ) 1.33 1 − 1.33 = 3.28 \pi _{T}^{*}=\left( 1-\frac{c_p{T_{1}^{*}}^{\prime}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right)}{c_{p}^{\prime}{T_{3}^{*}}^{\prime}{\eta _{c}^{*}}^{\prime}{\eta _{T}^{*}}^{\prime}\beta \eta _m} \right) ^{\frac{\gamma ^{\prime}}{1-\gamma ^{\prime}}}=\left( 1-\frac{1005\times 389.7\left( 8^{\frac{1.4-1}{1.4}}-1 \right)}{1158\times 1410.26\times 0.81\times 0.95\times 1.01\times 0.98} \right) ^{\frac{1.33}{1-1.33}}=3.28 πT∗= 1−cp′T3∗′ηc∗′ηT∗′βηmcpT1∗′(πc∗′γγ−1−1) 1−γ′γ′= 1−1158×1410.26×0.81×0.95×1.01×0.981005×389.7(81.41.4−1−1) 1−1.331.33=3.28

3.某单轴涡喷发动机在标准海平面 H = 0 H=0 H=0,马赫数 M a = 0 Ma=0 Ma=0以最大转速工作时，压气机的 π c ∗ = 12.0 \pi_c^*=12.0 πc∗=12.0，压气机效率 η c ∗ = 0.84 \eta _{c}^{*}=0.84 ηc∗=0.84、涡轮效率 η T ∗ = 0.92 \eta_T^*=0.92 ηT∗=0.92，涡轮后燃气总温 T 4 ∗ = 1100 K T_4^*=1100\ \mathrm{K} T4∗=1100 K，若该发动机采用转速 n = n= n=常数，喷管喉部面积 A 5 = A_5= A5=常数的调节规律，当 H = 11 k m H=11\ \mathrm{km} H=11 km（ T 0 ′ = 216.5 K T_{0}^{\prime}=216.5 \mathrm{K} T0′=216.5K）, M a = 2.0 Ma=2.0 Ma=2.0时， π c ∗ ′ = 8.0 {\pi _{c}^{*}}^{\prime}=8.0 πc∗′=8.0， η c ∗ ′ = 0.81 {\eta _{c}^{*}}^{\prime}=0.81 ηc∗′=0.81，涡轮效率 η T ∗ ′ = 0.91 {\eta _{T}^{*}}^{\prime}=0.91 ηT∗′=0.91，试计算:当飞行条件从 H = 0 H=0 H=0， M a = 0 Ma=0 Ma=0变到 H = 11 k m H=11\ \mathrm{km} H=11 km, M a = 2.0 Ma=2.0 Ma=2.0时，涡轮前燃气总温 T 3 ∗ ′ {T_{3}^{*}}^{\prime} T3∗′等于多少?假设涡轮导向器，喷管均处于超临界状态，两种情况下油气比 f = 0.01 f=0.01 f=0.01，引气冷却系数 v c o l = 0.01 v_{col}=0.01 vcol=0.01，定压比热容分别为 1.005 k J / ( k g ⋅ K ) 1.005\ \mathrm{kJ/(kg\cdot K)} 1.005 kJ/(kg⋅K)、 1.158 k J / ( k g ⋅ K ) 1.158\ \mathrm{kJ/(kg\cdot K)} 1.158 kJ/(kg⋅K);定熵指数分别为 1.40 1.40 1.40、 1.33 1.33 1.33；其它各种损失系数等于 1.0 1.0 1.0）

解： β = 1 + f − v c o l = 1 + 0.01 − 0.01 = 1 \beta =1+f-v_{col}=1+0.01-0.01=1 β=1+f−vcol=1+0.01−0.01=1

H = 0 H=0 H=0、 M a = 0 Ma=0 Ma=0，进气道内为绝能流动，可以得到： T 1 ∗ = T 0 ∗ = T 0 = 288.15 K T_{1}^{*}=T_{0}^{*}=T_0=288.15 \mathrm{K} T1∗=T0∗=T0=288.15K

采用转速 n = n= n=常数，喷管喉部面积 A 5 = A_5= A5=常数的调节规律

功率平衡： w c = w T β η m w_c=w_T\beta \eta _m wc=wTβηm，由于 β = 1 \beta=1 β=1、 η m = 1 \eta_m=1 ηm=1,可以得到 w c = w T w_c=w_T wc=wT，即：
c p T 1 ∗ η c ∗ ( π c ∗ γ − 1 γ − 1 ) = c p ′ ( T 3 ∗ − T 4 ∗ ) \frac{c_pT_{1}^{*}}{\eta _{c}^{*}}\left( {\pi _{c}^{*}}^{\frac{\gamma -1}{\gamma}}-1 \right) =c_{p}^{\prime}\left( T_{3}^{*}-T_{4}^{*} \right) ηc∗cpT1∗(πc∗γγ−1−1)=cp′(T3∗−T4∗)
可以算出涡轮前总温：
T 3 ∗ = c p T 1 ∗ c p ′ η c ∗ ( π c ∗ γ − 1 γ − 1 ) + T 4 ∗ = 1005 × 288.15 1158 × 0.84 ( 1 2 1.4 − 1 1.4 − 1 ) + 1100 = 1407.82 K T_{3}^{*}=\frac{c_pT_{1}^{*}}{c_{p}^{\prime}\eta _{c}^{*}}\left( {\pi _{c}^{*}}^{\frac{\gamma -1}{\gamma}}-1 \right) +T_{4}^{*}=\frac{1005\times 288.15}{1158\times 0.84}\left( 12^{\frac{1.4-1}{1.4}}-1 \right) +1100=1407.82\,\,\mathrm{K} T3∗=cp′ηc∗cpT1∗(πc∗γγ−1−1)+T4∗=1158×0.841005×288.15(121.41.4−1−1)+1100=1407.82K
涡轮功可写为：
w T = c p ′ ( T 3 ∗ − T 4 ∗ ) = c p ′ T 3 ∗ ( 1 − 1 π T ∗ γ ′ − 1 γ ′ ) η T ∗ w_T=c_{p}^{\prime}\left( T_{3}^{*}-T_{4}^{*} \right) =c_{p}^{\prime}T_{3}^{*}\left( 1-\frac{1}{{\pi _{T}^{*}}^{\frac{\gamma ^{\prime}-1}{\gamma ^{\prime}}}} \right) \eta _{T}^{*} wT=cp′(T3∗−T4∗)=cp′T3∗ 1−πT∗γ′γ′−11 ηT∗
解出涡轮落压比：
π T ∗ = ( T 3 ∗ η T ∗ T 3 ∗ η T ∗ + T 4 ∗ − T 3 ∗ ) γ ′ γ ′ − 1 = ( 1407.82 × 0.92 1407.82 × 0.92 + 1100 − 1407.82 ) 1.33 1.33 − 1 = 2.99 \pi _{T}^{*}=\left( \frac{T_{3}^{*}\eta _{T}^{*}}{T_{3}^{*}\eta _{T}^{*}+T_{4}^{*}-T_{3}^{*}} \right) ^{\frac{\gamma ^{\prime}}{\gamma ^{\prime}-1}}=\left( \frac{1407.82\times 0.92}{1407.82\times 0.92+1100-1407.82} \right) ^{\frac{1.33}{1.33-1}}=2.99 πT∗=(T3∗ηT∗+T4∗−T3∗T3∗ηT∗)γ′−1γ′=(1407.82×0.92+1100−1407.821407.82×0.92)1.33−11.33=2.99
涡轮导向器，喷管均处于超临界状态,可知涡轮落压比 π T ∗ \pi_T^* πT∗不变。

H = 11 k m H=11\ \mathrm{km} H=11 km、 M a = 2 Ma=2 Ma=2，进气道内为绝能流动，可以得到：
T 1 ∗ ′ = T 0 ∗ ′ = T 0 ′ ( 1 + γ − 1 2 M a 2 ) = 216.5 × ( 1 + 1.4 − 1 2 × 2 2 ) = 389.7 K {T_{1}^{*}}^{\prime}={T_{0}^{*}}^{\prime}={T_0}^{\prime}\left( 1+\frac{\gamma -1}{2}Ma^2 \right) =216.5\times \left( 1+\frac{1.4-1}{2}\times 2^2 \right) =389.7\,\,\mathrm{K} T1∗′=T0∗′=T0′(1+2γ−1Ma2)=216.5×(1+21.4−1×22)=389.7K
由功率平衡：
c p T 1 ∗ ′ η c ∗ ′ ( π c ∗ ′ γ − 1 γ − 1 ) = c p ′ T 3 ∗ ′ ( 1 − 1 π T ∗ γ ′ − 1 γ ′ ) η T ∗ ′ \frac{c_p{T_{1}^{*}}^{\prime}}{{\eta _{c}^{*}}^{\prime}}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right) =c_{p}^{\prime}{T_{3}^{*}}^{\prime}\left( 1-\frac{1}{{\pi _{T}^{*}}^{\frac{\gamma ^{\prime}-1}{\gamma ^{\prime}}}} \right) {\eta _{T}^{*}}^{\prime} ηc∗′cpT1∗′(πc∗′γγ−1−1)=cp′T3∗′ 1−πT∗γ′γ′−11 ηT∗′
可以计算出 H = 11 k m H=11\ \mathrm{km} H=11 km, M a = 2.0 Ma=2.0 Ma=2.0时，涡轮前燃气总温为：
T 3 ∗ ′ = c p T 1 ∗ ′ ( π c ∗ ′ γ − 1 γ − 1 ) c p ′ η c ∗ ′ η T ∗ ′ ( 1 − 1 π T ∗ γ ′ − 1 γ ′ ) = 1005 × 389.7 × ( 8 1.4 − 1 1.4 − 1 ) 1158 × 0.81 × 0.91 × ( 1 − 1 2.9 9 1.33 − 1 1.33 ) = 1564.64 K {T_{3}^{*}}^{\prime}=\frac{c_p{T_{1}^{*}}^{\prime}\left( {\pi _{c}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1 \right)}{c_{p}^{\prime}{\eta _{c}^{*}}^{\prime}{\eta _{T}^{*}}^{\prime}\left( 1-\frac{1}{{\pi _{T}^{*}}^{\frac{\gamma ^{\prime}-1}{\gamma ^{\prime}}}} \right)}=\frac{1005\times 389.7\times \left( 8^{\frac{1.4-1}{1.4}}-1 \right)}{1158\times 0.81\times 0.91\times \left( 1-\frac{1}{2.99^{\frac{1.33-1}{1.33}}} \right)}=1564.64\,\,\mathrm{K} T3∗′=cp′ηc∗′ηT∗′(1−πT∗γ′γ′−11)cpT1∗′(πc∗′γγ−1−1)=1158×0.81×0.91×(1−2.991.331.33−11)1005×389.7×(81.41.4−1−1)=1564.64K