航空发动机原理复习之计算题总结(二)

文章目录

类型三：压气机加级、换级、去级
- 03 知识回顾
- - 相似参数
- （1）压气机加级
- （2）压气机换级
- （3）压气机去级

类型三：压气机加级、换级、去级

03 知识回顾

相似参数

对于同一台压气机，无论压气机的转速 n n n、空气流量 q m q_m qm、进口总温 T 1 ∗ T_1^* T1∗和进口总压 p 1 ∗ p_1^* p1∗如何变化,只要保证压气机的相似参数相同，则压气机的增压比 π c ∗ \pi _{c}^{*} πc∗和效率 η c ∗ \eta _{c}^{*} ηc∗就不变。

1.转速相似参数：
n T 1 ∗ \frac{n}{\sqrt{T_{1}^{*}}} T1∗ n
2.空气流量相似参数：
q m T 1 ∗ p 1 ∗ \frac{q_m\sqrt{T_{1}^{*}}}{p_{1}^{*}} p1∗qmT1∗

（1）压气机加级

1.某压气机原有十级，在标准大气条件下（ T 0 = 288.15 K T_0=288.15\ \mathrm{K} T0=288.15 K、 p 0 = 101325 P a p_0=101325\ \mathrm{Pa} p0=101325 Pa），设计参数为： π c ∗ = 6.45 \pi _{c}^{*}=6.45 πc∗=6.45， η c ∗ = 0.82 \eta _{c}^{*}=0.82 ηc∗=0.82， n = 15100 r / min ⁡ n=15100\ \mathrm{r}/\min n=15100 r/min， q m , a = 13.2 k g / s q_{m,a}=13.2 \ \mathrm{kg}/\mathrm{s} qm,a=13.2 kg/s，为提高该压气机的性能，在保证原十级压气机增压比和效率不变的前提下，在第一级前面加上增压比 1.15 1.15 1.15，效率为 0.84 0.84 0.84的零级，试求加零级后压气机的转速、流量、总增压比和总效率。

解：

地面试车： T 1 ∗ = T 0 ∗ = T 0 = 288.15 K T_1^*=T_0^*=T_0=288.15\ \mathrm{K} T1∗=T0∗=T0=288.15 K；忽略进气道的总压损失： p 1 ∗ = p 0 ∗ = p 0 = 101325 P a p_1^*=p_0^*=p_0=101325\ \mathrm{Pa} p1∗=p0∗=p0=101325 Pa

首先求零级出口，即加零级后新十级压气机进口的气流参数：
p 10 ∗ = p 1 ∗ π c 0 ∗ = 101325 × 1.15 = 116524 P a p_{10}^{*}=p_{1}^{*}\pi _{c0}^{*}=101325\times 1.15=116524\,\,\mathrm{Pa} p10∗=p1∗πc0∗=101325×1.15=116524Pa
零级压气机功可写为：
c p ( T 10 ∗ − T 1 ∗ ) = c p T 1 ∗ π c 0 ∗ γ − 1 γ − 1 η c 0 ∗ c_p\left( T_{10}^{*}-T_{1}^{*} \right) =c_pT_{1}^{*}\frac{{\pi _{c0}^{*}}^{\frac{\gamma -1}{\gamma}}-1}{\eta _{c0}^{*}} cp(T10∗−T1∗)=cpT1∗ηc0∗πc0∗γγ−1−1
于是可以计算出：
T 10 ∗ = T 1 ∗ ( 1 + π c 0 ∗ γ − 1 γ − 1 η c 0 ∗ ) = 288.15 × ( 1 + 1.1 5 1.4 − 1 1.4 − 1 0.84 ) = 302.13 K T_{10}^{*}=T_{1}^{*}\left( 1+\frac{{\pi _{c0}^{*}}^{\frac{\gamma -1}{\gamma}}-1}{\eta _{c0}^{*}} \right) =288.15\times \left( 1+\frac{1.15^{\frac{1.4-1}{1.4}}-1}{0.84} \right) =302.13\,\,\mathrm{K} T10∗=T1∗(1+ηc0∗πc0∗γγ−1−1)=288.15×(1+0.841.151.41.4−1−1)=302.13K
根据相似准则
n T 1 ∗ = n ′ T 10 ∗ , q m , a T 1 ∗ p 1 ∗ = q m , a ′ T 10 ∗ p 10 ∗ \frac{n}{\sqrt{T_{1}^{*}}}=\frac{n^{\prime}}{\sqrt{T_{10}^{*}}}\,\,,\frac{q_{m,a}\sqrt{T_{1}^{*}}}{p_{1}^{*}}=\frac{q_{m,a}^{\prime}\sqrt{T_{10}^{*}}}{p_{10}^{*}} T1∗ n=T10∗ n′,p1∗qm,aT1∗ =p10∗qm,a′T10∗
可以求出加零级以后的流量和转速：
n ′ = n T 10 ∗ T 1 ∗ = 15100 × 302.13 288.15 = 15462 r / min ⁡ q m , a ′ = q m , a p 10 ∗ p 1 ∗ T 1 ∗ T 10 ∗ = 13.2 × 1.15 × 288.15 302.13 = 14.82 k g / s n^{\prime}=n\sqrt{\frac{T_{10}^{*}}{T_{1}^{*}}}=15100\times \sqrt{\frac{302.13}{288.15}}=15462\,\,\mathrm{r}/\min \\ q_{m,a}^{\prime}=q_{m,a}\frac{p_{10}^{*}}{p_{1}^{*}}\sqrt{\frac{T_{1}^{*}}{T_{10}^{*}}}=13.2\times 1.15\times \sqrt{\frac{288.15}{302.13}}=14.82\,\,\mathrm{kg}/\mathrm{s} n′=nT1∗T10∗ =15100×288.15302.13 =15462r/minqm,a′=qm,ap1∗p10∗T10∗T1∗ =13.2×1.15×302.13288.15 =14.82kg/s
总增压比： π c ∗ ′ = π c ∗ π c 0 ∗ = 6.45 × 1.15 = 7.42 {\pi _{c}^{*}}^{\prime}=\pi _{c}^{*}\pi _{c0}^{*}=6.45\times 1.15=7.42 πc∗′=πc∗πc0∗=6.45×1.15=7.42

压气机功可写为：
w c , s ′ η c ∗ ′ = w c , s 0 η c 0 ∗ + w c , s η c ∗ \frac{w_{c,s}^{\prime}}{{\eta _{c}^{*}}^{\prime}}=\frac{w_{c,s0}}{\eta _{c0}^{*}}+\frac{w_{c,s}}{\eta _{c}^{*}} ηc∗′wc,s′=ηc0∗wc,s0+ηc∗wc,s
于是总效率为：

2.某台压气机原有十级，在 H = 10 k m H=10\ \mathrm{km} H=10 km, M a = 1.5 Ma=1.5 Ma=1.5的飞行条件下，设计参数为： π c ∗ = 6.5 \pi _{c}^{*}=6.5 πc∗=6.5， η c ∗ = 0.83 \eta _{c}^{*}=0.83 ηc∗=0.83， n = 11000 r / min ⁡ n=11000\ \mathrm{r}/\min n=11000 r/min， q m , a = 26.0 k g / s q_{m,a}=26.0 \ \mathrm{kg}/\mathrm{s} qm,a=26.0 kg/s，为提高该压气机的性能，在保证原十级压气机增压比和效率不变的前提下，在第一级前面加上增压比 1.25 1.25 1.25，效率为 0.85 0.85 0.85的零级，试求加零级后压气机的转速、流量。（设 T 0 = 288.15 − 6.5 H T_0=288.15-6.5H T0=288.15−6.5H， H H H单位： k m \mathrm{km} km）

解： T 0 = 288.15 − 6.5 H = 288.15 − 6.5 × 10 = 223.15 K T_0=288.15-6.5H=288.15-6.5\times 10=223.15\,\,\mathrm{K} T0=288.15−6.5H=288.15−6.5×10=223.15K、 p 0 = 26436 P a p_0=26436\ \mathrm{Pa} p0=26436 Pa

压气机进气总温
T 1 ∗ = T 0 ∗ = T 0 ( 1 + γ − 1 2 M a 2 ) = 223.15 × ( 1 + 1.4 − 1 2 × 1. 5 2 ) = 323.57 K T_{1}^{*}=T_{0}^{*}=T_0\left( 1+\frac{\gamma -1}{2}Ma^2 \right) =223.15\times \left( 1+\frac{1.4-1}{2}\times 1.5^2 \right) =323.57\,\,\mathrm{K} T1∗=T0∗=T0(1+2γ−1Ma2)=223.15×(1+21.4−1×1.52)=323.57K
忽略进气道的总压损失，压气机进气总压
p 1 ∗ = p 0 ∗ = p 0 π ( M a ) = p 0 ( 1 + γ − 1 2 M a 2 ) γ γ − 1 = 26436 × ( 1 + 1.4 − 1 2 × 1. 5 2 ) 1.4 1.4 − 1 = 97047.4 P a p_{1}^{*}=p_{0}^{*}=\frac{p_0}{\pi \left( Ma \right)}=p_0\left( 1+\frac{\gamma -1}{2}Ma^2 \right) ^{\frac{\gamma}{\gamma -1}}=26436\times \left( 1+\frac{1.4-1}{2}\times 1.5^2 \right) ^{\frac{1.4}{1.4-1}}=97047.4\,\,\mathrm{Pa} p1∗=p0∗=π(Ma)p0=p0(1+2γ−1Ma2)γ−1γ=26436×(1+21.4−1×1.52)1.4−11.4=97047.4Pa

首先求零级出口，即加零级后新十级压气机进口的气流参数：
p 10 ∗ = p 1 ∗ π c 0 ∗ = 97047.4 × 1.25 = 121309 P a p_{10}^{*}=p_{1}^{*}\pi _{c0}^{*}=97047.4\times 1.25=121309\,\,\mathrm{Pa} \\ p10∗=p1∗πc0∗=97047.4×1.25=121309Pa
零级压气机功可写为：
c p ( T 10 ∗ − T 1 ∗ ) = c p T 1 ∗ π c 0 ∗ γ − 1 γ − 1 η c 0 ∗ c_p\left( T_{10}^{*}-T_{1}^{*} \right) =c_pT_{1}^{*}\frac{{\pi _{c0}^{*}}^{\frac{\gamma -1}{\gamma}}-1}{\eta _{c0}^{*}} cp(T10∗−T1∗)=cpT1∗ηc0∗πc0∗γγ−1−1
于是可以计算出：
T 10 ∗ = T 1 ∗ ( 1 + π c 0 ∗ γ − 1 γ − 1 η c 0 ∗ ) = 323.57 × ( 1 + 1.2 5 1.4 − 1 1.4 − 1 0.85 ) = 348.63 K T_{10}^{*}=T_{1}^{*}\left( 1+\frac{{\pi _{c0}^{*}}^{\frac{\gamma -1}{\gamma}}-1}{\eta _{c0}^{*}} \right) =323.57\times \left( 1+\frac{1.25^{\frac{1.4-1}{1.4}}-1}{0.85} \right) =348.63\,\,\mathrm{K} \\ T10∗=T1∗(1+ηc0∗πc0∗γγ−1−1)=323.57×(1+0.851.251.41.4−1−1)=348.63K
根据相似准则
n T 1 ∗ = n ′ T 10 ∗ , q m , a T 1 ∗ p 1 ∗ = q m , a ′ T 10 ∗ p 10 ∗ \frac{n}{\sqrt{T_{1}^{*}}}=\frac{n^{\prime}}{\sqrt{T_{10}^{*}}}\,\,,\frac{q_{m,a}\sqrt{T_{1}^{*}}}{p_{1}^{*}}=\frac{q_{m,a}^{\prime}\sqrt{T_{10}^{*}}}{p_{10}^{*}} T1∗ n=T10∗ n′,p1∗qm,aT1∗ =p10∗qm,a′T10∗
可以求出加零级以后的流量和转速：
n ′ = n T 10 ∗ T 1 ∗ = 11000 × 348.63 323.57 = 11418 r / min ⁡ q m , a ′ = q m , a p 10 ∗ p 1 ∗ T 1 ∗ T 10 ∗ = 26 × 1.25 × 323.57 348.63 = 31.31 k g / s n^{\prime}=n\sqrt{\frac{T_{10}^{*}}{T_{1}^{*}}}=11000\times \sqrt{\frac{348.63}{323.57}}=11418\,\,\mathrm{r}/\min \\ q_{m,a}^{\prime}=q_{m,a}\frac{p_{10}^{*}}{p_{1}^{*}}\sqrt{\frac{T_{1}^{*}}{T_{10}^{*}}}=26\times 1.25\times \sqrt{\frac{323.57}{348.63}}=31.31\,\,\mathrm{kg}/\mathrm{s} \\ n′=nT1∗T10∗ =11000×323.57348.63 =11418r/minqm,a′=qm,ap1∗p10∗T10∗T1∗ =26×1.25×348.63323.57 =31.31kg/s

（2）压气机换级

1.一台十一级的压气机，拟进一步提高流经压气机的流量和增压比，其办法是用一台增压比 π c 1 ∗ ′ = 1.96 {\pi _{c1}^{*}}^{\prime}=1.96 πc1∗′=1.96，效率 η c 1 ∗ ′ = 0.855 {\eta _{c1}^{*}}^{\prime}=0.855 ηc1∗′=0.855的压气机替换原压气机的第一级,并保证原压气机后面十级的增压比和效率不变。原十一级压气机在标准大气条件下的设计参数为: π c ∗ = 23 \pi _{c}^{*}=23 πc∗=23， η c ∗ = 0.82 \eta _{c}^{*}=0.82 ηc∗=0.82，$n=11950\ \mathrm{r}/\min ，，，q_{m,a}=64.0 \ \mathrm{kg}/\mathrm{s} ，其中第一级的参数为，其中第一级的参数为，其中第一级的参数为\pi _{c1}^{}=1.5 、、、\eta _{c1}^{}=0.83$，试求新十一级压气机的转速、流量，总增压比和效率。

解:

首先求原压气机第 1 1 1级出口的气流参数：

换级前
p 1 ∗ ′ = p 1 ∗ π c 1 ∗ = 101325 × 1.50 = 151988 P a {p_{1}^{*}}^{\prime}=p_{1}^{*}\pi _{c1}^{*}=101325\times 1.50=151988\,\,\mathrm{Pa} p1∗′=p1∗πc1∗=101325×1.50=151988Pa
第 1 1 1级压气功可写为：
c p ( T 1 ∗ ′ − T 1 ∗ ) = c p T 1 ∗ π c 1 ∗ γ − 1 γ − 1 η c 1 ∗ c_p\left( {T_{1}^{*}}^{\prime}-T_{1}^{*} \right) =c_pT_{1}^{*}\frac{{\pi _{c1}^{*}}^{\frac{\gamma -1}{\gamma}}-1}{\eta _{c1}^{*}} cp(T1∗′−T1∗)=cpT1∗ηc1∗πc1∗γγ−1−1
于是可以计算出：
T 1 ∗ ′ = T 1 ∗ ( 1 + π c 1 ∗ γ − 1 γ − 1 η c 1 ∗ ) = 288.15 × ( 1 + 1.5 0 1.4 − 1 1.4 − 1 0.83 ) = 330.79 K {T_{1}^{*}}^{\prime}=T_{1}^{*}\left( 1+\frac{{\pi _{c1}^{*}}^{\frac{\gamma -1}{\gamma}}-1}{\eta _{c1}^{*}} \right) =288.15\times \left( 1+\frac{1.50^{\frac{1.4-1}{1.4}}-1}{0.83} \right) =330.79\,\,\mathrm{K} T1∗′=T1∗(1+ηc1∗πc1∗γγ−1−1)=288.15×(1+0.831.501.41.4−1−1)=330.79K
同理，可以求出换级后新压气机第 1 1 1级出口的气流参数：
p 1 ∗ ′ ′ = p 1 ∗ π c 1 ∗ ′ = 101325 × 1.96 = 198597 P a T 1 ∗ ′ ′ = T 1 ∗ ( 1 + π c 1 ∗ ′ γ − 1 γ − 1 η c 1 ∗ ′ ) = 288.15 × ( 1 + 1.9 6 1.4 − 1 1.4 − 1 0.855 ) = 359.60 K {p_{1}^{*}}^{''}=p_{1}^{*}{\pi _{c1}^{*}}^{\prime}=101325\times 1.96=198597\,\,\mathrm{Pa} \\ {T_{1}^{*}}^{''}=T_{1}^{*}\left( 1+\frac{{\pi _{c1}^{*^{\prime}}}^{\frac{\gamma -1}{\gamma}}-1}{{\eta _{c1}^{*}}^{\prime}} \right) =288.15\times \left( 1+\frac{1.96^{\frac{1.4-1}{1.4}}-1}{0.855} \right) =359.60\,\,\mathrm{K} p1∗′′=p1∗πc1∗′=101325×1.96=198597PaT1∗′′=T1∗ 1+ηc1∗′πc1∗′γγ−1−1 =288.15×(1+0.8551.961.41.4−1−1)=359.60K
根据相似准则
n T 1 ∗ ′ = n ′ T 1 ∗ ′ ′ , q m , a T 1 ∗ ′ p 1 ∗ ′ = q m , a ′ T 1 ∗ ′ ′ p 1 ∗ ′ ′ \frac{n}{\sqrt{{T_{1}^{*}}^{\prime}}}=\frac{n^{\prime}}{\sqrt{{T_{1}^{*}}^{''}}}\,\,,\frac{q_{m,a}\sqrt{{T_{1}^{*}}^{\prime}}}{{p_{1}^{*}}^{\prime}}=\frac{q_{m,a}^{\prime}\sqrt{{T_{1}^{*}}^{''}}}{{p_{1}^{*}}^{''}} T1∗′ n=T1∗′′ n′,p1∗′qm,aT1∗′ =p1∗′′qm,a′T1∗′′
可以求出换级以后的流量和转速：
n ′ = n T 1 ∗ ′ ′ T 1 ∗ ′ = 11950 × 359.60 330.79 = 12460 r / min ⁡ q m , a ′ = q m , a p 1 ∗ ′ ′ p 1 ∗ ′ T 1 ∗ ′ T 1 ∗ ′ ′ = 64 × 198597 151988 × 330.79 359.60 = 80.21 k g / s n^{\prime}=n\sqrt{\frac{{T_{1}^{*}}^{''}}{{T_{1}^{*}}^{\prime}}}=11950\times \sqrt{\frac{359.60}{330.79}}=12460\,\,\mathrm{r}/\min \\ q_{m,a}^{\prime}=q_{m,a}\frac{{p_{1}^{*}}^{''}}{{p_{1}^{*}}^{\prime}}\sqrt{\frac{{T_{1}^{*}}^{\prime}}{{T_{1}^{*}}^{''}}}=64\times \frac{198597}{151988}\times \sqrt{\frac{330.79}{359.60}}=80.21\,\,\mathrm{kg}/\mathrm{s} n′=nT1∗′T1∗′′ =11950×330.79359.60 =12460r/minqm,a′=qm,ap1∗′p1∗′′T1∗′′T1∗′ =64×151988198597×359.60330.79 =80.21kg/s
原 10 10 10级压气机增压比： π c 10 ∗ = π c ∗ / π c 1 ∗ = 23 / 1.5 = 15.33 \pi _{c10}^{*}=\pi _{c}^{*}/\pi _{c1}^{*}=23/1.5=15.33 πc10∗=πc∗/πc1∗=23/1.5=15.33

新十一级压气机总增压比： π c 11 ∗ = π c 1 ∗ ′ π c 10 ∗ = 1.96 × 15.33 = 30.05 \pi _{c11}^{*}={\pi _{c1}^{*}}^{\prime}\pi _{c10}^{*}=1.96\times 15.33=30.05 πc11∗=πc1∗′πc10∗=1.96×15.33=30.05

压气机功可写为：
w c , s η c ∗ = w c , s 1 η c 1 ∗ + w c , s 10 η c 10 ∗ \frac{w_{c,s}}{\eta _{c}^{*}}=\frac{w_{c,s1}}{\eta _{c1}^{*}}+\frac{w_{c,s10}}{\eta _{c10}^{*}} ηc∗wc,s=ηc1∗wc,s1+ηc10∗wc,s10
可以算出后 10 10 10级压气机效率：
η c 10 ∗ = w c , s 10 w c , s η c ∗ − w c , s 1 η c 1 ∗ = T 1 ∗ ′ ( π c 10 ∗ γ − 1 γ − 1 ) T 1 ∗ ( π c ∗ γ − 1 γ − 1 ) η c ∗ − T 1 ∗ ( π c 1 ∗ γ − 1 γ − 1 ) η c 1 ∗ = 330.79 ( 15.3 3 1.4 − 1 1.4 − 1 ) 288.15 ( 2 3 1.4 − 1 1.4 − 1 ) 0.82 − 288.15 ( 1. 5 1.4 − 1 1.4 − 1 ) 0.83 = 0.8373 \eta _{c10}^{*}=\frac{w_{c,s10}}{\frac{w_{c,s}}{\eta _{c}^{*}}-\frac{w_{c,s1}}{\eta _{c1}^{*}}}=\frac{{T_{1}^{*}}^{\prime}\left( {\pi _{c10}^{*}}^{\frac{\gamma -1}{\gamma}}-1 \right)}{\dfrac{T_{1}^{*}\left( {\pi _{c}^{*}}^{\frac{\gamma -1}{\gamma}}-1 \right)}{\eta _{c}^{*}}-\dfrac{T_{1}^{*}\left( {\pi _{c1}^{*}}^{\frac{\gamma -1}{\gamma}}-1 \right)}{\eta _{c1}^{*}}} \\ =\dfrac{330.79\left( 15.33^{\frac{1.4-1}{1.4}}-1 \right)}{\dfrac{288.15\left( 23^{\frac{1.4-1}{1.4}}-1 \right)}{0.82}-\dfrac{288.15\left( 1.5^{\frac{1.4-1}{1.4}}-1 \right)}{0.83}}=0.8373 ηc10∗=ηc∗wc,s−ηc1∗wc,s1wc,s10=ηc∗T1∗(πc∗γγ−1−1)−ηc1∗T1∗(πc1∗γγ−1−1)T1∗′(πc10∗γγ−1−1)=0.82288.15(231.41.4−1−1)−0.83288.15(1.51.41.4−1−1)330.79(15.331.41.4−1−1)=0.8373
同理，新十一级压气机总效率：

（3）压气机去级

1.有一台十级压气机，在标准大气条件下，设计参数为： π c ∗ = 15.3 \pi _{c}^{*}=15.3 πc∗=15.3， η c ∗ = 0.83 \eta _{c}^{*}=0.83 ηc∗=0.83， n = 11950 r / min ⁡ n=11950\ \mathrm{r}/\min n=11950 r/min， q m , a = 60 k g / s q_{m,a}=60 \ \mathrm{kg}/\mathrm{s} qm,a=60 kg/s，其中第一级压气机的增压比为 1.20 1.20 1.20,效率为 0.84 0.84 0.84（空气 c p = 1.005 k J / ( k g ⋅ K ) c_p=1.005\ \mathrm{kJ/(kg \cdot K)} cp=1.005 kJ/(kg⋅K)，如果第一级出现故障，需要去掉第一级，同时保证后面九级压气机的增压比和效率不变，那么去掉第一级后的压气机的转速和流量应为多少？

解：

首先求原压气机第 1 1 1级出口的气流参数：
p 1 ∗ ′ = p 1 ∗ π c 1 ∗ = 101325 × 1.20 = 121590 P a {p_{1}^{*}}^{\prime}=p_{1}^{*}\pi _{c1}^{*}=101325\times 1.20=121590\,\,\mathrm{Pa} p1∗′=p1∗πc1∗=101325×1.20=121590Pa
第 1 1 1级压气功可写为：
c p ( T 1 ∗ ′ − T 1 ∗ ) = c p T 1 ∗ π c 1 ∗ γ − 1 γ − 1 η c 1 ∗ c_p\left( {T_{1}^{*}}^{\prime}-T_{1}^{*} \right) =c_pT_{1}^{*}\frac{{\pi _{c1}^{*}}^{\frac{\gamma -1}{\gamma}}-1}{\eta _{c1}^{*}} cp(T1∗′−T1∗)=cpT1∗ηc1∗πc1∗γγ−1−1
于是可以计算出：
T 1 ∗ ′ = T 1 ∗ ( 1 + π c 1 ∗ γ − 1 γ − 1 η c 1 ∗ ) = 288.15 × ( 1 + 1.2 0 1.4 − 1 1.4 − 1 0.84 ) = 306.49 K {T_{1}^{*}}^{\prime}=T_{1}^{*}\left( 1+\frac{{\pi _{c1}^{*}}^{\frac{\gamma -1}{\gamma}}-1}{\eta _{c1}^{*}} \right) =288.15\times \left( 1+\frac{1.20^{\frac{1.4-1}{1.4}}-1}{0.84} \right) =306.49\,\,\mathrm{K} T1∗′=T1∗(1+ηc1∗πc1∗γγ−1−1)=288.15×(1+0.841.201.41.4−1−1)=306.49K
根据相似准则
n T 1 ∗ ′ = n ′ T 1 ∗ , q m , a T 1 ∗ ′ p 1 ∗ ′ = q m , a ′ T 1 ∗ p 1 ∗ \frac{n}{\sqrt{{T_{1}^{*}}^{\prime}}}=\frac{n^{\prime}}{\sqrt{T_{1}^{*}}}\quad ,\quad \frac{q_{m,a}\sqrt{{T_{1}^{*}}^{\prime}}}{{p_{1}^{*}}^{\prime}}=\frac{q_{m,a}^{\prime}\sqrt{T_{1}^{*}}}{p_{1}^{*}} T1∗′ n=T1∗ n′,p1∗′qm,aT1∗′ =p1∗qm,a′T1∗
可以求出去第 1 1 1级后压气机的流量和转速：
n ′ = n T 1 ∗ T 1 ∗ ′ = 11950 × 288.15 306.49 = 11587 r / min ⁡ q m , a ′ = q m , a p 1 ∗ p 1 ∗ ′ T 1 ∗ ′ T 1 ∗ = 60 × 1 1.20 × 306.49 288.15 = 51.57 k g / s n^{\prime}=n\sqrt{\frac{T_{1}^{*}}{{T_{1}^{*}}^{\prime}}}=11950\times \sqrt{\frac{288.15}{306.49}}=11587\,\,\mathrm{r}/\min \\ q_{m,a}^{\prime}=q_{m,a}\frac{p_{1}^{*}}{{p_{1}^{*}}^{\prime}}\sqrt{\frac{{T_{1}^{*}}^{\prime}}{T_{1}^{*}}}=60\times \frac{1}{1.20}\times \sqrt{\frac{306.49}{288.15}}=51.57\,\,\mathrm{kg}/\mathrm{s} n′=nT1∗′T1∗ =11950×306.49288.15 =11587r/minqm,a′=qm,ap1∗′p1∗T1∗T1∗′ =60×1.201×288.15306.49 =51.57kg/s