Answer:
Pressure = 115.6 psia
Explanation:
Given:
v=800ft/s
Air temperature = 10 psia
Air pressure = 20F
Compression pressure ratio = 8
temperature at turbine inlet = 2200F
Conversion:
1 Btu =775.5 ft lbf,
= 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²
Air standard assumptions:
= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R
k= 1.4
Energy balance:
As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible
hence ![v_{a} ^{2} = 0](https://tex.z-dn.net/?f=v_%7Ba%7D%20%5E%7B2%7D%20%3D%200)
![h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }](https://tex.z-dn.net/?f=h_%7B1%7D%20%2B%20%5Cfrac%7Bv_%7B1%7D%20%5E%7B2%7D%20%7D%7B2%7D%20%3D%20h_%7Ba%7D%20%5C%5Ch_%7B1%7D%20-h_%7Ba%7D%20%3D%20-%20%5Cfrac%7Bv_%7B1%7D%20%5E%7B2%7D%20%7D%7B2%7D%20%5C%5C%20c_%7Bp%7D%20%28T_%7B1%7D%20-T_%7Ba%7D%29%3D%20-%20%5Cfrac%7Bv_%7B1%7D%20%5E%7B2%7D%20%7D%7B2%7D%20%5C%5C%28T_%7B1%7D%20-T_%7Ba%7D%29%20%3D%20-%20%5Cfrac%7Bv_%7B1%7D%20%5E%7B2%7D%20%7D%7B2c_%7Bp%7D%20%7D%5C%5C%20T_%7Ba%7D%3DT_%7B1%7D%20%2B%20%20%5Cfrac%7Bv_%7B1%7D%20%5E%7B2%7D%20%7D%7B2c_%7Bp%7D%20%7D)
= 20+460 = 480°R
= 533.25°R
Pressure at the inlet of compressor at isentropic condition
![P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}](https://tex.z-dn.net/?f=P_%7Ba%20%7D%20%3DP_%7B1%7D%28%5Cfrac%7BT_%7Ba%7D%20%7D%7BT_%7B1%7D%20%7D%29%20%5E%7Bk%2F%28k-1%29%7D)
=
= 14.45 psia
Answer:
B. To accurately measure spark advance, use a timing light that incorporates an
ignition advance meter. The spark advance cannot be determined by listening to the way the engine sounds.
Answer:
28.6 kg
Explanation:
The final weight can be calculated from the mixing data and formulae which is given as follows:
cement content = ![\frac{water content}{water - cement ratio}](https://tex.z-dn.net/?f=%5Cfrac%7Bwater%20content%7D%7Bwater%20-%20cement%20ratio%7D)
Computing the parameters and checking the tables gives 28.6 kg.
Answer:
![\Delta m = 102.25\,kg](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20102.25%5C%2Ckg)
Explanation:
The mass inside the rigid tank before the high pressure stream enters is:
![m_{o} = \rho_{air}\cdot V_{tank}](https://tex.z-dn.net/?f=m_%7Bo%7D%20%3D%20%5Crho_%7Bair%7D%5Ccdot%20V_%7Btank%7D)
![m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})](https://tex.z-dn.net/?f=m_%7Bo%7D%20%3D%20%281.25%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%29%5Ccdot%20%2825%5C%2Cm%5E%7B3%7D%29)
![m_{o} = 31.25\,kg](https://tex.z-dn.net/?f=m_%7Bo%7D%20%3D%2031.25%5C%2Ckg)
The final mass inside the rigid tank is:
![m_{f} = \rho \cdot V_{tank}](https://tex.z-dn.net/?f=m_%7Bf%7D%20%3D%20%5Crho%20%5Ccdot%20V_%7Btank%7D)
![m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})](https://tex.z-dn.net/?f=m_%7Bf%7D%20%3D%20%285.34%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%29%5Ccdot%20%2825%5C%2Cm%5E%7B3%7D%29)
![m_{f}= 133.5\,kg](https://tex.z-dn.net/?f=m_%7Bf%7D%3D%20133.5%5C%2Ckg)
The supplied air mass is:
![\Delta m = m_{f}-m_{o}](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20m_%7Bf%7D-m_%7Bo%7D)
![\Delta m = 133.5\,kg-31.25\,kg](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20133.5%5C%2Ckg-31.25%5C%2Ckg)
![\Delta m = 102.25\,kg](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20102.25%5C%2Ckg)