Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
Answer:
t=14ns
Explanation:
We make the relation between the specific access time and the memory percentage in each level, so
![60\% \Rightarrow 60/100 = 0.60\\35\% \Rightarrow 35/100 = 0.35\\05\% \Rightarrow 05/100 = 0.05](https://tex.z-dn.net/?f=60%5C%25%20%5CRightarrow%2060%2F100%20%3D%200.60%5C%5C35%5C%25%20%5CRightarrow%2035%2F100%20%3D%200.35%5C%5C05%5C%25%20%5CRightarrow%2005%2F100%20%3D%200.05)
![t= 0.6(5) + 0.35(5+15) + 0.05(5+15+60)\\t= 0.6(5) + 0.35(20) + 0.05(80)\\t= 3 + 7 + 4\\t= 14 ns](https://tex.z-dn.net/?f=t%3D%200.6%285%29%20%2B%200.35%285%2B15%29%20%2B%200.05%285%2B15%2B60%29%5C%5Ct%3D%200.6%285%29%20%2B%200.35%2820%29%20%2B%200.05%2880%29%5C%5Ct%3D%203%20%2B%207%20%2B%204%5C%5Ct%3D%2014%20ns)
Average Access Time is 14 nsec.
Answer:
0.1047N
Explanation:
To solve this problem we must remember the conversion factors, remembering that 1 revolution equals 2π radians and 1min equals 60s
![N\frac{rev}{min} \frac{2\pi }{1rev} \frac{1min}{60} =N\frac{2\pi }{60} =0.1047N](https://tex.z-dn.net/?f=N%5Cfrac%7Brev%7D%7Bmin%7D%20%5Cfrac%7B2%5Cpi%20%7D%7B1rev%7D%20%5Cfrac%7B1min%7D%7B60%7D%20%3DN%5Cfrac%7B2%5Cpi%20%7D%7B60%7D%20%3D0.1047N)
in conclusion, to know how many rad / s an element rotates which is expressed in Rev / min we must only multiply by 0.1047
Answer:
Chemical Engineers use chemistry, math and physics to design and use to make chemical products. The fibers in clothing are designed by chemical engineers.
Solution :
![$P_1 = 120 \ psia$](https://tex.z-dn.net/?f=%24P_1%20%3D%20120%20%5C%20psia%24)
![$P_2 = 20 \ psia$](https://tex.z-dn.net/?f=%24P_2%20%3D%2020%20%5C%20psia%24)
Using the data table for refrigerant-134a at P = 120 psia
![$h_1=h_f=40.8365 \ Btu/lbm$](https://tex.z-dn.net/?f=%24h_1%3Dh_f%3D40.8365%20%5C%20Btu%2Flbm%24)
![$u_1=u_f=40.5485 \ Btu/lbm$](https://tex.z-dn.net/?f=%24u_1%3Du_f%3D40.5485%20%5C%20Btu%2Flbm%24)
![$T_{sat}=87.745^\circ F$](https://tex.z-dn.net/?f=%24T_%7Bsat%7D%3D87.745%5E%5Ccirc%20%20F%24)
∴ ![$h_2=h_1=40.8365 \ Btu/lbm$](https://tex.z-dn.net/?f=%24h_2%3Dh_1%3D40.8365%20%5C%20Btu%2Flbm%24)
For pressure, P = 20 psia
![$h_{2f} = 11.445 \ Btu/lbm$](https://tex.z-dn.net/?f=%24h_%7B2f%7D%20%3D%2011.445%20%5C%20Btu%2Flbm%24)
![$h_{2g} = 102.73 \ Btu/lbm$](https://tex.z-dn.net/?f=%24h_%7B2g%7D%20%3D%20102.73%20%5C%20Btu%2Flbm%24)
![$u_{2f} = 11.401 \ Btu/lbm$](https://tex.z-dn.net/?f=%24u_%7B2f%7D%20%3D%2011.401%20%5C%20Btu%2Flbm%24)
![$u_{2g} = 94.3 \ Btu/lbm$](https://tex.z-dn.net/?f=%24u_%7B2g%7D%20%3D%2094.3%20%5C%20Btu%2Flbm%24)
![$T_2=T_{sat}=-2.43^\circ F$](https://tex.z-dn.net/?f=%24T_2%3DT_%7Bsat%7D%3D-2.43%5E%5Ccirc%20%20F%24)
Change in temperature, ![$\Delta T = T_2-T_1$](https://tex.z-dn.net/?f=%24%5CDelta%20T%20%3D%20T_2-T_1%24)
![$\Delta T = -2.43-87.745$](https://tex.z-dn.net/?f=%24%5CDelta%20T%20%3D%20-2.43-87.745%24)
![$\Delta T=-90.175^\circ F$](https://tex.z-dn.net/?f=%24%5CDelta%20T%3D-90.175%5E%5Ccirc%20%20F%24)
Now we find the quality,
![$h_2=h_f+x_2(h_g-h_f)$](https://tex.z-dn.net/?f=%24h_2%3Dh_f%2Bx_2%28h_g-h_f%29%24)
![$40.8365=11.445+x_2(91.282)$](https://tex.z-dn.net/?f=%2440.8365%3D11.445%2Bx_2%2891.282%29%24)
![$x_2=0.32198$](https://tex.z-dn.net/?f=%24x_2%3D0.32198%24)
The final energy,
![$u_2=u_f+x_2.u_{fg}$](https://tex.z-dn.net/?f=%24u_2%3Du_f%2Bx_2.u_%7Bfg%7D%24)
![$=11.401+0.32198(82.898)$](https://tex.z-dn.net/?f=%24%3D11.401%2B0.32198%2882.898%29%24)
![$=38.09297 \ Btu/lbm$](https://tex.z-dn.net/?f=%24%3D38.09297%20%5C%20Btu%2Flbm%24)
Change in internal energy
![$\Delta u= u_2-u_1$](https://tex.z-dn.net/?f=%24%5CDelta%20u%3D%20u_2-u_1%24)
= 38.09297-40.5485
= -2.4556