P1=P0(T0−0.0065hT0)5.35;P_1=P_0(\frac{T_0-0.0065h}{T_0})^{5.35};P1=P0(T0T0−0.0065h)5.35;
P1=101325Pa(291.15K−0.0065×3048m291.15K)5.35=69498.2032Pa;P_1=101325Pa(\frac{291.15K - 0.0065×3048m}{291.15 K})^{5.35} = 69498.2032 Pa;P1=101325Pa(291.15K291.15K−0.0065×3048m)5.35=69498.2032Pa;
ρ=PRT;\rho= \frac{P}{RT};ρ=RTP;
ρ=69498.2032Pa287Nm/kgK×290K;\rho= \frac{69498.2032Pa}{287Nm/kgK×290K};ρ=287Nm/kgK×290K69498.2032Pa;
m=ρV=283.9kg;m={\rho}V = 283.9 kg;m=ρV=283.9kg;
m=283.9kg60s=4.732kg/s;m = \frac{283.9 kg}{60s }= 4.732 kg/s;m=60s283.9kg=4.732kg/s;
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments