Answer:

Explanation:
Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

The initial pressure is:

The speed at outlet is:

![v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B%28250%5C%2C%5Cfrac%7Bgal%7D%7Bmin%7D%20%29%5Ccdot%20%28%5Cfrac%7B3.785%5Ctimes%2010%5E%7B-3%7D%5C%2Cm%5E%7B3%7D%7D%7B1%5C%2Cgal%7D%20%29%5Ccdot%28%5Cfrac%7B1%5C%2Cmin%7D%7B60%5C%2Cs%7D%20%29%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20%5B%281.125%5C%2Cin%29%5Ccdot%28%5Cfrac%7B0.0254%5C%2Cm%7D%7B1%5C%2Cin%7D%20%29%5D%5E%7B2%7D%20%7D)

The initial pressure is:


Answer:
Given that
Mass flow rate ,m=2.3 kg/s
T₁=450 K
P₁=350 KPa
C₁=3 m/s
T₂=300 K
C₂=460 m/s
Cp=1.011 KJ/kg.k
For ideal gas
P V = m R T
P = ρ RT


ρ₁=2.71 kg/m³
mass flow rate
m= ρ₁A₁C₁
2.3 = 2.71 x A₁ x 3
A₁=0.28 m²
Now from first law for open system

For ideal gas
Δh = CpΔT
by putting the values


Q= - 45.49 KJ/kg
Q =- m x 45.49 KW
Q= - 104.67 KW
Negative sign indicates that heat transfer from air to surrounding
Answer:
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Explanation:
Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.