Respuesta :
Answer:
[tex]P= 60.5 \frac{psia ft^3}{s} *\frac{1 Btu}{5.404 psia ft^3} *\frac{1 hp}{0.7068 Btu/s}= 15.839 hp[/tex]
Explanation:
Notation
For this case we have the following pressures:
[tex] p_1 = 15 psia[/tex] initial pressure
[tex]p_2 = 70 psia[/tex] final pressure
[tex] V^{*} = 1.1 ft^3/s[/tex] represent the volumetric flow
[tex] rho[/tex] represent the density
[tex]m^{*}[/tex] represent the mass flow
Solution to the problem
From the definition of mass flow we have the following formula:
[tex] m^{*} = \rho V^{*}[/tex]
For this case we can calculate the total change is the sytem like this:
[tex] \Delta E= \frac{p_2 -p_1}{\rho}[/tex]
Since we just have a change of pressure and we assume that all the other energies are constant.
The power is defined as:
[tex] P = m^* \Delta E[/tex]
And replacing the formula for the change of energy we got:
[tex]P = m V^* \frac{p_2 -p_1}{\rho} = V^* (p_2 -p_1)[/tex]
And replacing we have this:
[tex] P= (70-15) psia * 1.1 \frac{ft^3}{s} =60.5 \frac{psia ft^3}{s}[/tex]
And we can convert this into horsepower like this:
[tex]P= 60.5 \frac{psia ft^3}{s} *\frac{1 Btu}{5.404 psia ft^3} *\frac{1 hp}{0.7068 Btu/s}= 15.839 hp[/tex]
