Answer:[tex]\dot{m}=3.46lbm/min[/tex]
Explanation:
Initial conditions
[tex]P_1=15 psia[/tex]
[tex]T_1=60 F^{\circ}[/tex]
Final conditions
[tex]P_2=75 psia[/tex]
[tex]T_2=400F^{\circ}[/tex]
Steady flow energy equation
[tex]\dot{m}\left [ h_1+\frac{v_1^2}{2}+gz_1\right ]+\dot{Q}=\dot{m}\left [ h_2+[tex]\frac{v_2^2}{2}+gz_2\right ]+\dot{W}[/tex]
[tex]\dot{m}\left [ c_pT_1+\frac{0^2}{2}+g0\right ]+\dot{Q}=\dot{m}\left [ c_pT_2+\frac{0^2}{2}+g0\right ]+\dot{W}[/tex]
[tex]\dot{m}c_p\left [ T_1-T_2\right ]+\left [ -5hp\right ]=\dot{W} -5\times 746\times 3.4121[/tex]
[tex]-4\dot{m}-\dot{m}\times 0.24\times \left [ 400-60\right ][/tex]
[tex]-81.6\dot{m}-4\dot{m}=-4.949 BTU/sec[/tex]
[tex]\dot{m}=0.057821lbm/sec[/tex]
[tex]\dot{m}=3.46lbm/min[/tex]