Answer:44.61 KJ
Explanation:
Let h_1,V_1,Z_1 be the initial specific enthalpy,velocity&elevation of the system
and h_2,V_2,Z_2 be the Final specific enthalpy,velocity&elevation of the system
mass(m)=7kg
Applying Steady Flow Energy Equation
[tex]m\left [ h_1+\frac{v_1^2}{2g}+gZ_1\right ][/tex]+Q=[tex]\left [ h_2+\frac{v_2^2}{2g}+gZ_2\right ][/tex]+W
[tex]h_1-h_2=4 KJ/kg[/tex]
[tex]V_1=13m/s[/tex]
[tex]V_2=23m/s[/tex]
[tex]Z_1-Z_2=40m[/tex]
substituting values
[tex]7\left [ h_1+\frac{13^{2}}{2g}+gZ_1\right ]+7\times2[/tex] = [tex]\left [ h_2+\frac{23^2}{2g}+gZ_2\right ][/tex]+W
W=[tex]7\left [h_1-h_2+\frac{V_1^2-V_2^2}{2000g}+g\frac{\left (Z_1-Z_2 \right )}{1000}\right ][/tex]+Q
W=[tex]7\left [4+\frac{13^2-23^2}{2000g}+g\frac{\left (40 \right )}{1000}[\right ][/tex]+[tex]2\times 7[/tex]
W=44.61KJ