Given,
The enthalpy of the fluid = 3000 KJ/kg
Velocity of the fluid =60 m/s
At the time of discharge, specific enthalpy =2762 KJ/ Kg
a) Applying Bernoulli's Energy Equation,
"\\Rightarrow L_1+\\frac{v_1^2}{2}+gz_1+Q=L_2+\\frac{v_2^2}{2}+gz_2+W"
"\\Rightarrow v_2^2=2(h_1-h_2+Q)+v_1^2"
"\\Rightarrow v_2^2 =2(3000-2762-0.1)10^3+60^2"
"\\Rightarrow v_2=632.38 m\/s"
b) "A_1=0.1 m^2"
"v_1=0.19"
"m=\\frac{A_1V_1}{\\rho_2}"
"\\Rightarrow m=\\frac{0.1\\times 60}{0.19}=31.578 kg\/s"
c) "m=\\frac{A_2 V_2}{\\rho_2}"
"\\Rightarrow A_2=\\frac{31.578\\times 0.5}{692.38}"
"\\Rightarrow A_2 =0.0228 m^2"
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