Question

An insulated piston/cylinder device initially contains 8 L of saturated liquid water pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 55 min through a resistor placed in the water. Assume pressure of the Water to be constant during this process. If one-half of the liquid (by mass) is evaporated during this constant pressure process and the paddle-wheel work amounts to 800 kJ, then what will be the voltage of source.

$(Assumeatpressure{P}_{sat}=175kPa,v_f=0.001057m^3/kg,v_g=1.0036m^3/kg,u_{fg}=2038.1kJ/kg)$

• A. 320 V
• B. 287 V
• C. 394 V
• D. 425 V

Answer 1

The correct option is B 287 V


As per given information,

$V=8\times {10}^{-3}{m}^3$

$m=\frac{V}{v_f}=\frac{8\times {10}^{-3}}{0.01057}=7.568kg(x=0.5)$

$W_{electrical}=VI\times t$

$=\frac{V\times 8\times 55\times 60}{1000}=26.4VkJ$

From 1st law thermodynamics

$Q_{1-2}=U_{2-1}+W_{1-2}$

$Q=(U_2-U_1)+PdV-W_c-W_{shaft}$

$Q=\dot{m}[u_f+0.5u_{fg}-u_f]+P_{sat}[v_f+0.5v_{fg}-v_f]-W_c-W_{shaft}$

$W_g=7.5680[0.5\times 2038.1+175[0.5(1.0036-0.001057)\left]\right]-800kJ$

$26.4V=8376.054kJ-800kJ$

$V=\frac{7576.054}{26.4}$

$V=286.97\simeq 287V$