Question

A 4-L pressurecooker has an operating pressure of175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out ofliquid water for 1 h, determine the highest rate of heat transfer allowed.

Answer 1

To solve this problem, it is necessary to apply the concepts related to the Energy balance and the mass balance that allow us to find in each state the data necessary to find the total Power of the system through heat exchange.

From the tables of properties of the Water it is possible to obtain at the given pressures the values of the specific volume, the specific energy and the specific enthalpy.

Given these pressures then we have to

$P_1 = 174kPA$

$\Rightarrow v_f = 0.001057m^3/kg\\\Rightarrow v_g = 1.0037m^3/kg\\\Rightarrow u_f = 486.82kJ/kg\\\Rightarrow u_g = 2524.5kJ/kg$

$P_2 = 175kPa \rightarrow$ Saturated vapor

$\Rightarrow v_2 = v_g = 1.0036 m^3/kg\\\Rightarrow v_2 = u_g = 2524.5kJ/kg$

$P_e = 175kPa \rightarrow$ Saturated vapor

$V = 4L$

Considering the process performed, the kinetic and potential energy can be neglected as well as the work involved by specific interactions in the system. Although it is an unstable process it can be treated as a uniform flow process. Considering these expressions we can perform a mass balance for which

$m_{in}-m_{out} = \Delta m_{system} \rightarrow m_e = m_1-m_2$

Similarly through the energy balance you can get that

$E_{in}-E_{out} = \Delta E_{system}$

$Q_{in} -m_eh_e = m_2u_2-m_1u_1 \Rightarrow$ Since $W=ke=pe=0$

The initial mass, initial internal energy, and final mass in the tank are

$m_1 = mf_+m_g = \frac{V_f}{v_f}+\frac{V_g}{v_g}$

$m_1 = \frac{0.002m^3}{0.001057m^3/kg}+\frac{0.002m^3}{1.0036m^3/kg}$

$m_1 = 1.893+0.002= 1.895Kg$

At the same time the internal energy can be defined from the mass in state 1 as,

$U_1 = m_1 u_1$

$U_1 = m_fu_f+m_gu_g$

$U_1 = 1.893*486.82+0.002*2524.5$

$U_1 = 926.6kJ$

The calculation of mass in state 2 can be defined as

$m_2 = \frac{V}{v_2} = \frac{0.004m^3}{1.0037m^3/kg}$

$m_2 = 0.004Kg$

Then from the mass and energy balances,

$m_e = m_1-m_2\\m_e = 1.895-0.004\\m_e = 1.891Kg$

In this way the calculation of the heat of entry would be subject to

$Q_{in} = m_eh_e+m_2u_2-m_1u_1$

$Q_{in} = 1.891*2700.2+0.004*2524.5-926.6$

$Q_{in} = 4188kJ$

Therefore the Power would be given as

$\dot{Q} = \frac{Q}{\Delta t} = \frac{4188kJ}{3600s} = 1.163kW$

Therefore the highest rate of heat transfer allowed is 1.163kW