Question

g A 4-L pressure cooker has an operating pressure of 175 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 of liquid water for 70 min, determine the highest rate of heat transfer allowed. The properties of water are vf = 0.001057 m3/kg, vg = v2 = 1.0037 m3/kg, uf = 486.82 kJ/kg, ug = u2 = 2524.5 kJ/kg, and he = 2700.2 kJ/kg.

Answer 1

The highest rate of heat transfer allowed = 0.998kW

Step-by-step explanation:

Given

V = 4L = 4*10^-3m³

vf = 0.001057 m3/kg,

vg = v2 = 1.0037 m3/kg,

uf = 486.82 kJ/kg,

ug = u2 = 2524.5 kJ/kg,

he = 2700.2 kJ/kg

First, we'll calculate the initial mass in the cooker, m1.

This is calculated from the masses of the constituents.

m1 = V(vap)/α(vap) + V(liq)/α(liq)

m1 = ½V(1/α(vap) + 1/α(liq))

m1 = ½*4*10^-3(1/1.0037 + 1/0.001057)

m1 = 1.89414021479089337285534

m1 = 1.894kg

Next, we'll calculate the final mass.

This is given as;

m2 = V/α(vap)

m2 = 4 * 10^-3/1.0037

m2 = 0.00398525455813490086679

m2 = 0.00399kg

m(out) = m1 - m2

m(out) = 1.894 - 0.00399

m(out) = 1.89001kg

Then, we calculate the initial energy.

This is given as;

U1 = m1u1

U1 = (mu)vap + (mu)liq

U1 = ½V(u(vap)/α(vap) + u(liq)/α(liq))

U1 = ½ * 4 * 10^-3 (2524.5/1.0037 + 486.82/0.001057)

U1 = 926.165676118512874172562

U1 = 926.17kJ

Finally, we calculate the rate of heat transfer.

Rate = Q/∆t

Where Q = m2u2 - m1u1 + m(out)u(out)

Q = 0.00399kg * 2524.5 kJ/kg - 926.17kJ + 1.89001kg * 2700.2

Q = 4,187.307757

Rate = 4,187.307757/70 minute

Rate = 4,187.307757/(70 * 60)

Rate = 0.99697803738095238095238

Rate = 0.998kW

Hence, the highest rate of heat transfer allowed is 0.998kW