Question

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 75 min, determine the highest rate of heat transfer allowed.

Answer 1

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V$_f$ = V$_g$ = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v$_f$ = 0.001057 m³/kg

v$_g$ = 1.0037 m³/kg

u$_f$ = 486.82 kJ/kg

u$_g$ 2524.5 kJ/kg

h$_g$ = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V$_f$/v$_f$ + V$_g$/v$_g$

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v$_g$

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m$_{out$ = m₁ - m₂

m$_{out$ = 1.89414  - 0.003985

m$_{out$ = 1.890155 kg

so, Initial internal energy will be;

U₁ = m$_f$u$_f$ + m$_g$u$_g$

U₁ = (V$_f$/v$_f$)u$_f$  + (V$_g$/v$_g$)u$_g$

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E$_{in$ -  E$_{out$ = ΔE$_{sys$

QΔt - m$_{out$h$_{out$ = m₂u₂ - U₁

QΔt - m$_{out$h$_g$ = m₂u$_g$ - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW