Question

Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cooker has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of theexit opening is 8mm2. Determine
a)The mass flow rate of the steam and the exit velocity
b)The total and flow energies of the steam per unit mass
c)The rate at which energy leaves the cooker by steam.

Answer 1

Answer: a) Mr = 2.4×10^-4kg/s

V = 34.42m/a

b) E = 173J

Ø = 2693.1J

c) Er = 0.64J/s

Explanation: Please find the attached file for the solution

Answer 2

A pressure cooker at its operating pressure, will run at its best performance

• The mass flow rate is 2.37*10^-4kg/s and the exit velocity is 34.3m/s
• The total flow of energy is 0.583 KJ/Kg
• The rate at which energy leave the cooker is 0.638KW

(a) The mass flow rate

To calculate the mass flow rate of the steam, we start by calculating the mass using:

$m=\frac{dV}{V_f}$

So, we have:

$m=\frac{0.6}{0.001053}(1m^3/1000l)=0.57kg$

The mass flow rate is then calculated by:

$M=\frac{m}{dt}$

This gives

$M=\frac{0.57}{40min}$

$M =0.01425kg/min$

Convert to kg / s

$M=0.000237kg/s$

The exit velocity can be calculated as

$v = \frac{mV_g}{Ac}$

This gives

$v =\frac{(2.37*10^-4)(1.1594}{8*10^6}$

$v =34.34m/s$

Hence, the mass flow rate is 0.000237kg/s and the exit velocity is 34.34 m/s

(b) The total and flow energy

This is the sum of kinetic, potential and heat energy in the system

$pv=h-u=2693.1-2519.2=173.9kJ/kg\\\\θ=h+ke+pe=h = 2693.1kJ/Kg$

What this implies is that the sum of other energy is not significant compared to h.

Hence, the total and flow energy of the steam is 2693.1kJ/Kg

c) The energy rate leaving the cooker

This can be calculated by

$E=m\theta$

So, we have:

$E = (2.37*10^{-4}) \times (2693.1)$

$E=0.638kJ/s$

Approximate

$E =0.64kw$

Hence, the rate which energy leave by steam is 0.64 k w

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