Question

Determine the amount of heat, which should be supplied to 2 kg of water at 25°C to convert

it into steam at 5 bar and 0.9 dry​

Answer 1

Answer:

Total amount of heat to be supplied$=4864.52 \mathrm{~kJ}$

Explanation:

Given:

Mass of water to be converted to steam, $m_{w}=2 \mathrm{~kg}$

Temperature of water, $t_{w}=25^{\circ} \mathrm{C}$

Pressure and dryness fraction of steam = 5 bar, $0.9$dry

At 5 bar : From steam tables

$h_{f}=640.1 \mathrm{~kJ} / \mathrm{kg} ;\\h_{f g}=2107.4 \mathrm{~kJ} / \mathrm{kg}$

Find: The amount of heat

Enthalpy of $1 \mathrm{~kg}$of steam (above $0^{\circ} \mathrm{C}$ )

$h=h_{f}+x h_{f g} \\=640.1+0.9 \times 2107.4\\=2536.76 \mathrm{~kJ} / \mathrm{kg}$

Sensible heat associated with $1 \mathrm{~kg}$ of water

$&\mathrm{m}_{\mathrm{W}} \times c_{\mathrm{pw}} \times\left(t_{\mathrm{w}}-0\right) \\&=1 \times 4.18 \times(25-0)\\=104.5 \mathrm{~kJ}\end{aligned}$

Net quantity of heat to be supplied per kg of water

$=2536.76-104.5\\=2432.26 \mathrm{~kJ}$

Total amount of heat to be supplied $=2 \times 2432.26\\=4864.52 \mathrm{~kJ}$

Answer 2

Answer:-

Therefore $4864.52\:KJ$ of heat is supplied to $2\: kg$ of water at $25^ \circ C$ to convert it into steam at $5 bar$ and $0.9\: dry$​.

Given:

Mass of water$=2\:kg$

Temperature of water,$t=25^\circ C$

Pressure of steam$=5\: bar$

Dryness fraction,$x=0.9$

So, from the steam table,

For $5 \:bar$ pressure and $0.9$ dryness fraction

Enthalpy of saturated liquid,$hf=640.1\: KJ/kg$

Latent heat of vaporization,$hfg=2107.4\:KJ/kg$

(or the change in enthalpy when substance goes from saturated liquid to saturated vapor phase)

We know that the enthalpy of $1 \:kg$ of steam above $0^\circ C$ is

$h=hf + xhfg\\h=640.1 + 0.9 \times 2107.4\\h=2536.76\: KJ/kg$

Now, the  sensible heat associated with $1 \:kg$ of water is

$Q=m \times cp\times (t-0)$

Where $cp$ is the specific heat capacity  of  water at constant  pressure and its value is $4.18\: J/kg-K$

So,

$Q=1 \times 4.18 \times (25-0)\\Q=104.5\: KJ$

Hence, total heat to be supplied per kg of water is,

$H=h-Q\\H=2536.76-104.5=2432.26\: KJ$

Total heat supplied for $2\: kg$ of water is

$Hnet=H \times 2\\Hnet=2432.26 \times 2=4864.52\: KJ$

Therefore $4864.52\:KJ$ of heat is supplied to $2\: kg$ of water at $25^ \circ C$ to convert it into steam at $5 bar$ and $0.9\: dry$​.