Question

A liquid food with 12% total solids is being heated by steam injection using steam at a pressure of 232.1 kPa (Fig. E3.3). The product enters the heating system at 508C at a rate of 100 kg/min and is being heated to 1208C. The product specific heat is a function of composition as follows: cp 5 cpwðmass fraction H2OÞ 1 cpsðmass fraction solidÞ and the specific heat of product at 12% total solids is 3.936 kJ/(kg 8C). Determine the quantity and minimum quality of steam to ensure that the product leaving the heating system has 10% total solids.

Answer 1

Answer:

$m_{s}=20kg/min$

$H_{s}=1914kJ/kg$

Explanation:

A liquid food with 12% total solids is being heated by steam injection using steam at a pressure of 232.1 kPa (Fig. E3.3). The product enters the heating system at 50°C at a rate of 100 kg/min and is being heated to 120°C. The product specific heat is a function of composition as follows:

$c_{p}=c_{pw}(mass fraction H_{2}0)+c_{ps}(mass fraction solid)$ and the

specific heat of product at 12% total solids is 3.936 kJ/(kg°C). Determine the quantity and minimum quality of steam to ensure that the product leaving the heating system has 10% total solids.

Given

Product total solids in ($X_{A}$) = 0.12

Product mass flow rate ($m_{A}$) = 100 kg/min

Product total solids out ($X_{B}$) = 0.1

Product temperature in ($T_{A}$) = 50°C

Product temperature out ($T_{B}$) = 120°C

Steam pressure = 232.1 kPa at ($T_{S}$) = 125°C

Product specific heat in ($C_{PA}$) = 3.936 kJ/(kg°C)

The mass equation is:

$m_{A}X_{A}=m_{B}X_{B}$

$100(0.12)=m_{B}(0.1)\\m_{B}=\frac{100(0.12)}{0.1} =120$

Also $m_{a}+m_{s}=m_{b}$

Therefore: $100}+m_{s}=120\\\\m_{s}=120-100=20$

The energy balance equation is:

$m_{A}C_{PA}(T_{A}-0)+m_{s}H_{s}=m_{B}C_{PB}(T_{B}-0)$

$3.936 = (4.178)(0.88) +C_{PS}(0.12)\\C_{PS}=\frac{3.936-3.677}{0.12} =2.161$

$C_{PB}= 4.232*0.9+0.1C_{PS}= 4.232*0.9+0.1*2.161=4.025$  kJ/(kg°C)

Substituting values in the energy equation:

$100(3.936)(50-0)+20H_{s}=120(4.025)}(120-0)$

$19680+20H_{s}=57960\\20H_{s}=57960-19680 \\20H_{s}=38280\\H_{s}=\frac{38280}{20} =1914$

$H_{s}=1914kJ/kg$

From properties of saturated steam at 232.1 kPa,

$H_{c}$ = 524.99 kJ/kg

$H_{v}$ = 2713.5 kJ/kg

% quality = $\frac{1914-524.99}{2713.5-524.99} =63.5%$

Any steam quality above 63.5% will result in higher total solids in

heated product