Question

Liquid water at \(300 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) is heated in a chamber by mixing it with superheated steam at \(300 \mathrm{kPa}\) and \(300^{\circ} \mathrm{C}\). Cold water enters the chamber at a rate of \(1.8 \mathrm{kg} / \mathrm{s} .\) If the mixture leaves the mixing chamber at \(60^{\circ} \mathrm{C}\) determine the mass flow rate of the superheated steam required. Answer: \(0.107 \mathrm{kg} / \mathrm{s}\)

Short answer

The mass flow rate of the superheated steam required is 0.107 kg/s.

Step-by-step solution

Write down the mass and energy balance equations for the mixing chamber

The mass balance equation is given by $$\dot{m}_{water} + \dot{m}_{steam} = \dot{m}_{mixture} $$ and the energy balance equation is given by $$\dot{m}_{water} h_{water,in} + \dot{m}_{steam} h_{steam,in} = \dot{m}_{mixture} h_{mixture,out}$$

Calculate specific enthalpy of liquid water at \(20^{\circ} \mathrm{C}\)

To find the specific enthalpy (\(h_{water,in}\)) of the water entering the mixing chamber, we can use the formula $$h_{water,in} = C_p \cdot T_{water,in}$$ where \(C_p\) is the specific heat capacity of water at constant pressure (\(4.18 \mathrm{ kJ/(kg \cdot K)}\)) and \(T_{water,in} = 20^{\circ} \mathrm{C} = 293 \mathrm{K}\). Thus, $$h_{water,in}= 4.18 \times 293 = 1224.34 \mathrm{kJ/kg}$$.

Calculate specific enthalpy of superheated steam at \(300^{\circ} \mathrm{C}\)

Next, find the specific enthalpy (\(h_{steam,in}\)) of the steam entering the mixing chamber. We can use the formula $$h_{steam,in} = C_p \cdot T_{steam,in}$$ where \(C_p\) is the specific heat capacity of steam at constant pressure (\(1.996 \mathrm{ kJ/(kg \cdot K)}\)) and \(T_{steam,in}=300^{\circ} \mathrm{C}=573 \mathrm{K}\). Thus, $$h_{steam,in} = 1.996 \times 573 = 1143.27 \mathrm{ kJ/kg}$$.

Calculate specific enthalpy of the mixture at \(60^{\circ} \mathrm{C}\)

Now, find the specific enthalpy (\(h_{mixture,out}\)) of the mixture leaving the mixing chamber. We can assume the specific heat capacity of the mixture is the same as that of water, \(C_p = 4.18 \mathrm{ kJ/(kg \cdot K)}\), and \(T_{mixture,out}=60^{\circ} \mathrm{C}=333 \mathrm{K}\). Thus, $$h_{mixture,out} = 4.18 \times 333 = 1391.54 \mathrm{ kJ/kg}$$.

Substitute the known values in the energy balance equation and solve for \(\dot{m}_{steam}\)

Plug the values into the energy balance equation: $$1.8 \mathrm{kg/s} \times 1224.34 \mathrm{kJ/kg} + \dot{m}_{steam} \times 1143.27 \mathrm{ kJ/kg} = (\dot{m}_{steam} + 1.8 \mathrm{kg/s}) \times 1391.54 \mathrm{ kJ/kg}$$ Rearrange the equation and solve for \(\dot{m}_{steam}\): $$\dot{m}_{steam} = \frac{1.8 \times (1391.54 - 1224.34)}{1143.27 - 1391.54} = 0.107 \mathrm{kg/s}$$ So the mass flow rate of the superheated steam required is \(\boxed{0.107 \mathrm{kg/s}}\).