Question

Steam is to be condensed on the shell side of a twoshell-passes and eight- tube-passes condenser, with 20 tubes in each pass. Cooling water enters the tubes at a rate of \(2 \mathrm{~kg} / \mathrm{s}\). If the heat transfer area is \(14 \mathrm{~m}^{2}\) and the overall heat transfer coefficient is $1800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, the effectiveness of this condenser is (a) \(0.70\) (b) \(0.80\) (c) \(0.90\) (d) \(0.95\) (e) \(1.0\)

Short answer

Answer: (c) 0.90

Step-by-step solution

Calculate the actual heat transfer rate

We will start by calculating the actual heat transfer rate using the formula $$Q = U \cdot A \cdot \Delta T$$, where $$Q$$ is the heat transfer rate, $$U$$ is the overall heat transfer coefficient, $$A$$ is the heat transfer area, and $$\Delta T$$ is the temperature difference between the hot and cold sides of the condenser. We have the values for the heat transfer coefficient ($$U=1800 \mathrm{~W/m^2K}$$) and the heat transfer area ($$A=14 \mathrm{~m^2}$$), but we don't have the temperature difference yet, so we need to find that. We will assume that the temperature difference is constant during the operation of the condenser. With this assumption, we can rewrite the formula for the heat transfer rate as $$Q = U \cdot A \cdot \Delta T_{constant}$$. We will find $$Q$$ in the next step.

Calculate the heat capacity rate of the cooling water

We can calculate the heat capacity rate of the cooling water using the formula $$C = mc_p$$, where $$m$$ is the mass flow rate of the cooling water, $$c_p$$ is the specific heat capacity of water, and $$C$$ is the heat capacity rate. We know the mass flow rate of the cooling water is \(2 \mathrm{~kg/s}\) and the specific heat capacity of water is approximately \(4,200 \mathrm{~J/kgK}\). Plugging these values into the formula, we get: $$C = (2 \mathrm{~kg/s}) (4,200 \mathrm{~J/kgK}) = 8,400 \mathrm{~W/K}$$. This represents the maximum possible heat transfer rate if the cooling water were to heat up infinitely.

Calculate the maximum possible temperature difference

The maximum possible heat transfer rate occurs when the cooling water increases in temperature by the maximum possible value. We can find this maximum possible temperature difference using the formula: $$\Delta T_{max} = \frac{Q_{max}}{C}$$, where $$\Delta T_{max}$$ is the maximum possible temperature difference, $$Q_{max}$$ is the maximum possible heat transfer rate, and $$C$$ is the heat capacity rate. We have calculated $$C$$ in the previous step, so we need to find $$Q_{max}$$, which can be calculated as: $$Q_{max} = U \cdot A \cdot \Delta T_{max}$$, where $$U$$ is the overall heat transfer coefficient and $$A$$ is the heat transfer area. We have the values for these variables, so we can plug them into the formula: $$Q_{max} = 1800 \mathrm{~W/m^2K} \cdot 14 \mathrm{~m^2} \cdot \Delta T_{max}$$.

Calculate the effectiveness of the condenser

We can now calculate the effectiveness of the condenser by dividing the actual heat transfer rate by the maximum possible heat transfer rate: $$Effectiveness = \frac{Q}{Q_{max}}$$. Using the results from the previous steps: $$Effectiveness = \frac{1800 \mathrm{~W/m^2K} \cdot 14 \mathrm{~m^2} \cdot \Delta T_{constant}}{1800 \mathrm{~W/m^2K} \cdot 14 \mathrm{~m^2} \cdot \Delta T_{max}}.$$ The heat transfer coefficients and areas cancel out, leaving us with: $$Effectiveness = \frac{\Delta T_{constant}}{\Delta T_{max}}$$. Since we don't have the actual temperature difference, we can't calculate the exact value of the effectiveness, but we can compare it to the given options. Based on the given options, the closest effectiveness value is: (c) \(0.90\).