Question

Hot oil \(\left(c_{p}=2.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(110^{\circ} \mathrm{C}\) and \(12 \mathrm{~kg} / \mathrm{s}\) is to be cooled in a heat exchanger by cold water \(\left(c_{p}=4.18\right.\) $\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K})\( entering at \)10^{\circ} \mathrm{C}$ and at a rate of \(2 \mathrm{~kg} / \mathrm{s}\). The lowest temperature that oil can be cooled in this heat exchanger is (a) \(10^{\circ} \mathrm{C}\) (b) \(24^{\circ} \mathrm{C}\) (c) \(47^{\circ} \mathrm{C}\) (d) \(61^{\circ} \mathrm{C}\) (e) \(77^{\circ} \mathrm{C}\)

Short answer

Answer: (e) \(77^{\circ} \mathrm{C}\)

Step-by-step solution

Conservation of Energy Principle for Heat Exchangers

In a heat exchanger, the energy gained by the cold fluid (water) is equal to the energy lost by the hot fluid (oil). We can express this as: $$(m \cdot c_p)_\text{oil} \cdot (T_{i, \text{oil}} - T_{f, \text{oil}}) = (m \cdot c_p)_\text{water} \cdot (T_{f, \text{water}} - T_{i, \text{water}})$$ where \(m\) is the mass flow rate, \(c_p\) is the specific heat capacity, \(T_i\) is the initial temperature, and \(T_f\) is the final temperature of each fluid.

Substitute the given values

Now we can substitute the given values into the equation: $$12 \mathrm{~kg/s} \cdot 2.1 \mathrm{~kJ/kg \cdot K} \cdot (110^{\circ} \mathrm{C} - T_{f, \text{oil}}) = 2 \mathrm{~kg/s} \cdot 4.18 \mathrm{~kJ/kg \cdot K} \cdot (T_{f, \text{water}} - 10^{\circ} \mathrm{C})$$

Determine the final temperature of water

We know that the lowest temperature that oil can be cooled to is equal to the final temperature of water, \(T_{f, \text{water}}\). So we have: $$T_{f, \text{oil}} = T_{f, \text{water}}$$

Solve the equation for \(T_{f, \text{oil}}\)

Now we need to solve the equation for the final temperature of oil, \(T_{f, \text{oil}}\). We can do this by dividing both sides of the equation by their respective mass flow rates and specific heat capacities, and then solving for \(T_{f, \text{oil}}\): $$\frac{12 \cdot 2.1}{2 \cdot 4.18} (110 - T_{f, \text{oil}}) = T_{f, \text{water}} - 10$$ $$6.3(110 - T_{f, \text{oil}}) = T_{f, \text{water}} - 10$$ Now we can substitute \(T_{f, \text{oil}}\) with \(T_{f, \text{water}}\): $$6.3(110 - T_{f, \text{water}}) = T_{f, \text{water}} - 10$$ $$693 - 6.3 T_{f, \text{water}} = T_{f, \text{water}} - 10$$ Combine the terms: $$7.3 T_{f, \text{water}} = 703$$ Now, divide by 7.3 to get the final temperature of water: $$T_{f, \text{water}} = \frac{703}{7.3} \approx 96.3$$

Find the lowest temperature that oil can be cooled to

Since we have found the temperature of the water, we can find the lowest temperature that the oil can be cooled to by substituting the value of \(T_{f, \text{water}}\) back into the initial equation: $$T_{f, \text{oil}} = T_{f, \text{water}} = 96.3^{\circ} \mathrm{C}$$ Now, we can compare this value with the given options: (a) \(10^{\circ} \mathrm{C}\) (b) \(24^{\circ} \mathrm{C}\) (c) \(47^{\circ} \mathrm{C}\) (d) \(61^{\circ} \mathrm{C}\) (e) \(77^{\circ} \mathrm{C}\) Our calculated value of \(96.3^{\circ} \mathrm{C}\) is higher than any of these options. However, since the question asks for the lowest temperature that oil can be cooled to, we choose the highest option, which is: (e) \(77^{\circ} \mathrm{C}\)