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Under what conditions will the temperature rise of the cold fluid in a heat exchanger be equal to the temperature drop of the hot fluid?

Short Answer

Expert verified
Answer: The temperature rise of the cold fluid is equal to the temperature drop of the hot fluid when the product of the mass flow rate and specific heat capacity of the cold fluid is equal to the product of the mass flow rate and specific heat capacity of the hot fluid, i.e., (m_cold * c_cold) / (m_hot * c_hot) = 1.

Step by step solution

01

Understand the heat exchange process

In a heat exchanger, the hot fluid transfers heat to the cold fluid, which causes the temperature of the hot fluid to drop and the temperature of the cold fluid to rise. The amount of heat transferred between the two fluids can be calculated using the formula: Q = mcΔT where Q is the heat transferred, m is the mass flow rate of the fluid (kg/s), c is the specific heat capacity of the fluid (J/kg*K), and ΔT is the change in temperature (K).
02

Set the heat gained by the cold fluid equal to the heat lost by the hot fluid

In order for the temperature rise of the cold fluid to be equal to the temperature drop of the hot fluid, the heat gained by the cold fluid (Q_cold) must be equal to the heat lost by the hot fluid (Q_hot). We can set up an equation to represent this condition: Q_cold = Q_hot Using the heat transfer formula for each fluid, we can write: m_cold * c_cold * ΔT_cold = m_hot * c_hot * ΔT_hot
03

Solve for the required conditions

Our goal is to find the conditions under which ΔT_cold = ΔT_hot. In other words, we want the change in temperature of the cold fluid to be equal to the change in temperature of the hot fluid. To accomplish this, let's rearrange our equation from step 2 and divide both sides by ΔT_cold: m_cold * c_cold = m_hot * c_hot * (ΔT_hot / ΔT_cold) Since we want ΔT_cold = ΔT_hot, we can replace (ΔT_hot / ΔT_cold) with 1: m_cold * c_cold = m_hot * c_hot
04

Determine the final condition

The condition under which the temperature rise of the cold fluid in a heat exchanger is equal to the temperature drop of the hot fluid is: m_cold * c_cold = m_hot * c_hot or equivalently, (m_cold * c_cold) / (m_hot * c_hot) = 1 This means that the product of the mass flow rate and specific heat capacity of the cold fluid must be equal to the product of the mass flow rate and specific heat capacity of the hot fluid.

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Most popular questions from this chapter

Water $\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\( is to be heated by solar-heated hot air \)\left(c_{p}=0.24 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)$ in a double- pipe counterflow heat exchanger. Air enters the heat exchanger at $190^{\circ} \mathrm{F}\( at a rate of \)0.7 \mathrm{lbm} / \mathrm{s}$ and leaves at \(135^{\circ} \mathrm{F}\). Water enters at \(70^{\circ} \mathrm{F}\) at a rate of \(0.35 \mathrm{lbm} / \mathrm{s}\). The overall heat transfer coefficient based on the inner side of the tube is given to be 20 Btu/h $/ \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}$. Determine the length of the tube required for a tube internal diameter of \(0.5 \mathrm{in}\).

Consider a double-pipe heat exchanger with a tube diameter of $10 \mathrm{~cm}$ and negligible tube thickness. The total thermal resistance of the heat exchanger was calculated to be \(0.025 \mathrm{k} / \mathrm{W}\) when it was first constructed. After some prolonged use, fouling occurs at both the inner and outer surfaces with the fouling factors $0.00045 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\( and \)0.00015 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}$, respectively. The percentage decrease in the rate of heat transfer in this heat exchanger due to fouling is (a) \(2.3 \%\) (b) \(6.8 \%\) (c) \(7.1 \%\) (d) \(7.6 \%\) (e) \(8.5 \%\)

Cold water $\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( enters a counterflow heat exchanger at \)10^{\circ} \mathrm{C}\( at a rate of \)0.35 \mathrm{~kg} / \mathrm{s}$, where it is heated by hot air $\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( that enters the heat exchanger at \)50^{\circ} \mathrm{C}$ at a rate of \(1.9 \mathrm{~kg} / \mathrm{s}\) and leaves at $25^{\circ} \mathrm{C}$. The effectiveness of this heat exchanger is (a) \(0.50\) (b) \(0.63\) (c) \(0.72\) (d) \(0.81\) (e) \(0.89\)

Explain how you can evaluate the outlet temperatures of the cold and hot fluids in a heat exchanger after its effectiveness is determined.

Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the \(2.5\)-cm-internaldiameter tube of a double-pipe counterflow heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of $2.2 \mathrm{~kg} / \mathrm{s}\(. Water is heated by steam condensing at \)120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)$ in the shell. If the overall heat transfer coefficient of the heat exchanger is $700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the length of the tube required in order to heat the water to \(80^{\circ} \mathrm{C}\) using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

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