The radiator in an automobile is a cross-flow heat exchanger \(\left(U A_{s}=10 \mathrm{~kW} / \mathrm{K}\right)\) that uses air \(\left(c_{p}=1.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) to cool the engine-coolant fluid \(\left(c_{p}=4.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). The engine fan draws \(30^{\circ} \mathrm{C}\) air through this radiator at a rate of \(10 \mathrm{~kg} / \mathrm{s}\) while the coolant pump circulates the engine coolant at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The coolant enters this radiator at \(80^{\circ} \mathrm{C}\). Under these conditions, the effectiveness of the radiator is \(0.4\). Determine \((a)\) the outlet temperature of the air and (b) the rate of heat transfer between the two fluids.

Step by step solution

01

Determine the product of the heat capacity rates of two fluids

We need to find the product of the two heat capacity rates (for air \(C_{min}\) and for coolant \(C_{max}\)) to determine the outlet temperatures and the heat transfer between the two fluids. The heat capacity rates for air and coolant can be calculated as follows: \(C_{air} = \dot{m}_{air} \cdot c_{p, air}\) \(C_{coolant} = \dot{m}_{coolant} \cdot c_{p, coolant}\)

02

Calculate the effective capacity rates

First, we need to determine the heat capacity rate for air and the coolant: \(C_{air} = (10 \frac{kg}{s})(1.00 \frac{kJ}{kg \cdot K}) = 10 \frac{kW}{K}\) \(C_{coolant} = (5 \frac{kg}{s})(4.00 \frac{kJ}{kg \cdot K}) = 20 \frac{kW}{K}\) Since \(C_{air} < C_{coolant}\), we have \(C_{min} = C_{air}\) and \(C_{max} = C_{coolant}\). Now we can calculate the outlet temperature of the air:

03

Apply the effectiveness formula

The effectiveness is given by: \(Effectiveness = \frac{T_{air,out} - T_{air,in}}{(T_{coolant,in} - T_{air,in})}\) We can rearrange the formula and solve for \(T_{air,out}\): \(T_{air,out} = Effectiveness \cdot (T_{coolant,in} - T_{air,in}) + T_{air,in}\) Plug in the given values and temperatures: \(T_{air,out} = 0.4 \cdot (80 - 30) + 30 = (0.4 \cdot 50) + 30 = 20 + 30 = 50 ^{\circ}C\) So the outlet temperature of the air is \(50^{\circ}C\).

04

Calculate the rate of heat transfer

To find the rate of heat transfer \(Q\), we can use the formula: \(Q = C_{min} \cdot Effectiveness \cdot (T_{coolant,in} - T_{air,in})\) Plug in the values we have: \(Q = (10 \frac{kW}{K})(0.4) \cdot (50^{\circ}C) = 4 \cdot 50 = 200 \, kW\) So the rate of heat transfer between the two fluids is \(200 \, kW\). In conclusion, the outlet temperature of the air is \(50^{\circ}C\) and the rate of heat transfer between the two fluids is \(200 \, kW\).

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