A car radiator uses hot engine coolant (stream 1) at 95C to heat cold air (stream 2) entering at 20C. The desired outlet temperature for the air is 70C. The overall heat transfer coefficient (U) for the radiator is 150 W/m2K, and the total heat transfer area (A) is 0.8 m2. The mass flow rate of coolant (m1) is twice the mass flow rate of air (m2). Determine the effectiveness of the heat exchanger. Calculate the outlet temperature of the coolant (T, out).
The Correct Answer and Explanation is:
Given:
- Hot fluid inlet temp Th,in=95∘CT_{h,in} = 95^\circ C
- Cold fluid inlet temp Tc,in=20∘CT_{c,in} = 20^\circ C
- Desired cold fluid outlet temp Tc,out=70∘CT_{c,out} = 70^\circ C
- Overall heat transfer coefficient U=150 W/m2KU = 150 \, W/m^2K
- Heat transfer area A=0.8 m2A = 0.8 \, m^2
- m˙h=2m˙c\dot{m}_h = 2\dot{m}_c
Assume specific heats of both fluids are the same:
cp=1 kJ/kg⋅K=1000 J/kg⋅Kc_p = 1 \, kJ/kg\cdot K = 1000 \, J/kg\cdot K (reasonable for water and air).
Step 1: Determine Capacity Rates
Let:
- Ch=m˙hcp=2m˙c⋅cpC_h = \dot{m}_h c_p = 2\dot{m}_c \cdot c_p
- Cc=m˙c⋅cpC_c = \dot{m}_c \cdot c_p
Then:
- Cmin=CcC_{min} = C_c,
- Cmax=ChC_{max} = C_h,
- Cr=CminCmax=12C_r = \frac{C_{min}}{C_{max}} = \frac{1}{2}
Step 2: Determine Maximum Possible Heat Transfer
qmax=Cmin(Th,in−Tc,in)=m˙ccp(95−20)=m˙ccp⋅75q_{max} = C_{min} (T_{h,in} – T_{c,in}) = \dot{m}_c c_p (95 – 20) = \dot{m}_c c_p \cdot 75
Step 3: Actual Heat Transfer
q=m˙ccp(Tc,out−Tc,in)=m˙ccp(70−20)=m˙ccp⋅50q = \dot{m}_c c_p (T_{c,out} – T_{c,in}) = \dot{m}_c c_p (70 – 20) = \dot{m}_c c_p \cdot 50
Step 4: Effectiveness
ε=qqmax=5075=0.667\varepsilon = \frac{q}{q_{max}} = \frac{50}{75} = \boxed{0.667}
Step 5: NTU Calculation
NTU=UACmin=150⋅0.8m˙c⋅cp=120m˙c⋅cpNTU = \frac{UA}{C_{min}} = \frac{150 \cdot 0.8}{\dot{m}_c \cdot c_p} = \frac{120}{\dot{m}_c \cdot c_p} Cmin=m˙c⋅cp⇒NTU=120Cmin=120m˙c⋅cpC_{min} = \dot{m}_c \cdot c_p \Rightarrow NTU = \frac{120}{C_{min}} = \frac{120}{\dot{m}_c \cdot c_p}
From charts or equations for a cross-flow heat exchanger (Cmin mixed, Cmax unmixed), ε = 0.667 corresponds to: NTU≈1.1 for Cr=0.5NTU \approx 1.1 \, \text{for} \, C_r = 0.5
So: 120m˙c⋅cp=1.1⇒m˙c⋅cp=109.1\frac{120}{\dot{m}_c \cdot c_p} = 1.1 \Rightarrow \dot{m}_c \cdot c_p = 109.1
Step 6: Find Outlet Temperature of Coolant
q=m˙c⋅cp⋅(Tc,out−Tc,in)=109.1⋅50=5455 Wq = \dot{m}_c \cdot c_p \cdot (T_{c,out} – T_{c,in}) = 109.1 \cdot 50 = 5455 \, W Since m˙h=2m˙c⇒Ch=2⋅109.1=218.2\text{Since } \dot{m}_h = 2\dot{m}_c \Rightarrow C_h = 2 \cdot 109.1 = 218.2 Th,out=Th,in−qCh=95−5455218.2=95−25=70∘CT_{h,out} = T_{h,in} – \frac{q}{C_h} = 95 – \frac{5455}{218.2} = 95 – 25 = \boxed{70^\circ C}
✅ Final Answers:
- Effectiveness ε=0.667\varepsilon = \boxed{0.667}
- Coolant outlet temperature Th,out=70∘CT_{h,out} = \boxed{70^\circ C}
Explanation:
This problem involves a car radiator, a type of heat exchanger where hot engine coolant transfers heat to incoming cold air. The goal is to determine the effectiveness of the heat exchanger and the outlet temperature of the coolant.
The effectiveness of a heat exchanger measures how efficiently it transfers heat compared to the maximum possible. The effectiveness-NTU (Number of Transfer Units) method is suitable here, especially for a cross-flow configuration common in radiators. We begin by defining heat capacity rates for both fluids. Since the mass flow rate of the coolant is double that of air, the air stream (cold side) has the minimum heat capacity.
Next, we calculate the maximum possible heat transfer based on the minimum capacity rate and the temperature difference between the hot coolant and the cold air. The actual heat transfer is based on how much the air is heated (from 20°C to 70°C). The ratio of actual to maximum heat transferred gives the effectiveness, which is 0.667.
With this effectiveness and known area and overall heat transfer coefficient, we estimate NTU and determine the heat capacity rate of the air stream. This allows calculation of the actual heat transfer rate, which is then used to find the outlet temperature of the coolant. Because the heat exchanger is effective (ε = 0.667), and the capacity rates are such that energy is evenly split, the coolant also exits at 70°C, the same as the air.
This scenario illustrates fundamental heat exchanger principles and highlights how flow rates, area, and temperature goals affect system design and performance.
