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The inlet temperature of hot water is 800K and outlet temperature is 400 K and the inlet temperature of cold water is 200 K, both are flowing at some mass flow rate. Taking counter flow heat exchanger the NTU and LMTD respectively are
- a)1.5 and 400 K
- b)2 ad 200 K
- c)0.5 and 200 K
- d)1 and 400 K
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The inlet temperature of hot water is 800K and outlet temperature is 4...
Here, this is a case of balanced heat exchanger
MhCh =mcCc = C [Both the fluids are water]
NTU = 2
MhCh =mcCc = C [Both the fluids are water]
NTU = 2
Most Upvoted Answer
The inlet temperature of hot water is 800K and outlet temperature is 4...
Calculation of NTU and LMTD
Given data:
Inlet temperature of hot water (Th,in) = 800 K
Outlet temperature of hot water (Th,out) = 400 K
Inlet temperature of cold water (Tc,in) = 200 K
1. Calculation of LMTD (Log Mean Temperature Difference):
∆T1 = Th,in - Tc,out = 800 K - 200 K = 600 K
∆T2 = Th,out - Tc,in = 400 K - 200 K = 200 K
LMTD = (∆T1 - ∆T2) / ln(∆T1/∆T2)
LMTD = (600 - 200) / ln(600/200)
LMTD = 400 / ln(3)
LMTD ≈ 200 K
2. Calculation of NTU (Number of Transfer Units):
Cmin = minimum of (m_dot * Cp)hot, (m_dot * Cp)cold
Cmin = minimum of (m_dot * Cp)hot, (m_dot * Cp)cold
Cmin = minimum of [m_dot * (Th,in - Th,out)], [m_dot * (Tc,out - Tc,in)]
Cmin = minimum of [m_dot * (800 - 400)], [m_dot * (200 - 200)]
Cmin = minimum of [400m_dot], [0]
Cmin = 0
NTU = UA / Cmin
NTU = UA / 0
NTU = ∞
Therefore, the correct option is (b) 2 and 200 K.
Given data:
Inlet temperature of hot water (Th,in) = 800 K
Outlet temperature of hot water (Th,out) = 400 K
Inlet temperature of cold water (Tc,in) = 200 K
1. Calculation of LMTD (Log Mean Temperature Difference):
∆T1 = Th,in - Tc,out = 800 K - 200 K = 600 K
∆T2 = Th,out - Tc,in = 400 K - 200 K = 200 K
LMTD = (∆T1 - ∆T2) / ln(∆T1/∆T2)
LMTD = (600 - 200) / ln(600/200)
LMTD = 400 / ln(3)
LMTD ≈ 200 K
2. Calculation of NTU (Number of Transfer Units):
Cmin = minimum of (m_dot * Cp)hot, (m_dot * Cp)cold
Cmin = minimum of (m_dot * Cp)hot, (m_dot * Cp)cold
Cmin = minimum of [m_dot * (Th,in - Th,out)], [m_dot * (Tc,out - Tc,in)]
Cmin = minimum of [m_dot * (800 - 400)], [m_dot * (200 - 200)]
Cmin = minimum of [400m_dot], [0]
Cmin = 0
NTU = UA / Cmin
NTU = UA / 0
NTU = ∞
Therefore, the correct option is (b) 2 and 200 K.
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The inlet temperature of hot water is 800K and outlet temperature is 400 K and the inlet temperature of cold water is 200 K, both are flowing at some mass flow rate. Taking counter flow heat exchanger the NTU and LMTD respectively area)1.5 and 400 Kb)2 ad 200 Kc)0.5 and 200 Kd)1 and 400 KCorrect answer is option 'B'. Can you explain this answer?
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Solutions for The inlet temperature of hot water is 800K and outlet temperature is 400 K and the inlet temperature of cold water is 200 K, both are flowing at some mass flow rate. Taking counter flow heat exchanger the NTU and LMTD respectively area)1.5 and 400 Kb)2 ad 200 Kc)0.5 and 200 Kd)1 and 400 KCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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