Heat Transfer Miscellaneous Easy Questions and Answers | Page - 13

Heat Transfer Miscellaneous

Two infinite parallel plates are placed at a certain distance apart. An infinite radiation shield is inserted between the plates without touching any of them to reduce heat exchange between the plates. Assume that the emissivities of plates and radiation shield are equal. The ratio of the net heat exchange between the plates with and without the shield is

1. 1/2
1. 1/3
1. 1/4
1. 1/8

View Hint View Answer Discuss in Forum

q_{with shields} = 1
q_{without shields} n + 1

here, n = 1
q = 1
q_{without shields} 2

Correct Option: A

q_{with shields} = 1
q_{without shields} n + 1

here, n = 1
q = 1
q_{without shields} 2

A double pipe counter flow heat exchanger transfers heat between two water streams. Tube side water at 19 litre/s is heated frim 10°C to 38°C. Shell side water at 25 litre/s is entering at 46°C Assume constant properties of water, density is 1000 kg/m³ and specific heat is 4186 J/kgK. The LMTD (in °C) is ____.

1. 11.0206 °C
1. 11.06 °C
1. 11.0 °C
1. None of these

View Hint View Answer Discuss in Forum

Given: ṁ_h = 25 L/S; T_h,i = 46°C, T_h,o =?
ṁ_c = 19 L/S; T_c,i = 10°C, T_c,o = 38°C
ρ = 1000 kg/m³3, C = 4186 J/kg.k
Energy balance, ṁ_c = C(T_c,o – T_c,i) = m_n C(T_h,i – T_h,o)
19 (38 – 10) = 25 (46 – T_h,o)
T_h,o = 24.72 °C

LMTD = θ₁ - θ₂
ln θ₁
θ₂

θ₁ = T_h,i – T_c,o = 46 – 38 = 8 °C
θ₂ = T_h,o – T_c,i = 24.72 – 10 = 14.72
LMTD = 8 - 14.72 = 11.0206°C
ln 8
14.72

Correct Option: A

Given: ṁ_h = 25 L/S; T_h,i = 46°C, T_h,o =?
ṁ_c = 19 L/S; T_c,i = 10°C, T_c,o = 38°C
ρ = 1000 kg/m³3, C = 4186 J/kg.k
Energy balance, ṁ_c = C(T_c,o – T_c,i) = m_n C(T_h,i – T_h,o)
19 (38 – 10) = 25 (46 – T_h,o)
T_h,o = 24.72 °C

LMTD = θ₁ - θ₂
ln θ₁
θ₂

θ₁ = T_h,i – T_c,o = 46 – 38 = 8 °C
θ₂ = T_h,o – T_c,i = 24.72 – 10 = 14.72
LMTD = 8 - 14.72 = 11.0206°C
ln 8
14.72

For a heat exchanger, ∆T_max is the maximum temperature difference and ∆T_min is the minimum temperature difference between the two fluids. LMTD is the log mean temperature difference. Cmin and C_max are the minimum and the maximum heat capacity rates. The maximum possible heat transfer (Qmax) between the two fluids is

1. C_min LMTD
1. C_min ∆T_max
1. C_max ∆T_max
1. C_max ∆T_min

View Hint View Answer Discuss in Forum

The temperature difference is not for a given fluid but across the fluids and maximum heat transfer occurs for C_min and the temperature difference is equal to (T_hi - T_ci )

Correct Option: B

The temperature difference is not for a given fluid but across the fluids and maximum heat transfer occurs for C_min and the temperature difference is equal to (T_hi - T_ci )

Saturated steam at 100°C condenses on the outside of a tube. Cold enters the tube at 20°C and exits at 50°C. The value of the Log Mean Temperature Difference (LMTD) is ______°C.

1. 613.83 °C
1. 63.83 °C
1. 638.3 °C
1. 1163.83 °C

View Hint View Answer Discuss in Forum

Here T_ci → cold fluid which enters the tube (temperature)
T_co → cold fluid which exits at tube (temperature)
θ₁ = T_hi – T_ci = 100 – 20 = 80 °C
θ₂ = T_ho – T_co = 100 – 50 = 50 °C
θ_im = θ₁ - θ₂
ln θ₁
θ₂

θ_im = 80° - 50° = 63.83 °C
ln 80
50

Correct Option: B

Here T_ci → cold fluid which enters the tube (temperature)
T_co → cold fluid which exits at tube (temperature)
θ₁ = T_hi – T_ci = 100 – 20 = 80 °C
θ₂ = T_ho – T_co = 100 – 50 = 50 °C
θ_im = θ₁ - θ₂
ln θ₁
θ₂

θ_im = 80° - 50° = 63.83 °C
ln 80
50

Hot and cold fluids enter a parallel flow double t ube heat exchanger at 100 °C and 15 °C, respectively. The heat capacity rates of hot and cold fluids are C_h = 2000 W/K and C_c = 1200 W/ K, respectively. If the outlet temperature of the cold fluid is 45 °C, the log mean temperature difference (LMTD) of the heat exchanger is _____ K (round off to two decimal places).

1. 5.771 °C or K
1. 517.71 °C or K
1. 157.71 °C or K
1. 57.71 °C or K

View Hint View Answer Discuss in Forum

Hot fluid: (ṁ C_p)_h = 2000 W/K
T_{h_i} = 100 °C
Cold fluid: (ṁ C_p)_c = 1200 W/K
T_{c_i} = 15 °C, T_{c_o} = 45 °C
(ṁ C_p)_h (T_{h_i} - T_{h_o}) = (ṁ C_p)_c (T_{c_o} - T_{c_i})
⇒ 2000 × (100 - T_{h_o}) = 1200 (45 - 15)
⇒ T_{h_o = 82 °C
(∆T_i) = T_{h_i} - T_{c_i} = 100 - 15 = 85 °C
(∆T_o) = T_{h_o} - T_{c_o} = 82 - 45 = 37 °C}
(LMTD)_Parallel = 85° - 37° = 57.71 °C or K
ln 85
37

Correct Option: D

Hot fluid: (ṁ C_p)_h = 2000 W/K
T_{h_i} = 100 °C
Cold fluid: (ṁ C_p)_c = 1200 W/K
T_{c_i} = 15 °C, T_{c_o} = 45 °C
(ṁ C_p)_h (T_{h_i} - T_{h_o}) = (ṁ C_p)_c (T_{c_o} - T_{c_i})
⇒ 2000 × (100 - T_{h_o}) = 1200 (45 - 15)
⇒ T_{h_o = 82 °C
(∆T_i) = T_{h_i} - T_{c_i} = 100 - 15 = 85 °C
(∆T_o) = T_{h_o} - T_{c_o} = 82 - 45 = 37 °C}
(LMTD)_Parallel = 85° - 37° = 57.71 °C or K
ln 85
37