Open App

Mechanical Engineering Exam > Mechanical Engineering Tests > Heat Transfer > Test: Heat Exchanger Level - 3 - Mechanical Engineering MCQ

Test: Heat Exchanger Level - 3 - Mechanical Engineering MCQ

Test Description

10 Questions MCQ Test Heat Transfer - Test: Heat Exchanger Level - 3

Test: Heat Exchanger Level - 3 for Mechanical Engineering 2024 is part of Heat Transfer preparation. The Test: Heat Exchanger Level - 3 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Heat Exchanger Level - 3 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Heat Exchanger Level - 3 below.

Solutions of Test: Heat Exchanger Level - 3 questions in English are available as part of our Heat Transfer for Mechanical Engineering & Test: Heat Exchanger Level - 3 solutions in Hindi for Heat Transfer course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt Test: Heat Exchanger Level - 3 | 10 questions in 40 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study Heat Transfer for Mechanical Engineering Exam | Download free PDF with solutions

*Answer can only contain numeric values

Test: Heat Exchanger Level - 3 - Question 1

Dry saturated steam at 10 bars and 180^oC enters a condenser (counter flow heat exchanger) at the rate of 15 kg/s and leaves at 300^oC . The hot gas enters at 600^oC with mass flow rate of 25 kg/s. If the condenser tubes are 30 mm in diameter and 3 m long, make calculations for the heat exchange area correct upto two decimal places. Neglect the resistance offered by the metallic tubes Take the following properties for steam and gas For steam

t_sat = 180^oC (at 10 bar)

c_ps = 2.7 kj/kg - K

h_s = 600 W/m² - ^oC

For gas

c_pg = 1kj/kg -K

h_g= 250 W/m²-^oC

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 1

From steady state flow energy equation

Q = m_hc_ph (T_hi - T_he) = m_cc_pc(T_ce - T_ci)

⇒ 25 x 1 x (600 - T_he)

⇒ 15 x 2.7 x (300 - 180)

⇒ T_he = 405^oC

The overall heat exchange co-efficient is given by,

1/U = 1/h_g + r_o/r_i .1/h_s

(Neglecting resistance to heat conduction through interface wall and assuming d_o ≅ d_i)

⇒ 1/U = 1/h_g+ 1/h_s

⇒ U = 1/250 + 1/600 = 176.47 W/m²-^oC

For counter flow arrangement

ΔT_i = 300^oC; ΔT_e = 225.6^oC

∴ ΔT_m = ΔT_i =ΔT_e/ln (ΔT_i/ΔT_e)

∴ Q = UA(ፀ_m)

⇒ m_hc_pg(T_hi - T_he) = UA(ፀ_m)

⇒ A = 105.5 m²

*Answer can only contain numeric values

Test: Heat Exchanger Level - 3 - Question 2

Hot water having specific heat 4200 j/kg-K flows through a heat exchanger at the rate of 4 kg/min with an inlet temperature of . The cold fluid having a specific heat 2400 j/kg-K flows at a rate of 8 kg/min with inlet temperature . Make calculations for the maximum possible effectiveness if the arrangement of fluid flow is parallel flow.

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 2

Capacity rate for hot fluid, C_h = m_hc_ph

= 4/60 x 4200

= 280 W/^oC

Capacity rate of cold fluid, C_c = 8/60 x 2400

= 420 W/^oC

∴ C = C_min/C_max = 280/320 = 0.875

Effectiveness of a parallel flow heat exchanger is given as,

The effectiveness is maximum when BTU ➝ ∞

(i.e. UA >> C_min)

ε = 1/1+0.875

ε = 0.5333

1 Crore+ students have signed up on EduRev. Have you? Download the App

*Answer can only contain numeric values

Test: Heat Exchanger Level - 3 - Question 3

A counter flow heat exchanger is used to cool 2000 kg/hr of oil (c_p = 2.5kj/kg-K) from 105^oC to 30^oC by use of water that

enters the heat exchanger with a temperature of 15^oC. If the overall heat transfer coefficient is expected to be 1.5 kW/m²-.K , make calculations for the surface area required, presume that the exit temperature of the water is not to exceed 80^oC.

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 3

From steady state flow energy equation

m_cc_ph(T_hi -T_he) = m_cc_ph(T_ce -T_ci)

2000 x .5(105 - 30) = m_c 4.18(80-15)

M_c = 1380.2 kg/hr

∴ m_cc_pc = 1380.2 x 4.18 = 5769.24 kjhr-K

m_cc_ph = 2000 x 2.5 = 5000kh/hr-K

∴ C = 5000/5769.24 = 0.867

Effectiveness , ε = T_hi - T_he /T_hi - T_ci = 105-30/105-15 = 0.822

For a counter flow heat exchanger

NTU = 3.84

UA/C_min = 3.84

1.5 x A /5000/3600 = 3.84 ⇒ A = 3.55m²

Alternate method

T_hi = 105^oC T_ho = 30^oC

T_ci = 1.5^oC T_co M 80^oC

∴ Q_act = m_hc_h(T_hi - T_ho)

= 2000/3600 x 2.5 x (105 - 30)

= 104.166 kj/sec

For a counterflow heat exchanger

Q_act = UA θ_m

104.166 = 1.5 x A x 19.576

A = 104.166/15 x 19.576

A = 3.54 m²

*Answer can only contain numeric values

Test: Heat Exchanger Level - 3 - Question 4

Water (sp. heat = 4kj/kg -K) enters a counter flow heat exchanger (both fluids unmixed) at

15^oC and flows at the rate of 7.5 kg/s . It cools air (c_p = 2kj/kg-K) flowing at the rate of 10 kg/s from an inlet temperature of 120^oC. For an overall heat transfer coefficient 216.66 W/m²-K. and an exchanger surface area of 240 m², determine the LMTD for the heat exchanger?

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 4

Given For cold fluid [Water]

M_c = 7.5 kg/sec

C_c = 4 kj/kg-K

T_ci = 15^oC

For hot fluid [Air]

M_h = 10 kg/sec

C_h = 2 kj/kg-K

T_hi = 120^oC

U = 216.66 W/m²-K

A = 240 m²

For a counter flow heat exchanger,

Where C = C_min/C_max & NTU = UA/C_min

Based on given data

C_h = m_hc_h

= 10 x 2

= 20 kW/K

∴ C_h < />_c

∴ C_min = C_h &

C_max = C_c

Thus,

C = C_h/C_c = 20/30 = 2/3

NTU = UA/U_min

= 216.66 x 240/20000

= 2.599

≃ 2.6

Where

ε = Q_act/Q_th

⇒ Q_act = εQ_th

Also,

Q_act = UAθ_m

⇒ θ_m = Q_act/UA

⇒θ_m - εθ_th/UA = εC_h(T_hi - T_ci)/UA

⇒θ_m = 0.8 x 20 x 10³ x (120-15)/216.66 x 240

⇒θ_m = 32.3^oC

*Answer can only contain numeric values

Test: Heat Exchanger Level - 3 - Question 5

Steam in the condenser of a power plant is to be condensed at a temperature of 30^oC with cooling water from a nearby lake, which enters the tubes of the condenser at 14^oC and leaves at 22^oC. The surface area of the tubes is 45m² , and the overall heat transfer coefficient is 2100 W/m²-K. Determine the rate of condensation of the steam in the condenser. Take latent heat of vaporization of water as 2431 kJ/kg?

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 5

Since one fluid is changing phase, it is immaterial whether the heat exchanger is parallel flow or counter flow as both configurations will have equal LMTD and effectiveness

x = 0 x = L

∵ ፀ_o = 30-14 = 16^oC

ፀ₁ = 30-22 = 8^oC

∴ Q = UAθ_m

Q = 2100 x 24 x 11.55

⇒ Q = 108\ kW

For mass ofd steam condensing,

Q = m_steam x L

m_steam = Q/L 1087 x 10³/x2431 x 10³

⇒ m_steam= 0.45kg/s

*Answer can only contain numeric values

Test: Heat Exchanger Level - 3 - Question 6

A test is conducted to determine the overall heat transfer coefficient in an automotive radiator that is a water-to-air heat exchanger with both fluids (air and water) unmixed. The radiator has 40 tubes of internal diameter 0.5 cm and length 65 cm in a closely spaced platefinned matrix. Hot water enters the tubes at 90^oC at a rate of 0.6 kg/s and leaves at 60^oC. Air flows across the radiator through the inter-fin spaces and is heated from 20^oC to 40^oC. Determine the overall heat transfer coefficient (in kW/m²-K) of this radiator based on the inner surface area of the tubes. Take specific heat of water as 4.195 kJ/kg-K? Consider the heat exchanges to be counter flow.

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 6

Q = m_hc_ph(T_hi - T_he)

Q = 0.6 x 4.195 x (90-65)

Q = 62.93 kW

Tube side area for heat transfer, A_i = n π D L

⇒ A_i = 40 x π x 0.005 x 0.65

⇒ A_i = 0.408 m²

Again θ_i = 90-40 = 50^OC

Θ_e = 65-20 =45^oC

= 47.5^oC

Now Q = U_iA_iθ_m

⇒ U_i = Q/A_iθ_m = 62.93/0.408 x 47.5

⇒ U_i = 3.247 kW/m²-K

*Answer can only contain numeric values

Test: Heat Exchanger Level - 3 - Question 7

A counter flow double pipe heat exchanger is to heat water from 20^oC to 80^oC at a rate of 1.2kg/s. The heating is to be accomplished by geo-thermal water available at 160^oC at a mass flow rate of 2 kg/s. The inner tube is thin-walled and has a diameter of 1.5 cm. The overall heat transfer coefficient of the heat exchanger is 620 W/m²-K . Determine the length of the heat exchanger required to achieve the desired heating. Given

NTU = specific heat of water is 4.18 Kj/kg-K and for geothermal water is 4.31 Kj/kg-K.

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 7

The capacity rates are,

C_h = m_hc_ph = 2 x 4.31 = 8.62 W/K

C_h = m_hc_ph = 1.2 x 4018 = 5.02 kW/K

∴ C = C_min/C_max = 5.02/8.62 = 0.582

∴ Q_act = m_cc_pc [T_ce - T_ci]

= 1.2 x 4.18(80-20)

= 301 kW

Q_max = C_min (T_hi - T_ci)

= 5.02(160 - 20)

= 702.8 kW

∴ ε Q_act/Q_max = 301/702.8 = 0.428

⇒ NTU = 0.651

⇒ UA/C_min = 0.651

⇒ A = 5.11 m²

Now A = πDL

⇒ L = A/πD = 5.11/3.14 x 0.015

⇒ L = 108 m

*Answer can only contain numeric values

Test: Heat Exchanger Level - 3 - Question 8

A double pipe counter flow heat exchanger is to be designed to cool 12000 kg/hr of an oil of specific heat 1.95 Kj/Kg-K . from 85^oC to 55^oC by water entering the heat exchanger at 30^oC and leaving at 45^oC . If the overall heat transfer coefficient of heat exchanger is 400 W/m²-. K Calculate the surface area of the heat exchanger.

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 8

For the counter flow heat exchanger

M_h = 12000/3600 = 3.33 kh/s

C_ph = 1.95 kj/kh -K

T_hi - 85^oC; T_he = 55^oC

T_ci = 30^oC; T_ce = 45^oC

∴ θ₀ = 40^oC; θ₁ = 45^oC

⇒ ፀ_m = 31.9^oC

Q = m_hc_ph(T_hi - T_he) = 3.33 x 1.95(85-55)

⇒ Q = 194.805 kW

Again Q = UA(θ_m)

⇒ A = Q/u((θ_m)= 194.805/0.4 x 31.9

⇒ A = 15.27m²

Test: Heat Exchanger Level - 3 - Question 9

A counter flow, concentric tube heat exchanger used for engine cooling has been in service for an extended period of time. The heat transfer surface area of the exchanger is , 5 m² and the design value of the overall convection coefficient is 38 W/m². During a test run, engine oil flowing at 0.1 kg/s. is cooled from 110^oC 66^oC by water supplied at a temperature of 25^oC and a flow rate of 0.2kg/s. Calculate the fouling factor,

R_f”(m² - K/W).

Given for engine oil c = 2166j/kg-K;

& for water c = 4178 j/kg-K

A. 0.0019 m²-K/W
B. 0.0039 m²-K/W
C. 0.0029 m²-K/W
D. 0.00 m49²-K/W

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 9

From energy balance,

m_hc_ph(T_hi- T_he) = m_hc_ph(T_ce- T_ci)

⇒ 0.1 x 2166 x (110 - 66)

= 0.2 x 4178 (T_ce - 25)

⇒ T_ce = 36.4^oC

Θ_o = 73.6;Θ₁ = 41

⇒ θ_m = 55.7^oC

Q = m_cc_ph(T_hi - T_hi)

= 0.1 x 2166(110-66)

= 9530 W

Q = U A θ_m

⇒ 9530 = U x 5 x 55.7

⇒ U = 34.2 W/m² K

Overall resistance including fouling factor,

⇒R_f = 0.0029 m²K/W

*Answer can only contain numeric values

Test: Heat Exchanger Level - 3 - Question 10

The condenser of a large steam power plant is a shell-and-tube heat exchanger having a single shell and 30,000 tubes, with each tube making two passes. The tubes are thin-walled with 25 mm diameter and steam condenses on the outside of the tube with h_o= 11kW.m²-K. The cooling water flowing through the tubes at a rate of 30,000 kg/s and the heat transfer rate is 2 GW. Water enters at 20^oC while steam condenses at 50^oC. Find the length of the tubes in one pass, properties of water at are

c_p = 4.18 kj/kg-K, μ = 855 x 10^-6 N-s/m²

K = 0.613 W/m-K and Pr = 5.83.

Given that for laminar flow through pipes corresponding Nusselt number is 4.0 and for turbulent flow,

it is Nu = 0.023 Pr^0.4_.

(B) 4.55

Detailed Solution for Test: Heat Exchanger Level - 3 - Question 10

Q = m_cc_pc (T_c2 - T_c1) = 2 x 10⁹ W

T_c2 - 20 = 2 x 10⁹/30000 x 4.18 = 16

∴ T_c2 = 36^oC

Aslo, Q = UA(θ_m)_cf

Where A = N x πD(2L),

L = length of one tube pass

R_eD = Vd_p/μ = 4m/πDμ

= 4 x 1 kg/s / 3.14 x 0.02 m x 855 x 10^-6 Ns/m²

= 59567

The flow is turbulent,

∴ Nu_D = 0.023 Re_D^0.8 Pr ^0.4

= 0.023(59567)^0.8(5.83)^0.4 = 308

h_iD/k = 308

h_i = 308 x 0.613/0.025 = 7552 W/m² K

1/U = 1/h_i + 1/h_o = 1/7552 + 1/11000

U = 4478 W/m² K

(θ_m)_cf = = 21^oC

L = Q / U x (N x ?D x 2 )(θ_m)_cf

= 2 x 10⁹ / 4478 x 30000 x 2 x ? x 0.025 x 21

= 4.51

	Heat Transfer 57 videos\|77 docs\|86 tests

Heat Transfer

57 videos|77 docs|86 tests

Join Course

Information about Test: Heat Exchanger Level - 3 Page

In this test you can find the Exam questions for Test: Heat Exchanger Level - 3 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Heat Exchanger Level - 3, EduRev gives you an ample number of Online tests for practice

Top Courses for Mechanical Engineering

	Heat Transfer 57 videos\|77 docs\|86 tests

Heat Transfer

57 videos|77 docs|86 tests

Join Course

Top Courses for Mechanical Engineering

Important Questions for Heat Exchanger Level - 3

Find all the important questions for Heat Exchanger Level - 3 at EduRev.Get fully prepared for Heat Exchanger Level - 3 with EduRev's comprehensive question bank and test resources. Our platform offers a diverse range of question papers covering various topics within the Heat Exchanger Level - 3 syllabus. Whether you need to review specific subjects or assess your overall readiness, EduRev has you covered. The questions are designed to challenge you and help you gain confidence in tackling the actual exam. Maximize your chances of success by utilizing EduRev's extensive collection of Heat Exchanger Level - 3 resources.

Heat Exchanger Level - 3 MCQs with Answers

Prepare for the Heat Exchanger Level - 3 within the Mechanical Engineering exam with comprehensive MCQs and answers at EduRev. Our platform offers a wide range of practice papers, question papers, and mock tests to familiarize you with the exam pattern and syllabus. Access the best books, study materials, and notes curated by toppers to enhance your preparation. Stay updated with the exam date and receive expert preparation tips and paper analysis. Visit EduRev's official website today and access a wealth of videos and coaching resources to excel in your exam.

Online Tests for Heat Exchanger Level - 3 Heat Transfer

Practice with a wide array of question papers that follow the exam pattern and syllabus. Our platform offers a user-friendly interface, allowing you to track your progress and identify areas for improvement. Access detailed solutions and explanations for each test to enhance your understanding of concepts. With EduRev's Online Tests, you can build confidence, boost your performance, and ace Heat Exchanger Level - 3 with ease. Join thousands of successful students who have benefited from our trusted online resources.

Education Revolution