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4. A cylindrical storage tank with a diameter of 4 m and a length of 10 m will hold a fluid whose temperature must be maintained at 347 K. In order to size the heater required for the tank, the following worst-case scenario is considered: Ambient air temperature = -20 °C Wind speed = 20 m/s Tank wall temperature = 347 K What size (kW) heater will be needed?

4. A cylindrical storage tank with a diameter of 4 m and a length of 10 m will hold a fluid whose temperature must be maintained at 347 K. In order to size the heater required for the tank, the following worst-case scenario is considered: Ambient air temperature = -20 °C Wind speed = 20 m/s Tank wall temperature = 347 K What size (kW) heater will be needed?

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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please show how to derive h using nusselt equation and solve for qheatloss

4. A cylindrical storage tank with a diameter of 4 m and a length of 10 m will hold a fluid whose temperature must be maintained at 347 K. In order to size the heater required for the tank, the following worst-case scenario is considered: Ambient air temperature = -20 °C Wind speed = 20 m/s Tank wall temperature = 347 K What size (kW) heater will be needed? A = 2πrL = Theatloss De D = 4m r = 2m L = 10m Ttank = 347K Ta = -20C m V = 20- Twall S = 347K 2π(2m) (10m) = 125.6m² hAAT = k h = Nu De 4(Area) Wetted Perimeter (Ctrl) -
Transcribed Image Text:4. A cylindrical storage tank with a diameter of 4 m and a length of 10 m will hold a fluid whose temperature must be maintained at 347 K. In order to size the heater required for the tank, the following worst-case scenario is considered: Ambient air temperature = -20 °C Wind speed = 20 m/s Tank wall temperature = 347 K What size (kW) heater will be needed? A = 2πrL = Theatloss De D = 4m r = 2m L = 10m Ttank = 347K Ta = -20C m V = 20- Twall S = 347K 2π(2m) (10m) = 125.6m² hAAT = k h = Nu De 4(Area) Wetted Perimeter (Ctrl) -
Expert Solution
Step 1

Given:

Diameter, D = 4 m

Radius, r = 2 m

Length, L = 10 m

Temperature of fluid inside the tank, Ttank = 347 K

Ambient air temperature, Ta = -20∘C ⇒253 K

Wind speed, V = 20 m/s

Form table at wind speed 20 m/s, heat transfer coefficient, h = 34 w/m2.K

 

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