Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A water heater uses \(100 \mathrm{~kW}\) to heat 60 gallons of liquid water from an initial temperature of \(20^{\circ} \mathrm{C}\). The average density and specific heat of the water are $970 \mathrm{~kg} / \mathrm{m}^{3}\( and \)4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$, respectively. According to the service restrictions of the ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015, HG-101), water heaters should not be operating at temperatures exceeding \(120^{\circ} \mathrm{C}\) at or near the heater outlet. Determine the heating duration such that the final water temperature that would exit the water heater would be in compliance with the ASME Boiler and Pressure Vessel Code.

Short Answer

Expert verified
Answer: The required heating duration is approximately 913.55 seconds.

Step by step solution

01

Convert gallons to cubic meters

Begin by converting the volume of water from gallons to cubic meters for further calculations. 1 gallon is approximately equivalent to 0.00378541 cubic meters, so: \(60 \text{ gallons} = 60 * 0.00378541 \mathrm{~m}^{3} \approx 0.227 \mathrm{~m}^{3}\)
02

Calculate the mass of water

Next, calculate the mass of the water using its volume(0.227 m\(^3\)) and given density. \(m = V * \rho = 0.227 \mathrm{~m}^{3} * 970 \mathrm{~kg} / \mathrm{m}^{3} = 220.09 \mathrm{~kg}\)
03

Calculate the heat required

Now, calculate the heat required to heat the water to the maximum allowable temperature (120\(^{\circ} \mathrm{C}\)) using the mass, specific heat, and initial temperature (20\(^{\circ}\mathrm{C}\)): \(Q = m * c * \Delta T = 220.09 \mathrm{~kg} * 4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} * (120^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C}) = 91355.124 \mathrm{~kJ}\)
04

Calculate the heating duration

Finally, determine the heating duration by dividing the necessary heat by the rate at which the heater provides it (100 kW). \(T = \frac{Q}{P} = \frac{91355.124 \mathrm{~kJ}}{100 \mathrm{~kW}} = 913.55124 \mathrm{~s}\) The heating duration that would ensure the final water temperature to be in compliance with the ASME Boiler and Pressure Vessel Code is approximately 913.55 seconds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 3 -m-internal-diameter spherical tank made of \(1-\mathrm{cm}\) thick stainless steel is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located outdoors at \(25^{\circ} \mathrm{C}\). Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus the thermal resistance of the tank to be negligible, determine \((a)\) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). The emissivity of the outer surface of the tank is \(0.75\), and the convection heat transfer coefficient on the outer surface can be taken to be $30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assume the average surrounding surface temperature for radiation exchange to be \(15^{\circ} \mathrm{C}\). Answers: (a) \(23.1 \mathrm{~kW}\), (b) \(5980 \mathrm{~kg}\)

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be $2.5 \mathrm{Btu} / \mathrm{h} . \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$.

A cylindrical resistor element on a circuit board dissipates \(0.8 \mathrm{~W}\) of power. The resistor is \(2 \mathrm{~cm}\) long and has a diameter of $0.4 \mathrm{~cm}$. Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a \(24-\mathrm{h}\) period, \((b)\) the heat flux, and \((c)\) the fraction of heat dissipated from the top and bottom surfaces.

The inner and outer surfaces of a \(4-\mathrm{m} \times 7-\mathrm{m}\) brick wall of thickness \(30 \mathrm{~cm}\) and thermal conductivity $0.69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( are maintained at temperatures of \)20^{\circ} \mathrm{C}\( and \)5^{\circ} \mathrm{C}$, respectively. Determine the rate of heat transfer through the wall in W.

An electric current of 5 A passing through a resistor has a measured voltage of \(6 \mathrm{~V}\) across the resistor. The resistor is cylindrical with a diameter of \(2.5 \mathrm{~cm}\) and length of \(15 \mathrm{~cm}\). The resistor has a uniform temperature of \(90^{\circ} \mathrm{C}\), and the room air temperature is \(20^{\circ} \mathrm{C}\). Assuming that heat transfer by radiation is negligible, determine the heat transfer coefficient by convection.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free