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 refrigeration system is being designed to cool \(\operatorname{eggs}\left(\rho=67.4 \mathrm{lbm} / \mathrm{ft}^{3} \text { and } c_{p}=0.80 \mathrm{Btu} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\right)\) with an average mass of \(0.14 \mathrm{lbm}\) from an initial temperature of \(90^{\circ} \mathrm{F}\) to a final average temperature of \(50^{\circ} \mathrm{F}\) by air at \(34^{\circ} \mathrm{F}\) at a rate of 10,000 eggs per hour. Determine \((a)\) the rate of heat removal from the eggs, in \(\mathrm{Btu} / \mathrm{h}\) and \((b)\) the required volume flow rate of air, in \(\mathrm{ft}^{3} / \mathrm{h}\), if the temperature rise of air is not to exceed \(10^{\circ} \mathrm{F}\).

Short Answer

Expert verified
\(\dot{Q}_{eggs}\) = 10,000 x 0.14 x 0.80 x 40 = 44,800 Btu/h #tag_title# (5) Calculate the required volume flow rate of air#tag_content# Using the formula derived in step 3: \(\dot{V} = \frac{N m_e \Delta T}{\rho \Delta T_{air}} = \frac{10,000 \times 0.14 \times 40}{0.075 \times 10} = \frac{56,000}{0.75} = 74,667 ft^3/h\) Therefore, the required volume flow rate of air is 74,667 ft^3/h.

Step by step solution

01

(1) Calculate the total heat removal from the eggs

To find the total amount of heat that must be removed from the eggs, we can use the formula: \(q = mc_p \Delta T\) where \(q\) is the heat transfer, \(m\) is the mass, \(c_p\) is the specific heat, and \(\Delta T\) is the temperature change. The mass is equal to the number of eggs multiplied by the average mass of an egg, and the temperature change is equal to the initial temperature minus the final temperature.
02

(2) Find the rate of heat removal

To find the rate of heat removal in Btu/h, we need to divide the total heat removal by the time it takes, which is 1 hour. Let's denote \(N\) as the rate of eggs being cooled and \(m_e\) as the mass of an individual egg. Then the rate of heat removal from the eggs is: \(\dot{Q}_{eggs} = N m_e c_p \Delta T\)
03

(3) Calculate the required volume flow rate of air

The rate of heat removal by air can be expressed as: \(\dot{Q}_{air} = \dot{V} \rho c_p \Delta T_{air}\) where \(\dot{V}\) is the volume flow rate, \(\rho\) is the density of air, and \(\Delta T_{air}\) is the temperature rise of the air which is given to be not more than 10°F. Since the rate of heat removal from the eggs and by the air must be equal: \(\dot{Q}_{eggs} = \dot{Q}_{air}\) Substitute the values for the rates of heat removal from the eggs and by air, we have: \(N m_e c_p \Delta T = \dot{V} \rho c_p \Delta T_{air}\) Now, we can solve for \(\dot{V}\), the required volume flow rate of air: \(\dot{V} = \frac{N m_e \Delta T}{\rho \Delta T_{air}}\) Now, let's plug in the given values and calculate the rate of heat removal from the eggs and the required volume flow rate of air.
04

(4) Calculate the rate of heat removal from the eggs

Using the given values and using the formula derived in step 2: \(\dot{Q}_{eggs} = N m_e c_p \Delta T = (10,000 \, \text{eggs/h})\times(0.14\, \mathrm{lbm})\times(0.80\, \mathrm{Btu/lbm\cdot^{\circ}F})\times(90^{\circ}F-50^{\circ}F)\)

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

The air in an insulated, rigid compressed-air tank whose volume is \(0.5 \mathrm{m}^{3}\) is initially at \(4000 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) Enough air is now released from the tank to reduce the pressure to 2000 kPa. Following this release, what is the temperature of the remaining air in the tank?

Air enters the evaporator section of a window air conditioner at 14.7 psia and \(90^{\circ} \mathrm{F}\) with a volume flow rate of \(200 \mathrm{ft}^{3} / \mathrm{min}\). Refrigerant- \(134 \mathrm{a}\) at 20 psia with a quality of 30 percent enters the evaporator at a rate of \(4 \mathrm{lbm} / \mathrm{min}\) and leaves as saturated vapor at the same pressure. Determine (a) the exit temperature of the air and ( \(b\) ) the rate of heat transfer from the air.

Air enters a pipe at \(65^{\circ} \mathrm{C}\) and \(200 \mathrm{kPa}\) and leaves at \(60^{\circ} \mathrm{C}\) and \(175 \mathrm{kPa} .\) It is estimated that heat is lost from the pipe in the amount of \(3.3 \mathrm{kJ}\) per kg of air flowing in the pipe. The diameter ratio for the pipe is \(D_{1} / D_{2}=1.4 .\) Using constant specific heats for air, determine the inlet and exit velocities of the air. Answers: \(29.9 \mathrm{m} / \mathrm{s}, 66.1 \mathrm{m} / \mathrm{s}\).

Steam enters a nozzle with a low velocity at \(150^{\circ} \mathrm{C}\) and \(200 \mathrm{kPa}\), and leaves as a saturated vapor at \(75 \mathrm{kPa}\). There is a heat transfer from the nozzle to the surroundings in the amount of \(26 \mathrm{kJ}\) for every kilogram of steam flowing through the nozzle. Determine ( \(a\) ) the exit velocity of the steam and (b) the mass flow rate of the steam at the nozzle entrance if the nozzle exit area is \(0.001 \mathrm{m}^{2}\)

It is proposed to have a water heater that consists of an insulated pipe of 7.5 -cm diameter and an electric resistor inside. Cold water at \(20^{\circ} \mathrm{C}\) enters the heating section steadily at a rate of \(24 \mathrm{L} / \mathrm{min}\). If water is to be heated to \(48^{\circ} \mathrm{C}\), determine \((a)\) the power rating of the resistance heater and \((b)\) the average velocity of the water in the pipe.

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