External link to An ideal double-reheat steam turbine receives 300,000 kg/h of steam at 10 MPa and 400°C. After expansion from high pressure turbine shell, it is reheated twice from saturated vapor condition back to its initial temperature. The steam entering the condenser is 97% dry. Determine engine work, engine thermal efficiency, and steam rate.

# An ideal double-reheat steam turbine receives 300,000 kg/h of steam at 10 MPa and 400°C. After expansion from high pressure turbine shell, it is reheated twice from saturated vapor condition back to its initial temperature. The steam entering the condenser is 97% dry. Determine engine work, engine thermal efficiency, and steam rate.

| No Comments An ideal double-reheat steam turbine is a sensitive machinery that requires utmost care and precision in terms of its operation. In this article, we will discuss a theoretical case where an ideal double-reheat steam turbine receives 300,000 kg/h of steam at 10 MPa and 400°C. After expansion from high pressure turbine shell, it is reheated twice from saturated vapor condition back to […]

External link to A box girder 12 ft long is given the section shown. Determine the maximum deflection if it is supported on a simple support and exposed to a transverse load of 380 Lbf/ft uniformly distributed in the direction shown. A-36 steel. Determine the factor of safety.

# A box girder 12 ft long is given the section shown. Determine the maximum deflection if it is supported on a simple support and exposed to a transverse load of 380 Lbf/ft uniformly distributed in the direction shown. A-36 steel. Determine the factor of safety.

| No Comments The maximum deflection of a box girder 12 ft long with A-36 steel can be determined by using the following equation: δ = (5wL^4)/384EI, where δ is the maximum deflection, w is the transverse load per unit length, L is the length of the beam, E is Young’s Modulus for A-36 Steel and I is Moment of Inertia.Get the Complete Custom Written […]

External link to 1 kg of fluidis compressed reversible according to law pv=25, where p is in bar and v is in m^3/kg. The final volume is 1/4 of the initial volume. Calculate the work done on the fluid , and sketch the p-v diagram

# 1 kg of fluidis compressed reversible according to law pv=25, where p is in bar and v is in m^3/kg. The final volume is 1/4 of the initial volume. Calculate the work done on the fluid , and sketch the p-v diagram

| No Comments The work done on a fluid is the integral of pressure with respect to volume, W=∫p dv. Since we are dealing with a reversible process according to the law pv=25, it follows that v/p^2=constant (1/25). We can then use this information to determine the work done on the fluid.Get the Complete Custom Written Paper Written by Real Humans Who have exceptionally Excelled […]

External link to Two blocks of mass 50 kg and 40 kg are connected by a light inextensible string as shown in Fig. Find, from first principles, the acceleration of the system and tensions in the cable. Take Friction coefficient = 0.3

# Two blocks of mass 50 kg and 40 kg are connected by a light inextensible string as shown in Fig. Find, from first principles, the acceleration of the system and tensions in the cable. Take Friction coefficient = 0.3

| No Comments The problem at hand involves two blocks of masses 50 kg and 40 kg that are connected by a light, inextensible string as shown in Fig. The objective is to determine the acceleration of the system and tensions in the cable using first principles. It is also assumed that there is a friction coefficient 0.3 between block 2 (40 kg) and the […]

External link to In soccer practice Thato runs 74m in a direction of W30°S, he turns around and walks 22m in a direction of 30° North of East. Graphically determine the resultant displacement of Thato by means of addition of vectors

# In soccer practice Thato runs 74m in a direction of W30°S, he turns around and walks 22m in a direction of 30° North of East. Graphically determine the resultant displacement of Thato by means of addition of vectors

| No Comments In order to graphically determine the resultant displacement of Thato from soccer practice, it is necessary to first identify and break down his movements into two components: a 74 meter run in a direction of W30°S and a 22 meter walk in a direction of 30° North of East.Get the Complete Custom Written Paper Written by Real Humans Who have exceptionally Excelled […]

External link to Steam at 7 bar and dryness fraction 0.9 expands in a cylinder to a pressure of 1.5 bar and 300oC. Calculate the change of internal energy and the change of enthalpy per kg of steam.

# Steam at 7 bar and dryness fraction 0.9 expands in a cylinder to a pressure of 1.5 bar and 300oC. Calculate the change of internal energy and the change of enthalpy per kg of steam.

| No Comments When steam at a pressure of 7 bar and dryness fraction 0.9 expands in a cylinder down to 1.5 bar and 300oC, the change of internal energy and enthalpy can be calculated by using the first law of thermodynamics.[1] This states that when heat (Q) is added or removed from an isolated system with constant volume (V), then the change in internal […]

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